A School Course of Mathematics - Forgotten Books

A SCHOOL COURSE OF

MATHEMATICS

DAV ID MA IR

OXFORD

AT THE CLARENDON PRESS

HENRY FROWDE , M.A .

PUBLISHER TO THE UN IVERSITY OF OXFORD

LONDON, EDINBURGHNEW YORK AND TORONTO

P R EFA C E

THE material for mathematical education may be chosen

for the value of the knowledge of the material itself, or forthe value of the training got in the course of acquiringthe knowledge. The knowledge-value is greater the moreclosely the thing studied is related to human life and

interests, and less the more remote it is from these. As

to training-value, the appropriate progress from concrete toabstract is most possible in connexion with things of

concrete human interest ; so that a selection of materialmade for its knowledge-value is confirmed by the criterionof training-value .

In range this book is in general agreement with the

practice of our schools ; containing the Geometry, Algebra,and Trigonometry usually read by pupils that do not

specialize in mathematics. In detail there are a few

difl'

erences. Thus the addition theorem in Trigonometryis of no great use for the solution of triangles, nor has themanipulation involved suficient training-value to justifythe inclusion of the theorem . On the other hand, a certainamount of Solid Ge ometry is included from a belief in itsvalue for both knowledge and training.

The more the pupils can develop the subject for themselves and without help from the teacher, the truer is theirknowledge of it and the more valuable the training receivedin the process but care must be taken that this intensivemethod does not too greatly restrict the range of knowledge . This book has the form of a summarized discussionbetween teacher and pupils. In the earlier chapters thediscussion is given in some detail. In later chapters itismore condensed, so that tos the end of the book thereply put in the pupil

’

smouth is the final formal conclusion,

264673

iv PREFACE

which is reached only after a good deal of discussion.

The order of the development of the subject must dependto some extent on the suggestions made by the pupils indiscussion, and so must vary from class to class. Thisbook shows one possible development more immrtance is

attached to the method than to the order of development.In order that the discussion may be a real one, each discussion should take place before the pupils look at the cor

responding part of the book.

Some teachers may for the sake of drill and mechanicaldexterity desire an increase in the number of exercises ;

they can easily multiply them by numerical alteration of the

data. But I venture to think that this country pays toomuch attention to dexterity, and that to work honestlythrough the problems of the text and the exercises willresult in as great a degree of dexte rity as any one shouldrequire.

Occasional reference is made to principles. For a dis

cussion of principles readers are referred to Some Primariesof Mathematical Education by Mr. Ronchara Branford

,

which is being issued by the Clarendon Press. The

writing of my book has been in part due to readingMr. Branford

’

s manuscript.Valuable help in suggestions and proof reading has been

given by Messrs. Leonard Blaikie, Benchara Branford,J. G. Hamilton, and E . L. Kearney .

Most of the exercises are taken from examination papersof the Civil Service Commission, by the permission of the

Controller of His Majesty’

s Stationery Office .

Criticisms and suggestions will be gratefully received.

I shall be especially glad to hear of pupils’

suggestionswhich (whether right or wrong) have proved fruitful forthe development of the subject.

DAVID MAIR .

En cru o, Sow r

November, 1906.

C ON TE N T S

CHAPTER I . THE CAOHR

Determination of position of cache. Points of the com

pass right angle degree circle. Conclusions fromexperiment and from general reasoning. Isosceles triangle.

Chord properties of circle. Work is to be done by pupils,not by teacher. To bisect a line to draw a perpendicular.

Converse propositions.

CHAPTER II. POSITION or CHAIR ON FLOORPosition of a point given by distances from walls.

Parallels. Loci. Symbols, and their right use.

CHAr'rRR III. Anu s

Meaning of word ‘Geometry’. Units of area. Use of

brackets ; precedence of x over Suitable degree of

approximation. Area of rectangle, of right-angled triangle,of any triangle, of a curvilinear figure. Graphs. Algebraicexpressions.

CHAPTER IV. Vow uns

Air-space of a room . Volume of a cylinder (gallonmeasure) . Circumference of circle diameter. Areaof circle square on radius. Averages. Volume of

reservoir. Money lent at interest ; place and value of

abbreviations.

CHAPTER V. To Corr A MAP, SIZE UNOHANGRD

Copy made by measurement of distances ; number of

measurements needed . Case of three places, other methods.Plane table. To make an angle equal to a given angle ; to

bisect an angle. Conditions that fix a triangle . Variationof angles of a triangle that has two sides given. Triangle

fixed by three sides, by two sides and an angle, by a side andtwo angles corresponding congruences. Slider crank pair.

vi CONTENTS

CHAPTER VI. FURTHER MENSURATIONTo find the dimensions of a standard gallon measure .

Comparison of Fahrenheit and Centigrade thermometers

simple equations. Money and debt ; negative numbers.

Sheep enclosure quadratic equations.

CHAPTER VII. ON CRANE PROBLEMSPupil

's rate of advance. Wall crane ; surface circle ;

plane. Shear-legs ; degrees of freedom conditions thatfix position of load . Traveller crane ; parallels ; per

pendicular to a plane. Scotch crane. Derrick crane.

Cc-ord inates. To make a plane surface.

CHAPTER VIII. ANGLE-sun PROPERTY. PARALLELs

AND AREASVarious approaches to the fact that the three angles of

a triangle together make Prope rties of parallels.

Construction of parallels ; ruler and compasses, set-square ,T-square. Parallel rulers properties of parallelogram .

Area of parallelogram , triangle, polygon ; to make triangleor square equal in area to any polygon.

CHAPTER IX. ON DRAW ING To SOALEPlans of playground to various scales by chain-surveying

comparison of plans with one another and with play

ground. Plans by means of plane table ; comparison ;meaning of ratio. Horse-dealing indexnotation propertyof indices. Comparison of ratios involving incommen

surable numbers. Conditions that fix the shape of a

triangle ; sine of an angle. Areas of similar figures ;volumes of similar solids.

CHAPTER X. ON SIGNALLINGMorse Code ; summation of series. Semaphore signalling.

Flag signalling. Rosettes ; combinations of n things p at

a time . Binom ial theorem. Interest on a loan .

CHAPTER XI. ON CARRYING WATERPerpendicular and slant distances of a point from a line.

The greater side of a triangle is opposite the greater angle

and converse . Each bisector of the vertical angle of a triangle divides the base in the ratio of the sides. Calculationof the figure trigonometrical expressions.

CONTENTS

CHAPTER XII. ON SHORTENING CALOULATIONs.

SLIDE RULETable of powers of 2used to avoidmultiplication powers

of more useful. Table of powers replaced by graph.

Graph replaced by graduated line ; slide rule . Laws ofindices ; zero and negative indices ; square root from graphy-POI' ; index

CHAPTER XIII. DANGER ANGLE. PROPERTIEs or

CIRCLEs

Danger angle ; the an

gle in a segment Of a circle is con

stant ; angles greater an Tangent to a circle as

limi t of line cutting circle. Relations among angles of

cyclic quadrilateral ; limit when two vertices coalesce .

Rectangle properties of a circle to draw a tangent.

CHAPTER XIV. LOGARITHIIs

Index-power table with 10 01 as base advantages of base10 028. Negative and fractional indices. Logarithms withbase 10 checking table of logarithms. Calculation withlogarithms root extraction laws of logarithms.

CHAPTER XV. THEOREM or PvTHAGORAs

Theorem arrived at in variousways ; converse. Determination of a right-angled triangle h

'

om various data. Quadraticequations. Calculation of right

-angled triangle. Conditions thatfix shape of right-angled triangle. To find right

angled triangle with given area and perimeter ; simultaneousquadratic equations.

CHAPTER XVI. CALCULATION or A TRIANGLECalculation from a A B ; trigonometrical relations ; case

of LB obtuse ; sine and cosine of an obtuse angle. Calcula

tion from a b B ; various cases. Calculation from b c A

tangent of obtuse angle. Calculation from a b c ; case ofobtuse angled triangle proj ection on a line. Extension

of theorem of Pythagoras. Area of a triangle. A quadraticequation. Calculation of a triangle from the cc-ordinatesof the vertices ; and from other data.

CHAPTER XVII. SOLID GEOHETRY

Existence of perpendicular to a plane. Perpendiculars

to a plane are parallel ; converse . Number of points thatfix a plane. Existence of parallel planes. Angle between

PAGE

CONTENTS

two planes. Position of electric lamp fixed by lengths ofthree support ing strings discussion of the figure by drawing and by calculation. Volume of a pyramid. Positionof solid body fixed by six conditions. Conditions that fixa framework in form and in position. Congruence of

solids. Images. Similarity. Circle and sphere.

of similar solids.

TABLE or WEIGHTs AND MEAsUREs

AnswERs To CERTAIN NUMERICAL QUEsTIONs

INDEx To EXPLANATIONS or TERIIs AND PROPERTIEs

or FIGUREs

PAGE

A SCHOOL COURSE OF

MATHEMATICS

CHAPTER I

THE CACHE

1 . A BOY sometimes makes in a garden a cache or hidingplace for treasures, and covers it over so as to be indis

tinguishable. By what measurements can he fix the spotso as to find it againThe pupils should be asked for suggestions, and their

proposals discussed till a possible method is found If the

discussion can take place in a garden, it will go better.They will suggest measuring its distance from trees, posts,or other known points. Suppose the cache 0 is 7 feet froman apple-tree A,

12 feet from a plum-tree P, 4 feet froma cherry-tree T, and so on.

2. The cache is 7 feet from the tree A . Mark all the

points at this distance from A in the garden if possible,with a piece of string, or, failing that, on paper with stringor compasses, using 1 centimetre to represent 1 foot. Thesepoints make a curve that is called a circle, the apple-tree isat the centre of the circle

,and the distance of every point

of the circle from the apple-tree (7 feet) is called the radiu s

of the circle . The cache lies somewhere on this circle . Nowtake the next measurement , 12 feet from the plum-tree P

,

and mark all the points that satisfy this condition, that is,all the points that are 12 feet from the plum-tree. Whatkind of curve does this give What is its centre and its

radius ? The cache lies on this line also.

From these two measurements what do you know of the

position of the cache —It must be at one of the pointswhere these two curves cross, andwe could find it bydiggingat these points. But how could you avoid diggingat the wrong point —We could use another measure

ment Any other way —We could note that the

2 THE CACHE

cache lies north or east of the line joining the apple and

plum-trees, or note in some other way on which side of the

line it lies.

8. Suppose the trees A and P to be 9 feet apart, and thecache 1 1 feet from A and 7 feet from P. Mark the trees onpaper, 1 cm. representing 1 foot, and mark the two possiblepositions of the cache. W ith the same two trees, and thecache supposed to be 10 feet from A and 8 feet from P

,

mark its possible positions.Suppose that the boy noted the position by fastening one

end of a string at A , stretching it to C and making a knotthere and fastening the string there, and stretching it fromthere to P

,and cutting it of at P . How could he use this

string to find the cache ? Do it, taking the trees 9 feetapart and the two parts of the string 11 and 7 feet long.

How many points might he find if he forgot whether 1 1feet was the distance from A or from P ? Mark them all.

Suppose that when he tried to use the string he found theknot had slipped, what would he know of the position of

the cache Use the same lengths as before, and mark allthe positions he would get by supposing the knot at different points along the string. Use the string or yourcompasses, whichever you like.

4 . Are there any positions of the knot on the string thatwill not do f— Clearly 1 foot from an end will not

do. How does it fail Distinguish between the

positions that will do and the positions that won’ t. (Thepupils will find by experiment that the knot must be at

least feet from an end. )The two distances from the trees have been supposed to

up 18 feet, or in the drawing 18 cm . , and we havefound that the distance from each tree must be at least

45 feet, or in the drawing cm. Suppose this restrictionthat the two distances make up 18 feet or cm. removed,and find by experiment what relation there must be

between the two distances of the cache from the trees,the trees being 9 cm . apart on your drawing. Take 3cm .

as the distance from A, and take in succession 3, 4, 5,cm. as the distance from P. Then take 4cm. as the dis

tunes from A, and the same succession of distances from

CHAP. I, ARTS. 8-5 3

P ; and so on,till you see what the relation is, and till you

can prove it without any drawing at all — The sum of the

distances must be greater than 9 cm . and the difi'

erence lessthan 9 cm.

If you take any three points, and rule straight lines withyour straight-edge to join them, the figure you make iscalled a triang le, and the three lines you have ruled are

its sides. Make a triangle having sides 8 cm ., 10 cm ., and

13cm . long, another with sides 8, 10, and 17cm . long,and, if you can, four more with sides (1) 8, 10, 8 cm . ; (2)8,10, 19 cm (3) 8, 10, 7cm . ; (4) 8, 10, l cm . Try

making other triangles with different sides till you can give

a rule to distinguish the cases in which the triangle can

be made ; and justify your rule. The rule is importantenough to be called a proposition, and may be statedthus — Any two sides of a tr iangle are togethergreate r than the th ird .

5. Points of the Canvass. The terms north and easthave been used. How do you know which direction isnorth and which east Point to the south. How do you

know — It is the direction in which you must walk to gotowards the sun at midday. Where does the sun

rise and where set — Roughly, it rises in the east andsets in the west. If you face south and thenturn in the opposite direction, which way are you

facing — North. Does any one know the PoleStar and how it is used to fix directions (The Pole Staris not necessary, and need not be followed up if no pupilknows it. )If you face south and then turn halfway round to the

north,whichwayare you facing — East or west. In

turning halfway round to the north you have turnedthrough an angle called a r ight angle . Point out anyright angles you see about the room, the tables, desks, &c.

Through how many right angles do you turn in turningfrom south to north And in turning from south back tosouth again What fraction is a right angle of the wholeangle you turn through in turning from south back to

south again What fraction is a right angle of the angleyou turn through in turning round from any position tillyou again face in the same dire

2

ctionB

4 THE CACHE

Fold a sheet of paper, and call two points on the crease

A and B. Open the shoot out and fold it again so that Aand B fall together. Open it out again. Howmany anglesdo the creases make —Four. And are they all

equal — Yes, because they fit together. Then whatangles do they make with one another -Right angles.

Take two creases that form a right angle and fold themon one another. What angle does the new crease make withthe former two -Half a right angle. By foldingmake an angle a quarter of a right angle.

Nate for the Teacher. If any pupil raises the questionwhether the size of an angle depends on the length of the

arms, the question must be discussed. But the questionshould be left alone till a pupil raises it. It is a generalprinciple not to point out logical difiiculties to a pupil. Itis a waste of time to explain a dificulty that the pupil doesnot feel.6. Halves and quarters of a right angle have been men

tioned. In shopping we avoid fractions of a shilling byusing a smaller unit, the penny. In measuring we avoidfractions of a metre by using a smaller unit, the centimetre .

So with angles it is convenient to have a smaller unit thanthe right angle. The smaller unit is called a degree, and

FIG. 1 .

90 of them make a right angle. Look at your protractors;they are marked or graduated in degrees. Draw half a

dozen angles at random, and measure them in degrees.

How many degrees are there in the first two anglestogether Make an angle containing this number of de

grees. Make another angle as big as the first three anglesthat you made.


In turning from north to east, through howmany degreesdo you turn And from north to south And from northto north again And in turning from north halfway toeastAn angle may be indicated by a single letter or by three

letters, as A, w, pqr, in Fig. 1 . A straight line may be indicated by a single letter or by two, as 8, BC. Care mustbe taken that the sign used singles out one angle or line.

7. The Circle. Draw a circle and fold the paper so thatthe crease cuts the circle . Open it out again. Call thetwo points of cutting C and D, and call the centre of the

circle A . Draw the radi i AC and AD,and fold again so

that these radii lie along one another. How do 0 and Dfall Open out again and call the point where the creases

cross 0 . What are the angles at 0 — Right angles, forthey all fit when folded together. What do you

know of the lengths COand OD —They are equal becausethey fit together.

8 . The straight line joining two points on a circle iscalled s ehord. Draw any circleand lay in it a chord CD (Fig.

Bisect the chord at Oby folding 0 on

D . Join 0 to A,the centre of the

circle, and measure the angles AOGand AOD. Repeat with several otherchords.

Again draw a circle and lay in ita chord CD . Draw a line bisectingthe chord at right angles by foldingC on D and creasing the paper. How FIG . 2.

does the centre A lie with respectto this perpendicular bisector (When two lines meet atright angles they are said to be perpendicular to one

another. ) Repeat with several other chords.

9 . Can you tell from your drawing whether the centre Alies within a millimetre of the perpendicular bisector of thechord Within 01 of a mm. Within 0 01mm. And in

the former case, when the centre A was joined to the middlepoint 0 of the chord CD, can you tell whether the angleCOA difiers from a right angle by 1 degree By O

' l of a

6 THE CACHE

degree There is a limit to the accuracy of a conclusionmade from drawing. And can you predict whether thesame result would be true for additional chords if you drewthem —Not with certainty, though the result seemslikely. If wewant an accurate conclusion, and oneweknow to be true for all cases, we need some other methodthan this experimental one. Have we in this case anothermethod Return and compare the discussion in Art. 7.

In that therewas no measuring, nor did anything depend onthe accuracy of our folding or on the particular circle or

radii chosen and we may rely on the general truth of the

conclusion. Thus we know that— the straight l ine j o ining the centre of a circle to the m idd le point of a

chord is perpendicu lar to the chord ; and the per

pendicular b isector of a chord passes throu ghthe cen tre of the circle .

10 . The Isoseetes Triangle. Draw any triangle ACD,

having the sides AC and AD equal. A triangle with twosides equal is called an isosceles triangle, and the thirdside is called the base . Measure the angles at 0 and D.

Repeat with three or four other isosceles triangles. Whatdo you observe about the angles 0 and D in each case

Again, draw any straight line CD, and with your protractor draw lines from 0 and D making the same anglewith 0D and meeting in A , and so forming a triangle.

Measure the lines AOand AD. Repeat with three or four

other figures. What experimental result have you reached11 . If in Art. 7we began without a circle, but with only

two equal lines AC and AD, would there be any differencein our conclusions What do you know of the triangleACD made by joining C and D ? —It has two sides

equal. Consider the triangle when AC and AD are

folded together. What can you tell about it, and what ofthe angles at 0 and D — They are equal because theyfit together. State this as a proposition— Th eangles at the base of an isosce les tr iangle are equ al .Now begin with a straight line C’D (Fig. and draw

lines CE and DF from C and D on the same side of CDand making equal angles with it. Fold 0 on D. Howdo CE and DF fall —Together, since the angles are


equal. If,then, these lines are produced till they

meet the crease, what hap ne —They meet the crease init G. Then we have a triangle

FIG . 3.

GC’D in which we made the angles 0’and D equal . What

do we know of the sides —GC is equal to GD becausethey fit. This equality may be written

the sign being a short way of writing is equal toState your result as a proposition.

— If a tr iangle hastwo of its angles equal , the two sides Opposite tothese angles are equal .12. Let us return to the trees A and P , and the two

circles W i th these as centres, at one of whose intersectionsthe cache must lie. Call these intersections C and D

(Fig. Draw the

figure and fold C on

D , so as to get the

crease that bisectsCD at right angles.

What do we knowof this crease in con

nexionwiththecirclewhose centre is AAnd in connexion

with the circlewhosecentre is P ? —We

knowthat the crease FIG 4°

passesthrough A and

through P . So that the perpendicu lar bisector of a

chord that is comm on to two circles passes th roughthe centres of th e circles .

13. If we had found the middle point 0 of CD and joined

THE CACHE

it to A and P — Then AO is perpendicular to CD (Art.and so lies along the crease made by folding C on D . SO

does P0, so that AO and OF lie along the same crease

and make a straight line,the straight line that joins

A and P . So that the st raight l ine j oin ing the

centres of two c ircles b isects the comm on chordo f the two circles at r igh t angles .

*

When you see anything in a mirror the reflection or

picture or image seems to be behind the mirror rightopposite to the original and at the same distance from the

mirror. Now In Fig. 4 C and D are on opposite sides of

the line AP, right opposite to one another, and at the samedistance from it ; so we call 0 the image of D, or D the

image of C. And generally, if two points lie so that theline joining them is bisected perpendicularly by anotherline, they are called the images of one another in that line.

Now draw one circle on ordinary paper and another on

tracing paper. Put the tracing paper on the top of the

other and move it about so that one circle is sometimes inside

the other, sometimes outside, and sometimes they overlap.

How does the line of centres lie with respect to the common chord In every position And

'

In a position when thecircles are just ceasing to overlap, how does the common

chord lie — It has become shortened to nothing,and the

two points of intersection of the circles have become

one. And how does this single point lie withrespect to the centres of the circles — It is in line withthem, since the middle point of the chord is always in linewith them . So we have found that when two circlesmeet in only one point (or, in other words

,when they

tou ch one another) this one point (the point of con tact)is in line with the two centres.14. Suppose, now, that Fig. 2 (p. 5) is drawn in copying ink

and folded along the chord CD so that every point of thefigure prints itself ofi'

on the other side of CD ; or what isthe same thing, suppose the image in CD of every point ofthe figure drawn. What kind of figure have we then? -Two

On the Euclidian plan reference would be made here to theaxiom that two straight lines cannot enclose a space , but suchrefinements are out of place with beginners.

https://www.forgottenbooks.com/join

10 THE CACHE

distances from A and B,in the garden if possible. If the

garden Is impossible, take B on the wall of the room, or on

a desk, and A right below it on the floor. What do youknow of the position of 0 from the measured distancefrom A (say 7 feet) ? Mark all the points 7 feet from A .

These points altogether make what is called the locus of

points 7 feet from A. Then take the measured distancefrom B and mark all points on the floor at this distancefrom B, another locus . How do these two loci lie ?Assume another position for the cache and proceed as

before. Assume additional positions for the cache till the

pupils discover and experimentally satisfy themselves thatin each case the two loci are the same, so that the second ofthem gives no further information (Fig.

Taking B as before 5 feetabove A, assume 0 In succes

sion 1 , 2, 3, 10 feet from A ,

measure the distance GB in

each case, and make a tableshowing (1) the length of AOin feet for each position,(2) the length of BC corre

sponding, and (3) the numberof feet in AO divided by thenumber of feet in BC given

FIG“ 7°

as a decimal to two places.

17. Note. The work should throughout be done by thepupil, not by the teacher. In drawing and measuring thepupil cannot avoid doing the work ; in the matter of con

clusions from the drawing and measuring the teacher istempted to tell the conclusion instead of leading the pupils,by suitable questions, to discover the conclusion. The

general reasoning should also be that of the pupils witha minimum of leading bythe teacher. A necessary preliminary to the drawing of conclusions and to general reasoningis a certain amount of concrete mathematical experience.

If the pupils are slow at reasoning itwill be well to increasethe amount of drawing and measuring, and to continue thecourse chiefiy experimentally, and then to return to reason

out the results with the help of the concrete experience thus

CHAP. I,ARTS . 17, 18 11

gained. To repeat the words of another’

s reasoning is not

to reason.

18 . Ewnmle. A recluse lives in hiding, and communicates

with his friends through two post offices, A and B, of whichB lies 14miles north of A . The hiding-place H is supposedto be at the same distance from these two post ofices.

Make a map on a scale of 1 cm .

to 1 mile, showing A and B ; and

supposing H to be 8 miles fromeach of these points, mark whereit might be (Fig. Do the same,supposing H 9 miles from eachpoint, then 10 miles and so on .

By taking a great number of

pointswe get a line, and we knowthat if H is to be equidistant fromA and B it must lie on this line .

What sort of line is it and how FIG. 3.

does it lie ? If the pupils do

not yet know,take different positions for A and B,

andagain find the line on which H would lie, until the position of the line is made obvious.

Now consider H to be at any one of these positions and

join it to 0 , the middle point of AB . What do you know of

H0 Can you prove it by folding -Fold HA and HBtogether, and the crease made lies along H0 while A and B

fall together ; as we saw before. So that H0 lies along the

crease made by folding A on B ,that is, H0 is the perpen

dicular bisector of AB. And this is true no matter whichposition we take for H ; so that H always lies on thisperpendicular bisector. Can you state the result inmathematical form ; what are the conditions and what theconclusion ?— The locu s of poin ts equ idistan t fromtwo ' fixed points is the perpendicu lar b isectorof the l ine jo ining the points .

It is also known that the recluse returns home from A bytravelling north-north-west. NOW the direction midwaybetween north and west is called north-west (Fig. and

the direction midway betweennorth and north-west is callednorth-north-west. Find these two directions on your map

12 THE CACHE

by folding. On what line do we know that the hiding-placeH lies from the fact that he travels NNW. to get home

Rum.

W-N.W.

E.N.E.

FIG. 9.

Lastly,from the two conditions

,mark on your map the

spot at which the recluse’

s friends should look for him .

1 9 . In the preceding article itwas found that the locus ofpoints equidistant from two given points A and B is the

perpendicular bisector of the line AB . How could you

draw this perpendicular bisector — By folding A on

Could you do it without folding —Yes, byfinding enough points on the perpendicular bisector and

joining them. How would you find a point on

it -It has to be at the same distance from A and B, so

we open our compasses to any convenient radius and drawtwo circles with centres at A and B . The point H, wherethey cut one another, is a point on the perpendicularbisector. How many points do you need — Two,because two are enough to settle the position of a line

CHAP. I, ARTS. 19,20 13

drawn through them .

point K -In the same way as H take thesecond intersection of the firstpair of circles. Then join HEand we have the perpendicularbisector of AB (Fig.

If you had a line AB and

wanted to bisect it, that is, tofind its middle point, whatwould you do —Draw the per

pendicular bisector ; the point0 where it cuts AB is the

middle point. Or we couldmeasure AB ; suppose itcm. from this find half thelength, namely cm . , andmeasure off this distance fromone end of the line.

20 . Could you draw a perpendicular to froma point E that does not lie on the line —We might foldthe paper so that the crease passes through E , while one

part of the line 0D falls along another part. Then all thefour angles where the crease crosses CD will fit if folded onone another, so that they are right angles and the crease

is the perpendicular we want.Could you draw the perpendicular without folding

Could youfindon CD two points, P and Q, such that E would

FIG. 11 .

lie on the perpendicular bisector of P0 —With E as

centre, draw any circle that cuts CD in two points, and call

14 THE CACHE

these P and Q. Then E is equidistant from P and

Q so that the perpendicular bisector of PQ passes throughE (Fig. How many points on this bisector doyou need in order to draw it — One in addition to E .

Draw two equal circles, with centres at P and Q ,and call

one of the points where they cut one another H. Then EHis the perpendicular bisector of PQ, that is, it is the perpondicular from E on CD.

Howwould you draw at E a perpendicular to CD when

FIG. 12.

E lies on CD W ill the same method do —Yes, find

P and Q equidistant from E , then find H not on CD and

equidi

l

st

)ant from P and Q ; and EH is the perpendicular

(Fig. 2

How would you use your protractor to draw a perpen

dicular to CD at a point E in it Do so. Some protractorscan be used to draw the perpendicular when E is not on

CD. Can yours be so used

21 . Converse Propositions. We have found properties of

circles so related that a condition and the conclusion of one

become respectively the conclusion and a condition of the

other. Thus we had the trIo The perpendicularbisector of a chord of a circle goes through the centre.

(2) The line joining the middle point of the chord to the

centre is at right angles to the chord . (3) The perpendicularfrom the centre on the chord bisects it.Any one of these is said to be the converse of any other.

Another pair of converse propositions we had was (1) Theline joining the centres of two circles bisects their common

chord at right angles. (2) The perpendicular bisector of

CHAP. I, EXERCISES 15

the common chord of two circles passes through the centresof the circles.

The following are statements related in the same way.

Discuss which of them are true and which false. Drawa circle of radius 5cm . and centre A and take a point Bsomewhere inside the circle (Fig. Is it true that everypoint more than 5cm . from A liesoutside the circle ; and that everypoint outside this circle is more than5cm. from A Is it true that everypoint more than 5cm . from B liesoutside the circle ; and that everypoint outside the circle is more than50m. from B ? Is it true thatevery point more than 10 cm . fromA is outside the circle ; and thatevery point outside the circle is more

F 13than 10 cm. from A Is it true m’

that every point more than 10 cm.

from B is outside the circle and that every point outside thecircle is more than 10 cm. from B

Does it appear whether, from the truth or falsity of a

statement, you can tell whether a converse of the statementis true or false

Exmm sns.

1 . The positions of four points are specified thus : from 0 to Ais miles north and east ; from A to B is north andwest ; frot O O is 2 north end west ; and from 0 to D is

south and east . Show the points OAB OD on the scaleof 1 inch to one mile .

2. In Fig. 14 the thick line AB represents a level railwaywhich is carried partly through cuttings and partly on an embank

16 THE CACHE

ment. The irregular line represents the section of the ground. If

the scale Of horizontal distances is 1 inch to a mile, and the scale

of vertical distances is 1 inch to 100 feet, draw up a table showingthe height of the ground above the level of the railway at intervalsof half a mile starting from A, and indicatin points where theground is below the level of the railway by prot

einng the sign

3. Prick ofi‘

Fig. 15; in which A and B are two forts. A bodyof soldiers at Y are just out of range of each fort and want to

get as near X as possible without going any nearer to either fort .

how the point theywould go to, and draw the shortest path bywhich they could go . Drawthe straight of the path bylayingyour straight-edge against the circle and ough their destmation.

v9

FIG. 15.

4. AB is a tube which revolves uniformly on about its

middle point 0 , that is, turning through the same angle everysecond. If a marble moves from one end of the tube to the other

speed (that is, travelling the same distance everysecond) , whilst the tube makes a complete revolution, draw the

path of the marble . Represent the tube by a line 5inches long.

5. From a corner of a sheet of paper mark of 20 cm. along one

edge to A and cm. alo the other to B . Fold over the corner

along the line AB, and mar on the page the int 0 where thecorner falls. Open out the fold again and join to the cornerb a

str

iiilIt line . Prove shortly that this straight line is perpendic

to

18 POSITION OF CHAIR

four angles at either intersection all fit together.

0

Now

fold the crease B on itself in several places, maki ng new

across B and 0. Measure and tabulatethe lengths of theselast creases that liebetween B and C’.

with B How do

you know Measureand tabulate the an

make with C’. WhatFIG. 1 .

ments give —That the crossesD, E , cut B and 0 prettyexactly at right angles

,and that the distances between

B and 0’measured along them are pretty exactly the same.

The creases B and 0’are said to be paral le l. To be

made mathematically parallel in this way they would need

to have the angles between them and the creases D,E ,

made exactly right and the distances apart along all the

creases D, E , made exactly equal.Repeat this experiment by drawing, that is, draw any

line A, draw two others B and 0 at right angles to it, andat various points on B draw perpendiculars to meet 0.

Measure the lines and angles as before.

Again draw any straight line, and at a number of pointson it draw perpendiculars, all of the same length and all on

the same side of the line. Join their ends. What is theresult -The ends all lie on a straightMention any cases from everyday life of parallel lines.

The edges of boards of the floor ; the two Side pieces of

a ladder the two rails of a straight railway track ; thecourses of brick in a wall.NOTE . The existence of parallel lines is an experimental

fact. All systems of demonstrative geometry assume it in

some form , though often not explicitly.8. To return to our problem : it is now known that the

chair-leg lies on a parallel to the north wall feet from it.

CHAP. II, ARTS. 3, 4 19

How would you draw this parallel —We have had one

method, by finding a number of points on it ; a simplerwa is to measure feet along the neighbouring walls andto join the points thus found.

Is this enough to fix the leg—We need the distance

from a second wall. Is the leg then fixed Will anysecond wall do — It must be the east or the west wall, andthe leg is then fixed, for this gives a second locus meetingthe former locus in one point. The south wall would notdo, for the locus it would give is the locus we have already.

So that the position of the leg is fixed by its distances fromtwo walls that meet.

4. Having fixed one leg, say the right-front-leg, how willfix the chair — By fixing another leg, say the left-front

How — In the same way. Is the chairnow fixed — Yes, for you can place it in position from

these data.

How many measurements have youmade to fix the chairCould you have done with fewer When you have fixedone leg do you know anything about the position of the

second leg— If the first leg is set down on the proper Spot

P (Fig. the second liessomewhere on a circlewith this spot P as centreand with the distance between the legs as radius ;so that we need only one

more measurement. If

we know the distance of

the second leg from the FIG . 2.

north wall, its positionmust be Q or R, one of the points where the two loci meet.And if we know also that the second leg is nearer than thefirst to the east wall, its position is fixed without ambiguity.

What information then do we need to fix the chairThreemeasurements, in addition to a means of distinguishingbetween Q and R .

Ex. 1 . You are given that the right-front-leg is feet from thenorth wall and feet from the east, the other front leg is feetfrom the north wall, and the distance between the legs is 1

°5feet .

0 2


Suppose 1 cm. represents 1 foot and a e of your book repre

sents the floor ; and mark the position 0 the chair.

If you had been given that the left-front-legwas68 feet from thenorth wall, could on have marked its position ? Why not ? Whatcondition must ho d among themeasurements ?

5. When you have the chair in position can you move it

at all without shifting the two logs from their positions onthe floor — We may tilt the chair. Then statemore exactly what you have found— P rovided the chairstands on the floor the three measurements and the

m eans of d istingu i sh ing between Q and R se tt lethe position of the chair .

What measurements would be necessary if the chair isnot to be compelled to stand on the floor Think this over,and we will return later to the discussion of such a problem.

6. Is any other measurement possible for the second leg,instead of the distance from a wall, when you have the firstleg fixed The angle the line joining the two

FIG . 3.

with a wall or with a parallel to a wall would do. Andvarious other suggestions may be made.From the sketch given (Fig. 3) draw the chair in position

on the same scale as before.

7. A boy oncemade a cache in the corner of a rectangularlawn, and to keep a note of the position he made the

CHAP. II, ARTS. 5—8 21

drawing in Fig. 4. But, as boys and even men sometimeswill, he forgot to make a note of the scale of his drawing,whether an inch represented a foot, or every distance was115 of the true distance, or whatever else it might be. Whatdoes his drawing tell himof the position of the cache ? North odes cf lawn

distance of the cache fromthe two edges of the lawn.

He knows only that thecache is at the same distancefrom the two edges. Drawthe lawn to some definitescale

,this time marking the

scale,and draw the locus of

all points where the cachemight lie. How do you

mark the point that is 8 cm .

(say) from each edge -By

along the north edge to A, and there drawing a per

pendicular AB 8 cm. long. Or by measuring along theeast edge to C and drawing 0B perpendicular and 8 cm.

long. Or by measuring 8 cm. along to A and to 0, and

drawing perpendiculars long enough to meet.When the last method is used

,do we know that the

point B where the perpendiculars most is 8 cm. from eachedge -Yes, from our experimental result we know thatif 0A and CB are both perpendicular to 00, their distanceapart is the same all along as between 0 and C, that is, it is8 cm. If now the two edges OK and 0Y are foldedtogether, how do A and 0 fall And how do the per

pendiculars at A and 0 fall Then how does the point Bwhere they meet lie —On the crease.

8 . Does this result depend on the particular length 8 cm .

that we used —No, whatever the length B would lie on

the crease. What then is the locus of possible positions of the cache — It is the crease. Compare theangles the crease makes with OK and 0I by measuring.

They are equal as far as measurement can tell. Can

you give any other reason —They are equal because they


fit when the paper is folded. So that the locus of

possible positions for the cache is the bisector of the anglebetween the edges of the lawn.

State the geometrical property without reference to lawnsor caches— If two straight l ines OK and OY are at

r ight angles th e locu s of all po in ts equ idistantfrom them is the b ise ctor of the angl e XOY.

9 . Now produce X0 , Y0 , and B0 so that we have KOZand YOW crossing at right angles, while BOQ bisects theangle XOY (Fig. What can we say of this figure ?Are points on 0Q equidistant from 0W and OZ, and does

0Q bisect the angle ZOW ?

And do all points equidistantfrom XZ and YW lie on BQ ?Measure all the angles at 0

with your protractor and writetheir values on the figure.

When you stand facing northand turn about to face south,through how many degrees do

you turn ? And if from anyposition you turn about to facethe opposite way Does it

matter whether you turn toFIG . 5. the right or to the left Sup

pose that in Fig. 5 you standat 0 facing along OK and then turn about to face the Opposite way ; Write down the angles that make up the 180degrees you have turned through.

— They are X0W, WOQ,and QOZ,

if we turn to the right. Do this for everyline of the figure.

You found the sizes of all these angles bymeasuring. Butthe value of your results depends on the accuracy of yourdrawing and of your measuring

,which can never be very

great. Can you by reasoning tell what results you wouldget if you could draw and measure the figure with perfectaccuracy What would the angle X0Y be on a perfectfigure And how did you get the line 0B —By foldingOK on What, then, do you know of the anglesX0B and YOB on a perfect figure

— They are equal, and

CHAP. II, ARTS. 9, 10 23

together they make a right angle or 90 degrees ; so thateach is 45degrees.Can you tell the size of the angle XOW ? — It is 90

degrees, for the turn from OY to OX is 90 degrees and theturn from OY to OW is 180 degrees.

And what do you know of the sum of the angles BOX,

XOVV, and WOQ -It is 180 degrees. And of

these three angles you know two ; what, then,is the

WOQ —It must be 45 degrees to make the sum180 degrees.

10 . To write this down in the ordinary way takes a longtime, and some contractions or symbols are used to shortenthe writing. Thus

angle is denoteddegre eequa ls or is equa l totogether w ith or plusless or m inus

perpendicu larpara ll e ltr ianglether efore

and by the help of these symbols our reasoning may be

LXOY= 90°

LXOB LBOY= 90°

and LXOB LEOF

LXOB LBOY 45°

LYOX LXOIV »= 180°

and LYOX 90°

LXOW 180° —90° 90

°

and so on. But these contractions should be sparinglyused at first as there is danger of concentrating the attentionon the contractions instead of on the reasoning. The value

of symbols is best realized if the pupils are allowed to do theirwriting in longhand till they begin to realize how tediousit is ; they may then be encouraged to invent symbols forthemselves and to use their own inventions for a time.


11 . Now find the size of each angle of the figure bysimilar reasoning, and compare these results with the

measured results. Can you now answer any of the questions at the beginning of article 9 Are all points on OQequidistant from OZ and OW —Yes ; OQ bisects the angleZOW and folding along OQ brings OZ and OW together,so that if any length is measured along OZ and OW, and

rpendiculars drawn, they meet on OQ and have the same

ength just as we found already for OB, OX and OY. SO

that we now know that all points on BOQ are equidistantfrom KOZ and YOW.

The question whether OQ bisects LWOZ is alreadyanswered.

DO all points equidistant from XOZ and YOW lie on the

line BOQ Find a number of such points by drawing, andthen answer the question.

— No, all such points do not lie

on BOQ ; some lie on the line that bisects the angles XOWand YOZ. Here, then, we have two converse pro

positions. Are they both true, or both untrue, or whatCan you sum up what we have found in a single proposi

tion about a locus of points A little discussion shouldlead to the statement that : The locus of poin tse qu idistan t from two lines in tersec t ing at r igh tangles is the two l ines that b isect the fou r anglesmade by the intersect ing lines .

Ex. 2. Use the result of Article 8 to bisect a right angle by thehelp of ruler and compasses.

12. Suppose now that the lawn in Art . 7 is not rootangular, and that as before you know the cache is equidistantfrom two edges, the distances being measured perpendicularto the edges. Go over the work of Arts. 7—1 1, altering itto suit the altered condition.

13. Draw two straight lines AOB and COD crossing at O,and measure all four angles. The angles AGO and BOD

are called vert ical ly opposi te angles. The angles AODand B00 also are called vertically opposite. You have metwith several such pairs of angles. Can you state and

prove any proposition about a pair of vertically oppositeangles ? — The ver tically opposite angl es made bytwo in te rsecting lines are equa l .


the ends of the other move along the circumferences of the twocircles one end on each.

11 . A point P is known to be at least cm . from a point A ofa straight line shade the area in which P may not lie . P is alsoto be at least cm . from another point B of the line shade theadditionalprohibited area . Take more points 0, D,

on the line,and suppose P to be at least cm . from each, and shade theprohibited area. Finally, find experimentally and approximatelythe form of the prohibited area if P is to be at least cm. fromevery point of the line .

CHAPTER III

AREAS

1 . THE determination of areas is a problem of greatantiquity. Many races have passed through a stage of

development in which the land of a community was

periodically divided into a number of equal parcels ; and

the name Geom e try itself means land-surveying.

Do you know any units in terms of which areas are

measured —Acre, hectare, square foot, square decimetre,square inch, square centimetre

,and others. Draw as

many of these units as your paper is big enough to hold,and cut them out to use for measuring. Take a sheet ofpaper of any form, and see how many square inches you can

mark out on it. Does this tell the area — Yes, except forthe i r

regular pieces left at the edges. Are there nopieces left over except at the edges

—We can mark out

the inch-squares so that there will be no pieces left over

except at the edges. Could you measure the playground in the same way with a square foot as unit — Yes,but it would be somewhat laborious. Let us see if

we cannot find easier ways.

2. In measuring the sheet of paper, do you need to laythe unit-area on the sheet every time to mark out the unitsthere — We could simmy rule the sheet into squares, orwe could place it on squared paper and trace its boundaryon to the squared paper. Is there any choice as to

where you mark out the first square —We could use anystraight side of the sheet as the side of squares, and if thesheet was rectangular one square Should be placed in a

corner of the sheet.

The sheet is rectangular when its four angles are all rightangles.

28 AREAS

Take a rectangular sheet and rule it out as far as you can

into inch squares. Cut off the strips left over, and state therelation between the area

,the length, and the breadth of

the rectangle you now have .— The area in square inches

is the length in inches multiplied by the breadth in

inches. Does this result depend on the particularlength and breadth, or would it be true for others And

if the length is 1 inches and the breadth b inches, what isthe area —It is Ixb square inches, provided I and b are

whole numbers.

3. Let us now replace the strips that were cut away fromthe rectangle and try to find the area of the original sheetof paper. No inch-squares can be marked out on thesestrips, and we must use a smaller unit. Let us take as

smaller unit a strip such that ten of them placed side byside form an inch square. Mark out this strip-unit as oftenas you can on the strips that were cut away. If any area

still remains that has not been included take a still smallerunit, say a little square

,ten of which will make a strip

Thus, for instance, if the sheet of paper measures 127inches by inches, the area consists of

12x 8 inch squares,12x 2 8 x 7 strip-units,7x 2 of the smaller squares.

How many strip-units are there — 80.

were given the expression

12x2+ 8 x7

without knowing how it was arrived at, would its meaningbe clear — No

,unless there is an agreement about the

order in which the addition and multiplication are to beperformed ; if we began by adding the 2 and the 8, the

expression would mean 840. The expression then maymean the sum of the two products 12x2 and 8 x7, or itmay mean the product of the three quantities 12, 2 8, and7. Brackets may be used to avoid ambiguity ; thus

CHAP. III, ARTS. 3-5 29

means the sum of the two products 12x2 and 8 x7, while12x (2 8) x7

means the product of the three numbers 12, 2 8, and 7.

There can be no doubt about the meaning of either of these,but we can save ourselves some labour in writing if weagree to confine the brackets to one of the two cases. Let

us agree that in the absence of brackets or other indication mu l t ipl icat ion and division are to p re cedeaddi tion and subtraction. Then when we meet12x2+ 8 x7we know that we must first multiply 12 by 2and 8 by 7 and then add the results. So that with thisagreement the expression as it originally stood means 80

without any ambiguity, and if we want to make it mean840 we must use a bracket. 7

4. Now express in square inches and fractions the area ofthe sheet of paper.

—It is

l2 x 8 (12x 2 8 x 7) x 116 7x 2x r3-osquare inches. (Note the use of the bracket ; whatwould this expressionwith the bracket left outmean Thisexpression, which is now recognizable as 127x has the

value For most purposes it is sufiicient to give thevalue as 104, and if the length and breadth of the sheet areonly approximate , being measured to the nearest tenth of

an inch, the form has a misleading appearance of

accuracy, and even 104 is not reliable in the unit figure.

If you adopted as unit from the beginning the smallsquare of an inch in the side , how many of these unitswould you have in the rectangle And what is theirequivalent in square inchesEx. 1 . Make a loo of string, lay it on uared paper, and find

the area it encloses y counting squares. ove the loop aboutand pin it out, to make it enclose as t an area as possible.

What does the shape of the loop then 100 like ?

5. Now generalize the formula already obtained for thearea of a rectangle in terms of the length and breadthThe length being 1 inches and the breadth b inches the areais lxb square inches, where t and bmay contain fractions.Are these fractions limited to be tenths Or to be

80 AREAS

decimals And what is the formula if the dimensions are

given in centimetres6. What is the area of a rectangle measuring a inches

and b tenths by 0 inches and d tenths —It consists ofa x c inch squares,a x d b x c strip-units,b x d smaller squares,

a x e a x d x lla b x e x l

lo b x d x rh

square inches.This form is clumsy, and a shorter way of writing it is

desirable. One a little better isd b d

a x e

a x

130 ,

but even this may be improved on. Would it do to dropthe sign of multiplication andwrite ac to mean a multipliedby c — It would not do for numerals ; we cannot dropthe Sign in 12x8, for it would then be 128

,that is, one

hundred and twenty-eight. Maywe drop it except between numerals —Yes. A system is a little illogical that adopts difi'

erent conventions for letters and numerals,and this difierence of convention may cause dificulty.

However, these conventions are universal and very useful,and pupils that realize the difi‘

erence, and realize that bothconventions are no more than conventions, will find no

difficulty. We agree, then, that the Sign x may be droppedwhen no ambiguity results from dropping it, and we writethe number of square inches in the rectangle as

00 +

Area of a right-angled triangle.

7. Suppose that the playground that we want to measureis shown in Fig. 1, on a scale of 1 to 1000. We saw one

way of measuring its area, bymarking out a foot square as

CHAP . III, ARTS. 6-8 31

Often as possible . Can we now shorten this work -We

could mark out a few big rectangles on the playground,measure their length and

breadth, and so find theirBut some pieces

shallwe dowith these? Whatis their shape —They are

triangles, two of them righ tangled (i. 9 . having one anglea right angle). So thatto find an expression for thearea of a triangle would beusefu

Take a rectangular sheet of paper and cut it along a

diagonal , that is, along the line joining two oppositecom ers. What kind of figures have you now -Two rightangled triangles. Which of them is greater -Fit

ting together shows them equal experimentally, that Is, equalas far as observation will tell us. We will discusslaterwhether two triangles obtained by a cut accurately alonga diagonal of an accurate rectangle will fit accurately, thatis, whether the triangles are mathematically equal. Whatdo you know of the area of each triangle —It Is half thatof the rectangle.

Now draw any right-angled triangle . Is there any rectangle of which this triangle Is half Draw this rectangle.

Express the area of the rectangle In terms of the sides of the

triangle.-If the two sides of the triangle that meet at the

right angle aremp cm. and g cm . long, the area of the rectangle

is p xqsq. c And

the area of the

m

triangle itself? 9

-It isp xq-z-2 sq. cm.

8 . Take again one of the

triangles into which you cut

the sheet of paper. Call thecorner at the right angle Aand the othersB and FIG . 2,

Fold B on toA and fold C on

to A, forming creases DE and RE. Measure the sides and

angles of the figure ADEE. What Is the figureADEE

32 AREAS

As far as observation tells it is a rectangle. And

mathematicallywhat do you know of the figure ADEFThat the folding makes the angles at D and F right angles,and makes AD half of AB and AF half of AC, while A wassupposed a right angle from the beginning. Later wewill discuss whether this is enough to show ADEE a rectangle. Meantime, the sides AB and AC being 1) and g cm.

long, what is the area of the rectangle ADEF ?— It is

gxgsq. cm. i pqsq mm.

When the corner B of the triangle ABC is folded in onA,

is the piece of paper still a triangle —NO. y— Because the three boundaries are no longer straight,and a triangle is bounded by three straight lines.When B and C are folded in on A, how does the piece of

paper lie —By observation it is made up of the rectangleADEF and the folded-in parts which just cover the rootangle. Whether this is so mathematically we shallsee later. Then what is the area of the triangle -It is

twice the rectangle, or in sq. cm . Thus our previousresult is confirmed.

9 . Howmuch of the playground in Fig. 1 arewe now in a

position to measure — All but the triangle Q. Can

you suggest a method of dealing with the triangle Q Can

you divide it into right angled triangles — Yes, by a perpendicular from a corner on the opposite side.

You are now in a position to find the area of the playground of Fig. 1. Make the necessary measurements of thefigure, and, remembering that every length on the figure is

00 01 of the true length, findits area in square feet or insquare metres.

If your own playground issimple enough to treat Similarly measure it up and findits area.

FIG . 3. 10 . Could you now findthe area of a playground

such as Fig. 3 shows? Could you divide it into tIiangles

34 AREAS

12. Again,take a rectangular sheet of paper (ABCD in

Fig. 4) and make two cuts, ED and EC, from a point inone side to the ends of

B the opposite side. Can

you fit the corners cut

OH on to the remainingtriangle CED Whatthen is the area of the

triangle —It is half theC rectangle

,that is, half

FIG . 4 . the length of the rect

angle multiplied by thebreadth ; the length and breadth and the area being ex

premed in corresponding units, e . g. inches and squareinches or centimetres and square centimetres. And

in terms of the parts of the triangle -The area is halfthe base CD multiplied by the height EF, as we found

You assume that EF is equal to AD or BC.

Howdoyou know — By the experimental result that parallellines are at the same distance apart all along.

How would you cut the rectangle to give an obtuseangled triangle Cuts along AF and AC woulddo. And how far (1098 the discussion difier forthis case

13. Again, take any paper triangle CDE ; Fig. 4 willserve. Fold the Side DC on itself so that the crease

passes through E ,as BB in Fig. 4. Straighten out, and

then fold the corners CDE down on to F. What figureresults

— It is a rectangle exactly covered by the foldedIn corners. What are the length and breadth of therectangle —They are half the base and height of thetrIangle . SO that as before the area of the triangle is halfthe product of the base and the height.Repeat this for the case in which the angle D is obtuse

,

notIcing that till the folding is done an extra piece of paperto carry the point F must be left.14. Once more draw any triangle, acute or obtuse

angled, and draw about it a rectangle of which it ishalf, and so proceed to the expression for the area of a

CHAP. III , ARTS. 12—16 35

In this discussion we have chosen any side of the triangleand called it the base, while we called its distance from the

opposite corner the he igh t of the triangle. And as a

triangle has three sides, the expression we have found givesthree ways of finding the area of a triangle. Draw a

tIiangle, and use the formula in the three ways,and com

pare your three results. Further,check your results by

drawing the same triangle on squared paper and countingSquares.

15. The daily newspapers give curves showing the heightof the barometer for the preceding day or the precedingweek. In these, distances measured from an initial pointalong a base line running parallel to the lines of print(running horizontal ly, we may say by an obvious metaphor) represent time elapsed ; and distances perpendicularto the base line, that is, parallel to the columns of print

,

or ve rtical ly, represent the height of the barometer. Thelength of the perpendicular from any point of the base

line to the curve gives the height of the barometer at the

moment denoted by that point of the base line . A secondcurve gives the height of the thermometer. To save

space it is usual to use the same diagram for the twocurves, and to Show only the upper part of the diagram ,

thus omitting the bass lines.

Such curves can be more quickly read than a table of

numbers giving the same information. They are calledgraphs . The scales used are of course Shown on the

figure ; without them the curves would lose much of theirmeaning.

18 . Take a sheet of paper 20 cm. in breadth, and placea ruler across it at one end. Slide the ruler along, keepingit all the time parallel to the end of the Sheet, till it hasmoved 20 cm . along the paper . Find and Show in a tablethe area of the paper that the ruler uncovers by moving2cm. ,

4cm. ,6cm .

, 18 cm . , 20 cm.

Distance ruler has moved, in cm . 2 4 6

Area uncovered, in sq. cm 40 80 120 160

36 AREAS

Now mark a line OX 20 cm. long on squared paper (ifyou have no paper squared in centimetres, use any con

venient scale), and mark along it from O 2cm. , 4cm. ,

At the point marked 2cm. raise a perpendicular to OX to

represent on a scale of 1 cm. to 40 sq. cm. the area

uncovered by the ruler in moving 2cm . , that is, raise a

perpendicular of length 1 cm . At the point marked 4cm.

raise a perpendicular to represent on the same scale the areauncovered by the ruler in moving 4cm.

,and so on.

If you wanted to Show on your diagram the area

uncovered when the ruler had moved 1 cm . or 3cm.,

how would you proceed -Erect a perpendicular as be

fore . And for any other distance, for instance,cm .

Through the ends of all these perpendiculars draw a

smooth line. This line Shows graphically how theuncovered area varies as the ruler moves.

17. Again, suppose the ruler to be returned to its positionat the end of the Sheet of paper, while a second ruler is placedalong the Side of the sheet. Let the two rulers now move

away, each remaining parallel to its original position, insuch a way that the area uncovered is always a square,that is, has its angles right angles and its sides all equal.As before, Show in a table and in a graph the variation of

the uncovered area as the rulers move. From your graphread off the area of a square whose Side is cm.

,and the

length of the side of a square of area 178 sq. cm .

18 . Two rods 13cm . and 10 cm . long are hinged togetherat one pair of ends, and the other two ends are joined by an

elastic string. Find by drawing, and Show in a graph, howthe area of the triangle formed by the two rods and the

string varies as the angle between the rods increases, or,

in mathematical language, Show the area as a function of

the angle .

A lgebraic expressions.

19 . In finding the area of a rectangular piece of groundwe treated each dimension as made up of two parts, anintegral part and a fractional part. Let us now suppose the

CHAP. III , ARTS . 17—20 37

length made up of two parts a cm. long and 0cm . long, wherea and b may be fractional ;and the breadth of twoparts 0 cm . and d cm . long.

Sketch the rectangle, anddivide it into rectangleswhich have sides of dlengths a b c d cm . , as

in Fig. 5. Mark on eachrectangle its area in sq.

cm ., and so find an expression for the whole area.-It

Is In sq. cm .

a x e a >< d b x c b x d,

or, written in the shorter form,

ac ad be bd.

By considering the undivided rectangle you can giveanother expression for the area. It is the product of a b

and c+ d, which we agreed to write in the form

(a d),

(a b) (c d).

What then do we know of these two expressions —Thatthey are equal, Since each is the number of sq. cm in a

certain area.

20 . Is it true that whatever numbers a b c d are

Is it true only when these numbers are the lengths of lines,or is it true also when you buy a+ b loaves at c+d penceeach, or, as a particular case, 5white and 3 brown loaves at2; pence each -We are at liberty to buy the white loavesfirst and then the brown, and we are at liberty to pay down2 pence on every loaf and then the remaining halfpennywhich gives us as the price in pence

5x 2 5x ;i 3x 2 sq

in addition to the other form , 8 x21} pence . Or we couldrepresent the whole price by a figure, such as Fig. 5,representing a loaf by a centimetre across the page and a

38 AREAS

penny by a centimetre down the page the partitioned area

would then represent the four parts of the price . And the

same reasoning applies whatever numbers of loaves are

bought and whatever the price per loaf.May you then conclude that

ac + ad + bc + bd

for all numbers, whatever may be the concrete problem in

hand -It looks as if other cases could be represented inthe same way by Fig. 5.

It will be well for you to watch as to the applicability of

this formula, when you have occasion to use it. There are

kinds of numbers (negative numbers, complex numbers,vector quantities, some of which you may meet later,and which you have not yet considered in connexion withthis formula. Beware of applying to these without cautionresults you have found true for ordinary arithmeticalnumbers.

21 . In the sameway discuss the area of a square measur

ing a+ b each way. In future the units of length and area

will not always be explicitlymentioned it will be assumed

that they correspond,so that if the lengths are supposed

given in centimetres the unit of area will be the square

centimetre, and so on.

The result of the discussion may be written

(a b) x (a, b) a xa b x b a x b x 2,

(a b) (a b) aa bb abz,

(a t) (a b) aa bb 2ab,

and the same caution is necessary about the applicability of

this formula, and of all such formulas.

Ex . 5. In the same way discuss another expression for (a b c)x (a + b+ c) by considering a square measuring a + b+ c in the

side .

22. How would you calculate the price of 8 tons 7cwt. ofcoal at 193. a ton, and the price of 8 tons 18 cwt. at 193.

a ton ? Direct calculation is possible,but a little foresight

will often lighten the work .

In the first case the work is lightened by reckoning the

price at £1 a ton and then subtracting ls. a ton. In the

CHAP . III , ARTS . 21—23 39

second case we may in addition reckon first the price of

9 tons and then subtract the price of 2 cwt. Further,we may make formulas useful for such cases, in which weuse letters instead of numbers ; SO that any particular casemay be treated by giving the letters numerical valuesinstead of reasoning out the particular case for itself.As in Arts. 19 and 20 such formulas are closely con

nected with formulas about rectangular areas, whichnow be discussed.23. Draw four rectangles measuring

a b by c

a b by c

a b by a

a b by a

and express their areas in terms of rectangles having Sidesof lengths a b c d.

FIG . 6.

FIG . 7.

Figs. 6 and 7need little additional explanation. In eachthe shaded part is the original rectangle. In each expressthe areas a x e, a x d, bx c, bxd in terms of the rect

angles P Q B. S. By this means it is seenthat in Fig. 6

(a b) x (c d) = a x e b x c a x d

or, more Shortly,

— ad

and in Fig. 7

b) x (c b x d

(a

40 AREAS

Now draw figs. 6 and 7 over again, but making c

and d b. It results that

(a b) x (a b) = a x a b X b,

(a t) (a b) = aa bb,

and (a b) x (a b) = a x a b x b 2x a x b,

(a b) (a b) = aa bb 2ab.

Ex. 6. Calculate to three significant figures 997 x 1014 and

997x 997.

Is there any limitation on the values that a, b, c, d mayhave -a must be greater than b, and c greater than d, tomake the rectangles possible .

Note. The notation of the index is better not introducedat this stage, not until the need is felt of shortening thewriting ; and when that need is felt, the pupils may withgreat advantage be asked to invent contractions.

EXERCISES .

7. Draw any closed curve on squared paper, and find its area bycounting squares.

8 . A triangle ABC, right-angledat B, has side AB 6 inches 10and side BC 4 inches long. A

is divided into a number Of equal

parts, on which rectangles are

constructed as in Fig. 8 . Find

the sum of the areas of the

rectangles (i when 9 in number

(AB divide into 10 (ilgwhen 99 in number B divide

into 100 arts) .FIG . 3,

In eac case find the differ

ence between this sum and thearea of the triangle .

9 . Draw a straight line I Q, 9 inches long, and mark the positionof a point Z, 1 inches distant from the line and near the middleof the line . raw a straight line AB on tracing

-paper ; mark

a point N on this line 24} inches from A . Move the tracing-paper

about so that AB always passes through Z, and A is always on theline I Q. Find in thisway the locus of N that is, for each positionof A determine the position of N by ricking through the tracingpa er, and then drawa smooth curve t ough the points thus found .

ive reasons for thinking that the curve would never meet XQ.

42 AREAS

(that is, the side 0 posits the right angle) AB 10 cm. , and a lineKCL of indefinite en h at right angles to AB. On tracing permake a Copy aob of A C and lace it as shown in the rough etchin Fig. 10. Now bring a to and starting from this int slidethe t rac

'

per to the left till ab coincides with A o alwa s

being and ab perpendicular to KL. If M and N are t e

points of intersection of the two pairs of Sides, find from measurements (i) the area of CMoN when 0 has moved 5cm. , (11) the rateof movement OfM if 0 moves uniformly at 1 cm . per sec.

15. Find the area of a regular hexagon whose side is 1 inch .

(A regular hexagon has

six sides all equal and haseach of its six angles

A piece of wire nettinconsists of hexagomeshes an inch in the side(see the rough sketch inFig. Each horizontalside (such asAB or CD) ofa mesh is formed of twowires, and each of theother

FIG 1 1 sides, such as BC or CE ,

is formed of one wire .

Find, approximately, the length of wire contained in a square foot

of the netting.

16. Construct a triangle ABC, given A AB 76 cm . ,

57 cm . On BC, CA, AB as bases, and exterior to thetriangle, describe squaresK, L, M. Given that there is a simplerelation between the areas of K, L, M, find it by measurementand calculation.

Repeat with equilateral triangles K, L,M, instead of squares .

17. A recreation ground was surveyed by running a line throughit . From the line ofi

'

sets to the fence were taken at right angles

at regular intervals of 50 links, the length in links of the successiveoffsets on one side of the line being 30 (at the starting

-point) ,

37, 61 , 45, 72, 80, 40, 60, 69, 48, 19 . Give, as the decimal of anacre to the nearest hundredth, the area included between the linerun through and the fence . 100 links 1 chain . 10 squarechains 1 acre .

18. In a chemical experiment I require to dissolve one gram of

copper in acid . I have a reel of copper wire 100 metres long,which weighs 3} kilograms. What length of wire must I cut off

for the purpose ? Mark ofi'

the length in question, writing yournumerical answer below the measured length, calculated to the

degree of approximation attainable in measuring.

CHAP. III , EXERCISES 48

19 . Theflo

prilation curves, 1 , 2 and 3 (in Fig. l

a),are those of

England, e d, and Scotland respectively hat

population of each of those countries in 1841 When was thepopulation of Ireland decreasing most rapidly, and that of England

1821 1831 1841 1851 1861 1871 1881 190 1 191 1

FIG. 12.

increasing most rapidly ? Forecast from the curve the populationof England in 1911 .

(Examples of readi1 The population Of Scotland in 1831 was2millions, that of Ire and in 1821 was 68 millions, and that ofEngland in 1871 was 225millions . )

44 AREAS

20 . The Navy expenditure is given below for certain years.

Plot these on a diagramand drawa broken line through the pointsDraw also a straight line to lie as near this broken line as possible,and taking this straight line to indicate what the e nditure would

be in 1905-6, and in 1906—7, if there was no c ange of lens,estimate the expenditure for each of these years If the op tion

maybe estimated as 44 millions in 1906-7, whatwould t 9 cost per

head be in that year ?39

1897-8

1898—91899-1900

1900-1

1901-2

1902-3

1903-4

1904-5

21 . One man walks up:

street 600 yards long at the rate of 200

yards in three minutes, ut stops for five minutes in a shop halfway up the street. Another man, starti from the same end as

the first, but seven minutes later, walks r him at the rate of

100 yards or minute . Draw lines on the squared paperwhich willindicate t eir respective distances from the end of the street at

given times, using 1 inch to represent two minutes, and 1 inchto represent 100 yards.Hence, or otherwise, find where and when the second man will

overtake the first.

CHAPTER IV

VOLUMES

1. How shallwe measure the air-space of the schoolroomhow much oil an oil-drum holds ; how much coal can bemined from a given seam ; or howmuch water a swimmingbath or a reservoir holds What sort of unit do we need

Let us begin with the schoolroom .

Our unit must be a piece of space, such as a cubic metre ,or a cubic foot. These are suitable for measuring the

volume of a room. For smaller volumes the litre,gallon,

cubic inch, cubic centimetre, &c. ,are useful.

Make a decimetre cube and an inch cube of paper bytting out suitable

pieces of paper, foldingthem, and pasting themtogether (see Fig. 1)leave margins for the

pasting. If you can get

pasteboard make a footcube from it. In a corner

of the room mark withchalk how a metre cubewould stand.

Measure in metres thelength

,breadth, and FIG. 1 .

you

are in, which we will suppose gular. Suppose themto be 13 metres, 10 metres, and 4 metres, neglectingfractions for the present. (The teacher will of course use

the actual dimensions in place of these.) If you covered

46 VOLUMES

the floor with a layer of metre cubes,how many would

there be, and what would be their volume —130 cubes,

and their volume 130 cubic metres. How manylayers would fill the room

,how many cubes are there

then, and what is the volume or air-space of the room4 layers, 520 cubes, and the air-space is 520 cubic metres,or, in terms of the measured dimensions, 13x 10 x4 cubicmetres.

The building bricks that children use, or clay or plasticineshaped into cubes, should be used as models.

2. Now measure the room more accurately, and supposewe find the dimensions inmetres and Whatadditions must be made to the volume calculatedWith the ‘

building bricks a model, which is not to scale,may be made by increasing each dimension by a unit, or

a model to scale may be made by using pieces of board alongwith the bricks. Either model shows that we can fill upthe space left by

(a) three Slabs measuring (in metres) 13x 10 x

10 x4x0°6,4x 13x0°2 ;

(b) three pillars measuring 4x x 13x x

10 x xO°6

(c) a little block measuring O°6x x

What is the volume of a slab, say the horizontal one

How many metre cubes would ten such slabs togethermake -They would make 13x 10x4 cubes, so that one

Slab has a volume of13x

1

1

:x4

, or 13x 10x0°4 cubicmetres.

What is the volume of a pillar, say the vertical pillar

How many such pillars would you put together to be ableto cut them up intometre cubesAnd howmany copies of the little block would you put

to

gbe

e

th

?

er to be able to cut the figure so made into metrecu S

We can now calculate the volume of each piece, and wesee that to find the total volume we carry out operationsthat could be used to give the product x x44 , or

CHAP. IV,ARTS. 2, 3 47

Give separately the volumes of the original piece,the slabs, the pillars, and the little block. They are

Original pieceSlabsP illarsLittle block

610368

and, assuming the dimensions of the room to have beenmeasured to the nearest decimetre, we give as an appropriateresult for the air-Space 610 cubicmetres. What is the orderof importance of the variousvolumes you have given And

how many of them affect the result if it is wanted to one

significant figure If to two significant figuresIf you had begun with the decimetre and cubic decimetre

as units, what would you have had for dimensions and

volume of the room And what is this volume expressedin cubic metres

8 . Give a formula for the air-space in terms of the

dimensions of the room . The air-Space and the dimensionsbeing expressed in corresponding units, we have

volume length xbreadth xheight,or, if we denote volume

, length, breadth, and height byc, l, b, h,

Need the length, &c. , be whole numbers or involve onlydecimals of a particular kind —No ; for decimals to anynumber of places we can proceed by using successive units,smaller and smaller ; or we can begin with a unit smallenough for the smallest volume that occurs and in eithercase we arrive at the formula

v= lx bxh

Are vulgar fractions also included Vulgar fractionscould be considered for themselves in the same way as

decimals, but anyvulgar fraction can be expressed decimally

48 VOLUMES

—by a terminated decimal— to any degree of accuracy wechoose ; and therefore, to any degree of accuracy we choose,vulgar fractions are already included.

Laterwe shallmeetwith numbers that cannot be expressedwith absolute accuracy by either decimal or vulgar fractions ;for instance, the length in inches of the diagonal of an inchsquare (that is, the line joining opposite cornare). But thesenumbers also can be expressed decimally to any degree ofaccuracy we choose. So that for any rectangular piece of

Space of length l, breadth b, and height h, the volume (inthe corresponding unit) is given by

v lx bxh.

Ex. 1 . In Huxley’

s opinion the air in a room ought to berenewed at the rate of litres an hour for every person

present. On this reckoning, how often would complete renewalof the air be necessary in the room you are in ?

The standard gallon measure.

4. Get a standard gallon measure, and try to determineits capacity in cubic inches. How will you set aboutit — We could try how many inch cubes can be packedin. Or we could try how many go to the bottom layer andhow many layers there are. And what is the rela

tion between the number of cubes in a layer and the area of

the bottom of the gallon measure —The number of cubesin a layer is the number of inch squares that could bemarked out on the bottom. Measure the inside

diameter of the gallon measure (71 inches), draw a circle

of this diameter to represent the bottom, and mark out on

it as many inch squares as you can.

In packing-in the inch cubes you left some corners unfilled.

What will you do with these -We need smaller units.

These might be obtained by supposing an inch cube sliced

into ten equal slabs for the first smaller unit, a slab slicedinto ten equal pillars for the second unit, and a pillar cut

into ten little cubes measuring inch each way for thethird unit. Of these three kinds of smaller unitshowmany kinds doyou use to complete approximately the bottomlayer, and how are they related to areas on the bottom of

the vessel On your drawing of the bottom of the vesselshow the bases of all the units used for the bottom layer.

50 VOLUMES

curve) by ste ping round it with dividers. He also measures thediameter, an calculates the quotient

circumference

diameter.

Then all the results from the different circles are collected andtabulated in three columns, under the heads

Circumference Circumferencein cm. m cm . diameter.

Is there noticeable about the results ? Would the thirdcolumn affected if the measurements were taken ininchesEx. 3. Take a coin and measure its circumference by rolling it

along a graduated ruler . Measure its diameter and calculate

circumference diameter .

Ex. 4. Each pupil again draws a circle of any size on squared

paper, and draws a uare on a radius of the circle . Then hefindsthe area of the circ e and of the square by count

'

squares,°

udgingby eye what to allow for the squares partly inclu ed . Then0 calculates

area of circle area of square on radius.

All the results are collected and tabulated under the headsArea of circle

IArea of square Area of circle

in sq. cm. on radius in sq. cm area of square .

Is anything noticeable about the results -It is noticed thatthe numbers in the last column are pretty nearly the same, and

pretty nearly the same as in the last column of the precedingtable. Would you expect any of these results tobe quiteaccurate —No ; the measurement of a length is not hkely to bemore exact than to the nearest millimetre, or at the outside to thenearest tenth of a millimetre the eye is no great Judge of thearea of the squares partl included ; each line has some thicknessand in step ing the divi are along the circumference it is drficult

to step fair y on the line .

7. Further developmentsofmathematics show that ifall theoperationsmentioned in Exercise 4 could be accurately carriedout, all the numbers in the third columns would be the same

number, and they show how this number can be calculated toas many decimal places as you please . It has been calculatedto a great many places, but the value 31 416 is as accurateas you are likely to need, and for most purposes is

onearenough. This number occurs so often that (to savewriting)

CHAP . Iv, ARTS. 7,s 51

a contraction is used for it, namely the Greek letter 1: so

that we may say the circumference of a circle is 81 4 times

the diameter, or (if we need to be more accurate) 81 416times the diameter, or we can say it is 1: times the diameter,leaving over the decision which of these values is to be

“

223for 1r

, or whether a still more accurate value is to beu

And, calling the radius of the circle r, we may writedown the area of the circle in various ways, a few of whichare

times the square on the radius,31 416xradiusxradius,wxrxn

rr,

a n .

8 . Averages. There were various reasons why the valuesin the last columns of the tables in Exercises 2 and 4 werenot likely to be accurate . Are the values likely to be toogreat or too little — They might be either, some are likelyto be too great and others too little. If you arrangethe values in order of size

, greatest first, those at the

beginning are likely to be too great, those at the end too

small, while one from the middle of the list is likely to benearer the true value than one from either end. How willyou select a likely number from the list We mightchoose the middle number. Or we might take a numbermidway between the two extremes. Or we might (and thisis the usual way) add together all the values and divide thissum by the number of these values. The result of this iscalled the average of the values. Calculate the averageof the values in the third column of each table and compareit withEx. 5. In a company of soldiers

33 were23inches in height

14 7010 711 73

What was the average height ?0

If there were A soldiersof height a, B of height b, C'of height c,and D of height d, what would be the average height ?

E 2

52 VOLUMES

Ex . 6. A boy made in four successive examinations the marks73, 84, 90, 100. What was his average mark;3

As

fier the second

4

examination the boy calculated his average as 2785, after

85

the third examination be reckoned his average as7

290

after the fourth he reckoned itw 92-1 . Criticize his

method, showing to which examinations he gave too much weightand to which he gave too little .

9 . We have seen that a circle of radius r has an area (incorresponding units) of xrxr. Use this expression tocalculate the area of the bottom of the standard gallonmeasure, and the capacity of the vessel in cubic inches, andcompare with your previous results. Also give an expression for the capacity of such a vessel whose height is h andthe radius of whose base is r.

Ex. 7. Take a standard litre, measure its height (17“2cm. ) andits diameter (86 and find in two ways its capacity in

011

1

1

30 centimetres. Compare with the value given in arithmetical

ta es.

Volume of a reservoir.

*

10 . A reservoir has been made by damming up a valley.

The bottom may be taken as level ground, 26 hectares inarea. The area of the water surface as the water rises is

as followsDepth in metres

urf

5 10 15 20 25 30

Area of water 8 ace

in hectares37 49 66 85 104 120

and the reservoir is full when the depth is 30 metres.

How will you determine the quantity of water it will holdCan you divide thewater approximately into horizontal layersbounded by vertical sidesSuppose a wall built round the 26 hectares of level

bottom,carried up 5 metres

,and the space behind filled

The method of finding limits between which a quantity lies,discussed in articles 10-12 should be first introduced in connexionW ith some simpler problem than the reservoir. For instance , theclass might, for the area of a circle drawn on squared r, etan up or limit by counting the squares partially inclu£ofi€

e

an§ a

lower mit by leaving out of count the partially included squares.

CHAP . IV, ARTS . 9—11 53

up flush with the wall. Then when the water stands at 5

metres it is of a shape that we know how to deal with ; it hasthe same shape as the gallon measure with a horizontalbottom and vertical sides, a shape called cylindr ical .What is the volume, the height being 5metres and the areaof the base 26hectares Is it 26x5with any unit — No,because metres and hectares are not corresponding units ;we had better first express the 26 hectares as

square metres. Then we know the volume is in cubicmetres 5x or

11 . If now another wall is built round the water’

s edgewhile it stands at a depth of 5 metres, and then the reservoir filled to a depth of 10 metres, we shall have anothercylindrical piece of water of volume 1,850,000 cubic metres.

Treating the other layers in the same way, we have thevolumes of the six layers, in cubic metres,

or in all cubic metres. We know that this isnot exact, that it is less than the true volume, but we do not

know how much less. Could you find a result that ism ore than the true volume, and thus have some idea howfar wrong our results may beWe could obtain cylindrical layers by cutting away the

banks of the reservoir instead of filling up with walls.

Thus we begin at the line of the edge the water has when5metres deep and cut down all round to the bottom level ;this gives, when the reservoir is filled again to 5metres

,

a layer 37hectares in area and 5metres in thickness. Treating each layer in the same way, we have now for the six

layers, in cubic metres,

54 VOLUMES

or in all cubic metres. And we know thevolume of the actual reservoir to be less than this. Comparing the two results,we may say that the volume is about20

,

Could this work have been shortened — As each of the

areas had to be multiplied by 5 and these products added,we could have added the areas and then multiplied by 5.

12. If the area of the water surface of a reservoir at

different levels is given by

Level ofwater inmetres 0

Area ofwater surface insq. metres e f g

give two approximate expressions for the volume of water.

One of these is, in cubic metres,p x a p x b a x a p x d p x e p ,

and the other, written in the shortest of these forms,

Is there any limitation on the numbers that these lettersmay represent The discussion assumed the area of thewater surface to increase as the water rose, so thata b c d e f 9 must be in order of increasing magnitude

,

that is, each is greater than any preceding one.

13. Suppose there is a series of boxes, the bottom of eachmeasuring 80 cm. by 60 cm . , whi le the heights difi

‘

er.

Calculate the volume for various heights, and show in a

graph how the volume varies as the height changes.

Suppose another series of boxes all of the same height,

70 cm . , and with the bottoms square and of varying size.

Calculate and show in a graph how the volume varies as

the side of the square bottom varies.

Suppose a third series of boxes, all cubes, and varying insize. Show in a graph how the volume varies as the

length of the edge varies.In the first of these three cases the volume depends on,

CHAP . IV,ARTS . 12-14 55

or varies with, the height of the box, or, in mathematicallanguage, is a funct ion of the height of the box. In the

second the volume depends on, or varies with, or is a

function of,the length of a side of the square bottom. In

the third the volume depends on, or varies with, or is a

function of,the length of an edge of the cubical box.

14 . One man A lends another B and B agreesto pay at the end of every year for the use of this money£4 for every £100 he has borrowed, or, in the usual semiLatin phrase, 4 per cent. How much does he pay annuallyon the loan of —He pays £400.

If A and B agree that this £400 need not be paid over

but considered as an additional loan, what does B oweat the end of the year —He owes For

the second year B has now to pay on his loan of

What is the payment on this ; and, if this also is added tothe loan

, what is the amount of the loan at the end of the

second yearOn this plan,

by what do youmultiply the original loan tofind the amount of the loan at the end of the first yearAnd by what do you multiply the amount of the loan at

the end of the first year to get the amount at the end of the

second -In both cases the multiplier is 10 4.

On this plan, to what does a loan of £1 increase in a

year -It increases to And in two yearsx And in three , or four, or five years?

To x x or x x x or

£ 10 1 x x x x

These expre ssions become tedious to write. The pupilsshould be encouraged to express this tedium , and then askedfor suggestions of shorter ways of writing the expressions.The usual contraction is not immediately necessary, and thework may go on for a while without it, or with the help ofcontractions invented by the pupils.

Now calculate what a loan of £1 would increase to infour years, for different rates of payment for its use, say for1, 2, 3, 10 per cent. Tabulate these, and also show ina graph how the total to which the loan increases dependson the rate per cent , that is, show the total for 4 years as

a function of the rate per cent.

56 VOLUMES

Give an expression for the amount to which a loan of £1will increase in four years at r per cent. It is

-3What will it increase to in ten years ‘

P

If in the expression for the amount in four years the

brackets were obliterated, what would the expression mean

Compare this meaning and the original meaning when10 and when r 50.

Exnaorsxs.

8. Measure, in centimetres, a rectangular block of wood, andcalculate its volume .

Weigh it and find howmuch a cubic decimetre would weigh .

9 . A mahogany pattern, made from wood weighing 762gramsper cubic decimetre, weighed kilograms. A gun

-metal castinof the same size and sha e weighed kilograms, the gun

-meemployed weighing 8° kilograms per cubic decimetre . Showthat there must be a cavity in the casting, and find what weight of

gaotal would fill it . All the numbers are given to three significant

res.grll‘ake the wei ht of the wood as a grams per cubic decimetre,

of thegun-me as b kilograms per cubic decimetre, of the patternas c kilograms, and of the casting as d kilo ms, and give an

expression for the weight of metal that would the cavity.

10. Draw any rectangle and suppose it to re resent the bottomof a cistern on a scale of 1 inch to 1 foot . If allons of waterare run into the cistern, to what level will they it ? A cubic

foot is gallons.

11 . A room 460 square feet in area is to be flooredwithwood blocks,the flooring to be 2 inches thick . What will the blocks cost at

33. 6d. per cubic foot ?A room p square feet in area is to be flooredwith wood blocks, theflooring to be q inches thick . What will the blocks cost at r shillings a cubic foot ?

12. The ends of a workshop are 10 metreswide, the height at theridge of the roof is 8 metres and at the eaves metres. Theworkshop is 18 metres long. What is the floor area in square

metres, and the air-space in cubic metres ?

13. A ton of lead is rolled into a sheet g-inch thick . Determinethe area of the sheet in square feet, given that 1 cubic foot of leadweighs 712lb .

A ton of lead is rolled into a sheet m inches thick. Determine

58 VOLUMES

h 3

53“ the lowest point of the bottom, were measured

to

Explain the method of calculation you adopt.

20. If a square post has to be cut from a cylindrical tree trunk,find roughlywhat percentage of the wood is wasted.

21 . Fig. 3 shows the section ofa smallwatercourse.

The surface velocity is 20

inches per second, and themean velocity the sur

face velocity x Find

the discharge in gallons

FIG‘ 3° 22. The population of a

town increases uniformly,and in each period of three years the increase is 20 per cent. of the

pgpulation at the beginninfib

of that period. If thepo alation was

,000 in January 1903, w t will it be in January 956 and 1912,and what was it in January 1897By drawing a gra

ph, find the ulation approximately in

January of each year cm 1900 to 1

23. The table below shows the distances (in miles) from Londonof certain stations, and the times of two trains, one up and one

down. Suppose each run to be made at constant speed, and showby a graph the distance of each train from London at any time,

1 mch to represent 20 miles, and 3 inches to represent

London p.m p .m .

5} Willesden, arrivedo 4. (No intermediate

66 Northampton, arrive 5.50depart

113 Birmingham p .m .

At what point do they page

!

one another, and how far is eachfrom London at

'

oh of the three runs made by thestopping train is the fastest ?

CHAPTER V

TO MAKE A COPY OF A MAP .

1 . Ln us now discuss the problem how to make a copyof the map shown in Fig. How will you proceed Let

us begin with Kingston. Where will you place it — Any

0m.

FIG. 1 .

A local map should be used instead of that and the

places on it chosen so that the distances to measure are greater .

60 TO MAKE A COPY OF A MAP

where. Put it then at K (Fig. And where willyou putWimbledon What conditionmust it satisfy — It

must be at the same distance from Wimbledon as on the

given map, namely, cm., and anywhere at that distance

Put it down then at IV, anywhere on a circle

with centre K and radius cm. ; and consider Epsom .

What do you know of its position -Epsom is cm .

from Kingston, so that it must lie somewhere on a circlewith centre K and radius cm . Will it do any

at random on this circle — It is cm . from

Wimbledon which gives another circle on which Epsommust lie. Do these two circles fix a certain pointwhere Epsom must lie How will you decide between thetwo points in which the circles cut, or are you at liberty tochoose one of them at random

The pupilswill easily see that onlyone of the pointswill do,butnot so easily how to distinguish it. The usual conventionthat the top of the map is the north, so that Wimbledon iseast of Kingston, will enable them to place their copies withW east ofK, and to distinguish Epsom as the intersection

CHAP . v, ARTS. 2, a 61

that lies south of the lineKW. Orwithout using the pointsof the compass, the pupils may be led to notice that a mangoing from Kingston toWimbledon has Epsom on the right.Having now K, W, and E on themap, howwill you place

Esher —Bymeasuring its distances from two of the placesalready marked and proceeding as before. And theremaining places

2. Now count up how many measurements you havemade to copy the map.

—We had

0 measurementsWimbledonEpsomEsherRichmondTwickenhamHampton Court 2

a total of 11 measurements for 7 places.If there had been 20 places on the map, how many

measurements would have been needed —That means 13more places at 2measurements each, or in all 11 26, thatis, 37 measurements. In the whole number ofmeasurements how many less are there than 2 each for all20 places — 8 less, 2 being saved on the first place putdown, and 1 on the second place put down.

How many measurements are wanted for a map of Nplaces — 2 for each, except for the 3 saved on the first twoplaces, that is, 2xN 3 measurements.How would you copy the river How would you fix a

number of points on it -We could fix as many points as

we like in the sameway aswe fixed the places, and then drawthe river freehand through them .

8 . Return now toconsider the three placesK, W,E . Sayagain how they were fixed, and how many conditions or

measurements were needed to fix them. Would it make

any difference if we began by putting down W first of all

instead ofK ? Or if we had taken E second instead of W ?

Can you suggest anything else we could have measured inplace of the three lengths KW, KE , WE Could you

make use of your protractor -We could measure the

Mwwwl—l


angles at K, W, and E . Do 80 , and measure the

corresponding angles on your copy of the map.

We have now 6 measurements, 3 lengths, and 3 angles,and we know that the 3 lengths are enough to enable us

to make a map of K, W, E . If we had only 2 lengths, KWand KE , what more is needed Find by trial. —An anglein addition will do. Does it matter which —Theangle K will do ; whether one of the other angles woulddo is not clear. Measure the angle at K,

and use it

to draw the triangle KWE .

If you had only the length KW, could you make up withangles —With the angles at K and W we could draw the

Measure these angles and do so.

Suppose that you measure only one length KW, and for

the rest measure angles only. Could you in this way copythe whole map Do so.

Suppose again that you copy the whole map in this way,but by an error make KW of length 6cm . instead of cm.

Draw the whole figure that you get in this way and compareitwith the original map . Does it look like the originalmap

4 . The P lane Tabla— A man takes a sheet of paper on a

drawing-board to a high point in Wimbledon from whichhe can see a wide stretch of country. He fixes the boardhorizontal, and marks a spot W on the paper to representWimbledon. Then from W he draws lines pointing toKingston, Epsom, &c., and labels them to Kingston &c.

He now removes to Kingston, where we must suppose himto go up in a captive balloon to command a wide stretch ofcountry. On the line labelled to Kingston he marks a

point K to represent Kingston. Then he sets up the boardso that the line KW points toW imbledon, and draws from Klines toEpsom, &c.

,and labels them to Epsom &c.

,asbefore .

There are now two lines labelled to Epsom and he marksEpsom at their point of intersection and so with all theother places. In this way he makes a map of the district.Select two high points in your neighbourhood, and use

this method to make a map of the district.A board made with certain attachments so that it can

be levelled and fixed for accurate use in this way is called a

plane table.

CHAP . V, ARTS. 4, 5

5. Angles—In order to lay downW at the proper distancefrom K, you measdistance. Could you have laidoff the distance without a centimetre scale — Yes, by openingthe compasses to reach from Kto W and laying off this dis

tance. You measured theangles with your protractor, andlaid them 03with the protractor.

Could you have done without aprotractor Lay down K and

W without using a centimetrescale, and consider how you

will make the angle WEE of

the right size.— We could cut

or fold a piece of paper (Fig. 3)so as to fit the angle atKingston.

Then K’ being laid on Kand K’

P along KW, K’

Qp

'

vcs the direction of the

line KE . Use thismethod to laydownEpsom,

Richmond, &c. on yourmap, considering all the

time whether you can in

any way simplify the procedure.

By repetition of the cut

ting out of pieces of paperto fit the angles the pupilswill discover, or be led todiscover, that the pieces Aand B (Fig. 4) of the scrapof paper used need not be

cut away that it is sufiicientto lay K

}onK of the origi

nal map and rule K’P and

K'

Q in linewith theparts ofKW and KE that projectbeyond. Thenitisseen that FIG . 4.

63

KW as cm. , and laid off this


it is sufficient to mark the pointsK ’P Q without ruling any

lines. Thenext step is that ifwe know the lengthsK’P,K

’

Q,and PQ we can always place the pointsK

’P Q in proper rela

tive position without using the scrap of paper at all. That is,P and Q are taken anywhere on KW and KE on the originalmap, KP is laid ofi of the proper length on the new map,and Q is found as the intersection of two circles, one withcentre K and known radiusKQ, and the other with centre Pand known radius PQ. By carrying this out with as fewoperations as possiblewe arrive at the following construction.

6. To draw from Q (Fig. 5) a line making with QP an

FIG . 5.

angle equal to the angleK with the pointK as centre drawa circle cutting the two arms of the angle in A and B, and

with Q as centre draw an equal circle cutting QP in 0 . Thentake the length AB on the compasses and lay ofi this lengthas a chord CD in the second circle, and join DQ. DQP is

an angle equal to K.

How many points D are there7. Could you bisect a given angle K (Fig. that is, draw

a line dividing it into two equalparts — One way is to fold the

legs KL and KM together. The

crease line KN bisects the angle,since the two parts LKN and

MKN fitwhen folded together, andare therefore equal. Couldyou adapt the method just discovered for copying an angle to

the bisection of an angle —If weFre . 6 can choose A on KL , B on KM;


Make a triangle having a cm . , b cm . , c cm. ,

and measure the angles. Try to make a triangle havinga cm. ,

b cm . , 6 cm . What condition mustthe given lengths satisfy to make a triangle possible — Anytwo sides together must be greater than the third.

9 . If two triangles are made with the same values fora b c, can they be fitted together —Lay down the side a

(Fig. then a knowledge of the length b tells us thatA lies

FIG . 8.

somewhere on a circle L with centre 0 and radius b, and the

length 6 tells us that A lies on a circle Mwith centre B and

radius 0. So that A must lie at one of the two intersectionsof these circles, sayat the pointmarked A. Suppose anothertriangle A’B

’C

’ made, havingR C a, O'

A’

b, A’

R’

c,

and lay it on the other so that B ’ lies on B , and O"on 0 .

Then, since C'A’

b, A’ lies on a circle with centre 0 and

radius b, that is,on the circleMplaced A ’

B’C

’

so that A ’ lies on the same side of BC as A ,

then A ’falls on A and the triangles fit. Or we could alloweither way and if A

’falls at P , we know from

a former discussion (chapter I) that folding along BC bringsP on A, so that in this case also the triangles fit together.

10 . You know that 2 sides won’t fix a triangle. Couldyou in place of the third side take any other measurementor measurements —We could measure an angle

,or, if need

CHAP . V, ARTS. 9—1 1 67

Try to make nine triangles inwhich (all the lengthscentimetres)

1 . a 69,2. a 69,3. a ar

4. a

5. a

6. a

7. a

8 . a

9 . a 69 ,

The pupils should attempt these constructions with littleor no assistance . Half-successful and unsuccessful attemptsare of value for the following discussion.

For the fixing of a triangle by measurement of 2 sidesand

an angle or angles we may take as an angle to be measuredthe one between the 2measured sides, or one not between,that is, one opposite to one of the sides. Let us take theangle between the sides, and call the sides at and b thenthe angle is 0. Are these 3 measurements enough to fixthe triangle — Yes, for we can make an angle 0 with theprotractor, andmeasure the lengths a and b along its two legs.If two triangles are made with the same values of a b C,

will they fit together — Yes,as we see by fitting the 2

angles together so that the 2 longer legs fall together, andthe 2 shorter together.

Copy the triangle of Fig. 7 by means of its a b O, andwithout using a scale or protractor.

Can a triangle always be made from given values of a b C ?— The lengths of a and b do not matter. 0 must be lessthan Draw two lines 3 2cm . and cm. long,and inclined to one another at and join the ends of thelines. Has this triangle an angle of 200

° —No,the angle

of the triangle is 160° at the corner where we made an angleof 200

°

(Fig.

11 . We know that 2 sides alone are not enough to fixa triangle. Suppose, for instance, that we know a 6cm.

F 2


b 10 cm. , but nothingmore. Let us lay downB0 6cm . ,

Then what do you know of the position of A -That it

Gem. 0

FIG. 10.

lies on a circle with centre 0and radius 10 cm . (Fig. 10

,a

sketch drawn on a reduced

scale) . If now you are

given also that A how

will you find the position of A

Try drawing the triangle withA in various positions on the

circle . Or better, by drawingthe triangle in various positionsand measuring the angles A and

Cdraw a graph giving the angleA as a function of the angle C’

FIG. 11 .

CHAP . v, ARTS . 12,13 69

(Fig. From this graph read ofi the value of C thatcorresponds to A -There are two values, namely,0 18

°and C' 1 15

°approximately. Make a tri

angle having a 6cm . , b 10 cm . , 0 and anotherhaving a 6cm . , b 10 cm . , 0 and measure theangle A in each.

What is the greatest value A can have with the givenvalues of a and b — The graph shows it to be about

Make the triangle having A — Thegraph gives 0 and we make the triangle a 6cm . ,

b 10 cm . , 0 48° Measure LA of this triangle,

measure also LB.

12. Consider again a triangle having a 6cm ., and b 10

cm . and lay down CA 10 cm.

Where,then,

does B lie — It

lies somewhere on a circle withcentre 0 and radius 6 cm . (Fig. 12,drawn small). By taking AB in various positions, see howLA varies, and again draw a graphgiving LA as a function of LC.

What is the position of B whenLA is greatest And howwould Fm . 12'

you use Fig. 12 to find B for a given value of A

Use figures like Fig. 10 or Fig. 12 to make a graph givingLB as a function of L C (Fig.

From Fig. 10 or Fig. 12,or from the graphs of A and

B,make a graph of A B as a function of C and also a

graph of A +B + 0 as a function of C'.

13. We discussed the making of a triangle with 2 sidesand 1 angle given, the angle being opposite a given side, firstby laying down the Side opposite the given angle, and thenby laying down the Side next the given angle . Let us now

try laying down the angle first. Take a cm . ,b 6 9 cm . ,

The degreewas introduced as a unit of angle to avoid fractions

of a right angle . We now come to a fraction of a degree, whichwe may write decimally as above. Or we may now introduce

another unit called a m inu te 60 minutes make a degree, so that3

;°5 degrees is 37 degrees 30 minutes, which may be written

30


B Lay down the angle B equal to What is thestep — Mark ofi

'

BC equal to cm. along one legof the angle (Fig. 13,small). Where mustthe point A lie — It mustlie on the other leg BR of

the angle, and it must alsolie cm . from C', that is, ona circle with centre 0 and

radius cm . A must therefore lie at one of the intersec

tions of the circle and the

line. Are the two in

tersections equallysuitableThe one marked A gives a triangle ABC having a

cm. , b cm .,B so

that suitable. The point marked A ’

gives a triangle A'

BC' havingthe two sides of the right

length, but the angle is not 55 but 180°

or so

that this intersection will not do,and we can make only one

kind of triangle with the given values.

If you had another triangle DEF in which EF cm . ,

FD cm ., E could you fit it on to the triangleyou have drawn —P lace F on C’ and E on B, which willmake ED lie along BR. Then D must lie on the line BRand also on a circle with centre 0 and radius DF, that is,

cm . So D lies at one of the intersections of this line and

circle and A’is unsuitable, for this would make the angle

DEF 125°and not 55

°so that D lies at A and the

triangles fit.

14 . Would these results be the same when a b B are

given, no matter what values they have Take a cm .

and B and give b different values. Are the resultsthe same as before when b is 14 cm .

— Yes as before, A’

is unsuitable,and there is only one triangle.

you give any other values of b for which the results are thesame as before —Any value greater than cm . , that is,greater than a.

15. Are the results the same when b cm. ?

CHAP . V, ARTS. 14—16 71

Having laid down the side a and the angle B , we have, asbefore, two loci on which A must lie, a straight line and a

circle (Fig. 14) but in thisthe drawing shows that the twointersections of the line and thecircle lie on the same side of B .

In consequence we have two triangles of difi

'

erent shapes, bothof which satisfy the conditionsin the triangle ABC BCcm. , CA cm .

,LABC 55

°

and in the triangle A ’BC BC

cm .

, CA’

cm . ,LA

’BC

If you had another triangleDEF in which EF 5-4cm . ,

1°10 . 14.

FD cm . , E could you fit it on to the triangleABC? -We fitE to B andF to C, andfind thatDmust be atone of the intersections of the circle and straight line, namely,atA or A

’butwe have nomeans ofknowingwhether itwill

fall at A or A’. So that the triangle DEF will fit on to one

of the triangles ABC and A’BC, but we cannot tell which.

Angles like BA’C that are greater than 90

°are called

obtuse angles ; angles like the other angles of the triangleA

’BC, and like all the angles of the triangleABC, that are less

than are called acu te angles. If we knew whether thetriangle DEF has any angle obtuse, we could tell on to

which of the triangles ABC and A’BC it could be fitted.

16. Can you give any other values of b for which the

results are the same aswhen b cm .— As long as b is

less than cm. but great enough to make the circle withcentre C and radius b cut the line BR,

the results are the

As b gradually dim inishes, what happens tothe points A and A

’ where the circle cuts the line BRThey come closer and closer together till there is onlyone point, and then if b becomes still less the circle no

longer meets the line. Find by trial the valueof b for which A and A

’run together into one point.

It is about cm.

cm . When a straight line and a circle meet in only one


point, as in this case, they are said to touch one another, andthe line is called a tangent to the circle. The point wherethey touch is called the point of contact.

17. Draw the triangle ABC (having a and B

for the case in which A and A’run together so that there

is only one point A and measure the angle BAC. Drawother figures with other values for a, and measure the angleBAC of each.

Consider again the case of Fig. 14. If we draw a per

pendicular CN from C on BR ,how does N lie with regard to

A and A’ —It liesmidway between, since we know (chap. I

,

art. 14) that the perpendicular from the centre of a circle on a

chord bisects the chord. Verify this bymeasurement.

If you drew circles with radu cm .,

all with C for centre, how would N lie with regard to thetwo intersections of any circle and the line — It lies midway between. And if a circlewith centre C graduallydiminishes in size till it only touches BR ,

how do its intersections with BR lie with regard to N

— Always at the

same distance on opposite sides, so that when they run

together they must run together at N .

So that a tangent to a circle is perpendicu lar to theradiu s to the p oint of contact .Ex . 2. Draw a number of angles of From any point C on

one leg BC of each draw a perpen

dicular CN to the other leg BN .

Measure CN and CB on eachfigure, calculate ggfor each, andtabulate your results .

1 8 . Draw a triangle ABC inwhich a cm . , b cmB 55

°

(Fig. 15, drawn small) ;so that the construction gives

a line and circle meeting in

only one point A . If you

made another triangle DEFFIG 15 with the same values of a b B,

could it be fitted on to this triangle ? Suppose LEEF cm . ,

FD cm .—We lay EF along BC and


at B equal to 62° instead of so that only one trianglecan be drawn with the given values. When b is less than

cm . (say 11 cm. ) the circle with centre C and radius b

does not meet BR at all, but only BR produced ; so that thetriangles it would give have the angle at B equal to 62°instead of and no triangle can be drawn with the given

Case 6 is similar to case 5.

If you have two triangles in which a cm.,

b= 17cm. , B == can they be fitted together so as to

coincide — In the sameway as before it is seen that they can.

21 . Wherein do the data of cases 5and 6 on the one handand cases 1, 2, 3, 4 on the other difl‘

er, to account for thesedifferent results — In cases 5and 6 the angle B is obtuse,that is

,greater than a right angle, while in the others it is

acute, or less than a right angle.

If instead of a b B you were given b c C or c or

a c A , how far would the results difi’

er — To label thetwo sides and an angle opposite to one of them with theletters a b B is only a convenient shorthand, and anyother way of giving two sides and an angle opposite one

of them amounts to the same thing.

Ex. 3. Take B 40°and a in succession equal to 6, 9, 12, 15,

18 cm . In each case find the least value of b for which a trianglecan be drawn. In each case calculate b as a fraction of a, and

compare your results.

Repeat this, taking B = 30° instead of equal to

21 . The results we havereached as to making a tri

angle when two sides and

an angle Opposite one of

them are given can now be

summarized. We continueto call these Sides and anglea b B. Lay down the SideCB or a and the angle CBR

c or B , and from C let fall

FIG 17ON perpendicular to BB

(Fig. When the angleB is acute, and b is greater than a, there is one triangle

CHAP . V, ARTS. 21-23 75

satisfying the conditions ; when b is less than a but greaterthan CN, there are two triangles , when b is less than CN

there 1s no triangle . When the angle Bgreater than a, there is one triangle, but when b is less thana there is no triangle .

The results may be tabulated in the following form,

the symbol cc denoting as great a number as you can

imagineValue of B b lying between Number

0 and ON

ON a

a co

obtuse

22. We have discussed the making of a triangle from 3

given sides, and from 2 given sides and 1 given angle. Nextsuppose only 1 side given and consider whether we can

make up the datawith angles. In copying the map (Fig. l )we had such a case. What was it — Wemade the triangleKWE by means of the length KW and the angles atK and

W Then you can, out of a, b, c, A , B, 0, give such aset of data tomake the triangleABC.

— The seta, B, Cwoulddo. Make a triangle having a=6°9 cm. , B

C If you had a second triangleEF== cm . , E F can you fit the two trianglestogether to coincide P lace EF on BC. Then the angleE fits the angle B, and F fits C ; so that ED lies along BAandFD along CA , and D lies where BA and CA meet, thatis at A so that the two triangles coincide.

Taking a cm. and B find by experiment forwhat values of C a triangle is possible— Cmust be lessthan a certain angle .which is about Whena C for what values of B is a triangle possible— B must be less than 120

°or so. Experiment

with a number of other cases and try to state the conditionin a general form .

—The sum of B and 0 must be lessthan 180

°or so.

28 . We have treated the case when the given side lies


between the given angles. Take a difierent case, and try tomake a triangle having a cm . , A C

Draw BC equal to cm. , and make an angle ECK of 50°

any point L on CK and

make an angle CLMequalto Take L in a number of positions and for

each make LCLMContinue to try in thiswayuntil the line drawn at 76

°

passes through B. Denotethe point L for this position by A , and we havethe triangle ABCwith the

Fro . 18° given values of a, A , C.

Draw another figure of BOK with BC cm . andL C On a piece of tracing-paper make an angle of

lay it on your figure so that one leg points towards Cfrom some point in KC, and move it about till the other legpasses through B . This gives another way of making thetriangle.

Measure the angle B of the triangle. Measure also theangl

l

l

e

jgghich the line LM in its various positions makes

wr

24 . Taking a cm . , C find experimentally forwhat values of A a triangle is possible.

— A must be lessthan about Experiment with other values of aand C, and try to reach a general statement of the con

dition.—A C must be less than about

If you have two triangles ABC and DEF in which BC

EF= cm. ,LC= LF = LA LD can theybe

fitted together to coincide — Suppose that in making ABCwe used the angle of 76

°drawn on tracing-paper, and made

the point L (Fig. 18) slide along from C toK,the first leg of

the angle lying all the time along LC. Then the pointMwhere the second leg cuts BC travels all the time from C

towards B and then beyond B . The pointA is the positionoccupied by L at the moment whenM reaches B, and since

L travels all the time away from 0 there is only one such

CHAP . V, ARTS . 24-26 77

point. Now lay EF on BC and the triangle DEF on ABCso that the angles F and C fit together. If now the tracingpaper with the angle of 76

°is moved as before, the position

of D is found just as A was, and is at the same point.

25. We have tried to make a triangle from 3 sides, from

2 sides and an angle, and from 1 side and 2 angles. Can it

be made from 3 angles ? Try with the values AB C Make an angle of 20

°and call the legs

AK and AL (Fig. No length is given for AB ,so try

FIG . 19.

taking B at random on AK,and make the angle ABM

How will you complete the triangle - C must be whereAL and BM cross, so that the triangle is already complated. Is the angle C equal to as it was tobe —No, it is about See if you get any betterresult by taking B at other positions onAK, or by beginningwith another angle than A .

Trymaking triangles with other values for the anglesA = B = C = 60

°°

A B = C = 80°

A B = C 110

What is the result —Sometimes the third angle has thegiven value, sometimes not, but the third angle is alwayssettledwhen the first two aremade. Can you (withoutdrawing) predictwhat 0 will bewhen A 30

°and B 40

°

Experiment with difi’

erent values of the angles until youcan predict the third angle from the given values of two.

26. You made several triangles having A 20°and

B Call the sides of one of them a b c, and of


another a’b

’c’ Measure all the six sides, calculate as

c

c” and comparedecimals the three quantities g; 5,

the three results. What do you noticeRepeat with a different pair of the triangles that you have

drawn.

Repeat again with two triangles, both of which haveA 43

°and B or any other values.

Draw any triangle and measure its sides a b 0. Makeanother triangle with Sides equal to a, b

,0.

Measure the angles of the two triangles and comparethem .

Repeat, using a different number in place of

These results will be discussed later.

27. Let us summarize our results. Of the 6 elements of

a triangle— the 3 sides and 3 angles— any 3 that include at

least one side are enough to know in order to make thetriangle. In all cases but one the triangle is unique ; thatis, with the given values only one shape of triangle ispossible. The one exception is that in which 2 sides and

the angle opposite one of them are given, the angle is acute,and the side opposite this angle is less than the other givenside ; in this case two shapes are possible.

When two triangles are such that one could be placed on

the other so as to fit exactly, the two triangles are calledcongruent triangles.

28 . The Slider Crank Pain— Ia many machines thereoccurs

‘

an arrangement of three pieces known as slider,crank

,and connecting rod. In Figure 20, the slider C may

slide to and fro along the line OP, the crank BA is a rod

that turns about a fixed point B in the line OP, and the

CHAP . v, ARTS. 27,as 79

connecting rod AC is a rod connecting the slider C to theend A of the crank.

Take the crank 7 cm . long, and the connecting rod 16 cm.

long. What more must you know to fix the position of the

system Give the additional condition in as many differentways as you can.

Show what part KL of the line OP the slider C is able totraverse. Find the position of the system (1) when C is

12 cm . from B, (2) when C is 6 cm . from its end position K,

(3) when the crank makes 40° with BP , (4) when crank andconnecting rod are at right angles, (5) when the connectingrod makes 20° with CB . In each case see how many solutions there are, that is, how many positions satisfy theconditions.

By the help of measurements of the figures you havedrawn and of other figures, make graphs giving the anglesA B C of the triangle ABC as functions of the distanceKC. Also make a graph of A B C as a function of the

Bu rmese.

4. If you were told there were 3 towns A B C such that fromA to B was 5miles, B to C was 6 miles, and C to A was 12miles,what would you say ?

If you were told that the distance BCwas 6miles, CA 12miles,and the angle BCA 25degrees, what would be the distance AB ?5. Draw triangles all having a 6 cm. , and b 8 0m . , while thelo C has in different triangles various values between 0

°

andand measure 0 in each .

Draw a gra h giving c as a function of C.

Calculate t e area of a square on the side 0 of each triangle youhave drawn, and show this area as a function of C. On the same

graph show the values of the squares on a. and b, namely 36 and 64sq. cm .

6. (1) Construct a triangle PQR , with base PQ 9 cm. andBBQ(2) On the same base construct a triangle PQS with P8cm.

Draw the locus of R in (1 and of S inOn

fyour drawing mark e vertex T of a triangle PQT, which

has L PQ = 42°and PT-7cm.

State a set of data suficient to determine both the shape andsize of a triangle and explain why the 3 angles are insufiicrent .


7. A square reservoir has been made in a field and is to becovered up. It is, therefore, necessary to take measurements fromwhich the positions of its corners may be found again. Supposingthe field ular, give the least number of measurements (inaddition to the nown length of a side of the reservoir) that mustbeme . Give two sets of measurements either of whichwon 0 .

ngle ABC is supposed to be cut out of paper and laiddown on a sheet of aper, the corners A B C being marked on

the triangle . The tu angle is first turned over about the side BCit is next turned over about the side CA, and finall it is turnedover about the side AB. Draw any triangle for the t position,and then draw the triangle in its final position.

9 . Construct the quadrilateral ABCD having AB cm. ,

BC CD DA LABC Find by measurement the distance of D from the diagonal AC and determine thearea of the figure .

Now suppose AB and BC to be fixed, CD to revolve about C,and AD to be of variable length . Show the position of CD whenthe area of ABCD is amaximum, and give the maximum area.

10. Figure 21 shows a piece of ground, divided by hedges into

FIG. 21 .

triangles. How many measurements would have to be made in

order to drawa plan of the triangle ABC ? And how many for the

figure ABCD ? And howmany for the figure ABCDE ? In eachcase give one set of measurements (not more than are needed) andtell how you would use them to draw the plan.

If the figure was made n

apin the same we of n triangles, how

many measurements woul have to be made ? And how manysides would the outer boundary of the piece of ground have ?


0

fr. 0 .

ompregne 6 35

Noyon 8 15

Chauny 9 35

Tergnier 9 90St Quentin 1 1 65

From these data draw on squared per a diagram showing therelation between the distance travelfgl from Paris and the railwayfare Examine whether the railway fare is proportional to the

distance travelled, assuming distances to be given to the nearest

half mile, and 5centimes to be the smallest coin used for paymentof railway fares .

N .B .— A diagram is meaningless unless the scale of measure

ment is stated and the positions of the lines from which distancesare measured are indicated .

14 . In a field ABCD AB is chains, AC chains, AD109 chains, the angle BAC 27

°

and the angle CAD Drawa diagram to a scale of 2 chains to the inch, and find the area to thenearest rood .

15. Find the discharge per minute in gallons from a full

cylindrical pipe of 6 inches diameter, the mean velocity being12 inches per second .

CHAPTER VI

FURTHER PROBLEMS IN MENSURATION

1 . A GALLON is 277 cubic inches, and the standard gallmeasure is a vessel with a circular bottom and vertical sides,the height of the vessel being equal to the diameter of thebottom . Suppose we want to make a gallon measure of the

standard shape, and let us consider how we may determineits height and diameter from these data.

What do you know of the capacity of a cylindrical vessel— That the capacity is the product of the height h by the

And when the base is circularwhat do you know of its area — That it is x r x r if r

is the radius, or x d x d if d is the diameter.What then is the capacity of a cylindrical vessel h inches

high and (1inches in diameter? —It is x d x d x 71 cubicinches. And the capacity of a cylindrical vesselwhose height and diameter both measure (3 inches —It is

x d x d x (3 cubic inches, or more shortly written0-785ddd or 7

1, 1rddd.

Now how will you find out what (1 must be to make thiscapacity 277 cubic inches — We could try various valuesfor d and calculate the corresponding values of 0°785xd x

d x d. Do so for a number of integers 1, 2, 3,andfind the twovalues of 0°785ddd between which 277lies.

They are given by d ==7and d 8 . You could nowcalculatethe capacity for d equal to and thus get stillnearer to the capacity 277 c. in. but few pupils are fond

enough of arithmetic to make thisworth doing.2. Could you use a graph to find the value of d that will

make ddd equal to 277 -Yes, we could draw thegraph of ddd and read 03 the value of d forwhich it isequal to 277. As it takes some time to draw a graph,perhaps you could use one you have already drawn ; lookthrough them and see if anyof

2

them will help.—There is a

e

84 FURTHER PROBLEMS IN MENSURATION

graph giving the volume of a cube for various values of the

edge of the cube, that is, giving ddd for various values of d ;and the value of d that makes ddd will make

ddd 277 ; so that we Simply read ofi the value of d

that makes ddd 353.

Such a relation as ddd 277 in which two thingsare stated to be equal is called an equ ation ; a value of dwhich makes the two things equal is called a roo t of theequation ; and the process of finding such a value of d iscalled so lving the equation .

3. Find the height and diameter of a standard litremeasure, that is, of a cylindrical vessel that holds 1000 cubiccentimetres and whose height is twice its diameter.

4 . Look at a Fahrenheit thermometer. It tells us howwarm the room is. The mercury (or other liquid) insidethe tube stands opposite a mark on the scale. Thesemarks all have numbers, some of which are shownthere is not room to Show all of them . If it stands opposite55we say the temperature is 55degrees, or more completely55degrees on the Fahrenheit thermometer, which is writtenshortly 55°F. If in the morning the mercury stands lower,for instance at it is too cold, and the room is artificiallyheated till the reading of the thermometer is about 55

,that

is, till the temperature is about 55 If the temperature is70

° before the room is heated, it is already warmer than weneed, and we do without further heating.A man called Fahrenheit invented this thermometer and

called the lowest temperature he could reach bymixing snowand salt while he called the tem erature of his bodyHe marked with 0 the point atwhi ch the liquid stoodwhenthe thermometer was plunged in the mixture of snow and

salt, and with 100 the point at which it stood when at the

temperature of his body. The interval between these hedivided into 100 equal parts and marked with numbers from0 to 100. The graduation of the thermometer was carried

beyond 100 by marking ofi further divisions equal to thosefrom 0 to 100.

Itwas found that water froze at the temperature called 32

CHAP . VI, ARTS. 3—6 85

on this thermometer and boiled at 212. The freezing and

boiling points are more definite than the temperature of the

snow-and-salt mixture and the temmrature of the body, anda Fahrenheit thermometer is now made by fixing thesefreezing and boilingpoints, calling them 32and 212, dividingthe interval between into 180 parts, and continuing thesegraduations at both ends of the interval. It thus happensthat on the Fahrenheit thermometer the temperature of thebody is not exactly 100 but somewhat lower.

5. Look now at a Centigrade thermometer. On it thefreezing point of water is called 0° and the boiling point

How many degrees was it on the Fahrenheit thermometer from the freezing to the boiling point — It was180 degrees. And how many Fahrenheit degrees areequal to a Centigrade degree —8—gor or gFahrenheitde 8.

Blood heat or the temperature of the body is 37 degreeson the Centigrade scale, or more shortly is 37

°C. On the

Fahrenheit scale how many degrees above freezing point isthis temperature which is 37degreesabove on the Centigradescale —It is 37x or degrees Fahrenheit abovefreezing point. The Fahrenheit scale beginning 32degrees below freezing point, what is this temperature calledon that scale —It is called 32 or degrees.If a temperature is 23°C. ,

that is , is 3: degrees above freezingpoint on the Centigrade scale, how many degrees is it abovefreezingpoint ontheFahrenheit scale —It is xx degreesabove freezing point on the Fahrenheit scale. And

what is the temperature on the Fahrenheit scale, that is,when the mercury stands at the point marked a: on the

Centigrade scale,what is themark 31 atwhich it stands on the

Fahrenheit scale — It is a: x 32. So that inmathematical shorthand we write

y a: x 32.

6. Seawater boils at a temperature of C., turpentineat a temperature of 156

°C. ,

mercury at a temperature of

350°C. lead melts at a temperature of 325

°C. , and zinc at

a temperature of 391 C. Give these temperatures on the

Fahrenheit scale .


7. Themelting point of tin rs 442°F. Howwill youfind the

temperature Centigrade at which it melts — The temperature 1s 442 32 or 410 Fahrenheit degrees above the freezing

point of water. And each of these degrees 1s1—1—8

Centigrade, so that it re 410 1 °8 or 228 degrees Centigradeabove the freezing point, that is, the temperature is 228

°C .

Could you have found this result from the relation3, 1 °8x+ 32

by putting 442 for y and then trying to find a value of a:

that will make the relation true That is, can you find avalue of a: thatwill make 442= 1 °82: 32 — The value of a:

that does this will make 442—32 or 410 equal to 1 °

,8x or

will make 410 1 °8 or 228 equal to x. Whereindoes this difl'

er from the previous method of finding themelting point of tin on the Centigrade scale — It difi’

ers

only in treating all the quantities as mere numbers, withoutreference to their physical meaning.

8 . The process of finding a value of a: that will make442 1 °8z+ 32 is called solving the equation

442 32,

and the value 228 is called the root of the equation. Can

you solve the equation 31 1 °8x+ 32 as an equation in 23,

that is, can you put the relation in such a form as to Showat once the operations to be performed to give a: as soon as

we know what 31 is In other words, can you express theCentigrade temperature a: in terms of the Fahrenheittemperature 31 —The relation 31 32 will be true if31—32 is equal to 1 °8x. And these are equal if (y—32)is equal to x. The relation is now in the required formx (y—32) -5 which directs us to subtract 32 from yand then divide by

9 . Reason out this expression for the Centigrade temperature in terms of the Fahrenheit temperature from the waythe thermometers are made and without reference to equations.

of a degree

10 . On both thermometers low temperatures are shownby continuing the uniform divisions of the scales below the

CHAP . VI, ARTS. 7-12 87

mark 0, and numbering them 0,—1

,—2

,-3

, and so on.

What is the temperature Centigrade of the mixture of snowand salt which is marked 0 on the Fahrenheit scale — It is

32 degrees Fahrenheit below the freezing point. And 32

degrees Fahrenheit are or degrees Centigrade, so that it is degrees Centigrade below 0

°C. ,

that is, the temperature is 17°8°C.

11 . Could you have used to find this temperature eitherof the relations y= 1 °8x+ 32 and x= (31—32) whichwere found for positive readings, that is, for temperaturesabove 0 on the two scales — In the present case 31 is 0,so that the relations are 0 1 °8z+ 32 and x= —32)

Now we have not hitherto met with anyoperation requiring the division of a quantity such as

32, and it has at present no meaning. Can you assign

it a meaning that will give 2: its proper valueIf we suppose 32) to have the same meaning as

that is, if we take —32 + l °8 to mean that32 is to be divided by and then to have the minussign put before it, this supposition will give 2: its propervalue (32 or

Again look at the thermometers. We may take 32 to

mean that from the freezing point wemust, on theFahrenheitscale, go 32 degrees down to get to the point in question, and— 32) to mean that we must go 32 degreesdown on the Centigrade scale. Which agrees with making—32) mean the same as (32

12. On the Centigrade scale the freezing point of mercuryis Find it on the Fahrenheit scale — It is

degrees Centigrade below the freezing point of water ; whichis x or degrees Fahrenheit below. And the

Fahrenheit zero is 32 degrees below, so that the pointrequired is —32 or degrees below the Fahrenheitzero. That is, the freezing point of mercury is F.

Trywhether the relation 3; 32 gives this result.Here a: so that y x 32. Here

again you have anew form x Letus takeas a direction to go degrees down the Centigrade scale

from the freezingpoint. Thismeans going x or

degrees down the Fahrenheit scale, so we may appropriately


make x mean x or Then32means go 32 degrees up again, so that we arrive at the

point —32 or degrees below the mozing point, orat the point called 32 or

13. If the same temperature is p degrees below zero on

the Fahrenheit scale and qdegrws below zero on the Centigrade scale, find the re lation between 19 and q, giving it inthe two forms, one suited for the calculation of p when q isknown and the other for the calculation ofqwhenp is known.

As a particular case consider the freezing point of mercury,for which 19

14. The freezing point of me has p and

q so that for this case pWhat temperature has p and q exactly equal Use the

relation you have found, namely, q== (p 1 °8 or

p 1 °8 q—32.

—Let us suppom p and q to belong to thetemperature we are in search of. We separate the term

x q into two parts l°0 x q and 0

°8 x q. The former is

equal top ; leaving x q to be made equal to 32, in order

to And x q will be -32 if q is32 or 40. Reasoning back againwe have x 40

32 = 0 and 40 0°8 x 40 32 40,that is

,1 °8 x 40

32 40. So that —40° F. and —40°C. mean the same

temperature.

Or we could proceed more formally to solve the equation1 °8q

—32 p when p q. We have to find a value of pthat willmake 1 °8p

—32 p . The two quantities 1 °8p—32

and 19 being equal will still be equal if we subtract 1) fromeach of them . They then are o°8p

—32 and 0, so that wewant to make 0°8p

—32 equal to 0, or to make 0°8p equalto 32

, and these last are equal if p 32 or 40.

15. We had the relation 31 1 °8r + 32 between the

Fahrenheit reading y and the corresponding Centigradereading 93. When these readings a: and y are the same

what is their value That is, what value of 2: makesa: 32

Let us subtract a: from each of the two equal quantitiesa: and 1 °8x+ 32. They became 0 and 0°8x+32. Now

x 32 is positive for all positive values of a: and even


17 A farmer has sheep hurdles of a total length of

100 yards, which he sets up to enclose a rectangular pieceof ground. How will the area enclosed vary as the shape of

the enclosure varies And when will the area be greatest ?If one side is 10 yards long, what are the lengths of the

other three, and what is the area — The sides are 10,40

,

40 yards, and the area 400 square yards. Calculate thearea, supposing in succession that one side has the lengths10 , 15, 20, 40 yards, and draw a graph showing thearea as a function of the length of this side . From the

graph read ofi the length for which the area is greatest ;what are then the lengths of the other three sidesIf one side is 2: yards long what are the other sides, and

the area —The other sides are x,50 x

,50 2: yards and

the area a: x (50 as) square yards. Sketch the

enclosure (Fig. and from your sketch give another form

FIG . 1 .

for the area— It is 50 x 2: x x x or 5022—233: squareCan you draw the graph of 50x xx? —It

is done already, it is the graph of the preceding paragraph.

Can you find avalue of a: that will make 5023—5132: equal to520 — From the graph we see that a: 35 will approxi

matelymake 5023—232: 520. By putting a: and

Show that the value of a: lies between these twovalues. What problem connected with our sheep enclosureis this that we have solved — We have found the shapeof the enclosure when its area is 520 square yards.

What other value for a: does the graph give for an area of

520 square yards, and how is it related to the former, andwhat is the shape of the enclosure for this value

18 . When the enclosure is square what is the length of

each side — 25 yards. And if one side is 31 yards

CHAP . VI, ARTS. 17-21 91

more than 25, what is the length of a side next it -It is25—31 yards. And the area ? x (25—y)

Sketch the enclosure (Fig. and givethe area in another form .

The area is 25x 25 -y x y

square yards or 625—yysquare yards. Showin a graph the area as a

function of 31. For whatvalue of y is the area greatest Forwhat value of y is

the area 520 square yardsSome time ago we had a

F13 ° 2°

graph for the area of a square in terms of the length of the

side. Could you use this tofindwhat value of ymakes 625w 520 —Yes, it is the same value that makes 3131 625—520, that is, 105

,and from the graph we find this

value to be

1 9 . On the same diagram draw the graphs of and

5023—520. Give the distances of a point where the twographs cross from the two axes

,that is, from the two lines

from which all distances are measured. Use these graphsto solve the equation 502: xx 520.

- y — p

-25

20 . By means of any of the methods discussed, can youfind how the hurdles must be arranged to give an area

(1) of 100 square yards, (2) of 400 square yards, (3) of 700square yards ? One of these areas is impossible ; in thiscase

1

apply each method and Show why it fails to give a

rosu t.

21 . In discussing enclosures made with 100 yards of

we used a: for the length (in yards) of a side and yfor the length by which a side exceeds 25. What is therelation between a: and y, when they refer to the same

enclosure —It is a: 25+31. What other re lationdo you get between a: and y by expressing the area of

the enclosure in terms of a: and of y— We get xx

625- yy.

Suppose we have 200 yards of hurdles and use a: and y in

the same way. What is the meaning of y and what is the


relation between x and y-A square made with these

hurdles has a side of 50 yards, so that y is the excess of a sideover 50 ; and x y 50. Andwhat relation betweenx andydowe get from the expressions for the area? —We get

100x—xx 2500—yy.

What are these relations ifwe have 4 x a yards of hurdles— They are x a+y and 2cx—xx aa—yy.

22. We have seen how to use the graph of 3131 to finda value of x that will make 50x—xx 520. Can you

use the same graph to find a value of x that will make3x—xx 1

This will be our old problem if we make the area 1, and

the distance round the rectangle 2x 3,in any correSponding

units, for instance in square yards and yards or in square cm.

and cm . Theny is taken so that x or y= x and

the area-relation is 3x xx x yy. We have thento make x —

31g or 225—3131 equal to 1, or tomakeyyequal to The graph of yy shows that yy has this valuewheny Then for xwe have

x 31+

Therefore is a value of x that makes 3x—xx 1 .

Ve

g-

fig7 this result by calculating the value of 3x xx when

x

23. Ifyou tryin the samewaytofind avalue ofx thatmakes3x— xx 4 (that is, to solve the equation 3x—xx 4

what difficulty do you meet

24. We have seen that approximately 50x xx 520 whenx 35. Calculate the value of 100x—2xx when x 35.

Can you give a value of x that will approximatelymake 100x 2xx 1040 — We have it already, namely,35. Can you solve the equations

l5ox 3xx 1560

200x 4xx 2080 ;500x 10xx 5200 ;5x xx 10 52

that is, can you find for each a value of x that will make therelation true


the value also for x and so show that the exactvalue of x lies between and

27. Of the methods used above to solve equations some

give two roots, that is, two values of x that make the relationtrue, some only one. In all cases the rectangle (when thereis a rectangle at all) has two unequal sides. In each case

where only one side of the rectangle is found as root, findthe length of the other side and see whether it is also a root.

Em ersas.

1 It is known that a jam tin a inches high and b inches indiameter will hold x a x b x b pounds of jam . Find whethera tin inches high and inches in diameter will hold a pound

of jam .

2. I want to buy about 20 ards of certain iron wire , but I findit is sold only by the poun The wire is of an inch indiameter, and iron weighs 470 pounds a cubic foot . The area of

the section of a

fiiece of wire of inches in diameter is 85x d x d

square inches ow many pounds must I ask for ?3 The weight in pounds of a foot of iron piping is given by the

if:(b t) when the thickness is t inches, and the bore isb inches ; w is the weight in pounds of a cubic foot of the iron .

With b 3, t g, the weight of a foot is found to be lb .

find w . Hence calculate, to the nearest pound, the weight ofa foot if

4. If vand V are the volumes of a certain quantity of gas, underconstant pressure at 0

°and t

°

Centigrade, respectively,t

7 17X (1 271-

3)Find, to the nearest tenth of a cubic centimetre, the volumeoccupied at 18

°

Centigrade by some gas which occupied 100 c .c at

0°

Centigrade .

5. If an oak beam L inches long, B inches wide and D inchesdeep is fixed at one end and a weight W cwt . is placed at the other

end, it will break should W exceed

15xB xD xD

L

Find with as little work as possible whether such a beam 10 feet

8 inches long, 1 foot 10 incheswide, and 4; inches deep will breakif a weight of 2 tons is placed at the free end .

CHAP . VI, EXERCISES 95

6 The effective horse-power of a certain kind of water-wheel isx Q xh, where Q is the supply of water in cubic feet per

min . and h is the head of water in feet .

Find the effective horse-power in a case where the head is 5feetand the water is sufficient to make a stream 10 feet wide, 2 feet

deep and flowing at the rate of 5feet per second .

7. (1 horses are bought at £x each, b horses at £31 each, and c at£2 each What is the ave e cost per horse ? If , when re-sold,the gain on each is one-tent of its cost price , what is the totalgain and the total sellingprice ?

Give results, both for the general case, and for the special case in

which a = 7, b-5, c = 8, x = 20, y = 36, z= 45.

8 A steam plough ploughs an acre in a minutes Express, in

hours, the time it will take to plough a field b yards long and0 yards wide .

9 . If a man walk p miles in qhours, find how far he will walk in5hours how far he will walk rn x hours ; how long heto walk 5miles how long he will take to walk y miles ?10 . A steamer goes 0.miles up stream in b hours, and then 0 miles

down in of hours . The stream flows e miles an hour and the boatcould

nffmiles an hour in still water . Express a and c in terms

of b, e

If a =’

14, b = 2, c = l4, what are e and f ?

11 In order to find the diameter of a tube of uniform bore

some“gai

n was

h

ured

f into it and the height of the column

meas e werg t o t e merc was 25°6 grams the hei htof the column cm . What was

ury g

the diameter of the tube ? A cubic

cm . of mercury weighs grams .

12. A pint of water weighs l °l25

unds, and a pint of milk between°155and unds .

of the two weig s 1 1 48 pounds per

pint Between what limits does

the percentage of milk by volume

lie

13. A shot into whose barrelwill exactly t a sphere of lead

weighing 117th of apound, is termed

an N-bore gun . The diameterFIG:3°

of the barrel of a 12-bore gun is 29 inch . What rs the drameterof the barrel of a 20-hore gun ?

14. ABDO is a square field (see rough sketch Fig. each


side being 120 yds . AB runs east, and AC north . A foot th

runs across the field, and one point of it is 20 yds. distant th

from AB and from AC . If by following the foot th you find at

any moment that the distance you have advene north is 3thedistance you have advanced east, show by a diagram drawn on

squared

Piper to a scale of 20 yds. to an inch, the exact position

of the ootpath, and give the distance from B at which thefoo th leaves the field.

x yards is the distance of any point of the path from AC and

y yards its distance from AB, what relation between anand y willhold true for all points of the path ?15. Find in sq. feet the area of the kite shown in Fig. 4 on

a scale of an inch to a foot .

FIG . 4.

Find a formula for the area of a kite of the same shape whosebreadth is a feet and state

what the area would be of a

kite of the same shape and

of breadth 3(1 feet .

16. ABCD (Fig. 5) is a plan

(on a scale of l inch to 2

chains) of a corner plot of

buildingland sellingat£2200

an acre . Find the price of

the plot . A chain is 66 feet.

17 The sides of a rect

angle are measured as 0.

inches and b inches. Write

down(I) the area of the

FIG 5° rectang e, (2) the distance

round the rectangle, (3) the difference between the lengths of the

sides.

CHAPTER VII

ON CRANES : PROBLEMS IN THREE DIMENSIONS

1. a s chapter will treat solid geometry from the experi

mental and intuitional standpoint in order to bring out andadd to the subconscious experience on which geometricalknowledge rests. Then we shall return to two-dimensionalquestions and so give the three-dimensional ideas time to

arrange themselves before the logical discussion of them is

undertaken.

Special carewith regard to the rate of advance isnecessaryhere. For the best result with regard both to knowledgeand to training the method should be mainly that of discovery by the pupils, and this is possible only with a slowadvance. If the advance is quicker the teaching can onlybe dogmatic. Still greater speed will make the pupil’s mind

hn

ezl

le

thya confused blank, and may even be a danger to

A simme form of crane

FIG . 1 .

turn round till it lies

rs the wall crane (Fig. BAand CA are two rods fixedrigidly to a wall at B and Cand to one another atA . Theload L is raised bymeans of arope passing over a pulley atA . Make a model crane of

bamboo rods and find experi

mentally how the load moves

as the rope AL is lengthenedor shortened— It moves ina straight line.

A wall crane is sometimes

hinged at B and C so that itflat against the wall. If the

CHAP . VII, ARTS. 1-4 99

length of rope AL is kept constant while the crane turns,howdoes the load move To find out arrange the length ALso that the load nearly touches the floor, and mark itsvariouspositions on the floor. And how does the point A move—By experiment both paths appear to be circles.3. How does the load move if the crane turns and

the rope is varied in length at the same time How does

the point A move if the rod AB is removed How does Amove if AB remains while AC is removed In each of

these cases the locus traced out is called a surface. Give a

few phrases from everyday life inwhich ‘surface occurs.

The surface of a pond, the surface of the earth, the surface ofthe body, surface deep.

The meaning of surface 1s not identical throughout thesephrases, and in some is not very definite . A thing that issurface-deep affects to some slight depth the body to whichthe surface belongs ; the surface of the earth may mean the

top layer of the earth or it may mean where the earth and

air meet ; the surface of a pond may mean a top layer orskin of water, or it may mean where water and air meet.In mathematical usage the surface of the earth or of a pondmeans where the earth or the water meets the air.

In the case of the crane the locus of the load is hardly asurface in the mathematical meaning because the load has a

certain size. When the load is very small the locus is morenearly a surface in the mathematical meaning. Ifwe couldhave a load that did not measure anything In any dire ction,its locus as the crane turns and the rope varies in lengthwould be a mathematical surface.

Is the wall of the room a surface Is the paper on

the wall a surface What is the surface of the wall4 . What path does the point A trace out as the crane

turns Lower themodel crane till C andA are close to thefloor (keeping BC upright) and see howA moves -A movesin a circle. What property of a circle has the path of

A —A rs always at the same distance from C. Is

this enough to show that the path of A Is a circle Remove

the rod BA ; then as CA turnsA need not stay close to thefloor. Does it still describe a circle How does it difierfrom the circles we had in chapter I -In chapter I a

H 2

100 CRANE PROBLEMS

circle was the path of a point that was always at the same

distance from a tree and was always on the ground. In

the present caseA will describe a circle if it stays on the floor.Is the g round a surface Is the floor a surface —No,

but when a point moves on the ground it is moving in a

surface, namely, in the surfacewhere ground and air meet, it

moves (in language that is at once general andmathematical)on the surface of the ground The floor is not a surface,it is about an inch thick ; but ‘

the surface of the floor on

which the point A moves, is a surface.

5. Try to fit your straight edge against the floor ; alsoagainst a ball ; and against the side of a standard gallon or

litre measure.—It fits ainst the floor in any position, it

will not fit against a bi at all, in some positions it fitsagainst a gallonmeasure and In others itwill not. Thesurface of the floor is called a plane because the ruler fits

in any position the other surfaces are not planes.Let two pupils stretch a string and try to fit it against the

ground. On some ground it will fit in any position, andthat ground (or rather the surface of it) is called plane.

Other ground may be lumpy so that the string will not fit ;such ground 1s not plane.

How re the word plain used In geography (It IS a pitysuch slight differences of meaning of the same word are

marked by different spellings, plane and plain. )

8 . When the bar BA of the crane is removed, and A

moved,

to all possible positions, what locus does it trace out-A surface, all points of which are at the same distance

a surface is called a sphere and Cis called its centre. Can you name any objects of the same

shape —Cricket ball, association football, the earth.

When the bar BA rs replaced and CA removed, how does

A move And what positions of A lie on this sphere and

at the same time on the above sphere with centre C —All

points of the circle on which A moves when the crane is

complete lie on both these spheres.

7. The wall-crane is of use to raise bodies from the

ground to upper stories. The shear-legs (Fig. 2) is used

102 CRANE PROBLEMS

G to and fro, the other by lengthening or shortening thechain ; we say that the load has two degre es of freedom.

In the case of the wall-crane, when the crane is fixed and

will not turn flat against the wall, the load has one motiononly, it moves only up and down. In this case we say the

load has one degree of freedom .

8 . Measure the distance DO in several positions. Whatdo you find Can you prove that the distance DO is alwaysthe same Measure in several positions the angle that DOmakeswith EF. Canyou prove this angle constant -E and

F are fixed points so that the distance EF does not change,and the lengths DE and DE (the lengths of the beams) ofcourse do not change ; so that the triangle DEF is

d only changes in position ; and DO, the linejoining a vertex to the middle point of the opposite side is

fixed too. SO, in our model, we could replace the rods DEand DE by a pasteboard triangle DEF. And 0 beingmidway between E and F, CE is equal to OF, while thelengths of the beams DE and DE are the same ; so that thetriangles DEO and DEOare congruent, that is, they wouldfit if folded together ; so that LDOE and LDOF are right

,

and D0 is perpendicular tDo you know enough to prove that D moves in a circle—D is always at the same distance from 0 but this is not

enough, the locus of all points at the same distance from 0

is a surface that is called a sphere. We know further thatthe constant length CD stands out always at right angles toEF. It may be that a line of constant length turning aboutanother and being alwaysperpendicular to it traces outa circlewith its extremity but it is not yet proved. Noticethat in the wall-crane the locus ofA (Fig. 1) is that of theend of a rod CA that turns about BC and is always perpondicular to it.

9 . If the bar GD is removed, how does D move —Asbefore. If the bar DE also is removed —D moves

on a sphere. If the bars DG and DE were removed,how would D move

10 . Suppose ashear-legshas GD metres long, DE and

DE metres long, and the feet E and F set metres

CHAP . VII, ARTS . 84 1 103

apart. Find the distance DOby drawing the triangle DEF(Fig. 3, which is only a sketch) on a scale of 1

—3—0 . Now

(Fig. 4,sketch) use this distance and the

known length GD to find where G mustbe when the shear-legs is to pick up a

load 2 metres in front of 0, and whatlength of chain is wanted to reach downto this load.

If the load is to be set down again3metres in front of 0, how far forwardmust G move And if it is to be set

down on the deck of a Ship 3metres infront of 0 and 1 metre below the groundlevel, how long must the chain then beGive measurements of the position of

the load that you could take so as afterFIG 3wards to replace the load in its position.

-We could give the distance in front of 0 and the distanceabove or below ground-level ; or the angle that CDmakeswith

FIG. 4.

OH and the length of the chain ; or the distance of G behind0 and the length of the chain ; or other sets ofmeasurements.

Traveller emue.

11 . Another form of crane is the traveller crane sometimesused in an engineer’

s yard. Figs. 5 and 6 Show twosketches of this crane, Fig. 5after the manner of a pictureFig. 6 shows it as seen from a greater distance, and is

technically termed a proj ection . ABCD is a rectangularframe supported on 4 pillars. A beam EF rests on 2 sides

of the rectangle and travels along them,

remaining

104 CRANE PROBLEMS

always parallel to the other 2 sides. The crane G travelsto and fro along EF and raises or lowers the load H. A bar

across a box without lid, and carrying the load by a string,will serve as model. Or two desks may be placed so that

FIG . 5.

their edges will serve as the sides BC and AD of the frame,

to carry the bar EF.

How does the crane move when the beam travels Howdoes the load move when the chain is lengthened, when the

FIG. 6.

crane travels, and when the beam travels How may theload move when any two of these motions take place at the

same time How when all three motions take placetogether — The locus for any single motion is (experimentally) a strai ht line, for any two motions a plane, and forall three motions at the same time a block of space. W hen

106 CRANE PROBLEMS

points and lines were assumed to lie on the sheet of paper,that is, in the same plane.

You have seen experimentally that when the crane G

travels the load H moves in a straight line. Can you

describe the position of this line —It is parallel to EF and

at the distance GH from it. For the length of HG is the

same all the time, and (experimentally) HG is always .L to

EF,and lies all the time in a certain plane .

14 . When the beam and the crane travel at the same time

how may the crane move — It may move to any pointin the plane ABCD. And how may the load move,

the length GH remaining constant —It also moves in

a plane,and the distance GH between the plane in which H

moves and the plane in which G moves is the same everyWhat angles doesGHmakewith lines in these

two planes drawn from G and from H ? To measure the

angles made with lines in the plane ABCD lay a board over

the model to cover ABEE or EFDC tomeasure the anglesmade with the plane H moves in, take the stringGH of sucha length that H almost touches the bottom of the box.

—Theangles are all right angles.

When a numberof lines are drawn in aplanethrough apointG (Fig. 7) andanother line GH stands so astobe perpendicular

FIG. 7.

to all these lines the line GE is said to be perp endicu larto the p lane . And when a number of lines (such as GH)are perpendicular to a plane at different points, and all have

CHAP . VII, ARTS. 14—16 107

the same length, the plane in which the ends of these lineslie is said to be paral le l to the first plane. SO that, in thepresent case, the locus of the load H (when the beam and

the crane travel) is a plane parallel to the plane of the frameABCD.

15. Give examples of parallel planes from the school ora house — The floor and the ceiling are parallel planes, thenorth and south walls are parallel planes, so are the east andwest walls. All the floors in the school or house are planesparallel to one another. Of course floor means

here not the boardswhich may be 2 or 3cm . thick, but their

upper surface on which we walk ; wall does not mean thestone or brick mass 30 or 40 cm. thick

,but the surface of it

that we see ; and so with ceilingThe line in which the rope or chain of a crane hangs is

called a plumb line or vertical l ine . Any rope or chainhung by one end hangs plumb orvertical. A planeperpendicular to a plumb orvertical line is called a horizon tal plane ;thus the floor, the ceiling, and the rectangle ABCD of the

traveller crane are horizontal. A plane against whicha plumb linewill hang (without pressing against it) is calleda vert ical plane ; the walls of a room are vertical planes.

18 . We saw in Article 12 that we can fix the position of

the load H (Figs. 5, 6) by stating how far the beam FEmustmove from the position AB, how far the crane must movefrom F, andwhat length of chain must be paid out, to bringthe load to the position in question. And the order of thesemotions is indifierent.In more mathematical language we may say that the

position of the load isfixed by its distances from three planes,

namely, the distance AF or BE from the plane ABQP, thedistance FG from the plane ADSP , and the distance GHfrom the plane ABCD. Or, again, we may say that a pointcan travel from A to H by going suitable distances in thedirection of (that is, along or parallel to) the three lines,AB, AD, and AP, namely, a distance AF in the direction of

AD,FG in the dire ction of AB, and GH in the direction of

AP . And the order of these movements is indifi‘

erent thusthe route may be along AFGH or AFWH or ATGH or

ATUH or AV'

UH or AVWH .

108 CRANE PROBLEMS

17 Arrange your model so that the distances travelledfrom A by the load are 80 cm . in the direction Of AD ,

50 cm. in the directionAB, and 40 cm . in the direction AP .

Measure how far the crane would have had to travel in a

straight line to get from A to its present position, and how farthe loadwould have had to travel in a straight line to get fromA to its position. Can you test these distances by drawing ?

—We lay down (Fig. 8,sketch) the length AF,

let us say on

a scale of g, so that we make AF 16cm . long ; and drawFG

FIG. 8. FIG. 9.

at right angles to it and 10 cm. long, to represent FG on

the same scale. We then measure AG ; it is cm. long,representing a distance on the model of 94 cm . We now

lay down AG cm. (Fig. 9, also a sketch) and drawGH at right angles 8 cm . long to represent 40 cm . We

thenmeasure AH and find it 205 cm. , so that the distanceof the load from A in the model is 102cm .

P lace the beam,crane, and load at random. Specify the

position of the load by the distances AF, AT, AV (Figs.

5and Measure the distances of the crane and the loadfrom A, and check by drawing as before.

Scotch crane.

18 . The Scotch crane (Fig. 10) has a movable partthat consists of the vertical post AB,

the j ib AC, and the

tie-rod BC. The post AB is hollow and a pillar or pivotfixed to the ground fits into the hollow, so that the cranemay revolve about this pivot. The load hangs from C bya chain that may be varied in length. AD is a platformthat carries the steam engine and tackle for working thecrane.

l 10 CRANE PROBLEMS

angles GFL and BAP are both right angles, and so willcoincide. We thus have EH coinciding with AC and FL

lying along AP . Further, we know (chap. V, art. 18) thattwo tri angles AGP and FHQ having AC,= FH,

LCAPLHFQ, and the angles at P and Q, right angles, will coin

cide. So that FQ AP ; or the point where the chainmeets the ground is always at the same distance from the

Repeat the reasoning by which you knowthe triangles FHQ and AGP to be congruent.Ex. 1 . For a crane that has a j ib 10 metres long and inclined at

40°

to the post, find, by drawing, the radius of the circle fromwhich loads can be picked up.

Howmany degrees of freedom has the load Howmanyhas the j ib-head C? The surface on which the load maymove is called a cylinder .

20 . The Scotch crane is sometimes mounted on a truck,which runs along rails, and gives a greater choice of positions from which loads can be picked up. Suppose a crane,

whose jib is 9 metres long, and makes an angle of 43°

with the post, to run on a straight railway track 30 metreslong. Find, by drawing, the locus of possible positions on

the ground for the load when the truck is still. —It isa circle of radius 6 1 metres. By considering this locusfor all possible positions of the truck, find and draw the

locus of all pOMble positions of the load on the ground;

FIG . 12.

scale rice— It is a straight piece of ground metres

wide ending in two semicircles (Fig.

2L When the truck is at rest and the j rb swingsround, what is the locus of the j ib-head —It is a circle of

radius 6°1 metres, at a height of metres above theIf at the same time the length of chain by

which the load hangs varies, to what positions can the load

CHAP . VII, ARTS. 20—22 111

be brought — It can be brought to any point verticallybelow a point on this circle, that is, to any point on thepart of a vertical cylinder between this circle and an equalcircle on the ground .

If at the same time the truck travels on the rails, to whatpositions can the load be brought —It can be brought toany point above the area shown in Fig. 12 not more thanmetres above the ground. Its locus is a block of space

roughly of the shape of a date-box, bounded by the area

shown in Fig. 12, by an equal area metres above it, andby the vertical lines joining these two areas.

Make a model in clay or plasticine of this block of space.

Derrick

22. The derrick crane has the post AB (Fig. 13) heldin position by two otherbeams or stays AE and AD

fixed to the ground at E

and D. Further, the j ib BCis supported by a chain ACwhich can be altered in lengthand so change the angle of

inclination of the j rb to the

vertical. Suppose the crane

post AB to be 10 metres long F16 4?»

and the j ib 18 metres. E and D are 10 metres from B

and the angle EBD is a right angle.

By drawing the triangle ABD to a suitable scale, sayhaving AB 10 cm. , BD 10 cm. ,

LABD find howlong AE and AD must be made. Use this result to makea bamboo model.Use the model to find the locus on the ground of possible

positions of the load when the chain AC is 13 metres

long.-It is part of a circle of radius 127metres, i

of the

circle,because the stays AE and AD keep the j i from

swinging right round. Check the radius of this

What is the inclination to the vertical of the j rb when thechain AC is 13metres long Find from the model and by

112 CRANE PROBLEMS

23. Find, from the model and by drawing, the locus of

possible positions of the load on the ground when the j ibmakes an angle of 28

° with the post, and when it makes an

angle of What is the len h of the chain AC in eachcase If the angle between t e j ib and the post may beanything between 28

°and 80

°what positions on the ground

are possible for a load Show the positions on a drawing.

—The positions all lie between two circles with centre B

and radii and 8 5metres,excluding the part boundedby the lines BE and BD

(Fig.

When the length of ACis 13metres, what positionsmay the load have for different lengths of the chain CLand for different positionsas the j ib swings round, the

length CL being unalteredWhen the length CL re

mains unaltered, and the j ibFIG 14

does not swing, but the

length AC is altered, howdoes the load move and how does the j ib-head C moveHow many motions (or degrees of freedom) has the j ib

head C? When these 2 motions take place what is thelocus of the j ib-head —A sphere of radius 18 metres and

centre B ,or rather the half of it above ground, excluding

the part cut 03 by the staysAD and AE . And if the

angle between the j ib and the post must lie between 28°

and what is the locus -Part of the half-sphere at thetop and part at the bottom are then impossible. A certainJapanese lamp-shade gives the form if we cut away thequarter made inaccessible by the stays AD and AE .

24 . How many degrees of freedom has the load Whatis the locus of the load for eachmotion -When the lengthCL changes the locus is the part of a vertical straightline between the j ib-head C and the ground. When the j ibswings round, the locus is 3 of a circle, the remainingquarter being cut off by the stays. When the j ib is raised

1 14 CRANE PROBLEMS

Fig. 18 . Model this block of space in clay or

plasticine.

And if the angle between the j ib and the crane-post mustlie between 28° and what is the locus For the locuswe must cut away from Fig. 18 all that lies within the

FIG. 18.

vertical cylinder standing on the circle of radiusmetres (on our reduced scale), and all that lies outsidethe cylinder that stands on the circle of radius 17°7metres

Model this block of space.

25. Suppose the j ib to make an angle of 40° with the

crane-post, and to lie at first against the stay AD and then

FIG . 19.

to swing round till it is in the same plane as the stay AE .

Suppose the load L to hang just clear of the ground all the

time . Draw a figure of the ground Showing the feet B D Eof the crane-post and the stays, and showing the positionof the load when LDBL is equal to 30° (Fig.

A sort of picture of the crane may be made by setting

CHAP . VII,ART. 25 115

up a sheet of paper against AB and AE ,and drawing from

each point of the crane a perpendicular to this plane . The

FIG. 20.

foot of the perpendicular represents the point from which itis drawn. Draw this picture .

— The perpendiculars frompoints on the stay AD fall on the crane-post so that oneline AB (Fig. 21) serves for this stay and the post. L being

FIG. 21 .

on the groundwe can draw the perpendicularLL’ in Fig. 20,and so find the point L’ that represents the load in our

picture. The line CL will have the same length in the

picture as in its actual position. By drawing (Fig. 22)the-j ib at 40

°with the crane-post we find the length of

CL to be metres so we draw (Fig. 21) at L’a line

L’C

’.L to EEL

’and showing a length of metres on

the scale we are using. Now by joining A to C’ we finish

our picture Such a picture is called a proj ection.

Call Q the position of the load when it has swung roundthrough 60° from BD, R the position when it has swung

I 2

1 16 CRANE PROBLEMS

through and make pictures of these positions. Thesepictures are shown in Fig . 21 by dotted lines.

28 . It was seen (Arts. 12 and 16) that the position of

the load on a traveller crane can be Specified by givingits distances north, east, and downwards from the corner

A (Figs. 5, The load can be moved only to pointsin the block of space enclosed by the frame that supportsthe crane ; but if we wish merely to specify a particularposition and not to move the load to that point, we neednot restrict ourselves to that block of Space. W e may statethe position of any point by giving its distance north or

’

FIG. 22.

south of A , its distance east or west of A , and its distanceor below A . When the position of a point is given

in this way these distances are called co-ordinate s, thelines AB , AD, AP are called cc-ord inat e axes, and thepoint

.

A is called the origin o f co-ordinate s. Of two

opposite directions, e. g. north and south, it is usual tochoose one

, say north, and to agree that a distance statedsimmy,withoutmention of north or south

, is to be measurednorth; while a distance to be measured to the south hasa m inus Sign attached. Thus if a point has a cc-ordinateof 3 metres for the axis AB, the point is 3 metres northof A ; another point with a cc-ordinate —3metres is 3metressouth of A .

27. Again, suppose that a derrick crane has a b BFig . 13) that can be shortened or lengthened as flinchSS

118 CRANE PROBLEMS

faces have they —If one is rubbed to and fro on the otherthey may come to fit as the axle of a wheel fits into the

ut we can prevent that by turning one of themabout, and rubbing till they fit in all positions.

Do you know any other surfaces than planes that fit in all

positions Do an egg and egg-cup fit in all positions

They fit if we only rotate the egg without trying to turn itupside down. We can turn it upside down only by taking itout of the cup. So that they do not fit in all positions.

Do the cup and ball used in the game called Cup and Ballfit in all positions If a ball is pressed into a heap of

sand so as to make a hole that it fits, will it fit in all

positions —Yes, in both cases.

So that if two pieces of sandstone are rubbed together tillthey fit in all positions it might happen that one had a surface

like a ball, a spherical surface, while the other had a hollowsurface to fit it, also a spherical surface. If we took threepieces of sandstone and rubbed them till they fitted two and

two in every position, could their surfaces be sphericalNo, forwhen two spherical surfaces fit, one is round like a balland the other hollow

,so that these cannot both fit the third

one. What kind of surfaceswill these three pieces of

sandstone have —They may have plane surfaces.

As a matter of fact,when a true plane surface is wanted,

three of them are made at once in this way. Three platesof metal are taken and ground away till their faces fit

,two

and two, in all positions.

Having three plane surfaces made in this way, how couldyou make a straight line —If one of the plates is cut across,and then the side ground till a second plate fits on the side

just as it does on the face, the edge where face and side meetis a straight line.

Exsncrsss.

2. In a room whose walls run north and south, and east andwest, an electric glow lamp is to be placed at a certain point. Thewire enters the room at the north-east corner of the ceilin and is

to run everywhere parallel to an edge of the room. 6 linewhere two walls, or a wall and ceiling, or a wall and floor meet isan edge. ) Show how the wire m

'

ht be run actually, and whatother ways are possible if you no not trouble about supporting

CHAP . VII, EXERCISES 119

the wire or about the convenience of people walking about the

room .

lat co-ordinates would you use to specify the position of the

lamp

3. How many degrees of freedom has a railway train ? Howmany has a steamship ? And an airship ?

4. A derrick crane has a j ib 20 metres long and post 9 metreslong. When the chain AO (Fig. 13) is 17metres long it picks upa load, and then without chan e of the length CL the chain AC is

shortened to 15metres. Find y drawing through what height theload is raised.

The crane picks u a load from the ground when the j ib makesan angle of 50

°wit the vertical. The jib is then raised till it

makes35°with the vertical, then the crane swings round throughand deposits the load by lengthening the chain CL b metres.

Whgt i

l

s

0211s

?

difi‘

erence in height between thefirst and t positions

of t e

5. In building a bridge materials are sometimes brought intot ion by running them along a cable whose two ends are madegift

!

to posts on the two banks. Suppose the tops A and B of

the posts 30 metres a and the cable34metres long, and supposeeach of the cab e from the load W to a post to be strai ht.Find yexperimentwith a model and by drawing, the path the oaddescribes as it travels along the cable.

The load will of course always be between and below the pointsA and B, and in the vertical plane through A and B . But imagineit to be able to take all positions in the vertical plane the sum of

whose distances from A and B is 34 metres. Draw the locus of all

points it could then reach. This locus is called an e llipse.

Again, eu pose the load not confined to this vertical plane, butable to reac any position in or out of the plane the sum of whosedistances from A and B is 34 metres. Investigate the form of the

it could describe, and make a model of the surface. Thissurface is called an e llipso id. If the load is made fast to a point

of the cable, so that its distances from A and B are always the

same, what kind of path can it describe ?

6. A board has cut in it two grooves crossing at right angles.A rod AB has on it two knobs at A and .Bwhich slide along theooves. Investigate the kind of path any point P of the rod

ascribes. For this purpose draw two straight lines at right anglesto represent the grooves, and drawon tracing

-paper a line with the

knobs A and B marked to represent the rod.

Draw the path described byP for a number of different sitionsof P on the rod, some between A and B and some outsi e. For

120 CRANE PROBLEMS

what positions of P on the rod are the ovals flatter and for whatpositions are they rounder ? All these ovals are called ellipses.

7. A shear-legs (Fig. 2) has a backstay GD 9 metres long, thehead D moves in a circle of radius 7metres, and the foot of thebackstay may not come within 3 metres of O (the centre about

which D turns) , nor may it go further from it than 7metres. Find

how much of its circle D may describe, and where a load must liefor the shear-legs to be able to pick it up.

8. A plumb-line hangs from a point A on the ceiling of a room ;

FIG . 23.

two knots on the line, respectively7°6feet and feet from the floor

BC, have their shadows thrown by an arc lamp at distances of

feet and feet from the point where the line meets

the floor. How high is the lam above the floor, and how far

is it from the plumb-line ? Fin out by drawing on a scale of

1 cm . to 1 foot.

CHAPTER VIII

THE ANGLE-SUM PROPERTY OF A TRIANGLE .

PARALLELS AND AREAS

1 . IN chapter Vwe saw thatwhen two angles of a trianglehad been made we had no choice about the size of the thirdangle, we saw that there was some relation among the threeangles. W e will now discuss the nature of this relation.

A knowledge of this relation may help us tomake a trianglefrom such data as a cm . , B A Let the

pupils use the methods of chapter V to make this triangletheyv

ill notice that they are clumsier than if we knewa,B ,

2. Make a right angle by folding a sheet of paper and

then folding the crease on itself. Cut along the creases,

and from one of the four

pieces of paper cut a largeright-angled triangle ABC

(Fig . Fold the corners

A and C on to the corner B,

B being the right angle.

How do the angles A and 0

now lie Does this experi

ment tell you anything about0 the size of the two angles

Fro . 1 ° togetherIn this and the following experiments each pupil should

make his own triangle. The triangles will thus be variedin shape, and the evidence of more value than if all usedtriangles of the same shape.

3. Now let each cut out a paper triangle of any shape .

You know already that one side of a triangle is often chosen

CHAP . VIII, ARTS . 1—4 123

out and called the base. When one side has been chosenfor base, the angle opposite to it is called the ve rtex. Choosea base in each triangle, in such a way that the obtuse angleof triangles that happen to be obtuse-angled is the vertex,and call the vertex A . Fold the base BC on itself in sucha way that the crease passes through A (0A in Fig.

FIG. 2.

Now open out the triangle and fold all three corners on to

0 . What do you observe Does the experiment tell youanything about the sum of the three anglesNow out along the crease 0A , andwhat have you —Two

right-angled triangles, each giving the case of Art. 2 over

4 . Let each cut out another triangle of any shape . Tearoff the three corners, and

place the angles side byside as in Fig. 3. Whatdo you observe Whatdoes it tell about the sum

of the three anglesAll these results are

subject to the drawbacksinherent in experiment,one of which is that the FIG. 3.

result may be true for theparticular triangle dealt with , but not for other triangles.

This objection is to some extent met by the variety of

triangles used,so that we know that the result is true for

a considerable number of particular triangles at least. A

more serious drawback is the unavoidable error in drawing,folding, measuring, and judging the fitting of the angles.

Even if we could be sure there was no error of a tenth of

a degree, we should still be in doubt whether there might

124 THE ANGLE-SUMPROPERTY OFA TRIANGLE

be errors of a hundredth of a degree. The following discussion is less subject to these drawbacks.

5. Let the whole class now draw triangles, each as helikes. Let each measure the three angles of his triangle,add them up, and report the result. These results willnot differ much, except that one or two may difi

’

er fromthe rest by ten or twenty degrees or more, and are clearlydue to blunders. Let these latter be discarded and the

average of the remainder calculated.

This average is likely to be nearer the true value thanany individual result. For some individual results will betoo great and some too small, and one kind of error is as

likely as the other, so that when we take the average wemay expect the errors to cancel one another to some extent.

6. A pavement is sometimes made up of a number of

triangular tiles, all of the same size and shape, that is, allcongruent to one another. Look at the drawing of sucha pavement 4

, in which the angles of one triangle

FIG . 4

are marked A B 0. Mark all the other angles of the pavement that are equal to A, all those equal to B, and all thoseequal to 0. How many of each kind are there at a corner

where the tiles join And through what angle do youturn if you first face in any direction and then turn till you


T+E + U= 540°or six right angles ; P + Q+R is the dif

ference,namely 180° or two right angles.

Any angle, such as S or T or U, between one side of

a triangle and another side produced is called an exteriorangle Of the triangle. What is the relation between S, B, QBetween T, B , .P Between U, P , Q — S together withP m akes and Q+ 12together with P make so thatS Q+R .

The pupils will now be able at once, or with a littlequestioning, to state the results formally :The 3 angles of a triangle toge ther make 1 80

d egre es or 2 right angl es .

Any exterior angle of a triangle is equal to th e

two inte rior and Opposite angle s tak en together(that is, to the two angles of the triangle that are not besidethe exterior angle).Discuss the results you arrive at by setting out in the

other direction round the triangle of Fig. 5.—The results

are as before, except that the angles vertically OppositeS T U take the place of S T U. Repeat thereasoning by which you saw that vertically Opposite anglesare equal (chap. 11, art.

8 . [Note for the teacher. The conclusion that in walking

round a room, or round a triangle, one turns through 360difi

'

ers from the preceding experimental conclusions. It isnot a necessary result of the fact that in turning round on

one spot one turns through in fact, it is possible tosuppose that the angle turned through in going round

a room or a triangle differs from 360°and to work out

a geometry on that supposition. It is only a conclusionthat is in keeping with all our Oxperience ; it is what wemay call an intuition. A geometry worked out on thisintuition represents our experience very well, while thegeometries worked out on other suppositions fail in some

respects to represent our experience.

This experimental or experiential result may be made thefoundation of the theory of parallels, in place of the ex

perimental result of chap. 11, art. 2. To this we now

These remarks are of course unsuitable for considerationby children ; they are for the teacher.]

CHAP . VIII, ARTS . 8-10 127

9 . Draw a straight line, and at two points A and B

on it erect perpendiculars AC and BD (Fig. At a

number of points on BD erect perpendiculars EF,&c. ,

to

BD. Measure the angles these perpendiculars EF, &c. ,makewith AC and the length of each perpendicular interceptedbetween BD and AC. We have thus the experimentalresult that any perpendicular to BD is also perpendicular toAC, and that the perpendicular distance between BD and

AC is the same all along.

We shall now see that these properties follow from the

fact that the three angles of a triangle together make tworight angles, which we will for shortness call the angle

sum pr oper ty of a tr iang le .

10 . In Fig. 6 draw AG making an angle of 70° with AB .

At G, where this line meets BD,draw GH perpendicular to

BD, meeting AC in H. Now,without measuring, find the

value of all the other angles of the figure— Since GAH

and GAB make GAE is The angle-sum propertyapplied to the triangle ABC ,

in which LA 70°and

LB gives LBGA Then since BGA and

ACE make ACE is 7 The angle-sum property of

the triangle AGH now gives LAHG And so

on. Check these values by measuring all the angles.Are there two congruent triangles in the figure, that is,

two triangles such that if one was cut out it could be fittedon to the other -If AAGH is out out and turned round

128 THE ANGLE-SUMPROPERTY OF A TRIANGLE

so that its angle of 70° comes to A and its angle of 20°to

G (which can be done, since the distance between theseangles is AG) ; then its side p will lie along AB and its

side q along GB,and the triangles will fit. SO that the

distance GH between AC and ED is equal to the distanceAB between these lines.Actually cut out the triangle AHG of your figure and fit

it on ABG.

11 . Make Fig. 6 over again, but this time draw the lineAG making any angle with AB, and suppose that this anglecontains 0: degrees. Express all the other angles of thefigure in terms Of z, and repeat Art. 10 with the necessarychanges. We find that in this case also GE is at rightangles to AC and is equal to AB . And in this case the

point G may lie anywhere on BD, so that we know that

any line perpendicular to BD is also perpendicular to AC,and that the perpendicular distance between AC and BD is

the same all along.

GE was drawn perpendicular to BD. Could you showthe same results by beginning with a line drawn perpendicular to AC

‘

P

The lines AC and BD are said to be paral le l. In

general two lines are called parallel when a perpendicularto one is perpendicular also to the other and the distancebetween them measured along the perpendicular is the same

12. Draw two parallel lines (by making both perpen

dicular to a third), and draw another cutting them in 0 and

Q (Fig. Measure all the 8 angles of the figure. How

many difi’erent sizes of angles are there ‘

P


The case of OQ falling on WX and YZ so as to make theother pair Of alternate angles (C and F) equal, and all thecases of corresponding angles are left as exercises.The conclusion is that if a line fal ls across two othe rs

so as to m ake a pair of al ternate ang les, or a pairof corresponding angles equa l , the two l ines are

paral le l. In this case, as always, the pupils should be ledto give the statement Of the proposition.

Constmctions for parallels.

14 . Are any of your results of use for drawing parallelsin a simpler way than by making the two parallels perpendicular to the same line —W e have already (Art. 13)drawn parallels by making a pair of alternate angles equal,or a pair of corresponding angles equal. Does

this method enable you to draw through a given point Ca line parallel to a given line AB (Fig. 8) —Draw any

FIG. 8.

line through C to meet AB in D. W ith D as a centredraw an arc EF of a circle from DA to DC ; and withcentre C and the same radius draw an arc GH , G lying

.

on

CD . Open your compasses to reach from E to F, and Withcentre G and this radius draw an arc cutting GH in K .

Join CK. The alternate angles ADC and DCK which thetransversal CD makes with AB and CK being equal, CK is

parallel to AB . Repeat the proof that the anglesADC and DCK are equal.Ex. 1 . Draw a parallel to a given line, using only a piece

of paper or a piece of tracing-paper to measure angles and to

carry angles .

15. Variousmechanical means are used to draw parallels.

In each case the pupils should be provided with the apparatus and left to discover how to use it.

CHAP . VIII, ARTS. 131

One means is a wooden triangle called a set-square.

One edge p of the set-square is placed against the given lineAB (Fig. A ruler is placed against a second edge q of

the set-square, and the set-square is slid along the ruler tillthe edge p comes against the point C through which theparallel is to be drawn . Show that the line CD ruled alongthe edge p of the set-square is parallel to AB .

—The edge of

the ruler is in this case the transversal, and the two linesAB and CD make with it equal corresponding angles, eachangle being the angle O of the set-square.

Another means is the drawing-board and T-square (Fig.

TheT-square is a rulerAB with a cross-piece CDscrewed on below one end,so that when the ruler lieson the board the cross-pieceis below the surface of the

board, and fits against theside of the board. Showthat all the lines ruled withCD in contact with the

same side of the board are

If a set-square is set on theT-square and a line ruled

FIG. 10.

against an edge of the set-square, and the T-square slid

132 THE ANGLE-SUM PROPERTY OFA TRIANGLE

along the board while the set square slides along the

T-square, will a second line ruled against the same edge

of the set-square be parallel to the first line16. A third means is a pair of paralle l ru l e rs (Fig.

A and C are two points on one ruler so placed that AC is

parallel to the edges, B and D are two points on the othersuch that BD is parallel to its edges, and the distance BD

'

is

equal to the distance AC. Two equal rods AB and CD are

hinged to these points. Now hold one ruler still, move the

other about and rule a number of lines against it. Can you

show that these are all el

Measure AC or BD and one of the rods, and draw the

our sided figure (or quadrilateral) ACDB'

in a number of itspossible shapes, keeping AC in the same position for themall. What do you observe about the various positionsof BD

Figg. 12 being one of the possible positions, and the

lengths AB and AC being of course known, what furthermeasurements do you need to

0 make to be able to copy thefigure —The lengthBC woulddo, or the angle BAC, or the

length AD,one measurement

more in every case . Suppose

BC measured and the figure

drawn, including BC. DO you

FIG . 12. see in Fig. 12 any pair of

congruent triangles, that is,any pair of triangles that would fit on to one anotherThe triangles ABC and DBC have AB CD,

AC BD,

and BC BC, so that they are congruent. Repeatthe proof by which it was shown that the two trianglescan be made to coincide, and name the equal pairs of


parallelograms, and measure the opposite Sides. Thendiscuss whether the proposition is true. —It is true, for ifBD AC and BA CD (Fig. drawing the transversalBC gives two triangles ABC and BCD in whichLABO= LDCB,

LACE ==LDBC, and BC = BC, that is,

in?congruent triangles in which AB CD and

BD .

19 . See whether a 4-sided figure ABDO, in which BDis equal to AC and also parallel to AC, is a parallelogram .

First, draw a few figures from these data.— This time

the 2 triangles (Fig. 12) have BD AC, BC BC,LDBC LBCA, so that they are congruent and LABOLBCD ; which shows AB to be parallel to CD.

See whether a 4-sided figure ABDO in which BD = ACand AB I I CD is a parallelogram. First, draw a few figuresfrom these data.

—We lay down BD and draw BP and DQparallel to one another. Then we take C anywhere on DQand with C as centre and a radius equal to BD we draw a

FIG. 14.

circle cutting BP in A (Fig. This Figure BDCA has

BD=AC and AB CD. Is the figure a parallelogram,

that is,is AC parallel to BD —Two positions are possible

for A ,namely the two intersections of the circle with BP .

In one of the figures (Fig. a) AC is parallel to BD as nearlyas we can measure ; in the other (Fig. b) AC is obviouslynot parallel to BD .

CHAP . VIII, ARTS. 19-21 135

20 . Draw DM and CN perpendicular to DCmeeting BAin M and N. What more do you know of DM and ON—That DM and ON are both .L to AB is shorthandfor perpendicular or is perpendicular to and thatDM CN, each being the distance between the parallels.

Are there any congruent triangles in the figure -Triangles BDM and CAN are congruent ; for they have theangles at M and N equal, the sides DM and ON equal, andthe sides DB and CA equal ; and the sides opposite theequal angles are greater than the other pair of equalsides. How do you know that BD is greater thanDM — We saw that, given the lines BD and BP,

the leastradius of a circle with centre D that would meet BP wasthe perpendicular from D on BP . Repeat the proofthat two triangles can be fitted on to one another if they havetwo pairs of sides equal and the pair of angles opposite thegreater Sides also equal (chap. V) .Can you now tell whether either of the cases of Fig. 14

is a parallelogram -From the congruence of DBM and

CAN we know that the angles DBM and CAN are equal,

or, in other words, BD and CA are equally inclined to BA .

In the case in which the equal angles are correspondingangles made by the transversal AB with the lines BD andAC (Fig. l4a) the figure BDCA is a parallelogram. In

Fig. 141) the equal angles are not corresponding angles, andthe figure BDCA is not a parallelogram .

Show that in Fig. 141) BD and AC are equally inclinedto CD also.

21 . Let us now summarize our results. One small pointnot explicitly proved is left as an exercise.

The diagonal of a paral le logram (that is, the linejoining two opposite corners) divide s the paral le logram into two congruent triangles ; each pair of

Opposite sides are equal ; each pair of oppositeangles are equal .A lso a fou r-S ided figure is a paral le logram and

has th ese prop e r t ies,prov ided each pair of Opposite

s ides are equal ; or provided one pair of oppositesides are both equal and para l le l.If a fou r-sided figure has one pair of opposite

sides equal and the o ther pair para l le l , the equal

136 THE ANGLE-SUM PROPERTY OFA TRIANGLE

pair are equal ly inclined to the para l le l pair , bu ttwo kinds of figure are possible , on ly one of wh ichis a paral le logram .

Ex. 2. A four-sided figure or quadrilateral having the four sidesa b c d, and the four angles P Q R S, as in Fig. 15; findwhether it is a parallelogram

22. Draw two linesAB and AD at right angles. ThroughB draw a line parallel to AD and through D a line parallelto AB, these lines meeting in C. What can you tell of theangles of this figure by general reasoning And if a

number of lines perpendicular to AB and CD and ending onthese lines were drawn Such a parallelogram,

with itsangles all right angles, is called a re ctangle . We saw a

good deal of rectangles in our discussion of area, and weused the terms l ength and breadth without inquiry intotheir exact meaning. We now see that the length is theperpendicular distance between one pair Of parallel sides

,

which distance is the same all along, and the breadth thedistance between the other two sides .

What name do you give the rectangle when its sides areall equal in length -It is a square .

Name a set of data that will determine a rectangleThe length and breadth will do. Name a set that willdetermine a square.

—The length of the side is enough.

23. Areas.— In chapter HI we found experimentally that

the diagonal Of a rectangle divides it into two parts Of

equal area. Can you now give a formal proof —It has

been proved for a parallelogram,and a rectangle is a particularkind of parallelogram. Repeat the proof for the caseOf the rectangle .

We found further, experimentally (chap. III, art. that

a c and

a II c and P + S =

a ll c and P ==R ,

c and

c and

a c and P = R ,

P = R and

P = R 8nd

138 -THE ANGLE-SUM PROPERTY OF A TRIANGLE

Ex . 5. In chap. III, art. 12, a triangle was made from a

rectangle by cutting off two corners, and these corners were ex

perimentally fitted to the triangle to exactly cover it. Provethis result by general reasoning, and thus justify the e ressionf

aun

ldfo

l

i

;the area Of a triangle, half the product Of the base by

t e eig t .

24. Cut out a paper parallelogram ABCD (Fig. 17) andsee if you can cut it

in two and fit it togetherinto a rectangle.

-Cut

it along a lineE FperpenH dicular to two opposite

FIG . 17,sides, and then fitBCandAD together. You

thus have the experimental result. Can you prove it to bemore than experimental

,to be more accurate than errors of

execution and observation make it possible for a mere

experiment to Show —Produce DC to G and AB to H.

AD and BC are equal and so can be fitted together. Thecorresponding angles ADE and BCG are equal, so that DEwill lie along CG. The angle DAF is equal to the corre

sponding angle CBH,so that AF will lie along BH. Thus

ADEF will lie as ECKL,and the angles made by KL with

CK and BL are the angles DEF and AFE , that is, rightangles. Therefore the fitting together makes a rectangle

Give an expression for the area of the parallelogram.—It

is the same as of the rectangle, namely FL xFF,or, Since

FL is simply the length AB,

it is AB xEF,the product of

a side by the distance be

tween it and the parallel side.Or, using the usual terms, it

is the product of the base bythe height.Does it matter which side

you choose for base -If in

FIG. 13.

the case of Fig. 17 we choseAD for base, we could not

draw a perpendicular to AD that would meet BC. DM(Fig. 18) is the best we could do. Make the parallelo

CHAP . VIII, ARTS. 24, 25 139

grain in paper, cut off the piece ADM and fit it in the

position DNO. What will you then do to make the figurea rectangle -Cut off the piece OCP and fit it in theposition MBQ. Prove that this gives a rectangle,and therefore, in this case also, the area is the product ofthe base and the height.Ex . 6. Cut out a long thin parallelogram that will have to be cut

into several pieces to fit again into a rectangle, and Showhow todo it .

Ex. 7. Cut out a ieee of r of the form called a trapezium

(Fig. that is, a our-Sid gure with one pair of oppOSite sides

FIG. 19.

parallel, and show how to cut it up and make it into a rectangle .

(Take E and F the middle points of AD and BC, and out alongfrom these points to AB. )Also give an expression for the area of a trapezium in terms of

the lengths of the parallel sides and the distance between them.

Calculate the area of Fig. 19 from measurements.

FIG . 20.

25. Draw any quadrilateral ABCD, and find its area.


We join BD, draw perpendiculars from C and A on BD, and

use for each triangle the expression half the base by theheight Draw a number of triangles on BD as baseand equal to CBD in area, as EBD, FDB, &c. (Fig.

What do you notice about these triangles, or what propertydid you use to draw them —They all have the same

height, so that EFCGH is a straight line parallel toBD. Can you make a single triangle equal to thequadrilateral ABCD —If BH is in line with AB ,

the twotriangles ABD and HBD form a single triangle AHD.

To make DE in line with AD would also do. Statehow to make a triangle equal in area to a quadABCD.

— Through C draw a parallel to BD to

meet AB produced in H. Then ADH is a triangle equal to

26. In the same way make a quadrilateral equal in area

to a given five-sided figure. Show how to make a triangleequal to anypolygon, that is, to any area bounded by straightlines.

27. Draw a triangle with sides of 14, 11, 7cm. , and tryto make a square equal to it. If the side of the square isx cm . what do you know of a: —We measure the heightof the triangle and so find the area to be 37'9 sq. cm. Thenxxx 379 , and our graph of xxx (chap . III

, art. 17) givesus a: and we draw the square.

This method is a trifle tedious and not very exact wemay find a better method later .

Show how to make a square equal to any polygon.

ExEROISEs.

8. Bisect the angles of a rectangular Sheet of paper by fold

ing adjacent sides together, thus Obtaining four crease lines

running across the paper from com ers to sides. At the middle of

the per a quadrilateral is thus formed bythe creases. Determineits 8 pa in any way you like, and justify your statement by a

geometrical proof.

9 . Use the angle-sum prop

erty to make a triangle havinga 11 cm. , A C y calculating B and then using thevalues of a , B, 0Draw any two acute angles and mark them A and C. Without

measuringA and C, make a triangle having a 11 cm. and A and C


crease between AD and BC and compare it with half the sum ofAB and CD.

Now draw perpendiculars from C and D to AB and from A andB to the line of the crease produced, and give a

proof of the relation between the lengths of AB, C and the

14. From a corner A of a rectangular sheet of paper measureOfl

‘

AB along one edge a 24 cm. , and AC along the other 19cm. Join BC and cut along this line . Fold the triangle to

make a crease AO through A and at right angles to BC, meetingBC in 0 . Open the triangle out and fold the corners A, B, Cover to meet at O. the triangle out again and, in any wayyou please, make a f -size dra of it, showing the creases bydotted lines. Then answer either (a) or (b) .(a) State whether the three corners meet so that the part

folded down just covers up the rest of the triangle, giving reasonswhy this should or should not be the case .

(b What conclusion may be drawn about the sum of the anglesof t e triangle, and about the area of the triangle ? Give an

independent proof of one of these conclusions.

15. Imagine a triangle cut out and placed on the desk before you in any definite position . State precisely in each case

the measurementsa

n would take (a) to draw a fi re of thesame shape ; (b) to w a figure of the same size an shape ; (c)to enable you to make a drawing of the in exactly the same

position if the original were removed. on are to give themum number of measurements necessary in each case.

16. The four bars of a bicycle frame taken in order have lengths23, 58, 62, 56 cm. Take four strips of cardboard and jointhem by eyelets or paper fasteners into such a frame, making eachside 1th of its true length the joints to be loose.

Is the form of the frame settled by the given lengths ? Howwould you add another bar to fix the form ?An actual bicycle frame is fixed in shape with no such additional

bar . How is that ?17. A four-sided field has two sides parallel and of lengths

260 and 200 yards, a third side J. to them and 220 yards long.

The fourth side is bounded by a footpath which is to be moved so

as to make the field rectangular Without loss or gain of area.

Draw the field, using 1 cm . to represent 20 yards, and show howthe fourth side must be altered .

What is the area of the field (in acres)18. ABCD is a rectilinear figure bavin AD parallel and

unequa

él to BC. Show, with proof, how to e a rectangle equal

to AB D .

CHAP . VIII, EXERCISES 143

AA'

, BB’

, 00 , DD'

, EE’are parallel straight lines. The

distance between each and the next being h ft . and their lengthsa, b, c, d, e ft. res ively, find an expression for the area of the

figure CBA. When a == 5 c 82 ,

d = 6°3, e = 2°4, h = 3°1 , draw the fi to scale, calculate

the area, and mark it on the figure . the figure completelyspecified ?

19. A 20-lb . steel plate originallymeasures 10 ft . , by 4 ft. 6 in .

at one end and 4 ft. 9 in. at the other, the width measurementsbeing square to the 10 ft . edge . It is first sheared parallel as faras possible to a width of 4 ft . 7 in . , next one uare corner

measuring 3 ft. x 2ft. is cut ofl'

, next a circular he 9 12 in. in

diameter is out in it, and, finally, a rectangular manhole measurin 23 in. b lb in. is cut in it.

t is t e weight in lb . of the remaining piece of the original

20 . Write down the values Of 1 a: for the following values of15 : 01 , 1 , 2, 5, 10. Show by a graph how 1 + m

varies as as increases from 01 to 10, taking one inch as unit .

From the gra

ph red ofl

'

the value of 1 Why is the graphnot very useful or reading ofl

'

the value of

21 . Water has to be diverted from a river through a six

inch diameter pipe running full bore at a velocity of one foot per

second to irrigate a field of 20 acres. How long will it take todeliver an inch ofwater over the whole area ?22. Suppose you are asked to make a of a four-sided

figure ABCD in which AB and figure beingon a sheet of paper of th are using. Howmeasurements would you have to make(1) when your copy is to have the same

sheet that the given figure has on its sheet ; (2)on the sheet does not matter ? Give, in each case, one set Of

measurements which would be just enough, and state briefly howyou would use these to draw the figure .

23. A pint of milk weighs 0. pounds, a pint of water weighsb unds, and a pint of a mixture of the two weighs c pounds .

at fraction (by volume) of the mixture is milk ?If b 1 1 25, a 11 28, and c 11 27, what fraction is milk ?

CHAPTER IX

ON DRAWING TO SCALE

1 . IN the course of our discussions we have frequentlyused figures showing objects not full size, but drawn to

some smaller scale. For instance, in chapter V we had a

map of a piece of country,and in chapter VII we had

models of cranes smaller than the real cranes, and drawingsof cranes still smaller. What was the relation betweenlengths on the map, or the model, or the drawing, and thecorresponding lengths on the original object — In eachcase all the lengths on the copywere the same fraction of the

original lengths. Andwhatwas the relation betweenthe angles of the copy and the angles of the originalEach angle of a copy was equal to the corresponding angleof the original object. This was assumed to be so,

without any special statement. We will now discuss thetruth of the assumption.

2. By the help of a measuring-tape only make a plan of

the playground or of a field, making every length on the

plan 331, of the actual length, that is, using a scale on

FIG . 1 .

which 2 mm. repremnt a metre.— We begin with (say)

a straight piece of fence AB,measure it and lay

.

down thecorresponding length . [Figure 1 and the following figuresin the text are mere Sketches, and not drawn to any

146 ON DRAW ING TO SCALE

halves of AB, AT, and BT ; that is, they are equal to AC,AD, and CD, and also e

qual to CB , CE, and EB .

know,therefore, that AA TB is congruent to AADC and

to ACBB (chapter V) ; and this gives what we wanted toknow,

that the angles at A and A’are equal, that the

angles at B and B'are equal, and that the angles at Tand

T’are equal since each is equal to the angle ADC.

4 . Ex . 1 . How does your proof tell you that the Is at A'

to that at A, and not to the angle ADC or t e angle

Ex. 2. Certain reasoning was quoted to show CE DT, to showA s ACD and CBE congruent, and to show A s A

'

B’T and AOB

cor

l

i

Lgruent

. Give the reasoning inx. 3. Show that AATB can be out up into four triangles,

each of which will exactly fit

5. Now make a plan of the fence and tree on a scale of3 mm . to 1 metre . Call the ends of the fence on the planX and Y, and the tree Z (Fig. Divide XI into three

FIG . 3.

equal parts by cutting offXP and PQequal toA'

B’

. Through

1

1

;s

aid Q draw parallels to XZ and YZ, and by the same

11 Of reasoning as before show the three an

gles of AXYZ

to be equal to the three 1three angles of AABT.

ang es Of AA B’and to the

6. Ex. 4. Compare the areas Of the triangles ABT, A’

B'

Tmpare them also with the area of th tria l

the tres end the two ends A and B of she ac

l

t

l

rsral

to Q, and

CHAP . IX, ARTS. 4—8 147

?h

éough A and B draw parallels to CQ. These parallels trisect

Ex. 6. Give some other way of trisecting the line PQ.

Measure P Q, divide the length by 3 arithmetically, and measure

this distance ofl’

along PQ.

7. Show that the angles of AATB are equal to those of

the triangle formed in the playground by the tree and thetwo ends of the fence.

— Suppose the length AB laid 03

along the fence as often as possible . Laid 03 500 times itwill just cover the fence. From all the points of divisionsuppose parallels drawn, just as in Pi 3. Then we shallhave 500 triangles all congruent to TB, and a greatmany parallelograms, and the proof runs as before.

Show that the angles of A T’E

’and XYZ are equal to

those of the triangle made by the tree and the fence— Theproof as to A ’

T'

E'

does not differ from that for ATB . In

the case of AXYZ if we lay ofl‘

XY repeatedly along thefence, 333 lengths will not cover the fence and 334 are too

long. Let us divide up AXYZ, as in Fig. 3, and use

AXPE. XP can be laid ofl‘

an exact number of timesalong the fence and an exact number of times along XY.

And the former proof holds as between AXPR and the

trian 19 on the playground, and also as between AXPRand XYZ.

8 . Suppose LMN (Fig. 4) a plan drawn to a scale of

P3mm. to 1 metre, L corresponding to A, M to B, and N

to T. Show that the

angles of LMN are equalto those of -We

want a length that can belaid ofi an exact numberof times along LM and

also along A ’B

’. Such a L K

length is A’F, of FIG. 4.

A ’B

’

,which is equal to

LK, of LM for each of these represents the fence on

a scale of mm . to 1 metre. Draw FG [I to B‘T’

, and KU

HtoMN . Then we Show, as before, that A’G is of

A’T

’

, FG is of HT , LU is of LN,and KU is

ofMN ; that As A’

FG and A’R

’T’ have their angles

equal, and that As LKUand LMN have their angles equal.

148 ON DRAWING TO SCALE

Also each of the As A’I’G and LKU represents the fence

and tree on a scale of mm. to 1 metre, so that thesetwo triangles have their angles equal. Therefore As LMNand A

’B

’T

’

also have their angles equal .

Ex. 7. Howwould you divide A'

B'

into 10 equal parts—We

could measure A'

B’

(we really know this length already we used

it to make the plan A'

B’

T divide the length by 10 arithmetically,and lay this distance Ofi

'

repeatedlyalong A'B

'. Or we could draw

any line through A'

, lay off any convenient distance along it 10

times, join the l0th division to B'

and draw parallels throughthe other 9 oints of division.

Howwo d you divide LM into 13 equal parts ?

Ex. 8. To compare the A s A’

B'T

ingofl'

of A'

B'and of LM, we could have produced A'

B'

to 13 times its length, and produced LM to 10 times its length.

Complete the comparison of the triangles in this way.

9 . Now consider another method of making a plan, inwhich we measure only one length, say the fence, and forthe rest measure angles only, so that the position of the treeis fixed by the angles TAB and TBA . In what cases isthe advantage of this method greatest over the method of

measuring only lengths by a measuring-tape or a surveyor’

s

chain -The advantage is greatest when the area consideredis large. Then restriction to the tape or chain would meanwalking many kilometres. On the present plan all the

measurements can be taken from a few points.10 . Let us by this method make plans of the fence and

tree to the same scales as before, scales on which 1 metre isrepresented by 2, 1 , 3, mm . in succession. In the firstcase we draw AB 1 500 of the length of the fence, ona sheet of paper pinne to a drawing-board. Then we set upour drawing-board (or plane-table) at A in a horizontalposition, and turn it round till the line AB points alongthe fence. We then draw a line from A in the directionof the tree the point T that represents the tree will liesomewhere on this line. Now we transfer our drawingboard to the other end of the fence, make BA point alongthe fence, and draw from B a line towards the tree. Thepoint T lies where this line crosses the former one (Fig.

The other cases difl'

er only in having A ’B

’

(Fig. 2)of the length of the fence , .XY (Fig. 3) of


AT A’T’

LZ BT B’T

’YZ MN

at’

at’

at’

at’

37’"

WV—b? bt

’

from these measurements and the values of AB, A’B

’

, &c. ,which were used to make the plans, calculate as decimal sAB A

’R

'

XY LMd

AT, H y ,

XZ, m , an compare them. DO the same

91 B’T’ YZ MN

BA’B

'

A”“

fi’ ML

'

The fractions dealt with in this article have a Special name

h its full title is ‘the ratio ofMN tOML

18. Draw any two lines intersecting in O and across themdraw a series of parallels, and letter the figure as in Fig. 5.

Measure up the figure and calculate as decimals the0A OB 00 0D

or fractions5a

—,

AOb 65

0

, mai

ze , and

gompare them .

0 a aCompare also

OB Ob CC Oc’&c. State

formally your experimental result. Can you prove the resultby general reasoning — Suppose that we find OA

CHAP . IX, ARTS. 13—15 151

and OB a 2°1, the centimetre being the unit. We mark OPalong OA equal to 0°l cm. and drawHi parallel to Ao and

Bb to cut Oa in p . Then OA is 36 times OP, and we can,

as before, compare the triangles OAa and OH) and show Outo be 36 times Op . Also OB is 21 times OP, and by comparing As OBb and OH) we can show Ob to be 21 times Op .

We had 9-4 Q; and nowwe have 91; 86

OB Ob 21

Since Ou is 36 times 019 and Ob is 21 times Op .

14. Suppose OA and OB not to be exactly of the lengthsgiven, but to be and when more accuratelymeasured. How would you then prove that the fractions

or ratios g; are equal — We should then have

to make (or to imagine made) OP cm . long, so thatOA would be 358 times OP and OB would be 213 times OP .

The proof runs as before, and from our more accuratemeasurements we get a more accurate value for the fractionsOA/OB and Oa/Ob, namely or Give

the proof in detail for this case.

If by greater refinement in measuring we could measure

OA and OB to 3 decimal places, or to 4 decimal places,how would you proceed with the comparison of these ratiosor fractions — W e choose OP of

\

an appropriate length,and the proof runs as before.

Is it possible to measure a length with absolute accuracy,or should you expect every increase of refinement to add to

the number of digits we must use for the expression of the

length — The latter seems likelier. Then to whatextent is your proof of the equality of OA/OB and Oa/Obvalid -It is valid only if it is agreed to pay no attentionto errors of a certain degree of smallness.

15. That it is unsafe to neglect small things is wellillustrated by an old story. Two men were negotiating thesale of a horse, and the seller ofl’

ered to g ivo the horseprovided the horseshoe nails were bought at the rate of1 farthing for the first, 2 farthings for the second, and so

on, the price doubling with each nail. The buyer jumped


at the offer. What was the price of the last nail, therebeing 8 nails in each shoe -The price of the last nailwas more than So fast does the smallsum Of one farthing multiply.

By unaided arithmetic this calculation is very heavy.

The price of the last nail in farthings is 2x2x2x x2,there being 31 2’

s in the product. A short way of writingsuch a product is a considerable help. The product inquestion may be wri tten 231 . W ith this notation the pricesof the lst, 2nd, 3rd, 4th nails are in farthings 1 , 2,22

,23

,The number written above and to the right to

indicate the number of 2’s is called the index, and the

product so expressed is called a powe r thus 231 is calledthe thirty-first power of 2 and its index is 31 .

16. Can you suggest any shorter way of calculating 231than to begin with 2 and double the necessary number of

times —We could divide the 31 2’

s into two lots of 16

and 15. If we then multiply together the 15 2’

s, we can

get the product of the 16 2’

s by doubling once more ; andthese two products multiplied together will give the product of 31 2’

s. With the index notation this relation maybe written 2l5x21° 231 . Can you shorten the

calculation any further —We could divide the 15 2’

s intobatches, say into three batches of 5 2

’

s ; we could cal

culate the product of 5 2’

s,and then three such products

multiplied together give the product of 15 2’

s. W ith our

notation 2“x25x25 215. You have now a fairlyquick way of calculating 231. Use it to calculate thisquantity roughly, say by Stopping in each product of theprocess at two significant figures.

25 2x2x2x2x2 32 ;

215 32x32x32 approximately ;216 33,000x2

231 33,000 x66,000

And this being the number of farthings, the price of the

last horseshoe nail is The second figure isunreliable, so the price may be given roughly as17. Perhaps the quickest way is to calculate first 232.We can divide the 32 2

’

s into two batches of 16 each,


no number of decimal places will express the ratio with complete accuracy. Let us for shortness denote 1

,

by m, and Then OA is greater thanm

+ 1 times OB . On OBA cut off

OP equal to times OB , and OQ equal to -1times

OB . Thus A lies between P and Q. And

and Qq being all drawn parallel (Fig. the point orbetween 19 and q.1 9 . If nowwe suppose OB divided into 71 equal parts, the

nth part of OB can be laid off an exact number of times

along OP, namely m times ; and also an exact number of

times along OQ, namely m+ 1 times. Our former reasoning

therefore applies and Shows that Op is 13 times Ob, andm 1

times Ob. Hence 00 , which is intermediate91

m+ 1in length between OP and OQ, lies between 2times Ob, that is, between 3°

and 3° 654

times Ob. In th is case then the small error has not

multiplied.

Further, we could have begun with a value giving OA/OBto a hundred decimal places, and the form of reasoningwould need no changego. So that whatever degree of accuracywe consider necessary for OA OB we know that to thatdegree of accuracy Oa/Ob has the same value.

Remembering that OA,OB, Oa, Ob are mere numbers,

namely, the numbers of centimetres in the lengths of the

lines, we can state the result approximately in any of the

orms

CHAP . IX,ARTS . 19-21 155

20 . In the preceding articlewe found that approximatelyOA/OB GalOb. Now we cannot by means of decimalsexpress the lengths OA, &c. , accuratelyas for the ratios we must use approximations. Do all

these small errors mount up so that the other forms

OA xOb OaxOB , &c. , which we deduce from OA/OBOa/Ob, contain serious errors, or are the errors in theseforms also small Suppose the lengths of all these linestaken in centimetres, to a millionth of a centimetre, so thatOA p approximately if not accurately, and that if p is

not the accurate value it is within a millionth of it. Thus0A is greater thanp and less than p + 0°000,001

or, if for shortness we write 15 for and use the

symbol to mean‘is greater than and to mean ‘is

less thanp k OA p k.

Suppose q, r, s the lengths of the other lines to the sameapproximation, so that

q—k OB q+k,

r—k Oa r+ k,

s—k Ob 3+k.

Let us try to find by how much £ can difl'

er fromOa p r

q-a p xs froq r

, &c.

21 . The greatest possible value of the fraction 3;be found by giving OA its greatest value p It, and OB its

This discussion (articles 20-24, 26, 27, 29) is too difficult for thepresent stage. Let the pupils assume that the neglect of a small

quanti will only cause a small error,in the result . Let them

proc on this assum tion to discuss the statements at the end

of art. 24 and to wor through arts. 25 and 28, and then letthem go on to art. 30. At the end of the course contained in

this book the pupils could with profit return to the discussion

given here .


least value q—k ; it is+

2} To find howmuch this difiers

from2 .we bring the two fractions to the same denominator

q x (q k) in the usual arithmetical way. The two numera

tors are then qx (p + k) and (q—k) Xp, or qp+qk and qp lp ,and their difference is qk pk or kx (q p) The difl

'

erence

between the fractions is thereforekx (p Q)

i that is, it isp q 9 X (g k)

one millionth of

In the same way Show that the least possible value of the94.

p mfraction

OBdiffers from

qby one millionth of

“3q")

Since q+ k is greater than q k, the fractionQM

is less

10+ 9 24 Pthan the fraction

M—k)so that

0Band9do not difier by

more than a millionth ofN;—

q

k)—r or, in the shorter form

already used, by more than1

222513)

In the same way show thatOb

andZdo not difier bymorek0+ 9

OA 0a22. Knowing that

OB Obare equal, what can you

tell about the pam'

hle difierence betm en the appmximl te

values { and2-Since { does not difl

'

er from these equal

fractions by more thanZ—(

é’

gandannot difier bymore

Mr a)

q(q- k)that is, than one millionth of

P

q (q—k)+

rnor -

8bymore than


two sides in the same ratio .

FIG. 7.

If two triangles have the th ree angles of the

one equal to the th ree angles of the other, corrZesponding sides of the two tr iangles have the same

FIG . 8 .

ratio . For M ple, in Fig. 8, in which LALE , and LO LF, we know that

b sc f

’

CHAP . ix, ARTS. 159

Th ree paral le l lines fal ling across two otherlines divide them in the same ratio . For example,in Fig. 9

Ex. 9 . The third form of statement just given has not beencompletely proved. S

qlythe missing pomt ; for instance, bydrawing

through A a paralle to BF.

25. Ex. 10. As a test of the statements of the preceding articlecalculate all the ratios mentioned as decimals from measurementsof the figures.

Ex. 11 . Fig. 10 shows a sextant, an instrument for measuringangles. The frame ABC carriesan arm AD pivoted at A.

arm carries at the end A a mirrorE set at right angles to the frame.

F is anothermirror ndicularto the frame and para el to AB .

To measure the angle betweentwo lines rimning from the observer

’

s position to two objects Pand Q at the same level as theobserver in otherwords,the anglesubtende by P and Q at the FIG' 10°

observer’

s position) , the observerholds the sextant up horizontally so that, with his eye at the pointG marked on the bar AB, he sees P over the mirror F. He thenturns the arm AD until he sees Q in the mirror F, the ray oflight from Q coming to his eye by reflection at E and then at F.

It is known that when a ta of light falls on a mirror and isreflected, the ray falling on t e mirror and the reflected ray are

in the same plane with the perpendicular to the mirror and makeequal angles with it. The point G is taken so that when the armAD lies along AB and the observer looks at P over the mirrorF, he sees P also tn the mirror by reflection at E and F.

Show how the point G may be fixed on the bar AB . And showthat the angle subtended by P and Q is twice the angle BAD.

Each obj ect is so far away that all lines from the frame to itmay be considered parallel.

Ex. 12. Use a sextant to measure the angles for the purpose ofdrawing the plan of the playground.

26. Let us now return to the plans of the playgroundmade by a measuring-tape or a surveyor

’

s chain only. W e

found corresponding angles of all the plans to be equal.


But our proof depended on the commensurab il ity of the

lengths involved, that is on the fact that a unit could bechosen in terms of which the lengths could be expressed

Suppose we have to do with two triangles ABC and DEF

(Fig. in which the lengths of the sides are incomm en

surab le , that is, such that no unit can be chosen in terms

FIG. 11 .

of which the lengths can be expressed. Suppose that, inwhatever units and to whatever degree of approximation thelengths a, b, c, d, c, f of the sides are taken, the fractions or

e

d

prove corresponding angles of these triangles to be equalLet us imagine BC divided into a great number n of equal

parts, for instance amillion, and imagine EF also divided inton equal parts. Let us imagine that BA contains more thanm and less than m+ 1 of these small parts of BC ; from the

equality of the fractions5andgwe know that ED contains

more than m and less than m 1 of the small parts of EF.

Imagine further that AC contains more than p and less than1) 1 of the small parts of BC, and consequently DE con

tains more than p and less than p+ l of the small parts

Let us imagine made on BC and EF triangles KBC and

REF for which (15are equal to$1, and gand 3equalto5; Their sides being commensurable we know how to

6ratios5and5are equal, and

5and are equal. Can we

prove their corresponding angles equal. And these trianglesdo not difl

'

er much from the triangles ABC and DEF.


Can you suggest any other circumstances that are enoughto prove the triangles similar or, in other words, can you

give any other set of data that are enough to fix the shapeof a triangle —Possibly the triangle would be fixed in

shape by one angle and the ratio of one pair of sides.

Possibly two triangles, with a pair of corresponding anglesequal, and one equal ratio among cormsponding sides, might

Discuss all the possible forms of theseconditions ; first draw as many as you can of the trianglesdetermined in shape by the angle B and one ratio

,giving B

in all cases the value and taking as the ratio each of

the following in succession0

4 ;

0-5,3 ;

2 2-0, o-4 ; o-e, M ,

1-5;b a

g re,21

,— Drawing gives the following

experimental results. There is one shape of triangle possible

when R and 2 Zare given. There is one shape

when B and the ratio of b to another side aregiven ratio is greater than 1 . There are twoshape when R and the ratio of b to another side are givenif the ratio is less than 1 .

29 . Show that if two triangles ABC and

Fro. 12.

have the angles B and E equal and the ratios 2 f;aequal, they have also the other corresponding angles equal,

CHAP . IX,ARTS . 163

and so are similar. Assume the ratios 3 to lie

between along BA mark 03 BK

BC, and BL times BC ; and so for ADEF. Then

proceed as before.

Show that if two triangles ABC and DEF have B E

g g, each ratio being greater than 1

, the two tri

angles are similar. Where does this proof depend on the

condition that the ratios2and are greater than 1 -In

the course of the proof we cut ofi small equal lengths BWfrom BA and BF from ED (Fig. and draw WX

FIG. 13.

parallel to AC, and YZ parallel to DE . The condition that

2and -

fi. are greater than 1, shows that WX WB and

YZ YE , and in consequence that the triangles EWKand EYZ are congruent.80 . Now state generally what conditions fix the shape of

a triangleA triangle is fixed as to shape(1) by a knowledge of two angl es, or(2) by a knowledge of two rat ios of its sides, or

(3) by a knowledge of one angle and of the rat ioof the two sides that form th is angle , or(4) by a knowledge of one angle and of the ratio

the opposite side bears to another side providedth is rat io is greater than unity .


Or, in other words, two triangle s are sim ilar(1) if the

yhave two pa irs o f corresponding

angles e qua or

(2) if they have the ratios of two sides to the

th ird in one t riangle e qual to the corre spond ingratios in the o ther ; or

(3) if th ey have a pair of corresponding anglesequal, and the rat ios of the s ide s form ing th eseangles also e qual ; or

(4) if th ey have a pair of corresponding anglesequal and the rat io of the side opposite th is angleof one triangle to a second side equal to the

corresponding rat io in the oth er triangl e , provided th ese ratios are greater than 1 .

81 . When a triangle is to be made from B 35°and a

value of g, for what values of Pbare there two shapes, for

what values one, and for what values none — We lay downthe angle B and put down the point A anywhere on

one arm (Fig. We then calculate the value of b,

FIG . 14.

and with this as radius draw a circle which cuts the

other arm BX in C. If is less than 1, this circle cuts

BX in only one point, and there is only one shape. If -5is greater than 1

,the circle (if great enough to meet BX

at all) cuts it in two points and gives two shapes. Let

fall the perpendicular AD or p from A on BX. If the

given ratio2is equal tog, the circle cuts BX in only one


angles whose sines are and verify your resultsby reference to the book of tables.

88 . Areas of similar figures. Consider again Figs. 1—4,which we now know to be similar. What is the ratioAB

and what is the ratio of the area ABT to the area

A’B

’T

’ that is, how many times does the area ABTcontainthe area A

’B

’

T’ Pair the four figures in all possible

ways, and for each pair give the ratio of the bases and the

ratio of the areas.

If two sim ilar triangles have a pair of correspondingsides in the ratio k, find the ratio of their areas.

—Suppose

the similar triangles of Fig. 15to haveg—g, 76. Draw the

FIG . 15.

perpendiculars AM and DN . The areas of the trianglesare i xAMxBC and i xDN xEF. BC = kxEF, and

AM= kxDN. So that the first area is ri xkxDk xEF ;or kxk timesfixDN xEF. The ratio of the areas is kxk,or with the index notation it is 152. Prove that

AM BCAH kxDN

'

,that 18

,that

DN EF

84 . If two plans of the playground are made by chainsurveying, are they similar —Yes, because each plan is

made triangle by triangle, and taking pairs of corresponding triangles in succession we prove their angles equal.Thus the whole figure conforms to our definition of similarity. And if the two plans were made by the

measurement of angles only in addition to one base lineThe proof that they are similar is much the same.

If the two base lines of the plans, made by either

CHAP . IX, ARTS. 33-36 167

method, have the ratio k, what is the ratio of the two areas

that represent the playground —Each triangle of one planis as to area kxk times the corresponding triangle. So thatthe whole area of the first is kxk times the area of the

second. Is the result afl'

ected if the playground hasa curved boundary -By taking triangles enough we can

reduce as much as we like the area along the curvedboundary with which we cannot deal. So that we can

prove the result true to any approximation we like.

85. Volumes of similar solids. If a schoolroom is p cm.

long, g cm. broad and r cm. high, what is its volume — It

isp x q xr cubic centimetres. And if a model of theschoolroom is made, each dimension being k times the

original dimension, so that its length, breadth, and heightare in centimetres kxp , kxq, and kxr, what is the volumeof themodel ?— It is kxp xkq e c. cm. , or (omittingthe signs of multiplication) kpq r or kkkpqr, or (withthe index notation) kq r. What ratio does the

model bear to the room as to volume — The ratio is ks.The original room and the model are said to be sim ilar ,

and in general two solid figures in which correspondingangles are equal and corresponding lines in the same ratioare said to be similar. If this ratio is k, what is the ratio ofthe volumes For instance, suppose a model of a mountainto be made

,every distance on the model being k times the

original distance, and the two being assumed similar.

Suppose the mountain cut up into rectangular blocks andthe model into corresponding blocks. Then each block of

the model is kkk times the corresponding block of the

mountain So that, apart from small pieces left over thatdo not make rectangular blocks, the ratio is kkk. And

by cutting out smaller blocks we can make the piece leftover as small as we like, so that the ratio of the wholemodel to the whole mountain is kkk or ks, to as close an

approximation as we please.

86. How many conditions fix the shape of a triangleTwo, as we saw inArt. 80. How many conditionsfixthe triangle as to size and shape -One length is needed

in addition, for instance the length of a side. Or considering size and shape together, we saw earlier (chap. V

,


art. 27) that three conditions fix the triangle.

many conditionsfix the triangbon a sheet of paper —This brings us back to the problemof fixing a chair on the floor (chap. II). To fix the positiontakes three conditions, so that shape, size, and position takesix conditions. Or we can consider shape, size

, andposition all at once . For instance, these are all fixed bygiving the distances of each corner from two adjacentedges of the sheet of paper, in all six conditions. Sug

gest other ways of fixing the triangle in shape, size, andposition, and see if the number of conditions is six in allcases.

Em crsss.

16. Draw a line AB 38 inches long, and AC of indefinite lengthmakingwith it an angle of On ACmark 06AK, KL, LMeachof length 15 inches. Join MB and draw KX and LY IIto MB,

cutting AB in X and Y.

Show by measurement that by this construction AB is dividedinto three equal parts.

Give also a general proof, showing that the construction wouldhold if AC were drawn at some other angle to AB, and some otherlength taken instead of 15 inches. For the general proof you mayfind it convenient to draw lines through X and Yparallel to AC.

17. Make a triangle, ABC, having LA 90°and LB 18

°and

AB 14 cm. long, and let fall AM perpendicular to BC. Measure

CA AM CMCA, CB, AM, CM, and calculate the ratios

CB’

A_

R’

CA

three siinificant figures.

It is nown that, if the figure could be perfectly drawn and

measurements taken with absolute accuracy, each of these ratios

would have the value “5;

1By means of a table of squares

evaluate this to three figures, and compare it with the average ofthe values alread found.Show briefly, t at if a number of triangles ABC are made, in all

of which A and B have the same given values, then the ratio

CA/CB is the same in all these triangles. State , with reasons,

whether the ratio of the angle B to the angle A is the same as

the ratio CA/CB .

18. An enlarged photograph of Fig. 16 is to be made, bc beinrepresented by a line BC 15centimetres long. CopyFig. 16, anthen draw the enlargement, denoting the points corresponding toa b c &c. , by A B C &c.


while remaining in contact with the paper ; also that whenr is in its first position a is made of theWhere must the drawing pin placed if it is

tracingyg

a

rr revolves, the tracing of ABC

presently coincides with X24. Compute to two significant figures, the diameter and cross

sectional area of a s crmen of copper wire, one mile of whichweighs 4 lb . One cubic inch of copper weighs lb .

25. Two quantities a: and y are so related that the value of y fora given value of a: may be found as follows z

—Along 0X (Fig. 19)

FIG . 19.

measure OP (in centimetres equal tow, and draw PQ perpendicularto OX to meet LM in Q ; t on P0 Copy the figure and findthe values of ywhen a: 2, 3, 4 and 5.

Measure 0L, and in each case calculatea:

y

0Lto one decimal

place. State the relation between a: and y algebraically.

CHAPTER x

ON SIGNALLING

1 . IN theMorse Code, which is a good deal used in postaltelegraphy and other methods of signalling, there are two

signs, the dot and the dash. Each letter of the alphabet isrepresented by some arrangement of dots and dashes. Howmany arrangements can be made that contain three signs orless —We set out the arrangements, 1 sign at a tima2 at a time

,and 3 at a time, and count them up

With one sign at a time there are 2 arrangements, with2 signs 4, with 3 Signs 8 ; in all 14.

2. How many arrangements are there with 4 SignsHaphazard counting soon becomes tedious and even inac

curate, and for expeditious andaccurate counting some systemis necessary. How would you convert a 3-arrangementinto a 4-arrangement, or a 2-arrangement into a 3-arrangement Does it matter (for the purpose of counting the4-arrangements) whether you add the sign at the beginning,at the end, or somewhere in the middle — By taking a

3-arrangement and inserting a dotwe get difl'

erent 4-arrange

ments according to the position of the inserted dot, but

if we treat all 3-arrangements in this way the same

4-arrangement turns up in more than one way. For in

stance, with a dot inserted at the beginning is thesame as with a dot inserted at the end. But if we

'

always add the dot at the end each 3-arrangement gives one

172 ON SIGNALLING

4-arrangement, and there rs no repetition. Byadding a dashat the end instead of a dot we get another 4-arrangementfrom each 8-arrangement. So that there are twice as manyarrangements of 4 signs as of 8 signs.8 . In this way tabulate the number of arrangements of

1, 2, 8, 4 signs.

The number of l-arrangements is 2.

2 2x2 or 4.

3 2x2x2 or 8.

4 2x2x2x2 or 16.

Howmany 10 arrangements are there — The number Is

ten 2’s multiplied together (which is also written and

is 1024. How many warrangements — The numberis n 2

’

s multiplied together, which we write shortly asHowmany signs are there In the longest arrangements of

a code that represent letters of the alphabet, the code beingmade with as short arrangements as possible — To stop at3-arrangements would give only 14 letters. To go to

4-arrangements gives us the 26 letters of the alphabet and4 arrangements to spare.4 . Consider again the building up of an arrangement of

4 signs. What choice have you for the first sign And

the first sign being chosen, what choice have you for thesecond In how many ways then can the first two signs

be chosen Continuing in this way discuss the possiblenumber of arrangements of 4 signs, of 10 Signs,and of n signs.

5. Howmany arrangements can be made without goingbeyond 10 signs for any one arrangement — We must addthe numbers for 10 signs, 9 signs, &c. It IS 1024+512+

or 2046. Por ar

rangements of more signs than 10,thismethodwould be very

heavy. Let us denote by N the number of arrangementsthat can bemadewithout going beyond 10 signs. Nowwritedown twice the number of 10-arrangements, twice the number of 9-arrangments, and so on, whose sum is 2N— Thesenumbers are 2048+4. Can you subtract the expression found for N

from the one now found for 2N, without adding up either ofthem What is the result — It is 2N N = 2048 2,

174 ON SIGNALLING

In these two expressions and 2 occur once ; but everyother number and so on to 22) occurs twice. So

that by subtracting we getN 1 2.

8 . Semaphore. Let uS use for signalling an arm hinged tothe top of a post so that it can be Set in any of the seven

FIG. 1 .

positions of Fig. 1,so that we have 7 distinct signs. By

using two signs in succession how many Signals can be

made —They areAA

,AB

, AC, AD, AE , AF, AG,

GA, GB, G049 in all. With 3 signs in succession, how manysignals — 7x 7x7. W ith 10 signs and with n signs?— The number with 10 signs is the product of 10 7

’

s, or

written shortly or multiplied out about 280 million, thenumber with or signs is the product of n 7

’

s or

How many signals can be made without going beyond3 Signs for each W ithout going beyond 10 —When 10

CHAP . X, ARTS. 8—10 175

signs are allowed, we may use 10 or 9 or or 1 , and so

get signals, say Nsignals. Then we calculate 7 times N. 7 times 71° is theproduct of 11 7’

s or is 711 , and so on, so that 7N is 71 1 710+

tr+ 7i Subtracting, we have 7N—N 711—7

, or

7 —7about 330 million. This calculation

is tedious. Later, we must try to find a shorter method.

Flag signalling.

9 . A ship has a variety of flags and makes signalsby arranging flags in order down a r0pe . W ith two

flags A and B, how many signals are possible, both flagsbeing used - Two, with A above or with B above.

And with 3 flags A , B, C, how many signals, using all3 flags -By actual trial there are 6 signals.How would you make a 3-flag signal from a 2-flag

say from g in which A is above — We could put a

above A, between A and B, or below B . In the same

way we can make three 8-flag signals from i so that

there are 6 in all. Using a fourth flag D in addition, how many 4-flag signals are possible -Beginning

any 8-flag signal we can insert D in the lst, 2nd,3rd, or 4th place, so that each 3-flag signal gives 4 4-flagsignals, and the 6 8-flag signals give 24 4-flag signals.10 . Tabulate these results and extend them.

A single flag gives 1

2 flags give8 2x3

4 2x8 x4

5 2x3x4x5

2x 8 x4x x 10 IO-flag2x3x4x xn n-flag

Roughly multiply out the number of 10-flag signals.

3}million.

ON SIGNALLING

11 . Suppose you begin with 4 flags without first discussing 3-flag arrangements. What choice have you for thetop flag

— It may be any one of the four. And the

top flag being chosen, in how many different ways can you

choose the 2nd -Three ways it may be any one of the

remaining three flags. Continuing in this way, findhow many 4-flag signals you can make— The number is4x8 x2xL

In the same way discuss the number of 10-flag signals,and of n-flag signals.

Ex. 1 . It has been supposed all along that we had only one flagof each kind . If, instead, we have asmany flags as we like of eachkind, what is the number of 4-flag signals with 4 kinds of flags,and of n-flag signals with n kinds ?12. Again, suppose we have 5 flags a, b, c, d, e, all dif

ferent. How many Signals can we make using 2 flags at a

time -The Signals area a a a c c c c

b c d c a b d e

e e e e

a b c d

If a is made the top flag, we have a choice of 4 for thebottom flag ; and it is the same for any other top flag. Forthe top flag we have a choice of 5. So altogether the

number of signals is 5x4, or 20. How many 8-dagsignals can be made with these 5flagsW ith at flags, all different, how many 2

-flag signals can

you make —For the top flag we have a choice of n flags.

And with each top flag we have n 1 possible bottom flags.

Altogether, nx (ri — 1) or n (n 1) Signals. Howmany 8-flag Signals — To any 2-flag arrangement we can

add one of the remainingn—2flags, and so get n- 2 signals.

80 that there are n (n—1) X (n—2) or n (n—1) (n—2) 8-flagsignals. Howmany 4-flag signals And howmany7-flag signals

18 . At a party all the girls are given difierent rosettes,each made up of one or more colours. Duplicates of theseare given to the boys, and each boy has to find as his

178 ON SIGNALLING

would, if order counted, make 24 rosettes. So that therewould be 24xN 4-colour rosettes if order counted, and weknow this number to be 5040, so that 24N 5040, and

N 5040 24 210. Or, if we do not multiply out, we

have the orm

15. With 100 colours, how many rosettes of 1, 2, 3, and

4 colours can be made, colour alone counting — Thenumbers are respectively

100 x99.

100 x99 x98.

100 x99 x98 x97.

2 2x 3 2x3x 4or approximately 100 ; 4950

W ith it colours, how many rosettes of 1, 2, 3, and 4

colours can be made, colour alone counting -The numbersare

n (u—l ) (n- 2) n (n— 1) (n—2) (n—3)2x 8 2x3x4

Any selection of 4 colours or other objects from amongn difl

’erent colours or other objects is called in mathematics

a combination of n things 4 at a time. In general,a selection of p things from among 10 things all different isa combination of at things 1) at a time.

Binomial Theorem.

16. To multiply 83 by 7we add together 7 times 80 and7 times 3. No matter what values a b e have, a times

b+ c is the sum of a‘ times b and a times 0, or in mathe

matical form a x (b+ c) a x b a xe, or a (b+ c) ab

ac. In this way multiply x+ a and m+ b together.— The

product is as as be: ab. Multiply this product byx-i-c and put first the term containing a: three times, thenthe terms containing a: twice, then those containing a once,

and last of all the term without a.— The new product is

containing a: twice are often grouped together and written(a b chew,orwith the indexnotation (a b c)a

”. Group the

terms containing a: once in the sameway, and write thewholeproduct, using the index notation — It is a“ (a+ b+ e) :c

2

CHAP . x, ARTS. 15-19 179

(ab be ca)a abc. In this expression the number ab c that multiplies 50

2 is called the coefficient of and

ab be ea is called the coeficient of a. Multiply this expression by (0+ d and group the product according to powers of a.

-It is

(abc bcd+ cda dab)x+ abcd.

17. In this product what is the relation between eachcoeflicient and the numbers a b c d —The coeflicient of

x3 is the result of selecting one of the numbers at a timeand adding the selections. In the coefiicient of x3 occurs

every possible combination of the numbers a b e d, taken 2at a time ; the coeflicieut is the sum of all possible products2 at a time. The coefiicient of a is the sum of all possibleproducts of a b c d, taken 3 at a time. The term that doesnot contain a is the only possible product of the fournumbers all together.18 . When the product (a a) (a: b) (a: e) (a: d) is multi

plied out, how many of the quantities a: a b e d occur in

each term —By looking at the multiplying out we see

that every term contains one letter from each factor, andthat every possible combination of one letter from each factor

How many terms then are there altogethermany groups when we group together all the

terms that contain a the same number of times —Thenumber of ways of choosing one letter from each factor is2x2x2x2, or 16 so this is the number of terms reckonedthis way. As to groups, a group may contain a

tor a

“or

x2 or a, or not contain a at all ; so there are five groups.

How is the group containing made up-It takes a:

from three factors and a difl'

erent letter from the remainingfactor. And the group containing a“ —It takes a:from two factors and other letters from the remaining two.

It will contain as multipliers of every possible combination, 2 at a time, of the numbers a b e d. How are

the remaining groups made up

19 . Suppose for a moment that a b c d all have thevalue 1. How will you writemore shortly And what are the groups of the multipliedout form -With the index notation the expression is

written (a: The term containing a four times occurs

180 ON SIGNALLING

once only, by taking a from each factor ; it is simply at .The term not containing a: occurs once only by taking 1from each factor ; it is 1 . Terms containing a: three timesoccur by taking 1 from any one factor and a from the

other three , four such terms occur, so the group is 4a“.

Terms containing a twice occur by taking 1 from any two of

the factors and a from the remaining factors ; the numberof such terms is the number of ways of choosing 2 factorsfrom the 4 factors, that

'

Is 4x8 2, or 6 ; so the group is

Terms containing a: once occur by taking a: from any one

factor and 1 from the remainder , this one factor can be

chosen 1n 4 ways, so the group is1 for a b 0 d in the

c) (a d), and alsoby simply multiplying o20 . Give the various groups of terms when a a, a b,

x+c, a+ d, aH-e, x+f x+g are multiplied together. Howmany terms are there In each groupAlso find the number of terms by multiplying out

(a: W, and check the result by actually multiplyingtogether x+ c, a+ b, &c.

When a factors x+ c, x+ b, . a+ i aremultipliedtogether

,howmany terms are there In the groups containing

a“, an” a”

? xvi-8 M"— The numbers of terms aren (n—1) (n—2) n (n—1) (n—2) (rt—3)

2x8 2x3x4

When (a: is multiplied out, write down the first five

groups.— They are

n n— l ) n (n—l ) (n—2)2 2x8

at” 8’

n (n—l ) (n—2) (n—3)2x8x4

The result that

The remaining articles of this chapter should be postponedtill they are wanfid In article 4 of chapter XII .

182 ON SIGNALLING

that bears interest in the fourth year, in the fifth year, andin the tenth year —The sum in the fourth year is

b b b b 3

or a (l + -

1—O—

O) ,

in the fifth “ (I 'll—CODY, and in the tenth d (l + -

i-g—O)

9

o

24. And what does the whole amount to at the end of

the tenth year, original loan and accumulated interestCalculate this amount when the sum lent is £740 and therate of interest is 3 per cent. , first by direct calculation, andthen by the help of the binomial theorem.

—The number of

pounds Is740x(1 T—go)

o

° We multiply £740 ten times

by and find £994 108. 0d. The binomial theoremgIves

8 3 1° 3 9 10 9

(100 4-1) -

1-

66) 10 x ( 100 ) 2

x

or,in ascending powers of

1 IOX10

2

X 9

(100 )2

+

3 10

(166) is only o-oooooooooooooooe, and the terms

beside it are not much bigger. Let us therefore begin at

the other end, that Is, with the series in ascending powers.

We calculate the successive terms to five decimaland place them in a column .

1

The successive terms become smallerthe seventh and following terms are too

CHAP . X,EXERCISES 188

fifth decimal place. The sum of the six terms is 134392,and the accumulated value of the loan is 740 x 18 4892

pounds, or £994 108. 0d.

In this case it is doubtful whether the use of the binomialseries saves any time over direct calculation. P resently weshall have calculations for which there is no comparisonbetween the methods.

Ba na n a.

2. A railway has 34 stations. How many kinds of third classsingle tickets does it take to supply the system ?

3. A school has 87 ls, and each of them sends a Christmascard to every other. owmany do they send ?4. The floor of a conservatory is to be paved with red, white ,

and black tiles according to thedesi

gn indicated in Fig. 2. If

the ength and breadth of the

floor are respectively 40 and 20times the length of the diagonalof a square tlle, find the numbf:

of n

ab?tile

l

s

lgf

egach colour

t t w'

ignorin

the smaller til

le

eg.g

5. In the game of dominoesa number of blocks are used .

Each block is marked off into

two parts, and on each of theseparts a number of pips, between

Tlimd 6 inclusive

édaremuarked.

ese are arra In

sible ways, andn

ghe ips onptlie FIG . 2°

two parts of a bloc may be the same or no twoblocks are alike. Howmany blocks are there ?6. A publisher uses sheets of paper 24 inches by 10 inches to

make a book each page of which measures 8 inches by 5inches .

He folds each sheet of paper so that it makes 6leaves or 12pages .

Sketch a sheet (

zi pper, back and front, mark the creases and

write on the sev pages the numbers 1, 2, 12, so that whenthe sheet forms part of the book these numbers will appear inro r order right side up. Mark corresponding corners of the

and front views of the sheet with the letters A, B, C, D .

7. Nina cards numbered 1 to 9 are arranged in three heaps as in

Fig. 3. The bottom card of heap 1 is taken away and placed on

184 ON SIGNALLING

top of heap 2; then the u rd now at the bottom of heap l is takenand put on hea 3. Similarly, the bottom card of heap 2 is placedon top of heap and the card left at the bottom of heap 2put on

heap 1. heap 2. heap 3.

FIG . 3.

l . the card at the bottom of heap 3 is laced on

1, and the card so left at the bottom on heap 2. S etch thein their position at this stage .

When this operation is re ted, name the positionssuccession by card No. 1. how many

return to its original°

tion ? Are the

their original positions ? ow do you know ?8. A marble statue is 5feet 6 inches h. A statuette which

is an exact co y of the statue on a er scale and ln

weighs lb. 0 statuette is 9

it woul have the statue,to the nearest hundredweight ? Note that

volume of statue

9. A wooden ball 25centimetres in diameter weighs 5°9 kilograms. Find, to the nearest hundredth of a gram, the weight ofa cubic centimetre of the wood . The volume of a ball ofd is imn, where 7r

10. The number of cubic feet of timber in an oak log, whichtapers uniforml

y;may be calculated from the formula 0°2618h x

w ere a and b are the numbers of feet in the

etere of the ends of the log, and h is the number of feet

in the length of the log. Find what is the weight of such a

log of oak whose length is 15 feet, and the diameters of whoseends are 3 feet and 2 feet respectively, assuming that a cubic footof the oak weighs 50 lb. Give your answer to the nearest cwt .

If you assumed the diameter to be 2 feet 6 inches throughout,

186 ON SIGNALLING

16. Draw two straight lines AOB and COD crossing at rightangles. On draw a line PQ 8 cm. long, and on P0mark R 1 cm so that P alwayslies on AB and Q on CD.

positions, and so draw the path of R .

Repeat, taking R in succession at distances of 2, 4, 6, 7 cm .

from P .

In the case when R is midway between P and Q, discuss bygeneral reasoning the nature of the path of R .

line and a curve between twopoints A and B .

Erect perpendiculars at intervals of l centimetre alongline AB . Join the points at which successive perpendiculars meetthe curve Estimate approximately the area of the figure byassuming It to be equal to the polygon so made.

18. Water runs at 4} miles an hourof which is 3 square feet. How (to the nearest

million) does it deliver in 24 hours ? is gallons.

draw lines 0X and CY, meeting at an

angle 0X 10 cm. and CY 75cm joiningthese lines to XYand cm. apart . Drawand mark T the centre and S any point boundary .

Place the tracing-paper on this semi-circle with 0 above theS, and pin it at this point. Now rotate the tracing,

point on OK (say X 1) comes over the boundary of the

prick through Y1 the corresponding int on OYon the boundary prick through Y

.2 an so on. Draw

of Yas X traces out the gIven figure .

locus of Ybe if 0 were pivoted at T?

CHAPTER XI

ON CARRYING WATER

1 . A W has to take a bucket from a cottage C, fill it atsome point P of a ditch D,

and carry it to a trough T.

Taking the distances as Shown in Fig. 1,find, by drawing

a =47m .

6-e6m .

FIG. 1 .

to a scale, the length of the journey for difl'

erent

positions of P,say at distances of 5

,10, 15, metres from

A and show your results in a table.

At every point P draw a J. to the ditch, of a length thatrepresents (on the scale chosen) the length of the journeyfor that position of P . By drawing a curve through theends of these .La, make a graph showing the length of

the journey for any position of P . From your graph findthe position of P that gives the Shortest journey.

2. Produce TB to U so that EU= BT, that is, take Uthe image of T in the line of the ditch. Suppose the manto go from C to U, crossing the ditch at some point P ,

188 ON CARRYING WATER

and going straight from C to P and from P to U. Give

a graph showing the length of the journey for difl'

erent

positions of P . From the graph find which journey is theshortest. Can you by any better method than the graphdecide which is the shortest journey from C to U — One

path is straight. It is therefore the shortest.

Note for the teacher. It is our constant experience thatthe straight path is the Shortest. Whether this connexion

between straightness and shortness is in the nature of

things does not concern us, as it is not suited for dis

ensaiou by the young.

Can you now find the shortest route from C to T bya better method than the graph — P being any point onthe ditch, folding our figure about the ditch brings T and Utogether, and brings PT and PU together, so that any route

CRTis equal to the route CPU. Therefore,for the shortest

route the bucket must be filled at Q, Where the straight pathCUcuts the ditch.

8 . So far, P has been assumed to lie on the stretch AB of

the ditch. How does the route CRTof Fig. 2 compare withthe route CAT — Observation shows that when P lieson AS and moves towards S, each piece, CP and TP, of the

FIG . 2. FIG . 3.

route continually increases. graph showing thedistance CR as a function of the distanceAR. Which pointof D is nearest to C Can you prove it Vbeing theimage of C in the line of the ditch (Fig. the straight


to r. We know then that t is less than r which is a part ofLODE ,

therefore LOED is less than LODE .

State these results formally.—The greater ang le of

a triangle has the greater side opposite , and the

greater Side has the greater angle opposite . Or,since the reasoning applies to any two of the three sides

or angles, the order of size of the angles of a triangleis the same as the orde r of size of the o pposi tesides .

5. Draw again the ditch D, the cottage C, the trough T,the image Uof the trough, the point Q that gives the shortestroute, and let the line joining C to Tmeet the ditch in Z,

as 6. Name all cases of equal can find

FIG . 6.

and calculating ratios. In particular,

8? Draw QW .L to

AB and see if it gives any more ratios equal to67

,

Confirm your conclusions by general reasoning.— QU

being QT, the ratio9; is

cg TBU being 0A, the

Q 9 Q0 AO QAAs BUQ and ACQ are srmrlar, and

Ac U 3

give all the ratios that are equal to

U QB’

BU BTThat TBU is ll CA gives As CAZ and

TEZ similar, and Since QW isHTBU,TB TZ BZ

CHAP . XI, ARTS. 5-7 191

2%(WP; Compare the anglesTQB and UQR, and

the angles CQW and TQW. Keep only enough of the figure

to show the equal ratiosUT

,

Wt and remove the

rest. What conditions with regard to this reduced figuremust hold to make then three ratios equal — It is enoughto know that QW bisects the angle CQT, and that QZ Is

FIG . 7.

.L QW or bisects the angle TQU (Fig. for this enablesus to restore the original figure. Show how to

restore the original figure.6. Let us call Q the vertex of the triangle QOT. Then

QW is the bisector of the vertical angle of the triangle.

The angle TQU, made by producing a side CQ, is called theexternal vertical angle , and QZ is the bisector of the

external vertical angle. The point W (in ordinary English)divides the base CT into two parts, namely, from W to C,and from W to T. By an extension of the language the pointZ is also said to ‘divide ’

the base CT into two parts, namely,the part from Z to C and the part from Z to T. By means

of these terms state the result concisely.— The bisector

of the vertical angle of a tr iangle divides the baseinto two parts that are in the same ratio as the

two s ides ; the bisector of the external verticalangle divides the base into two parts that are alsoin the same ratio.

7. Consider the propositions converse to those just stated.

State them, and see whether they are true. —If a point W


divides the base CT of a triangle CQT internally

83, then WQ bisects the vertical angle

Draw TU QW, and meeting CO in U. ThenCW C

WT QU’ because TUIIWQ. And

CW CQWT

1s givenQ—T

Therefore QT

QU. Now the angles CQW and

TQW are equal to the two baseangles of the isosceles triangle QTU,

for QW TU. Therefore LCQWLTQPV, that is, QW bisects the angle

Having given that a pointZdividesthe base CT of A CQTexternally so

Fm ' 8'

53 gg, r draw TY ZQ to

meet CQ in Y, and use a similar proof to show that QZbisects the external vertical angle of the triangle.

8 . Re turn to the original problem of carrying water bythe shortest route. In what ways can you Specify thepoint of the ditch at which the bucket should be filled

(1) It is the point Where the line from C to the image of Tcrosses the ditch. (2) It is the point Q such that the twoparts QC and QT of the path are equally inclined to the

ditch. (3) It is the point Q that divides AB internally inthe same ratio that the line CTdivides it externally.

9 . From the data in Fig. 1 calculate the distances of

Q and Z (Fig. 6) from A and B.— All lengths being in

metres, we haveZB TB

BA

ZB

“ X2678. Hence ZA 784-63 141 Again, in

21

the same wayAO CA 68_QB

and

1 so that or ZB


longest side of a right-angled Awill be called the hypotenuseand no longer a side . In a right-angled triangle you knowthat the Sine of an acute angle means the ratio of the side

opposite the angle to the hypotenuse. Another useful termis cosine . The cosine of an angle is the ratio of the side

beside the angle to the hypotenuse . Thus in ACQA the

cosine of m is %3; it is Written‘cos m Use this term

to give additional lengths on Fig. 9 in terms of r,3,

CA TBm.

— CoamCQ TQ

’ so that CA — r cos m and

TB 3 cos m.

11 . Join CT and let it meet QX in Z and QY in W.

Draw CE and TF .L to QY. Suppose the acute anglebetween QY and CT an angle of h degrees. Mark all anglesof 71 degrees on the figure , and express any additionallengths you can in terms of r

,s,m,

n.—All the ratios

ZA ZB CE TF ZQ75

,

Z—T

,

CW,

TW’

ZWare equal to Sln h . The thlrd

of these ratios gives CW sin n CE AQ r sin m,so

that CWm

, and the fourth ratio gives TW sinn

TF BQ s sinm,so that TW

’ 8 srn mAll the ratios

CA Q WE E WQ l to can TheCZ

,

TZ’

1170:

WT,

WZare equa c y glve

CA r cos m 3 cos mand other

cos it cos n

relations.

12. From the values found for CW, TW, CZ, TZ

CW CZ CWdeduce the ratIoS

TW TZ TW13 equal to

r sin m s sin m r CZ r cos m s cos m

sin 70. am 73 s TZ cos n cos n 8

These results are a further proof that the two bisectors of

the vertical angle ofAQCTdivide the base in the ratio of thesides QC and QT (or r and s).

CHAP . x1,ARTS 11-13 195

Ex. 1 . Givem the value make r==74, 8 35, draw the figure,measure 91 and calculate the lengths CW, TW, CZ,

TZ. Comparethese with the measured lengths. The positions of C and Tmaybe given by their distances from QX and QY calculate these and

compare with their measured lengths.

18 . One more ratio connected with a right-angled trianglemust be named. The ratio of the side opposite an acute angleto the side beside the angle iscalled the tangent of the angle.

(Here again the term side does not apply to the hypotenuse. )

Thus —3is the tangent of m, and iswritten ‘tan m

The values of m,Sin cosm

,tan m are all given when

we know the value ofm or of one of the ratios, so that theseratios are related.

Giveone relationbetweenSinm,cosm

,and tanm.—Sin 711

CA sin m“

Q—C'HSOS m

aar flO thfit

cosm QC CAtan m.

By expressing CT (or other length of Fig. 9) in difl‘

erent

r sin onways, find a relation between m and n.

—CW

and TW“3“ m

. so that CT = (” 2m m

SID ‘n 8111 73

r cos mandTZ

s cosm’ so that CT (r—s) cos m

.

n cos n

(r+ s) sin m (r-s) cos msin n cos n

that is, (r+ s) tan m (r 8) tan h .

By means of this relation and mathematical tables cal

culate 72 when r 74, s 35

, m 41 . Hence, calculater CW CZ

the lengths CW,TW, CZ, TZ, and the ratios 3

:

TW’

TZ.

ExEncISEs.

2. Draw an angle whose sine is and find the value of the

angle , of its cosine, and of its tangent .

3. An angle has a cosine of Find the angle, its sine,and its tangent .

4. The tangent of an angle is 4. What are the angle, its

sine, and its cosine ?


5. Can the cosine of an angle be equal to 2? What valuescan the cosine have ? And the sine ? And the tangent ?

6. Having given a line AB and two points C and D, find a

point E in AB such that EC and ED are equally inclined to AB .

7. Having given a piece AB of a straight line, show how to

divide it intemall or externally in a given ratio. Is the methodof calculation or 0 drawingmore convenient when the ratio is givenas a number ? Which is more convenient when the ratio is givenas that of two lines shown ?8. By drawing and measuring find the values of the sine, cosine,

and tangent of angles of 10, 20, 30, 40, 50, 60, 70, 80For each of these angles verify that the tangent is the quotient

of the sine by the cosine.

Draw the graphs of the sine, cosine, and tangent of an anglefrom 0 to 90 degrees.

9. With your protractor make an angle of By drawing findthe values of the sine and tangent of this angle .

Tabulate beside these the values given in the tables and accountfor the differences in the results .

10. A picture is hung with the top horizontal by a cord 2 feetlong fastened to the two top corners and passing over a nail, eachpart of the string making an angle of 60

°wrth the top of the

picture. Howmuch will the picture be raised by shortening the

flord by£1

i

?

nches and adjusting the picture so that the top is stillorlzon

l 1 . Find, by drawing and by calculation, the altitude of the sunwhen a man casts a shadow twice as longas himself. (The altitudeis the between the man

’

s shadow and the direction of the

sun’s

12. In a survey it was necessary to continue a strai ht line X0

past an obstacle which could not be seen over. A°

no OYwasmeasured at right angles to X0 , and from the point Y lines YPand YQ were drawn which cleared the obstacle the les 0 1

’P,

OYQ were 36°and 53

°respectively, the distance OYwas 50yards.

What lengths were set off along YP and YQ so that PQ was in the

same straight line with X013. A survey line is Tim chains 18° north of east from A

to B, chains due east from B to C, and chains 70° northof east from C to D. Find, b calculation, how far north and east

each straight piece goes, and to three significant figures) how far

north and east D is of A.

Check your results by a drawing to scale.

14. Draw a triangle ABC, havin AB 14 cm . ,‘

AC

11 cm. , LRAG Draw AD isecting the angle BAC

and meeting the base in D. DrawDE parallel to AC to meet AB


he graduated so that the angle of swing of the rod to right or leftof the vertical position, may be read di rectly on it . Where mustthe marks corresponding to swings of 10

°

20°

and 30°on each side

he placed, andwhat is the greatest swing we can read on the scale ?

20. There are four stations on a line of railway. How manyticketsmust be supplied for use

on the line ?

21. The area of the Trent basin is square miles, and theannual rainfall 31 inches. The river discharges cubic feet

per minute. What percentage (to the nearest integer) of the rainthat falls is discharged by the river ?22. A recent Blue Book gives the weekly expenses in summer of

a number of households as follows —The average weekly familyexpenditure in

261 cases was

289

416

382

596

How many cases are included, what is their to tal expense for

th

h

e

lweek, and what is (to the nearest penny) the average for the

w o e ?

23. How many shot, g-inch in diameter, will 0 to a

lgound if a

cubic foot of lead weighs between 708 and 712 l e volumeof a sphere is i n x

24. An air-

pump with a cylinder of volume A is used to exhaust

a vessel of vo ume B. After it strokes of the pump the pressure of

the air in the vessel is (AfB)n

x p where p is the atmospheric

pressure. Taking p as 30 inches of mercury, A as 17cubic inches,and B 130 cubic inches, find the pressure in inches of mercuryafter 15strokes and after 50 strokes.

25. Draw two lines QP and QR meeting at an angle . Find

whether every point on the line that bisectsfrom QP and QR ; and Whether every point equidistant from QPand QR lies on the bisector of LPQR .

CHAPTER XII

ON SHORTENING CALCULATIONSTHE SLIDE RULE

l . RECALL the meaning of the terms index and power(chap. IX

, art. Calculate the powers of 2 up to the

index 15 and exhibit them in a table — The product of

2 2’

s is 4, of 3 2’

s 8, of 4 2’

s 16, and so on. Call theindex a: and the power y, so that y 2x

, and the table is asfollows

This table enables us to some multiplicationswith little labour . Suppose, for instance, we want tomultiply 128 x 16. The table shows that 128 is 27 or 7 2’

s

multiplied together, and that 16 is 2‘ or 4 2'

s multipliedtogether. Thus 128 x 16 is or 11 2

’

s all multipliedtogether, or The table shows that 21 1 is which istherefore equal to 128 x 16.

In the same way find the products 256x 128 and 512x

Can you use the table to find the quotient 512—The table shows to be the product of 14 2’

s; and

200 ON SHORTENING CALCULATIONS

512 to be the product of 9 2’

s. So we have to divide the

pmduct of 14 2’

s by 9 2’

s, which leaves the product of

5 2’

s, and the table gives 32 as the product of 5 2’

s. So

that 512 32.

2. But we must deal with many numbers that do not

appear in this table. For some the table may be madeto serve. For instance, suppose we want to calculate328 We may take 327-68 as an approxi

mate value. Now lox 128 l0x27, and

32768 21532768

100 100

so that the quotient is (10 x 27) or which is256

10000 1 0 256.

W ith regard to the number 215, you know already thatthe number 15 that shows how many 2

’

s have to be

multiplied together is called the index. You know alsothat the number 2“ itself (or is called the 15thpowe r of 2

’The relation may also be

expressed by saying that 15 is the logar ithm of

the number 2 that forms each of the 15 factors being calledthe base of the logarithm . The relation is also written15 log2 or, when it is not necessary to show thebase of the logarithm,

15 log

3. For many numbers our table is useless. A more

useful one may be found by using another number insteadof 2. Let us take a number not much greater than 1, say

Calculate the powers of up to the index 10, eachto 3 decimal places.

Index Power

1

2 10 72

3 10 83

4 10 94

5

The series of powers increases so gradually that anynumber between 10 10 and 11 05 may, with tolerable


Why do you take only four terms of the series Whatare the values of the 5th and 6th terms The o therpowers of being also calculated by the binomialtheorem we have the table

Index 10, 20, 30, 40, 50,

Power 2705,

showing that the error of the former method mounts so

rapidly that it gives 1 °01 loo correct to only one decimallace .

You must then go beyond the index 100 to get a powergreater than 10. Give some idea how much further youmust go.

— Successive calculation is slow and inaccurate.

We use the binomial series and find

10 1200

10 1300

showing that we need not go so far as the index 300.

5. A class of 30 boys might easily complete the table,each of them calculating less than 10 entries. One wouldcalculate the entries for indices 11, 41, 71, stoppingat the first power that is beyond 10. The second wouldtake indices, 12, 42, 72, the third 13, 43 and

so on. How would the first boy calculate his entries, andhow may he check them —He calculates each by the

binomial series. As a check he calculates them without theuse of the binomial series. Thus 1 °01u being 1 °01 1°x

he takes the value of 10 1” and multiplies byfindin

lg 1

°01u Further, he knows 13 48,

and t at

1 °011 1 x x

so he finds the remaining entries by multiplyingrepeatedly by The results will be below the truevalues, but near enough to expose any serious error .

A few of these entries follow for the sake of illustrativecalculations.

CHAP . x11,ARTS. 5, 6 203

6. By means of the entries table already givencalculate as far as you can

(1) x (5) 107x793

(2) -5 (6) 793-5

(3) x (7) x

(4) (8) 780

(1) is approximately and is so

that 7°76x is multiplied by itself 252 times, or

The entry for index 252 is necessary for the evaluetion of this.

(2) and we havenot the corresponding entry.

(3) 1°616 x 1°0148 x and we have

not the corresponding entry.

1 0148 1 ° as

this time we happen to have the entry.

(5) 107x 793 x x x x

x to evaluate which we should lookout the power for index 215and multiply by 10

,000.

(6) 793-5 10 x l0x PM"10 x as we happen to have the entry.

(7) x x even our

complete table would not give this, as it does not containindices even so high as 300.


(8) 780= 10 x 7 10 x

10 POP “. If we had the entry for index 164 it would

not help here. The division would still remain to be

performed, and the sole object of our table is the avoidance

of multiplication and division.

7. Can you suggest any way out of the difficulty found inusing the table to calculate 7°40 x7°57 and 780 ?

Can you make use of the last entry for which the power is10 — If the last index on the table is k, sothat 10,

we can write 1-01404 in the form 1-01k x 1-olm-k

or lox

1 °01404' k

, and the table will contain 404—k.

You know that k lies between 200 and 300. To findmore exactly where it lies calculate the entries for indices210

, 220, They areIndex210

220

230

240

Now calculate the entries for 231, 232, They are

9959

1O°O59

The number 10 lies between these two powers, so we need

go no further. And of the two is nearer to 10, so

we take the approximate result,10.

Now calculate l °olm .— It is 1 °Olz31 x 1 °01 173, or lox

and the complete table would give this. Thus thetable serves for such products as x

Can you now calculate 780 -We found this tobe 10 x and we want the dividend to have agreater index than the divisor. We replace 10 byand have the form or for whichform the table is useful.

8 . The relation between power and index may be shown

on a graph. As data from which to draw the graph it is


(5) 107x793. By the graph x is and the

product required

(6) is by the graph, so that 793is

(7) 7°40 x We mark the points A and C (Fig. 2)

FIG . 2.

that represent these two numbers and draw the abscissaeAB and CD, which are the two indices to be added. Theirsum CE is too great to show in the graph. So instead of

considering the number we divide the index into twoparts CF and FE ,

FE being the part that extends beyondthe rectangle containing the graph, and CF the greatestindex of the graph, namely 231 ; we consider x

We know 10, and the graph gives

so that 7°40 x7°57 56°C.

(8) 780. In this case the indices of 1525 and

are GH and KL (Fig. of which the greater is to besubtracted. So we lengthen GH by a length HM equal tothe greatest absciss namely 231 . We then cut off a lengthMN equal to KL,

'eaving GN, to which index the power

corresponds. Now GN GH +HM KL, so that

$335= x 1 0 1“ 1525x 10

9 . It was seen that we need not read OR the indicescorresponding to numbers that we are multiplying or

dividing. We can leave the index axis ungraduated without loss of usefulness, as in Fig. 2.

We can even,by putting other numbers along the index

axis, do without the curve itself. Suppose, for definiteness,

CHAP . x11, ARTS. 9, 10 207

that our unit for the index scale is 1 millimetre, so that thegreatest abscissa is 231 mm. long. Read off from the graphthe indices corresponding to powers 1

,2,3,

10.

They arePower 1, 2

,3,

4, 5,

s,

7, s, 9,

10.

Index 0,70

, 1 10, 139, 162, 180, 199, 209, 221, 231.

Make dots at distances 69 , 110, mm. along the indexaxis ; but instead of marking them with these distances,mark them with the numbers 1

,2,3, 10 (Fig. 3, sketch) .

A i E D

1 2 4 5 6 7 8 9 10

Further, make dots on the index axis corresponding to

the powers but therewill be no room to write these powers beside the dots.

10 . Can you use this scale (shown reduced in Fig. 3) inplace

°

of the graph to calculate x — Denote by B thedot corresponding to the power then the length AB (inmillimetres) is the index corresponding to or

Denote by C the dot corresponding to the power so

that the length AO is the index of and

Now mark of BD of the same length as AC, and we havex l °o1‘3 x or x or

And we know that the point D is marked on the scale, notby its distance from A, but by the corresponding power.

D is close to the dot that corresponds to the power so

that and x

How would you calculate x — The graph is

practically straight from to so that the divisionsof Fig. 3 for would be practicallyequally Spaced if there was room to put them in.

We judge the position of B for by eye, and so for G.

Then if D lies between two dots we estimate by e 9 how itdivides the interval, and so get the second decim place inthe product.Ex. 1 . Show that if the graph was actually straight between the'

nts and the divisionsof Fig. 3 for wouldg: actually equally spaced.


11 . How will you mark of? BD equal to AC -It could

be done by opening dividers to the length AO, and so transferring it. Would it help to have a second copy ofthe scale —We could then place the point A of the second

scale against B (Fig. and C on the second scale wouldthen lie against the point D that we want.

A C

1 2 5 6 7 8 9 10

FIG. 4.

Make two such scales on strips of paper, and use them to

ga

nulate x2°68. Use them also to calculate

12. Use your two scales to .calculate 9°l9 x2°68.—B

being the point marked on the first scale, 0 the pointon the second, we put the beginning of the second

scale against B (Fig. We know that 9°18 x

C

r-

fi-r

. T-r*

but 0 is beyond the end K of the first scale, that is,AC is greater than 231 . l0 x

so we slide the first scale along till A lies where Kwas, to the position A ’K The point on the first scalenow opposite C is marked so that and

9°19 x

The sliding of the first scale to its new posi tion 1s hkelyto lead to inaccuracy. Can you suggest an improvement—We could make the first scale on a strip of twice thelength, and repeat the graduations on the added length.

The added length would then be ready to hand in the

Position A ’H ’ whenever itwas needed.


shortly written 1 °012, the 2 showing how many factorsthere are ; so l °o1 1°x 1 °01 lo may be written the 2

showing how often the factor 1°01 lo occurs. In the sameway (1

°01 means the product of 3 factors, each factor beingSo that we may write the relations in the form

any number of factors instead of 2 or 3. If we have a

factors, each being each of these a factors containsthe factor 10 times, so that the factor occurs

altogether a x 10 times ; so that

(1.01 l0)

a

It matters nothing that there are 10 factors in each batch.

The reasoning applies equally to any number b, so thatwe know

1°01ax".And nothing depended on the particular number thatforms each factor. Any number k, integral or fractional,does equally well, so that we know

(kbyz

What sort of numbers must a and b be —They are

whole numbers, as they show the number of factors. We

know what 1 °017 means and what 1 O° 16 means, butwe haveno meaning for 1

°

16. Can you generalize —Asbefore

,we Show that

-e

where of course a must be greater than 17. We showfurther that

1 °01b

and ka -z-kb ka‘ b

,

a being greater than b.How would you express simply — It is

17 factors multiplied together, and then divided by 30

factors. The result is unaltered if we omit the multiplica

CHAP . x11, ARTS. 15—18 211

tion by the 17 factors, and at the same time omit thedivision by 17 of the 30 factors. There remains simplydivision by 13 factors, which we express by 1 PM”

.

What becomes of the relationwhen a 18, and when a 17 -When a 18 we have18 factors divided by 17 factors, so that one is left, and thequotient is When a 17we have 17 factors dividedby 17 factors, so that the quotient is 1 . If you takethe relation and forget for a

moment that it has any meaning, and put 18 for a, whatdoes it become -It becomesHitherto, we have not met the index 1 . Can you give

it a meaning -The index shows how many factors thereare. The index 1 says there is only 1 factor, so thatmeans simply

17. Take again the relation and

ignoring its meaning write 17 for a.— The relation be

comes What do you makeof this, with a zero for an index —An index zero wouldseem to say that there are no factors. We know that

is 1 . Clearly, then, if 1 °O1O is to

have any meaning it must mean 1. Is it possible that1 0 1"might turn up somewhere where its meaning had tobe something else — In expressions such as 1 °01b,

can only turn up when a and b are equal, and then-1 °01 b 1, so that in all such cases 1 is the appro

priate meaning of

What meaningwould you give to It“ — This symbol willturn up from h“ k"or k“ k“, or such expressions, so itsappropriatemeaning is 1

18 . Once more write a 30 in the relation-It becomes

Here is another new kind of index, a negative index. Can

you give it a meaning — We know that multiplyingtogether 17 factors and then dividing by 30 factors amounts

simply to dividing 1 by 13 factors. So , while means

that 13 factors are to be multiplied together, mustmean that 1 is to be divided by 13 factors. Is it

possible that “ 13 might turn up where it would have tohave some other meaning

-It can only turn up when


there are 13more factors in the divisor than in the dividend,so that the meaning already given is appropriate.

What meaning would you give to If“

19 . We have sometimes wanted to find the square rootof a number

,that 1s, to find another number which, multi

plied by itself, gives theoriginal number. Hitherto, our only means has

been the graph of xxx

(which may also be written xx or Can you

use the graph y= l °o1x to

find the square root of a

number Find, for in

stance, the square root of-At the point A

Fm , 7,that marks the power

we draw the abwissaAB to the curve (Fig. We bisect AB in a and lay in

the curve the abscissa DE , whose length 1s AC. The power2°73 represented by D is the square root. For 2°73= 1

so that 2°73x2°73 1 °01” x 1 °014B

Could you use a complete table of powers of to finda square root — The table shows that 7°48 lies between1 °

and 1 °01203. New 1 °01 101 x 1 °01101 1 °01209, so thatis the square root of or and we can

look its value out in the table. We might have taken1 °01203 as the value of but we could not separate the203 factors into two equal batches. To find a square rootwe must use only even indices.

20 . We had such relations as x

which are true so long as a is a whole number. Let us for

a moment forget the meaning of this relation and put ,1

instead of a. What does the relation become -It be’

comes x 1 ° =01t If 1 °Olt°

1s to have anymeaningit must mean the number which, multiplied by itself,makes that 1a, the square root of 1 °01 .

Calculate the squares of Do thesegive you the square root of x

so that 1 °OO5is pretty accurately the square root of


EXERCISES.2. A, B, C, D are four stations on a railway the distance AB

is 10 miles BC, 10 CD, 8. The following 1s an extract from a

time-table

Down.

A dep. a.m . D dep. a.m .

B iarr. a.m .

dep. a.m.

arr. a.m .

dep. a.m.

D arr. a.m . A arr. e .m.

Plot in one diagram on squared paper graphs to Show the positionsof both trains at any time between those given, assuming thatthey run uniformly between the stations. When, and where,do they pass one another ?3. The water supply of a village of inhabitants is brought

by a pipe 150 uare inches in section, and the water flows alongthe pipe steadi y at 2 miles an hour. Give, to the nearest tenthousand gallons, the total daily water suppl and also, to thenearest ten gallons, the daily supply per hea A gallon is 277cubic inches.

4. A number of girls are to be distinguished bywearingdifferentrosettes. R ibbons of three colours are available to make therosettes, each rosette may be of one colour, of two, or of three ,and no two rosettes are to be the same . For how many will thesethree colours serve ?

If another colour was added, making four , how many more girlscould be included ?

5. Calculate the value of7 (1 where P

7°

n 24, the value of P being given to the nearest

hundredth, those of r and n exactly.

6. Calculate, to two decimal places, the value when a: 1 of

m7 11 3 1 3

31 7! 13:

where 3 ! means 1 x2x3, 5l means 1 x2x3x4 x5, &c.

7. A cask is 4 feet 6 inches deep, its greatest diameter is 2 feet3 inches, and the diameter of each end is 2 feet . Calculate thenumber of gallons which it will hold . The volume of a cask ofdepth h feet of which the greatest and least diameters are D and d

CHAP . XII, EXERCISES 215

D d 2

feet 18 approximately 854 x h x2

cubic feet, and a cubic

the error caused by takin this formula instead of

the followin more accurate formula olume of cask of de th11 feet of w

°

ch the greatest and least diameters are D an dfeet is

005236xhx (8 ><D2 4xDxd 3xd2) cubic feet.

8. An upright pole 10 feet high casts a shadow feet long atmidday on a certain day. Another upright pole of the sameheight, 100 miles further north, casts a shadow feet long at

the same time . Deduce the earth ’

s perimeter, supposing the

earth a sphere .

CHAPTER XIII

DANGER ANGLE. PROPERTIES OF CIRCLES

1 . A sm r’

s chart (Fig. 1) shows on a certain coast twopoints, a church 0 and a lighthouse L , with the notice thatthe danger angle is This means that if a ship Sgets so near 0 and L that the angle CSL is greater than

she is in danger of rocks or shoals. Draw the chart

on double the scale shown in Fig. 1, draw on tracing-paper

an angle of move the tracing-paper about so that thetwo legs of the angle always pass through 0 and L, and so

draw the curve that separates safe water from unsafe.

Repeat, taking 50°in succession as the danger

angle. What are the curves — They appear to be arcs of

circles (that is, parts of circles) that pass through C and L .

Supposing one of these curves a circle, how can you findits centre —We know that the centre lies on the perpendicular bisector of a chord of a circle. By drawing twochords we find the centre as the intersection of the two

bisectors. Find the centres of all these curves.

How many positions of S on a danger circle do you

need to lay down before you can find the centre of the

circle — One, for by joining it to C and L we get twochords. Find the centres of the curves in this way.

Howmany points on a circle fix it ? That is, how many

218 DANGER ANGLE. PROPERTIES OF CIRCLES

lie on the same side as S of CL . WhenLCSL is obtuse, we

make the angles CLO and LCO equal to LCSL 90, mak

ing 0 lie on the other side of CL from S. In each case 0 isthe centre and 00 or 0L the radius. Where is 0when LCSL is right8 . Suppose the positions of C and L kept fixed while S

is varied. Can S move at allwithout changing the centre 0(and in consequence the circle) — The position of 0 is

settled by the size of the angle CSL , as the constructionjust given shows ; so that S may move without changing0 so long as it does not change the size of the angle CSL .

That is, if we return to our original idea of a ship’

s chart,and suppose CSL the danger angle, the position of 0 is

unchanged so long as the church and lighthouse subtendthe danger angle at the Ship. That is, all positions whereC and L subtend the danger angle lie on the same circle.

Are there any points inland at which C and L subtendthe danger angle Do C and L subtend the danger angleat all points on the circle we have found — If the figure is

folded about CL, every point Son the danger circle falls on

another point S’ where the

same angle is subtended (Fig.

3) in other words, the imagein CL of every point S thatwehave found gives another pointS

’where CL subtends the dan

ger angle. Again, the anglesubtended at any point on thearc CTL is clearly not the

danger angle. The completelocus of points where the danger angle is subtended is thetwo arcs CSL and CS

’L .

In Fig. 3, draw theM n to

C, S, L ,T, join TC and TL

,

FIG . 3. and by expressing as manyangles of the figure as possible

In terms of Ls CTO and LTO, find the relation betweenLCTL and LCOL and between LCTL and LCSL .

LCTL +LCSL

CHAP . XIII, ARTS. 219

4. Draw any circle with centre 0 (Fig. Suppose a

line K to move up across the circle from X to Y. Let Kin any position meet the circle in P and Q, and let R be anypoint above K on the circle. By measuring a number of

positions find how the angle PRQ varies as the line K

FIG . 4.

moves across the circle— The angle increases from 0°to

And how doesLPOQ vary? It increasesfrom0°to and then diminishes again to What

is the relation between Ls PRQ and PDQ in anyposition — Until the line K passes the centre 0 ,

LPOQis twice LPRQ ; after that 360

° LPOQ is twice LPRQ.

Hitherto we have been concerned with angles less thanand in this article we have assumed LPOQ to mean

the angle less than 180° that is contained by OP and OQ.

Until the line K reaches the centre 0, the angle FOQmeans the number of degrees a line must turn through toget from OP to OQ, turning coun te rc lockwise , that is,opposite to the direction in which the hands of a clockturn. Let us continue this meaning for the angle POQafter the centre 0 is passed, and admit angles of more than

W ith this meaning how does the angle POQ vary as

the line K moves across the circle from X to Y — LPOQincreases from 0

°at first contact of the line with the circle

to 360°at last contact.

5. In Fig. 4 join R0 and produce it to any point T.

Draw the figure in all possible forms, LPOQ less thanor greater than the LPRQ enclosing O or not

220 DANGER ANGLE . PROPERTIES OF CIRCLES

enclosing O, or havi O on one leg. What is in each casethe relation between TOP and LTRP, between LTOQ and

LTRQ —The angle-sum property of a triangle showsOPR + ORP +ROP equal to and so equal to BOP -lPOT. So that POT is equal to ORP + OPR , or to twiceeither of them, since they are equal. In the same way

2LTRQ. And how is LPOQ related to— In every case by addition or subtraction of

TOP and TOQ, and of TRP and TRQ, we find LPOQL2PRQ ; LPOQ meaning the angle through which a lineturns counterclockwise to get from OP to 00.

6. Any piece of the circumference of a circle is called an

arc. The area enclosed between an arc and the chordjoining its ends is called a segm ent . Thus in Fig. 5,

ADB is an arc, and the space

bounded by the arc ADB and

the chord AB is a segment.Further, the angleAER is saidto be an angle in the segmentADB

,or an angle at the cir

cumference standing on the

arc ACB . The two anglesbetween 0A and OB , one lessthan 180

°and one greater

than may be distinguished as the angle at O

Fm . 5.

standing on the arc ACE and

the angle at O standing on thearc ADB . By means of these terms state the conclusionarrived at.— The angle at the centre of a c ircleis dou ble the angle at the circum fe rence thatstands on the sam e arc , and the angle at the c i r

cum fe re nce on a given arc is the sam e for a ll

po in ts a long the circum fe ren ce . Or, in other words,al l the angl e s in the sam e segm en t of a circle are

equ a l , each be ing half of the angle at the cen trethat stands on the sam e arc. Thus in Fig. 5 the

acute angle AEB and the angle AOB less than 180°stand

on the same arc ACB, while the obtuse angle AFB and

3111

5Engle AOB greater than 180

°stand on the same arc


Such a line that is .L to a radius of a circle and meetsthe circle in one point only, or touches the circle, is (inconveniently enough) given a name that you have metalre

pdy in a different sense. It is called a tangent to the

c1rc e.

9 . Itwas found that in Fig. 3 Ls CSL and CTL made uptogether State the result as generally as you can, and

prove it. —The sum of a pair of Opposi te angl es of a

quadr ilate ral inscribed in a c irc le is For eachangle is half that at the centre on the same arc, and the twoangles at the centre make upState and prove a converse proposition.

— If a quadrilateral has the sum of two opposite angles equal toa circle can be drawn about it, that is, passing through itsfour corners. Suppose that in Fig. 7Ls S and Q of PQRSare together Find the centre 0 of the circlethrough P Q R as the intersection of the .L bisectors of

PQ and QR , and draw the circle. The locus of pointswhere PR subtends an angle 180 Q is the arc PTR of thiscircle and the image in PR of the arc. So S lies on one

of these arcs. If S lay on the image PUR, say at U, weshould have a quadrilateral with the angle PQR greaterthan Let us exclude such angles, and S willlie on the arc PTR, and the proposition will be true .

Ex. 2. Draw any quadrilateral in a circle, measure its angles,and so find the sum of each pair of opposite angles.

Ex. 3. Draw a uadrilateral having a pair of o posits anglestogether equal to Measure the other two ang es and so findtheir sum . Draw a circle through three of the com ers of the qua

drilateral and see if it goes through the fourth com er.

10 . If the sides of the quadrilateral are produced, thusmaking exterior angles of the figure, how may these twoconverse prepositions be stated — A quadrilateral in a

circle has the angle between a pair of sides equal to theexterior angle between the other pair of sides ; and if a

quadrilateral has the angle between one pair of sides equal tothe exterior angle between the other pair, a circle can be

drawn about it.In a circle draw a quadrilateral PQRS (Fig. and pro

duce one side RS both ways. Join O, the centre of the

circle, to P Q R S. Suppose LOSP to contain a degrees,

CHAP . x m, ARTS. 9, 10 223

LORQ b degrees, and LSOR 0 degrees. Are all the otherangles determined when a b c are known —Yes, forthese angles are enough to enable us to draw the wholefigure except as to size. Draw a number of figuresof difi

'

erent forms and in each figure express all the other

FIG . 7.

in terms of a b c. In particular, what values haveSPQ, RQP , PSV, QRW — In the figure drawn (Fig. 7)

OSE 90 so that PSV 904-54 4 .

QRW = 180 ORQ ORS 90 +

POQ POS QOR—ROS (180—2a) — 2b) —c

2b—2a— c ; whence OPQ.

SPQ ore OPS

RQP RQO+ 0QP


Thus the particular results wanted are

QRW ==SPQ 90+-

2

‘f—b,

PSV= PQR

Ex. 4. Check these expressions by measuring a, b, c on one of

your figures, substituti their values in the expressions, and

comparing the results wit their values measured on the figure.

11 . Suppose now in each of your figures the angle 0 to

become so small that we may neglect it. Draw the figuresagain for this case. What do the expressions for the anglesbecome, and what is the form of the figure —The ex

pressions become QRW SPQ 90 b, and PSV

PQR 9o+u for thefigure drawn. S and R come so closetogether that OS and OR are indistinguishable from .Ls to

VW, and VW is indistinguishable from a tangent to the

12. Consider the case in which VW is actually a tan

gent. Draw any tri

angle PQR in a circle(Fig. and at R draw

VRW J. OR and therefore a tangent to the

circle. Join O to P , Q,R. Call LORP a and

LORQ b. Draw figuresof all possible forms,

and for each figure ex

press all the otherangles of the figure interms of a and b. In

particular, what are theanglesRPQ,RQP ,PR V,QRW —Fr0m among

FIG . 8. the expressions for theangles we pick out

PRV PQR 90+ a, and QRW QPR 90— b, for the par

ticular figure drawn (Fig. Thisagreeswith the result foundfrom the quadrilateral bymaking 0 very small. State


angle OLP equal to A (Fig. Draw the .L bisector of

CL , and the .L at L to LP. The intersection of these .La is

the centre of the circle.

FIG . 9 .

Ex. 7. Show how.

to draw a circle to touch a given line at a givenint and to pass through another given point. Is there any case

In which it cannot be done ?

Rectangle properties of circles.

15. Draw a large circle, centre 0, and take a point 0lying either inside or outside. Let a point P travel alongthe circumference and see how the length OP varies. Begin

with P where you please then let it travel round the circleand back to the initial point, so that the radius CP turnsthrough Give a graph showing the length OP as a

function of the angle through which CP has turned.

When is OP greatest and when least ‘

P

Ex. 8. How does the angle COP vary ? What difference according as O is inside or outside the circle ?

16. The line OP , considered as unlimited in length,usually cuts the circle in a second point Q. For eachposition of P measure OQ, and calculate OP x 00. Give a

graph showing OP xOQ as a function of the angle CP hasturned through. What is your experimental result

The product OP xOQ is constant.

CHAP . XIII, ARTS . 15-17 227

Ex. 9 . For each of a number of positions of CP, draw on squared

per a rectangle with sides equal to OP and OQ, and find the area

y counting squares.

17. Draw the diameter AB that passes through 0. Whatrelation have you between lengths made on PQ and

on this diameter by O and the circle —OP xOQState this relation as a proportion, that

is, as an equality of ratios. — It may be stated

OP OBrOP 0A OQ OB

0A 00’

on 00’

0A 0P'

What method have you hadfor proving the equality of

ratios Can you apply it hereCan you find any similar tri

angles ? - Join AQ and BP .

In Fig. 10, where O lies insidethe circle, the vertically oppositeangles AOQ and BOP are equal,the angles QAB and QPB in thesame segment QAPB are equal,and the angles AQP and ABP

in the same segment AQBP FIG . 10.

are equal. The triangles AOQAO PO

and POB are Similar, and How do you

k th t thQ0

aLBO

d tAO BO

?now 9. ese are equ an no

QO PO

FIG . 11 .

Proceed with the of 0 outside the circle — In thisQ 2


case (Fig. 1 1) the angle ABP between two sides of the

quadrilateral ABPQ in a circle is equal to the externalangle AQO between the other two sides, and in the same

way LBPQ LQA0 . So that the As OBP and OQAOA OP

s1m11ar,and the ratios0? OB

are equal. How

OB 0Ado you know It Is not

OPthat 1s equal to

OQ

Ex . 10. From measurements of each of your figures calculate

g; 3—2and so test the proposition.

Ex. 11. Prove the proposition for two positions of OP , neitherposition passing through 0.

18 . The length of the radius being r cm . and the dis

tance of 0 from the centre being 3 cm . , express OA , OB,

and 0A xOB in terms of r and a— Lengths being in

cm. and areas in sq. cm .,we have in Fig. 10,

AO OA—OO r—s, B0 r+ s,

0A xOB (r rr—ss, or —82

In Fig. 1 1

0A s—r, OB s+ r, and 0A xOB (s s

z—rz.

Express OP xOQ in terms of s and r.

Ex. 12. Measure r and s on each of your figures, calculate the

difference between r”and s’ , and compare it with the values you

found for OP x OQ for various positions of CP .

Ex. 13. Draw a rectangle with sides r 19 centimetres and r 41

centimetres, and show how to cut it up and rearrange it as a square

r centimetres in the side, with a little square 19 centimetres in the

side cut away.

19 . Cons1der the case of 0 inside the circle. On the

diagram that holds your graph of OP as a function of the

angle turned through by CF, draw a graph giving OQ as

a function of the angle turned through by CP . For whatpositions is OQ greater than OP, and for what positions isit less -Draw the chord MON ..L to AOB . When P lieson the greater arcMBN, OQ is less than OP ; when P lieson the less arcMAN, OQ is greater than OP ; and when P


this relation become at the points G and H where P and Qpass one another -It is 0GxOG 8

93—73

.

21 . How would you draw a tangent to a circle, centre C,from a point 0 outside it (Fig. 12) —We could lay a ruleragainst the circle and through 0 and so draw the

tangent. That method gives the tangent very well,but does not give the point of contact very accurately.

FIG . 12.

G being the point of contact we know that LOGO is a

right angle, so that G must lie on a circle on 00 as

diameter. We draw this circle and so find the two pointsof contact of tangents from 0 .

By considering similar triangles, find a relation betweenOG, OA and OB.

— Join GA and GB . The angle OGAbetween the tangent GOand the chord GA is equal to theangle GBA in the alternate segment. So the two trianglesOAG and 0GB have LOGA LOEG, and the angle at Ois an angle of each triangle. The triangles are therefore

OA OGslmflar

,and

0G OB, or OGZ OA xOB .

22. We saw earlier that the sign of multiplication x

could sometimes be omitted without loss of clearness. In

the present case confusion would msult from its omission,

but a dot may be used without loss of clearness. The

above relation is then written OG2 OA .OB .

An equality of two ratios, 9 . g. gs “(é inFigs. 10 and

CHAP . XIII, EXERCISES 231

11, is called a preportion , and the four lengths OA , OQ,OP, OB are said to be in proportion. When the pro

portion contains the same length twice, as a numeratorin one ratio and as a denominator in the other, this lengthis called a m ean proport iona l between the other two.

Th

éis

o

OG in Fig. 12 is a mean proportional between 0Aan B.

28 . Suppose the length of the tangent 0G to be tcm.

and give a relation between t, r, s.—OA s- r

, OB s

+ r, t"OA .OB sz—r

z. It may also be put in the

forms 32

r2+ t

2,and r

2sz—tz. What sort of

triangle do the three lengths r, s, t form —They form the

right-angled triangle 0GC.

24 . Sum up your conclusionswith regard to the rectangleproperties of a circle — If a l ine is drawn throu gh a

point 0 and cu ts a circle in P and Q, the area of

th e re ctangle con tained by OP and OQ is independent of the direct ion in wh ich the l ine is drawnt h rough 0. If the circle has a radius of r cm . and

O is s cm . from the cen tre ; th en the ar ea of the

rectangle is — 32sq. cm . when 0 is inside the

circle , and 43° when 0 is ou tside . A lso when

0 is ou tside , the area of the rectangle is equal tothe square on the tangent from O to the circle .

ExRRcIsns.

18. Draw a circle with a triangle ABC inscribed in it. Draw

the creases that a pear when (i) B is folded on to C ; sii) AB

is folded on to A8. Do the crosses meet on the circle Testby a geometrical proof.

19. A regular hexagon is inscribed in a circle . Express, in

degrees, the value of the angle at the centre subtended byeach side .

A paper circle is creased so as to show a diameter AB, and the

circle is then folded so that the intsA and B are brought togetherto the centre, two

’

other creasesllibing thus obtained at right angles

to AR. Prove that the six points now marked on the circum


ference by the three creasesare angular points of a regular hexsgm .

m nh but s genu al proof should also be given.

20. Three straight pieces of railway track meet at threeA, B, 0, thus forming a triangle. Where would you place a gun so

as to cover the three tracks ? If your gun won’

t carry far enoughto cover the three junctions at once, where will you place it tocommand some part of each track ?21 . A gasometer is in the form of a cylinder with a rounded top,

and the follo dimensions are

given diameter 28 feet,

height at edges 4 feet, height in t 0 middle 16 feet. Draw a

section of the gasometer on the scale of 1 inch to 10 feet, and calculate the number of cubic feet of gas that thegasometerwill hold.

Take the volume of the portion at the top above 14 feet) as halfthe volume of a cylinder of the same base an height, and the areaof a circle as times the square on its radius.

22. There are shoals off a coast on which stand two rominentobjects P and Q. A chart of the coast shows two as from

P and Q meeting at an angle a

What single observation might bepossib

le range of danger to test if the dangerzone

23. Take three points L, M, N , and suppose them to represent

the positions of three lighthouses along a stretch of coast LMN.

A ship 0 is 03 the coast, and the angles LOM and

to be 80°

and Construct the position of the ship, give no

proof, but point out why there is only one solution.

Prove the truth of a property of the circle that yo

24. A barometer tube, whose internal diameter is 2 centimd ires,is 120 centimetres long and is filled with mercury. Find theweight of the mercury, being given that a cubicmercury weighs 136 grams, and that the area of a circle is

x (diameter)2. Also find the cost of the mercury at 68. 3d.

a kilogram.

25. State how you would roceed if you were re

quired to draw

about a given circle a quagrilateral, which is to e such that a

circle may be drawn about it. Briefly justify your method.

To what extent is the shape of the quadrilateral at your

26. A canal-lock is to be made 100 feet long and 17 feet wide,and to let a boat through 11 stream the water in the lockmust rise 8} feet. e sluice to mit the wate r is to be designed


two other roads goes through the centre of thememorial enclosure,and is parallel to the front of the palace.

34. Draw a circle of radius 3 inches, and take any point P at a

distance of 5 inches from the centre of the circle. Mark fourpoints on the circle at distances 5, 6, 7, 8 inches respectively fromP , and join them to P bystraight lines. Then make a table likethe following and fill it 1n.

What conclusion can you draw from your results ? What isthe length of the tangent drawn from P to the circle ? Drawthe tangent.

CHAP . XIII, EXERCISES 235

Prove, for the case in which the cutting line passes through thecentre of the circle, that, if from a point outside a circle twostraight lines are drawn, one of which cuts the circle and the othertouches it, the rectangle contained by thewhole linewhich cuts thecircle and the part outside the circle, is equal to the square on theline which touches it .

35. Two straight lines AC and BD cut one another at K so

that LAKE AK BK CK and

DK A, B, C, D are the angular points of a quadrilateral .Measure the angles of the quadrilateral. Show from our

measurements that it is im ssible to describe a circle a out

the figure, and prove the t eorem by which you justify yourconclusion.

How may the same conclusion be arrived at, from a considerationof the given dimensions of AK, BK, CK and DX ?

36. From a point C outside a circle two straight lines CAB and CTare drawn, CAB passing through the centre and cuttin the circle inA and B, CT touching the circle at T. Prove that t 9 sqCT is equal to the rectangle CA . CB, and express thisalgebraically, denoting

bCA by h, the radiusof the circle by r, and the

length of the tangent y t.

Imagine your figure to represent a section through the centre ofthe earth, supposed to be a sphere of miles radius, and C thetop of a cliff. Show that, if the height of the cliff is k feet, the

distance of the horizon from C is approximate lyVF miles.

CHAPTER XIV

LOGARITHMS

1 . In discussing the Slide Rule and other methods of

speedy calculation (chap. XII) we considered two quantitiesa: and 31 so related that y How far would thatdiscussion be altered if instead of we used a difierent

number,if for instance we used — In this case also

we should have a series of indices and correspondingpowers. For the powers running from 1 to 10 the series of

indices would be different, there would be more entries inour table. The graph made with 1 °001 as base would servethe same end in just the same way as the graph made from

The curve would be the same kind of curve, and

would difl'

er only in the graduation of the index axis.

2. Calculate far enoughto give a rough estimate of the number of entries in our

table — The binomial series for

1 1000 x 0°001 {03951999m9

2x

3

in which we get the fourth term from the third by multi

plying by00

298

, the fifth from the fourth bymultiplying by

and so on, the further successive multipliers being

0

296

1 To multiply speedily by

we take them in the forms 1 1

The series is 1 1 01 66

each term calculated to three places ; the remaining terms

238 LOGARITHMS

The numbers have also been looked up in a tableof logarithms, and the logarithm there given for each has beenentered in the third column. Compare the first and thirdcolumns. Each entry in the last column multiplied by

is nearly equal to the corresponding entry in the firstcolumn. Hence, instead of calculating our table we can

make use of a table of logarithms.

For instance, let us calculate x The extractgiven above is from a seven-figure table of logarithms but

for the present calculation and for most practical purposesa four-figure table is enough. We look up the number 7°76

in a four-figure table and find given as its logarithm .

This tells us that As the logarithm of

we find so we know thatHence x x

x 10 x One thousandth of 89

is and we want from the table a number whoselogarithm is about The table gives

LogarithmO°O8S8

0 0892,

and from either of these we get approximatelySo that 7°76x

Calculate in the same way 1 °616x

107x793, 793 x 1525 780.

Negative and Fractional Indices.

6. Let us now recall when we first met with indices,what we have used them for, and what we have learnedabout them. W e first met indices in connexion with an

old problem of horse-dealing (chap. IX, art. In con

nexion with that problem we had to calculate the productof 32 2

’

s or 233. State again in ordinary English, as wellas in mathematical symbols, the method we adopted to

lighten the calculation.-We separated the 32 2

’

s into twobatches of 16 ; in mathematical symbols, we tookinstead of 232. Then we separated the 16 2

’

s .of each batchinto two batches of 8 or in symbols we took insteadof How will you write the whole product of

32 2’

s at this stage When several pairs of brackets occur ,

CHAP . XIV ,ARTS. 6-8 239

it makes the expression clearer to use different kinds of

brackets — The whole product of 32 2’

s is multipliedby itself, which we may write as P. The last step isto separate each batch of 8 2

’

s into two batches of 4 2’

s, to

put instead of 28. A batch of 16 2’

s is the product oftwo batches of 8, and so is looked on as x (2

4V; thiswe write more shortly as And the whole productof 32 2

’

s, being the product of two batches of 16, is

X or

The next process is the calculation of this number. The

statement in mathematical symbols of this calculation is

[awry [26sState this in English without the use of mathematicalsymbols or language.

— We take a batch of 4 2’

s and

multiply it out the product is 16. Putting 16 in place of

24 gives the second mathematical form used. The next stepis to multiply together the two batches of 4 2

’

s,that is, to

multiply 2 16’

s together the product is, to two significantfigures, 260. This corresponds to the third mathematicalform [260

2]2. We have now to multiply 260 by itself the

product is 68,000. Finally, we multiply by itself,and as the product of 32 2

’

s we get million.

State what operations the symbols (10902, (32V,310 direct you to carry out. Carry out the operations

and account for the results you obtain.

7. Indices turned up again in connexion with the MorseCode and the Semaphore (chap. X

, art. There we hadsuch relations as 710 x7 711 and 2"x2 State inEnglish what these relations mean, justify them, and

generalize them. State what operations each of the following expressions directs to be carried out, and carry themout — 103x 102, 10

5, 3

4x32, 32x33 x34, 39.

8 . Work again through the discussion of indices in chap.

XII on the Slide Rule, from article 13 onwards. We therefound that so long as m and n are positive whole numbers

A. am xa” MM ”

;

B . am

a” when m n

and when m n ;

(amp; a a

mxn.

240 LOGARITHMS

Once more state what operations each of these expressionsdirects to be carried out, and so justify the equations A B C.

What operations do the following expressions direct youto carry out ? Carry out the operations, giving each valueto two significant figures. When two expressions have thesame value, explain why

(2x 25x (3x

32x43, (2 25 35,

(3 35 25, (4-e

42 32, (3 32 42.

You find such relations as (2x 25x35 and (332 42. Generalize these relations and justify them.

More general forms areD. a

mxbm ;

E . (a-z-b)”

um-z

The expression (2x directs us to multiply 2 and 3

together, and then to take 5 such products and multiply

them together. So in the final product there are 5 3’

s and

5 2’

s. And the expression 25x35means simmy the productof 52

’

s and 5 3’

s, so the two expressions are equal. Nor

have the particular numbers 2, 3, 5 any influence on the

reasoning ; the index 5must be a whole number, the numbers 2 and 3 could be fractional So we know that, mbeing a whole number, (ax b)” a

'”x The relation(a-e a

’” b’” is justified for integral values of m in thesame way.

9 . In chapter XII we found meanings for such expressions as In the same wayshow what the following expressions should mean, and give

their values to two significant figures

1001 ,1 1 20+ 31 x2° 10x

10-2 102x 103

State again how we found a meaning for a negative integral index —When m 17, n 30, the relation B givesa17

a3° 1 a

“, but if we pay no regard to the original

meaning of the relation B and write it in the form appropriate to the case m n, we have a

"03° of ”

. We

242 LOGARITHMS

11 . We gave a meaning also to one fractional index or

05. It was agreed that was to be defined by theequation a

°°5x a, or in other words, that was tomean the square root of a. Can you find a meaning for

by assuming to obey the laws A , B, C, D, E

A gives x This merely gives us a newindex 0002 for which we have no meaning. But if wetake more factors, if we take factors, we get x

x al. If this is to hold must be

a number which, multiplied times by itself, makes a ;or in other words, is the 1

,000th root of a. For

instance, for we know that multiplied times by itself makes 10. Again, the relation 0gives a

, which is another way of stating therelation we have just dealt with.

What meanings will you give toa°°341

, a“'824 — W e have met already in the relation

x We will use this, which may also bewritten as the definition of In the

same way may be defined as or

and so for And in the same way is the

product of 74 factors, a°°34l the product of 341 factors ;

and a"824 the product of factors.

Find the values of these numbers when a 10, that is,the values of

12. Do the laws A,B, C, D,

E hold for all the indices

now defined, that is, for all fractions with as denominator — Suppose m and n 01 52. Then A

becomes xa°°152

am”

. If for shortness we write 74:for this is k3°xk1°2 km ; which in this form is

obviously true. B gives -sa°“152

or 1 am”

,

that is, k3°+ k1°2- 122

or 1 clearly true ; this introduces the negative fractional index and the

definition of as 1 0 givesao-oouso

.

This introduces a new fraction not among those con

sidered. We shall return to it. Meantime,is it true that

a“ ? More generally, is it true that

aP ? means multiplied by itselfp times, so thatmeans multiplied by itself 1) x 1000 times.

W e may arrange these 1000 xp factors in batches of 1000.

CHAP . XIV , ARTS . 11—14 243

Each batch is then or a so that the whole1000p factors make up , and so a”. Moregenerally (W, and we could have used this as the

definition ofMN.

13. Let us include all possible fractions among our indicesbydefining a

ll? by the equation (awl)? a ; and by definingu,m by this combined with aplq or by up ,

which two definitions we have seen to be equivalent.This gives two definitions of according as we take

as 3 —100 or Are these two definitionsequivalent —The question is whether a

sand

as°

are equivalent. Let us denote the 100throot of a byp , and the 1000th root byq. Now 1000 q

’

smultiplied together make a, and we can arrange the 1000 q

’

s

in 100 batches of 10. The product of the 100 batches is a,so that each batch is the 100th root of a, that is Q ”

p .

The first relation above is as

or a3or

p1°°

a. The second is 03°

or (q at“or

a. And we know q1°

19, so that and

the two definitions are equivalent.The 100th root of 10 is and the 1000th root is

Calculate independently and

Consider as an example of the more

general (a? é —We write P for and

Q for P °°1°2 By definition P 152

. Sothat Q1 P 152 100 P 100 152

a3x 162

or

am. But am is the definition of

ao-oom

.

State what operations each expression in the precedingparagraph directs to be carried out.

14. Any number, integral or fractional, positive or negative, has now a meaning as an index. The base alone isrestricted to a positive number, integral or fractional.The arrangement of a table of logarithms can now be

explained. It is a series of indices and correspondingpowers, but the base is not although we were able touse the table for that base. The base is 10 and the relationbetween the index a: and the power y is y Thisrelation is also spoken of as that between a number 3; andits logarithm x

,and is written in the form a: logy.

244 LOGARITHMS

In article 5we found that the table gave 7as thelogarithm of This we write either as

or as log To see whether thisentry tallies with our definition of a fractional index weshould have to calculate and and n o if

they are equal. This is laborious, and we can find easier

ways of checking the table.

15. As a first check on the table take as an approximation to 0 0009977The relation to be checked isthen or 10

which we alreadyknow to be true.

Again,by a geometrical constructionfind the square rootof 10.

— We draw a circleon a diameter 1 1 units long(Fig. and divide the diameter intopartsa of length 10and b of length 1 . At the

point of divisionwe draw the.L chord. Half its length 0 is x/ 10. The angles A and Cof the figure are equal, each being 90

°—B,and the angles

F and F are equal, each being right. So that we have two

similar triangles and g "

I? or cxc a xb or c =

J a xb m . We could also begin with a diameter oflength 7 divided into parts of 5 and 2. This will give a

better result. The size of the figure is also imwrtant for

accuracy. The square root of 10 18Use also your graph of y l °o1”c (chap. XII, art. 8) tofind

the square root of 10.

We have thus or log Comparethe entry in the table of logarithms.

16. Use the same methods to find the square root of— It is What is its relation to 10

so that 3°16a 10, or

or log a

Continue by extracting the square root of and then

246 LOGARITHMS

Rewrite this, using the logarithm notation instead of the

index notation— It is

log 7°76

log

1

therefore 7°76x 1 58

19 . State the equation 7°76 389° in terms of logarithms -0°8899 is the logarithm of 7°76 (to the baseor shortly 0°8899 == log 7

°76. Express 0°1987 and

1 °

0886 similarly, and then express the equation 0°8899

in terms of and0° log 1 °0886 log 12

°26 °

,log 7

°76+ log 1°58

log And if you substitute for its valuein terms of 7°76 and — The relation is

log 7°76+ log log (7

°76x

Calculate x in the same way, and express theresult in various ways— The product is and

being &c., the equation x

is simply another way of writing x

Again, we state the relations as 08 876 log 7°72

,&c.

,and

the equation may be writtenlog

-Hog log or log x

The calculation of 7°72x is usually more compactlywritten in such a form as

log

log

log

x

Prowed with the remaining calculations.

20 . We have had several equations likelog log log x

Give a general form for such equations and justify it as the

CHAP . XIV ,ARTS . 19—21 247

equation just written was justified— The general equation

is log a+ log b log (ax b) .

To justify it we suppose a number p found such that aor (what 18 the same thing) p = log a ; and a number q

such that b 10° or q log b. Then a xb is 10Px 10‘1 orand another way of expressing this relation 18 p q

log (a x b) ; and since 19 means log a and q means log b, theequation we began with is justified.

21 . Some time ago (chap. VI), we discussed what theheight and diameter of a standard gallon measure must be.

Such a measure has its diameter and height equal, and hasa capacity of cubic inches. Can you use logarithmtables to give these dimw sions — If d is the diameter in

incheswe know that ifxd xdxd 277-4,

277-4x4 2-774x4x 100d x dxd

31 42

log2 774 0 4431

log4

log 100 2

log3 142

25480

2 log 100

log 3532

so that dxdxd 3532

and therefore log d logd log d log

The left side of this equation is 3x log d, and we seen

that log so that3x log d 25480

log d 08 493

and the table gives 08 493 log7°068

so that d

248 LOGARITHMS

and both diameter and height of the galloninches.

Ex. 1 . Calculate the heigggoand diameter of a standard litre

1measure whose capac1ty 1s

its diameter.

22. Compare the method of extracting a cube root bylogarithm tables with that by means of the graph of a”

(chapter VI) , and with the ordinary arithmetical method.

The graph method is only a rough method. The arithmetical extraction of a square root is cumbersome, for a

cube root it is forbidding, and for higher roots the labour isso great that to calculate such a number as without logarithm tables is possibly beyond the patience of anyman. Logarithm tables are a great help for multiplicationand division but their true value as a labour-saving devicefirst appears in root extraction.

Ex. 2. Calculate the number

23. Can you generalize the relationlog (dx dx (1) log d+ log d+ log d,

or (more compactly) log (13 3 logd— If n is any whole

number log d” n log d, for this is only the short way of

writing log (dxdx d = log d+ log d+ therebeing n factors d on the left, and as terms log d on the

right. Can you deduce this relation from the mean

ing of logarithm —Suppose log d k, then d

and da (lok)s

so that log (33 3k, that is,

3 log d. And so for the case when 3 is replaced by n.

Is it true that log d'"n log (1 if n is fractional

Put log d It, then at 10"and no matter whether n isintegral or fractional d” so that log d

”

7110 n log d.

24. The quantities z and a being related by z lookout a number of corresponding values in the tables and

draw the graph of s as a function of u from a 1 to s 10.

What are the corresponding limits for u It was stated(in art. 1 of this chapter) that the graphs 3] and

y 1 °001 in difi'

ered only in the graduation of the m-axis.

Can you choose such a scale for u that the curve z 10“

250 LOGARITHMS

8. The population of the United Kingdom was estimated atin the middle of the year 1891 , and at ten

population increases (i) by the samenumber of individuals every year, (ii) by a constant percentage

mrysyégar, estimate the population in the middle of the years 1894

1

9 . The earth completes a revolution round the sun in 365days,and the planet Venus completes a revolution in 225 days. The

from the earth to the sun is 90 million miles . Find thedistance of Venus from the sun, assuming that the square of the

th

s revolution is proportional to the cube of its0 sun.

10. Obtain the quotients

(y2-3y+ 2) - (x2-3x +2)and y

5-x5

y-x y

-x

The limiting values of these quotients when 31 is put ual to 35

after division are called the differential coefficients of a: 33: 2and respectively. Show that the differential coeficient ofx2-3x-l-2 18 2413-3 and that of $ 5 is 5x4.

11 . A ladder79 metres long isplaced against awall, the foot of theladder being 38 metres from the bottom of thewall. If the he

'

htint on the wall reached by the ladder is x metres, fingz ,

given tfit x2+3°82 7

'92.

12. The area of the surface of the human body is proportionalto the square of the height, and the weight is re rtlonal to thecube of the height . The loss of heat by the y varies as thesurface area and this loss must be made up b food. Does a manof 150 lbs. or his son of 110 lbs. need more ood forin proportion to his weight ? If the father needs 7 ounces a dayfor this purpose, what does the son need ?

13. Given that the area of any triangle the lengths of Whosesides are a , b, and c isVs (8 -a) (8 “the“ ?

reduce this to its simplest form for a triangle in whi ch a2 b3 02,

and calculate the area when b mches, 0 inches.

14, The rvice ipe to a house 75feet from the main isinch inmm: For how man burners, each takiug 5cubic

feet of gas per hour, will th e serve The number of cubic feet

per hour delivemd by

CHAP . XIV, EXERCISES 251

d is the diameter of the pipe in inches, and L is the length of the

pipe in yards.

15. A ny measures inches across and a nny 1 inch .

If the t°

cknesses were in the same ratio, what tionma

decimal to the nearest hundredth) would the weight of the

penny be of the weight of the penny ?If a pennyweighed twice as much as a halfpenny, and was of

the same th1ckness, what would it measure across (to the nearesthundredth of an inch) , the halfpenny measuring 1 inch ?If a disc measures a inches across and b inches in thickness,

then its volume is x a x a x b cubic inches.

16. The time of swing of a pendulum is pro rtional to the

square root of its length. If a pendulum 39 inc es long swingsonce per second, how many swings per minute will be made by a

pendulum 5feet long ?

17. On a map drawn on a scale of an inch to a mile two points

are found to be inches apart . One point is at the sea level, theother at a height of feet. Lay off on squared paper to anysuitable scale the horizontal and vertical distances between thesepoints, and by measurement find the length of wire that wouldrun straight between these points.

Having given that(distance between the points 2 z (horizontal distance)?

(vertical

calculate the length of wire in feet to the nearest hundred.

18. In return for an immediate payment of £575178 . 6d. a Life

Assurance Society undertakes to pay at my death, reckoning that on the average a man of my age will live nineteen yearsmore . Find, to the nearest tenth per cent, what rate of interestthe Society pays on the average for money thus deposited with it,assuming that a sum of £19 invested at r per cent . amounts in 7;

years to £1) (1Investigate the correctness of the formula used.

19. Find the value of where a

b 24193 , c 5769 .

20. Explain the meaning of a“and a1 /5.

Between what integers does lie ? Give reasons for youranswer.

21 . The formula p P gives the pressure p of theatmosphere in pounds per square in c at a height of a feet abovesea level, Pbeing the pressure in pounds per square inch at the sealevel. Fmd the pressure at the top of Snowdon, 3570 feet abovesea level, when P 15.

252 LOGARITHMS

22. Give an account in continuous English of indices, explainingthe convenience of the notation for the integral index, and the

from in to no tive and fractional indices. Imattach to the orm as well as the matter of the

account.

23. A jet of water issues from the side of a tank with a horizontalvelocity of J2gH feet per second, where H feet is the headof water above the orifice, and g a certain constant. The waterretains this horizontal velocity unchanged while falling, and in t

seconds from issuing it falls2 feet. Taking H 10, and g

32, find whereabouts the jet str ike a horizontal plane 100 feetbelow the orifice .

24. The relation between 19 and 0 indicated by the table

is suspected to be given by the equation pr“

k, where a: and k

owing for slight errors of observation, find, asnearly as ou can, the values of a: and k, and give the value of pcorrespon g to the value of e.

25. If a“

b”, and b a°

, show that y =x/c. With the aid ofthe tables of logarithms ap ly this result to find the values of log. 5and log. 211 , where e 2518.

26. It is said that any angle ABC can be trisected by the following method

On tracin -pa r drawa circle, centre 0 . Produce any radius

OD to E so t at E 0D, and through D draw llflVof indefinitelength perpendicular to OE . Place the tracing per figure so

e whole of the circle lies within the angle BG', arrange itso that MN es through B, one arm of the an le passesthrough E an the other touches the circle . Then prio throughD and 0 , and join the points thus obtained to B .

’

Carry out this construction for an angle of using a circlewith a radius of 3 cm. test the result with your protractor, andeither justify or disprove the soundness of the method .

27. The weight of the model of a bell made of the same metal asthe bell on the scale of 5centimetres to the metre is 26 grams.What is the weight of the bell ?28. Drawa circle of

oradius cm.

SDrawa di

ameter AB and pro

duce it to O so that B 3°

cm . uppose a e ongmall‘

y lyi

along OBA to turn about 0 till it ceases to cut the c1rcle . 03P and Q its points of intersection with the circle. Make a table

LOGARITHMS

sent a rock 1n the ofing. Measure the anngipa

:t,, prove that she will certai be out of danger

the rock he is steered so that the angle is always less

Draw the course of the ship, supposing the angle LSM 1s con

stantly kept at State your method.

32. Once in ancient Greek times the problem of making a

cubical altar of double the volume of an existing cubical altarhad great importance . Take the edge of the original altar as

80 cm. and solve it (i) in the Greek fashion the intersectionof the

grepphs y

9 160mand m3 = (ii) e graph y x3

(iii by ogarithms and (iv) by the solution equation x3 2

as ollows -This equation has a root between 1 and 2 ; pu ga: 1 + 31 find from it another having its roots less 1 . By

flitting y z/lOfind an equation in z . This equation a root

tween 2 and 3 ; find from it an equation m u with its roots less

by 2, and so on .

CHAPTER XV

THEOREM OF PYTHAGORAS

1 . NAME any case you have had of a right-angled trianglein which you knew a relation between the three sides.

If C is the centre of a circle,

AB a diameter, and M0 the .Llet fall on AB from a point Mof the circumference, we knowthat CM3 002 OM2

(chap.

XIII, art. Also if C is thecentre of a circle and 0G is a

tangent to the circle at G we8 ' 0

know that 002 063 4-6402

(chap. XIII, art. In eachcase what condition must besatisfied in order to make sucha relation true — It is enough Fm . 1 .

to know that the triangle is

For suppose ABC a triangle in which B is

FIG . 2.

a angle (Fig. W ith C as centre draw a circlepassing through A ,

and produce BC to form a diameter

256 THEOREM OF PYTHAGORAS

DE of this circle ; for such a figure it has been proved thatAC2 ABM-BC”

. Or, again , with C as centre draw a

circle through B (Fig. and produce AC to form a diameter DE of this circle ; we know that for this figureA02 AB2+BC

’. Repeat these two proofs.

Ex. 1 . Redraw the triangle ABC on squared paper, and draw thesquares on AB , BC, and AC. Count the number of little squarescontained in each of these three squares, and compare with theprevious result .

Ex. 2. With l cm . or 2cm . as unit draw a triangle ABC havingLB BA 3 units, B0 4 units . Find the length of ACby measuring, and also by calculation. (This property of thenumbers 3, 4, 5has been known from very early tunes, and is stillused by builders to square the foundations of a house . )

2. Draw on squared paper a triangleABC (Fig. 3) havingB a 4

, c 3. Draw the squares on the sides of

the triangle, and count thenumber of unit squares included in each. In the case

of the square ACDE on b thereare fractions to take account of.Can you avoid the discussionof these fractionsbyconsideringwhere D and E lie —We go

from A toE by going4units inthe direction BA and 3 in the

direction BC ; let us say4 unitseast and 3 units south. From

F161. 3. C to D is also 4 units east and3 south. Prove that

if D and E are found in this square — AsABC and EFA have LB BA FE , so

that the As are congruent and AC AE . The congruencegives also LBCA LFAE , so that LFAE +LBACLBCA +LBAC leaving LCAE also In the

same way CD CA and LACD Thus CD is and

IItoAE and the figure is a parallelogram. Thus all the sidesare equal and all the angles right. What is the area ofthe square BEE C of Fig. 3 —BF is 7 units long and


tenuse, gives such a figure as Fig. 7, and says concisely,Behold I In modern Europe it is the custom to give

formal statement and proof. Supply this formal statement

and proof.

5 Take any right-angled triangle (Fig. 8) of which AB or

c is the hypotenuse. Let it move at right angles to the side

FIG. 8.

FIG . 7.

0 through a distance 0 to the the

motion the side 0 is continually covering up a new part of thepaper while the sides a and b are continually uncoveringparts previously covered. Since the area covered by thetriangle is always the same, the total area covered up byc is equal to the total area uncovered by a and b. Give

expressions for these areas, first ascertaining the .1. distance

between the two positions of a, and the J. distance betweenthe two positions of b. —Congruence of As shows the J.

distanceB0 tobe a, and the 1 distanceAH to be b. So thatside a traces out an area of a°, and side b an area of b2.

Hence once more we have 02

a2 b“.

Find the areas traced out by a and b by considering the .L

distances (on Fig. 8) between BE and CF and betweenAD and CE— If CF cuts AB in O the area traced out bya is BE x BO, and the area traced out by b is AD x AO.

Show how to divide the square on 6 into two rectanglesequal to a

2and b2.

6. Draw a right-angled triangle ABC with the 3 squareson its sides, as in Fig. 9. Draw CLM IIto AD, and join KB

CHAP . XV, ARTS. 5-7 259

and CD. Compare as to area the A KAB and the squareKACH which have the same base and the same height. Inthe same way compare ACAD andrectan lo ADMD. If AKAB isro about A in the directionof the arrow through a right angle,how will it lie ? Thus comparethe areas KACH and ADMD, andin the same way compare the areasBCGF and BEHL .

The discovery of the propertyof a right-angled triangle that thesquare on the hypotenu se ise qua l to the sum of th e

squares on the other twosides is ascribed to the Greek FIG . 9.

sketched in this article is given by the Greek geometerEuclid.

7. We have seen that if a A ABC has C a right angle,we can draw a circle with centre B and radius BA,

produceBC to form a diameter DE ,

and produce AC to form a

chordAF (Fig. and so showthat b2 AC . CF = DC . CE

(c 02- 4

2, so that

02

a2 b2. Suppose now that

we know that cit

afI 62, but 0

do not know whether the angleC is right. Make the same

construction, and see if you

can tell whether LC is right.-In this case we do not

know whether CF (Fig. 10) isequal to CA. But we knowthat CA . CF = DC.CE (cknow that o

s a2 b”, so that

FIG . 10.

cz—a

z. And we

CA .CF b“, and CF

bz/CA b. The chordAF is bisected at C, therefore BCis .Lto ACE, and the angle C is right .So that a triangle in which the square on one side

s 2

THEOREM OF PYTHAGORAS

is equa l to the sum o f the squares on the o th er twois a r ight-angled t riangle .

8 . Draw any triangle ABC right-angled at C, and drawCD 1 to AB (Fig. Suppose BC to be of length a

centimetres, and so on, as

marked in the

Howmanyof these lengthsdo you need to know to

fix the triangle in size and

shape Will one lengthdo, for examfle a -One

9 length is not enough, forFIG . 11 . we can make any number

of different triangles all ofwhich have the side a of the given length ; and so for anyother single length. Will two lengths do -They doin some cases at least, for example a and b. In howmany ways can you choose two quantities from the fivea, b, p , m, n ?

-The number of ways is 5x4+ 2 or 10.

Justify this formula 5x4 2. Try all these 10 ways.

9 . When the two given sides are a and b, orp and n, or

p and m, the method of drawing the triangle is obvious.

When a and m are given how do you proceed — We laydown BC of the given length a cm. And D is on the locusof points at which CB subtends a right angle, that is, on

the circle on CB as diameter ; D is also at distance m cm.

from B. The intersection of the loci fixes its position. CAdrawn . l. to BC cuts BD produced in A .

similar cases given by other pairs of the quantities a, b,p ,

m, n.

—Similar cases are given by the pairs a, p ; b, 71

b, p .

What cases remain to discuss —The remaining cases

are a, n ; b, m ; m, 01. When m and n are given we laydown ADB , and C is given as the intersection of the .Lat D

and the circle on AB as diameter.


want. Take C in a great number of positions. In eachcase produce BC, and draw the curve that all these positionsof BC touch.

Begin again with AD, and find the triangle ABC by the

11 . Can you, by considering similar triangles or therectangle property of a circle, give a relation between a

, n,

and any third length about our right-angled triangle ABC(Fig. 1 1 ) — The triangles BCD and BAC are similar, so

that :1m+ a

The circle about ADC has its centre on

AC, so that BC is a tangent and a2

at its value draw the graph of and so findthe value of m that makes this expression equal to a’

,that

is, 102. Use this value of m along with n to drawthe triangle

,and measure a on the triangle. Also use this

value of m along with a to draw the triangle and

measure a.

Multiply out (m+ — It is m2 3°2m+ If

you denote m+ by x, can you give an equation for a:

corresponding to the equation m (m 102 for m

Since m2+ 3°2m+ we know that x2

105. Now calculate a: andm.—Logarithm tables

give log 105 2 0212, so that log s and x

Hence m+ 1°6 and m

Calculate m for right-angled triangles that have

i a n

11 a 120, n

iii a n

draw the triangles and measure a on each. Is the trianglealways possible whatever values of a and n are given

12. We have seen how the value of m may be calculatedfrom the equation m2+ 3°2m 102. Of course m can onlybe calculated from m2 ma a

2 when n and a are given

numerical values. But possibly you can put the relation insuch a form as to indicate how the calculation is to be carried

out when n and a have numerical values— Put a: for

CHAP . XV, ARTS. 11—13

Then a” m2+ms + so that a}

?

When at has been calculated, m is given by

m a: Or, without mention of a, we can give m by

Z 2

Use the formula just given to calculate m when a

and n

log alog a

2a? 1873

n

log (a2+ 32942

C2

I

m

18 . By the help of the theorem of Pythagoras show

how from two lines a and2units long to draw another

units long. Take a and n


and use 2millimetres as unit . —We draw at right2

PR = 12

’

(Fig. Then em a2

so that

QB is the line required.

Show how,knowing a and n, to draw a line of length

m en

4 go— In Fig. 14,with centre Itand radius

RP,draw a circle cutting RQ in S. QS is the length m .

Does this suggest a means of finding m from given valuesof a and it without any calculation at

all —Draw a circle of diameter 91

units. Draw a tangent PQ of lengtha units (Fig. and draw through Qthe diameter QSRT. ThenQP

2QS.QT or a

2

so that QS is the length m that satisfiesthe equation a

2 m (m n) .

14. To sum up, we have seen that aright-angled triangle is determined byany two of the five lengths a, b, p,m,

n and we have seen how in eachcase the triangle can be actuallymade .

QuadraticEquations.

15. We have seen that m2+ 3°2m 102 if m or

in other words m2+ 3°2m—102 0 if m 0. Nowsubtract 971 times wn. from m2+ 3

°2m—102.—There

remains m— 102. Now subtract times m

from m 102— There remains or, to the degree

of accuracywe areworking to, there remainsnothing. Thus(m+ x (m m2+3

°2m—102.

Calculate thevalues of y m2+ 3°2m 102 for m 0, 4,

8,12, 16, 20, and draw the graph of y from m 0 to

m 20.— The values arem 0

,8,

12, 16,

20,

y—102

,12

,80

,205

, 362.


17. Now calculate the values of y or m2+ 3°2m 102 for

m= — 4,—8, — 12

,—16,

—20.-When m —4, y is (—4)

x ( —4) —102 or 16—13- 102 or —99 ; and so

for the others,the result being

m 4, 8, 16

,

y—99, —64, 103

,

and agreeing closely with the values of 3.

Use these values to continue the graph of 31 so as to

show it from m —20 to m 20. Read off from the

graph values of m for which y 0.— They are m and

m 1 1 °8. The first is the root we had before of the

equation m2 m 102 0. The other,the negative

value ofm, is meaningless for the original problem in whichm was the length of a line . The same equation mightturn up in some other connexion in which a negative rootmight have meaning.

Could you have obtained these two roots otherwise thanfrom the graph — Take the expression m2 32m 102 in

the form (m (ia This will be zero if eitherfactor is zero, that is, if m or if m

Ex. 4. You have frequentlyused the graph of :r2for positive valuesof x . B the aid of the assumption that a) x b) ab draw

the grap of for negative as well as posit1ve values of x.

In the case of sums of money, by attaching a meaning tonegative sums we extended the applicability of a formula,and so attained some economy of writing ; with practice thiseconomy becomes an economy of thought also. Here againcertain assumptions about negative quantities enable us to

treat m“711—102 and (mx ("t as equiva

lent for all values of m. In these and other cases of extension of meaning our ultimate justification is the economy of

thought that results.

18 . Return again to the original calculation of the root ofthe equation m2 m 102 and repeat it, rememberingthat —a) x —b) ab.

-We found that (m+ 102

105, and that 105 But since

x ( l0 °2x we know that 105.

Hence the equation (m 1 °6)it 105 is satisfied by making

Page 262.

CHAP . XV,ARTS. 17-19 267

either m+ or m+ which give thetwo values m and m

Generalize this for the case m2 nm—a2 0.

-We have(m a

°+ and since a

“+n

2/4 is equal to

x/m x Jm fi ,

and also to !M bd VFW ).

we can satisfy the equation by making m n/2 equalxto

either t/ a’or m ,

that is, by making an equal to either

s/M —nfl or J im19 . Solve the equations

102—3°2x—102

x2+ 3°2x+ 102

x2 102

x2+ 3°2x 10

x2+ 3

°2x 10

x2—3°2x l

de° 1

giving in each case as many values of a: as you can ; and in

any case when you can find no value of x, explain where themethod breaks down. In any case where you can find no

root, draw the graph of the expression, and see if the graphalso fails to give a root.What is the distinguishing feature of the equations among

the above seven that have no root — When the last termof the equation is positive and greater than 1 there is no

root. Find generallywhen the equation a?”

ax b 0

has roots— We find that a2/4—b. If a

2 4—bis positive so that we can find its square root, we sha 1 get

roots of the equation. But if b is greater than a2/4—bis negative and has no square root

,and there is no root of

the equation.


Calculation of tight-angled trianglefrom various data .

20 . We have seen that a right-angled triangle is fixed inshape and size by any two of the lengths a, b, p , m, n of

Fig. 11 on p. 260. Incidentally we have seen how m maybe calculated from given values of n and a. Show how mhaving beencalculated, bandp may clechecalculated.

—Theyare given by b2 n (a m) and p

? mu. Calculatethem for a 101 and n and compare with the resultof drawing and measuring.

21 . In Fig. l 1 how many right-angled triangles are

there Write down the relation between the three sides

of each in terms of a, b, p , m,

71. Through how manypoints can a circle be drawn From the four pointsA,

B, C, D,choose three in all possible ways to draw a

circle through, and in each case write down the relationamong a

,b, p , m, n that the rectangle property of a circle

gives.—The six relations are

(m+ a)z

a°+ b

2;

“2 m2

b2 ”2 ;

ma ;

m (m n)n (m n).

The triangle being fixed by any two of these lengths, itmay be possible to calculate the remaining three in terms

of those two. We have seen how to calculate m, p , b in

terms of n and a . See if,by means of the six relations

above,any other pair can bemade to serve instead of n and a.

22. The calculation of m,b, p from n and a was made

from the equations v, vi, iv. The lengths being determinedin this way, the equations 1, 11, 111 must also be true . Can

you prove that any quantities a, b,p,m, n that satisfy iv, v, vi,will also satisfy i

,11,iii —The equations v and vi give

a°+ b

2 iv andv give a2

m2+mn m2+p2; iv and vi give b2 712-H im 71

2

Can the truth of any three of the six equationsbe deducedfrom the truth of the other three —It can in some cases,


Ex . 8 . A right-angled triangle ABC is made by cutting a square

AGED along a diagonal . Calculate the angles of the triangle , and

the six ratios of its sides

24. Can you express the ratio?in terms of an angle

It is called sinB (chap. IX, art. Can you express

any other ratios in terms of B 2”is cosB, and a

tanB (chap. XI, arts. 10,

The triangle is fixed in shape by the one ratio or sinB .

So cosB and tan B are also fixed. Can you express them in

terms of sinB — The relation p2 m2 a

2gives

(gr l , or (sinB)"-(cosB)3 1.

which expresses cos B in terms of sinB . It gives alsom 2

a2 1 1

(15) (I?)r 1

(tan B)2 (BinBl

z’

(sinB)2

2or (tanB)

1 (sin B )?

For shortness of writing it is usual to write sin2B insteadof sin B xsin B, or (sin B)

2, and so with the other ratios ;

so that the relations given are writtencos

°B l—sinzB ;

sinzB2Btan

1 sinzB

Ex . 9 . Express sinB and tan B in terms of cos B, and sin B and

cosB in terms of tan B .

Ex 10 . Look up the tables for sin and use the expressionsjust given to calculate cos 40

°

and tan and compare with thevalues given by the tables . Further check all three ratios bydrawing a triangle with angles of 90

°and measuring its sides,

and calculating the ratios

25. Suppose you are asked tomake a right-angled trianglewith a perimeter of 30 cm.

, that is, with the sum of its threesides 30 cm. in length. How will you do it, and how far is

CHAP . XV, ARTS. 24—26 271

its form at your disposal -Draw two lines GK and CYat

right angles. Make a loop of threadmeasuring 30 cm . round.

Drive in a pin at C, and another at some point A of CK.

Put the loop over these two pins, and stretch the loop to thefurthest point on CY that it will reach. Many triangles arepossible, for A may be taken on OK anywhere less than15cm . from C. State the condition that the perimeteris 30 cm. in terms of a, b,m, n, p .

—It is 30.

Suppose the triangle is to have a perimeter of 30 cm . and

an area of 20 sq. cm.-By means of the loop of thread we

can find the length of CB for any position of A,and calcu

late the area. W e draw a graph giving the area as a functionof the length CA , and read off a value of CA that makes thearea 20 sq. cm .

Another way of finding the triangle of given area and

perimeter is to take any value of CA, to find the correspond

ing value of CB that gives the triangle the required area,and to measure the perimeter of the resulting triangle . We

thus get a graph of the perimeter as a function of CA , and

from the graph we read of the value of CA that gives therequired perimeter.

State in terms of a, b, m, n, p the conditions that the perimeter is to be 30 cm. and the area 20 sq. cm .

—They are

a+ b+m+n 30 and ab/2 20.

26. How would you, without any geometrical construction, calculate a, b, m, n

, p for the triangle that hasa + b+m+ n 30 and ab/2 20 ? It will be well bymeans of the relations that always hold among a, b, m, n, p ,to get rid of some of these quantities. T getting rid ofm and m— We know that (ir.-t n)2 a

2+ and the peri

meter condition a b m n 30 gives

(in-Mi)2 (30—a— b)2,

so that (30—a—b)2

a2+ b

“.

Can you now, without further regard for m,n,p , find from

the relations ab 40 and (30—a—b)2

a” b2 what values

a and b must have —W e could take any value for a, calculate from ab 40 the corresponding value of b, and repeatfor a number of values of a. We could then calculate thevalue of (30 a- b)

‘

-a2- b2 for each pair of values of a and


b, and show this quantity also in a graph as a function of a.

A value of a thatmakes this expression (30 a b)2- a

2 b2

eq

yal to 0 gives a triangle forwhich a+ b+m+ u 30 and

ab 2 20.

You could also take any value of a, and calculate from(30 a b)

2a2 b2 the corresponding value of b. You

could then calculate ab 40 for each pair ofvalues, and drawthe graph of ab—40 as a function of a. A value of a thatmakes ab—40 0 gives the triangle we seek.

Having in either way found a, how do you calculate theremaining quantities b, m,

n,p— The equation ab 40 0

gives b. Then a+ b+m+u 30 gives m+ a . Then19 m2 a

z—pz, n

? b2 -p2give p , m, n.

27. From the equation (30—a—b)2—a

2—b2 0 calculateb when a 2

,4,6, 8, 10,

-When a 2 the e

quation

is (28—b)2—4—b3 0. We multiply out (28—b) and

get 784—56b+ b2, so that the equation is 780—56b 0 ;

and b is that is, When a 4, the equa

tion is (26— b)°— 16—b2 0 ; it simplifies to 660 52b

and gives b 12 7. Similarly for other valuesofa. Can

you, without giving a a numerical value, simplify the equation (30—a- b)

2—a° b2 0, so that the one simplification

will be available for all numerical values of a (30 a b)2

multiplied out is 900 60 a 60 b a‘2 b2 2ab, so that

the equation is 900 60 a 60 b 2ab 0. This is equivalent to 60 b—2ab 900—60 a, and to

Using the relation in this last form we can substitute anyvalue of a and at once easily calculate the correspondingvalue of b.

Can you, from the equations a+ b m n 30 and

ab 40, calculate a and b without the help of graphsWe have seen that these equations can be put in the form

900—60 a 40

60—2aand b

Each of these equations gives b when we know a ; and

when a has the right value the two equations give the same


We conclude that any two independent conditions fix

a right-angled triangle, that each condition may give the

value of one of the quantities a, b, m, n, p, or may givea relation among these quantities.

Exxncrsss.

11 It is required to mark out with tape on a plot of grass

a rectangle of 26 yards by 12yards . What ought the length of itsbe to the nearest foot ?

of this length would help you to layand state the geometrical proposition of

which you have made use .

Drawa triangle ABC, having LA a right angle, LB and

AB 12 centimetres long . Bisect

BC at D and join AD . Measurethe length of AC, and calculate

the lengths of all the other linesand the sizes of all the otherangles in the figure

13. The frame represented inthe rough sketch in Fig . 15

supports a table ; ABCD is a

square 12 inches in the side, and

each prolongation AK, BL, CM,

DN of a side is 3 inches ; thelegs are at K, L, M, N, and the

com ers of the table are at the

same points . What is the shapeFIG 15. of the table, and what are Its

dimensions to the nearest tenthof an inch Justify your statement .

14 . With your protractor make an angle of and find byd i

e

l

lfithe sine and cosine of the angle .

Ch your results by calculating from them sin? 38°

cos2

15. Figs. 16 and 17 are two different arrangements of five

pieces of cardboard (four congruent triangles and a uare ) Showthat Fig . 17has the form of two squares placed side y side .

Hence or otherwise show how two squares may be joinedtogether, and the resultingfigure out up and re

°

oined into a single

square . What kind of triangle do the sides 0 the three squares

form ? Justify your answer.

CHAP . xv,EXERCISES 275

16. Draw two circles in contact and a common tangent (notthe one at the point of contact of the two circles) . Draw threemore figures, varying the radii. On each figure measure in centi

FIG . 16. FIG . 17

metres the radii R and r of the touching circles, and the common

tangent T. Make a table like the followin filling the first threecolumns from your measurements and caIculating the last twocolumns from them.

Area of Area of

square on T rectangle4Rr

What inference can you draw from the last two columns of yourtable ? Attempt to justify it geometrically17. Two ships A and B were steaming at 12 and 18 knots respec

tively, A going due east and B due north . B crossed A’s direction

east of A, and ten minutes afterwards A crossed the track of B,

south of B . If D represent the distance in nautical miles betweenthe shi at m minutes after the first event (m being less than 10)prove t

36+ (1°3m

For what value of m is (1°3m zero ? Is it ever negative ?

For what value of m is 36+ (1'3m least, and what is this

value ? What is then the length of D ?Make a diagram to a scale of 2 inches to a nautical mile, showing

the positions of the ships at the moment B crossed A’

s direction

and at intervals of 1 minute thereafter up to 10 minutes. Measure

the distances and plot them on squared paper, representing


1 minute by 1 inch and 1 nautical mile by 2 inches. From thisfind by inspection the time at which the ships were nearest to one

another, and their distance apart at that time, and compare withyour previous results.

A knot is a speed of one nautical mile per hour.

18 . Make a A having angles of 50°

and From it measurethe sine, cosine, and tangent of

Draw any right-angled triangle ABC right -angledex

press the sine, cosine, and tangent of B in terms of the sides

a, 0 From known relations between a, b, c express sin B andcos B in terms of tan B . In the case of B = 50

°

verify theserelations from the values found above .

19 . Make an angle whose tangent is Calculate the sineand codne, and check your results by measurement of your figure

20. A litre measure is cm. high . What must be the height(to the nearest millimetre ) of a half-litre measure of the sameshape ? Amon measures of the same shape the volume varies

as t 0 cube of t e height .

21 . A projectile whose weight is to pounds and diameter d incheswrought-iron plate when moving at the rate of 0 feet per

second. Assuming that the penetration p (in inches) is given bythe formula

find the penetration when d 135, w 1250, and v 2016.

22. A water main is 2 feet in diameter. How many gallons are

contained in 1 mile of it ? And at how many miles hour mustthe water flow in it to sup y a population of

head in 12 hours ? cubIc foot gallons. The area of

a circle is times the square of the diameter . )

23. Along a line AXB of indefinite length set off AX 8 cm. ,

XB 4cm. , and describe a circle with AB as diameter, namingthe centre 0 . Draw CXD at right angles to AB, cutting the circle

at C and D. Calculate the lengths of OC and OK, and show thatXC s/ 32cm . Use your figure to find to the nearest tenth thesquare root of 32.

If the len?h of AX be denoted bya and of XB by b, state in

its simplest cm the value of XC, j ustifying your statement by

geometrical reasoning.

24. The volume of a sphere varies as the cube of its radius. If

the mass of Saturn is that of the Sun, and its density

CHAPTER XVI

CALCULATION OF A TRIANGLE

1 . WE have seen (chap. V) that a triangle is determinedby any three of the six quantities a, b, c, A ,

B , C ; exceptthe three angles. We have seen that the sum of the threeangles is always so that a knowledge of two angles isas good as a knowledge of three the three amount only totwo independent conditions. So that we may say that anythree independent quantities of the six a

, b, c, A , B. C, fixthe triangle.

We have seen (chap.

2. Lot us suppose weknow a side a and two

angles A and B of the tri

angle ABC (Fig. Draw

the J. p . Suppose the

FIG . 1 . lengths of the various linesof the figures to be given

in centimetres or any other unit by the numbers a,b,

m, n, p , as shown in the figure. Write down all the

relations you know among a, b, p, m,n, A , B, and see if

you can by means of them calculate the parts b, c, C, m,

n, p, from the known parts a, A ,B .

—First, we know

C = 180—A—B . Then cosB -33 and sinB

a right-angled triangle thatany set of data that enableus to draw the triangleenable us also to calculatethe triangle. Is this trueof all triangles

CHAP . XVI, ARTS. 1-3 279

m andp . Knowingp we get b from sinA and 71 from

cosA . Finally, c m+ a gives 6.

Ex. 1 . Carry out these calculations for the case a 1472A = 58 B = 46°3, the length being In centimetres and the angles

in degrees. Check your results by drawing and measuring.

3. In general we do not want on, n, p , but only b, c, C.

Can you rearrange the equations found so as to give b

and 6without mention of m,n, 17

—These equations give

P a sinB

sinAand p a sinB, so that b

sinAFor 6

we have 6 m+ a , a cosB, n b ocsA , so thatc

’

a cosB + bcosA . If you wanted 0 without calculating b, can you express 6 in terms of a, A, B

— Putting

for b we have 6 a m B +a sc osA

am A S 1n B

sinA cosB + cosA sinB

sinA

This expression for c is heavy to use in calculation. Can

you find a bette r Try the relations that would resultfrom dropping a .L from A or B instead of from C.

— The.L let fall from B Is equal to a sin C, and also to 6 sinA, so

that ca sin 0

C being known at once from C 180sm A

A—B . The .L let fall from A gives 0 not so

good because b is not among the original data.

Use these relations, which may be written in the forma b 6

sinA sin B sin C’

to find b and 0 when a A B 463.

Looking out sin 58 °4 and sin 46 3 in the tables we getx

280 CALCULATION OF A TRIANGLE

we may

loglog

—11 0 270

log 0-9303- 1

log b 1 0 967, 1249 .

And so for c.

In this work you have looked up two entries for eachangle, namely for B

sin 46°3

log 0 7230 —l .

But in another place the tables give the logarithm of sinas a single entry. This is the table called, not very

appropriately, ‘logarithmic Sines. The actual entry is

log sin 98 591,

which is greater by 10 than the value already found. Toenter 1 would make the table clumsy ; for compactness the logarithm of every sine, cosine, and tangent isshown in the table with 10 added .

4 . To survey a of land two points A and B

FIG . 2.

metres apart are taken ; and to fix a point C' theCAB and CBA 123

°are measured (Fig. Cal


making OP ’10 cm. , and measure OM

’andMP ’

. Stilltaking as our definitions

sin 123°

cos 123°

OM’

[OPgive theirvalues— P

’M’is cm. and OM’

is cm. , so thatsin 123

°and cos 123

°

How will you find cos 123°from the tables — The tables

give angles from 0°to The angle P ’

OM’is 180

°123

°

The cosine of an acute angle is positive, and cos

57°is simply the length OM

’divided by the length OP

’

,

without regard to the fact that OM is negative. Cos57°is

or 054, by measurement ; or more accurately fromthe tables, cos 57

°05446. Thus cos 123

°

6. In view of these extended definitions, see what corresponds for an obtuse-angled tri angle (Fig. 2) to the relationsa/sinA b/sinB c/sin C and e a cosB bcosA for

an acute-angled triangle — SinB is CF/CB, and is positive,so that a sin B CF b sinA . Also AD is csinB and

is also b sin C, so that the relation a/sinA b/sin Bc/sin C is true for this triangle also. CosB

, however, is

negative, and is - BF/a, so that BF a cosB . Thus6 AF BF bcosA a cosB, the same relation asbefore.

Our extended definitions thus prove convenient. It savesspace in the memory, or in the notebook, to be able to use

the old formulas instead of keeping a second set.

Ex. 2. Draw a few triangles and measure their sides and angles.

a, b, 6 being the lengths of the sides of a triangle in centimetres,and A , B , C the values of the angles in degrees, calculate and

compare2

, {3for each triangle .

Calculate the angles and sides of one of the right-angled trianglesmade by cutting in half an equilateral triangle 10 cm . in the side,

and compare2, B’ 5. Do the same for the triangle made by

cutting a square 10 cm . in the side along a diagonal.

Are the sides of a triangle proportional to the opposite angles ?

7. In the case of triangles fixed by two sides and an

angle, what classes are to be distinguished (chap . V) Given

a, b, A, how will you calculate B, C, c— The equation

CHAP . XVI, ARTS . 6-8 283

a/sinA b/sin B gives B . Then A +B + C 180°

gives C,and c/sin C a/sinA gives 0.

Draw the triangle for the four cases aall four cases having A 20 and b 10. Then calcu

late the triangle for each case — Remembering that everyvalue of sinB that gives an acute angle for the value of B

gives also an obtuse angle, we have the following resultsB C

in degrees

12481641 impossible

impossible possible

In the case when B the sum of A and B is greaterthan 180, so that C is impossible and the only triangle possiblefor a has B and C 144°l . In the case

when a the value of sin B is greater than 1 , and no

angle has a sine greater than 1 so that there is no trianglewith a These results agree with the results of

drawing.

Of all possible values of a, which give two triangles, whichgive one, and which none, when b 10 and A 20

Generalize the result — When a is greater than or equalto 10 there is one triangle when a lies between 10 andthere are two triangles ; when a one triangle ; whena is less than no triangle . The general results are

a b 1 triangleb > a > b sin A 2 triangles

a b sinA 1 trianglea b sin A 0 triangle

8 . Given b, c, A ,how will you calculate a, B , C? Write

down again all the relations between a, b,p, and choosethe useful ones.

—Sin A p/b gives 19, and cosA n b

gives i t. Then m c—n gives m. Then tan B p m

gives B. And m/a cosB, or p/a sinB , gives a.


Solve the triangle A 20, b 14, c 6.

— We have9 14sin 20 n 14 cos 20 = This gives 7:

greater than 6, a case we didC not contemplate. Draw

a triangle (Fig. 4) in which n isgreater than 0, and see how

10 you would solve it. -Denotethe exterior angle at B by Q.

1) and n are given as before .

as is now n—c. Q is given bytan Q = p/m,

and B is 180—Q.

a 19 given by m/a cos QcosB or p/a sin Q

or pla sin B . For the case in question this gives m 7°2,

tan Q Q 34, a 7°2/cos 34

What meaning would you suggest for tan B, when B is

obtuse — In accordancewith the extended definitions givento the sine and cosine (art. let us suppose OP to rotatefrom OX through the angle B and drop PMJ. to OX (Fig. 3)then tanB is MP/OM the directions OX and OY beingpositive

,the opposite directions negative . When the angle

is obtuse MP is positive and OM negative, so that thetangent is negative . How will this definition applyto the solution of the triangle A 20

, b 14, c 6

Tan B —p/m and by drawing an

angle whose tangent is we find the angle to beOr the table gives tan 34, so that B 180—34 146.

We can thus solve the triangle without mention of Q.

Write down the formulas you would use to solve thistriangle, without mention of Q, and compare them withthose you gave for an acute -angled triangle.

— The formulasdiffer only in the sign of m. And if you reckon on

positive when F lies on the same side of B as A ,and nega

tivewhen it lies on the other side —The formulas are thenidentical in form .

Here, again, by our extended definition of tan B we have

the formulas for the solution of a triangle exactly the samefor acute and obtuseangled triangles.

9 . There remains the case in which the three sides a, b, care given, and the angles A , B , C are to be calculated. What


AF is k, n - Ic. We have now az—b2 771

2— 172 and

m = c+k, so that a3

or

FIG . 5.

If we 71 instead of k, the equations for m and n

e m+ a and m—n as before. Thus,m—n (a + b) (a—b)/c 22486x

m+ a c 6340,

whence m and n

and —cosA —n/b cos

Thus A is an obtuse angle that has the same numericalvalue of the cosine as so A 180

11 . Can you tell from the given values of a, b, 6 whether

a triangle is right , acute or obtuse-angled, without calculating the angles —We know already that if a2 0

2 b2,

the angle B is right. A lso we have seen that in all cases

az— b2 (m—n) c and c m+n. Hence, in all cases

,

112—b2 (2m—c) c or 2mc c°+ a

z—b2. Now F may lie

(see Fig. 6) to the left of B,in which case m is positive it

may lie to the right of B ,in which case m is negative ; and

it may coincide with B ,in which case m is zero. When

cit-t a

2 b2 we know from 27110 02+ az—b2 that m is

positive, that F lies to the left of B , and therefore theangle B is acute . When 0

2 a2 b2 we know that m is

negative, F lies to the right of B , and LB is obtuse . Whene

‘2as b° the formula shows that m 0, that F coincides

with B, and LB is right ; which we knew already.

Need you in this way consider each angle separately to

CHAP . XVI, ARTS . 1 1 , 12 287

determine the nature of the triangle -Only the greatestangle can be right or obtuse, and the greatest angle is

FIG. 6.

opposite to the greatest side. So let us call the greatestSide b ; then the triangle is acute right , or obtuse-angled,according as c”+ a

z is or . b2.

Ex. 3. Taking as the lengths of two sides of a triangle 5and10 cm . , and taking in succession as the length of the third side 16,14, 12, 10, 8, 6, 4, 2cm . , draw all the triangles you can , andfind bycalculation what kind of triangle each is. When no triangle can

be drawn, explain why.

Ex. 4. Two rods 5 and 10 cm. long are hinged together, andthe free ends

°

oined by an elastic thread . Show in a ph thelength of the is sad as a function of the angle between ti

l

: rods.

Give another gra h showing the area of the square whose side isequal to the thr as a function of the angle between the rods.

Show on the same diagram the sum of the squares on the two rods.

For what values of the angle between the rods is the single square

greater than the sum of the two, and for what values is it less ?

12. In Fig. 6 BF is called the proj ection of BC on

BA . In general, if J. s Pp and Qq are let fall from the twoends P and Q of a line on another line X (Fig. pg is the

proj ection of PQ on X . Also positive directions beingchosen for PQ and X, for instance, those marked by arrowheads in the figure, the projection pq is considered positiveif it runs in the positive direction of X and negative if inthe other . Can you express the projection in terms of the

angle between PQ and X — If A is the angle between thepositive directions and l the length of PQ, then pg t cosA,


when A is acute and negative when A is obtuse .

FIG . 7.

Express the projection n of b on c in terms of the angle A ,

and in place of the relation found between a, b, c, a give one

between a, b, e, A .

— ~Since n b cosA ,the relation 2m

b°+ cz—a2 becomes 2bo cce A b°+ c°— a

°. Write

down an expression connecting a, b, c, B,and one connecting

a,b, c, C.

13. Draw any triangle and the three J. s from the angleson the opposite sides . Find, by experiment, whether thethree i s are usually concurrent

,that is, meet in one point.

In any triangle ABC draw the .Ls BE and CF (Fig. 8)meeting in 0 . Pick out any set of four points through

FIG . 8.

which a circle will pass ; a number of points through whicha circle will pass are called concyclic.

—The angles BFCand BEC are both right , so that the circle on BC as dIameterpasses through B,

F, E, C. For the same reason the c1rcleon AO as diameter passes through A, F, O, E . Ja n


side and produce it to divide the square on that side intotwo rectangles. What rectangle property do you get fromthe circle through B , F, E ,

C That AF .AB

AF.AC. P ick out a rectangle drawn on your figureequal to each of these. Repeat for the two circles on AB

and AC as M eters. What area of the figure is equal toa” b2 0

” —We have seen that the two rectanglesmarkedP are equal, as are the twomarked Q, and the two marked R.

We have then to subtract P + Q which makes 02 from the

sum of Q+R (which is a”) and P +R (which is The

remainder is 2R . Express R in terms of sides and

angles of the triangle— One rectangle marked R is a xCD,

the product of a and the projection of b on a the other isb x CE , the product of b and the projection of a on b. The pro

jection CD is b cos C and the projection CE is a cos C.

So from either rectangle we have R abcos C.

Youthus arrive again at the equation 0° a2 b“ 2abcosC ;

which is an extension of the theorem of Pythagoras thatc2

uz+ b3 when C In Fig. 9 join 00 andHA , and

compare the triangles CBC and ABH with one another and

with the rectangles marked Q, and so give a proof after themanner of Euclid’

s proof of the theorem of Pythagoras, thatthe square on one side o f an acute-angled tr iangle isless than the sum of the squares on the oth e r two

sides by twice the rectangle contained by one of

these sides and the proj ection on it o f the oth er .

15. Reconsider the preceding article for an obtuse angled

triangle.

Calculation of the area of a triangle.

16. A set of data that fix a triangle, fix the area and

enable us to calculate it. The most useful expressions forthe area are in terms of the three sides and in terms of two

sides and the included angle. Express the area in terms of

b, c, A .-The area is half the product of the base and the

height, pc/2 (Fig. 1 or And p b sinA, so that thearea is 4} besinA .

Find an expression in terms of the three sides— We

must express p in terms of the sides. We have p” aa—m’

CHAP . XVI, ARTS . 15—17 291

while m is given by 20m ez+ u

z—b2. This expressionfor m shows that (a+m) 2c is equal to

—b2 or (e+a)2—b2 or — b) ;

and that (a—m) 2c is equal to2ao—c

2—as+ b2 or b”- (e -a)

2or (b -a) .

Therefore 402132 or - m) is

(e+ a + b) (e+ a —a).And the area or grep is

The symmetry of this expression in a, b, e is a check on

its correctness. We get the same result no matter whichside we choose as base. Use the formula to calculate thearea of a triangle whose sides are 13

,10

,8 cm . Also

measure one of the angles and calculate the area from 1}bexsin A . Also draw a perpendicular and use hoe. See howclosely your results agree.

A quadratic equation.

17. The relation a2 b2 0

2 2becosA between a, b, e,A ,

enables us to calculate any one of these fourquantitieswhenwe know the other three. Use it in the following cases, andcompare it for rapidity with methods already given :

(1) a b e

(2) b c A 1 18 ;

(8) a b A 28.

Consider how, when a, b, A are given, the different casesarise of two solutions, one solution

,and no solution. The

equation for e is 02—2becosA a

2 b2 or

(e- b cosA)2

a2—b2+ b2 cos2A .

Since sin2A 0082A 1

,the right hand side of the equation

may be written a2— b2sinzA and the equation itself

(e—bcosA)2

a2 - bzsin2A .

For this to give any value at all for e, the right sidemust be positive, that is, a must be b sin A .

* For suchvalues of a the equation always gives two values of e. But

Discuss the oase a = b sin A.

U 2


a negative value of e is useless as the side of a triangle .

Both values of e will be positive when

that is, when b’ cos3A a"‘- b2sin3A

or b’ (oes3A + sin3A ) a

2

b’ a”

or b a.

And when a b there is only one positive value of e and

18 . The position of each vertex of a trianglemay be givenby its distances from two .1. lines or axes OK and Of (Fig.

These distances are

called co-ordinates.

In Figure 10 the co

ordinates ofA are

and the co-or

dinate measured Hto0X being given first ;the co ordinates of Bare and the

co-ordinates of C are

and Theco-ordinate is

not measured in the

direction OX, but in

the opposite direction. Similarly, any co-ordinate measuredin the opposite direction to OI’ is negative.

Can you calculate the length AB and the angle that ABmakeswith OK (or with a parallel to OK) — IfAM is drawnI I to OK we have a right-angled triangle in whichAM and BMWhence AB2 40824-55542 and AB And

tan BAM so that LEAM is

Calculate the engths of the other sides and the anglesthey make with OX, and so the angles of the triangle.

Check by drawing.


OX =p ; drawXZ and OY perpendicular to OX, making XZ qand OY unity ; join YZ, and on it describe a semicircle, cuttingOX in A and B . OA and OB are the required roots.

Solve the equation x3 x+2 0 in this way.

8. For printing a circular, a printer’

s estimate is 18 . 9d. for 50cepies, or 28 . 8d. for 100. Presuming each of these estimates toconsist of (i) a charge for setting up the type, independent of the

number of copies printed, (ii) a charge for rinting and paper,

proportional to the number of copies printed didwhat his estimatefor printing 1000 copies would be .

9. The given numbers are known to follow approximately thelaw y a bx. Determine the probable values of a and b.

y_.-

. o-65, 0-91, 1-77, 2-13.

x = l °5, 3-0, 5-0, 7-5, s, 10.

10. In a certain experiment the following values of a: and 3,wereobtained

Determine the probable values of y when a: was 3, 5 and 8, andexhibit them in a table .

Assuming that there is a simple law connecting a: and find thelaw,

and check your results by means of it .

11 . The sides of a rectangle are p and q inches in length .

Froma oint within the rectangle four straight lines are drawn per

mdicu ar tothe sides find the sum of these lines in terms ofp andg.

ve that the sum is the same for all positions of the point withinthe rectangle .

If the point is outside the rectangle, and perpendiculars are

dropped from it on the sides, produced if necessary, show that thesum of the perpendiculars can still be expressed ,

in terms ofp and gby regarding some of the lines as positive and others as negative.

12. Multiply together x+ a, w+ b, a: c and arrange your result

by bracketing together like powers of x .

Describe your result in words and deduce from it the result of

multiplying x a, a: b and a: e together.

CHAP . XVI, EXERCISES 295

25 -36, that ia, 53-2x5x § = 42-2x4x §therefore

that ia, (5therefore 5 1} 4 gtherefore 5= 4.

14. Show that

£02+pz +q (a: g1 9 a If! )2 2 2

and give the values of :c for which x3+pa: +q 0.

What values of :c satisfy the equation whenp

-o°l and q==

15. The tangent of an angle which lies between 90° and 180°out a construction forfinding the angle, andcosine.

16. Draw a field ABCD in which AB 80 metres, AC 94m. ,

AD 78m . , LCAB LOADCould you find the area of a quadrilateral field if you knew the

lengths of the sides ? Why‘

I

17 A tunnel has to be bored through a mountain connecting two

places A and B. To get from A to B over the mountain I have towalk4 kilometres upwards at an angle of 23

°and then 2kilometres

downwards at an angle of 34°with the horizon, the path being all

in one vertical plane. Find the length of the tunnel, the difference

iflevel (

i

f its two ends, and the inclination of the tunnel to theorizonta

18. It is noticed that when the elevation of the Sun is 22°the

shadow of the summit P of a mountain falls at Q on the seashore .

OR of 800 feet is measured horizontally at right anglesto POand the angle PRO is then found to be Calculate theheight of P above sea level.Also find the height of P by drawing.

19. From the top of a hill of height 874metres above the levelof an adjoining lake the angles of depression of the to of anotherhill and of its reflection in the lake were found to be I and

(the hill to in the reflection is vertically below the actual hill topand is as ar below the lake surface as the actual top is above) .

fezd the height of the second hill from a drawing to a suitablee .

Also calculate the height of the second hill, having°

ven that itsratio to the height of the first hill is sin (d

’

d) -sin (d d), whered and d

’

are the two angles of depression

20. Solve the equations y 0°4x :e+0°6y 12, and verify


the solution byplotting, on squared paper, the graphs of the equa

tions.

21. Plot on squared paper the following sets of values of a: and

y, which are in mehes za: 38 6°l

y 8°l .

There being a relation of the form y as b which is approxi

mately satisfied by these values, find what the relation is.

22. When a body is thrown up in the air with a velocity of 96feet per second, its height above the ground t seconds after it was

thrown is 96t 16t2 feet. Find its height after 1, 2, and 3seconds .

When is it again at the same heights as at these times 723. When a photograph is to be taken, the plate on which theicture is made must be placed at a distance a: from the lens

,this

sum cc being connected with the distance 3; of the object from the

lens bythe relation £471

,wheref isa length dependingon the

lens used. For a s 4 inches, find the distance of the late fromthe lens when t e distance of the object is (1) 6 feet, 2) 20 feet,(3) 100 feet, and show that the distance of the plate for 100 feet is

24. At sea level water boils at 212°Fahrenheit . At a he

'

ht Hfeet above sea level the boiling point is lowered B degrees low

where B and H are connected by theH 520B B3. At what temperature will water boil at the topof a mountain feet high25. A road is to be cut in the side of a bill that slopes at A

de es to the horizontal. The breadth of the road is to be cyards,an the slope of the ground behind the road is to be B degrees tothe horizontal. Find an expression for the amount of earth to beremoved in making 1yards of the road .

Also calculate the number of cubic yards of earth removed peryard of road when c 30, A 40, B 75.

26. The weight of rails suitable for a given railway is obtainedfrom the formula

W 17(L 3

where W weight in pounds per yard,t load on one driving

-wheel in tons,o greatest speed in miles per hour .

Find the value of W when L 8 and o 55, these numbersbeing given to the nearest integer.

27. A pint measure is cm. high to the nearest mm. Find

what must be the height of a litre measure of the same shape.


34. Draw any triangle . Taking any side as base, make on it anisosceles triangle equal in area to the original triangle. Measure

the two perimeters, and see which is greater, and establish the resultbygeneral reasoning.

ake now the isosceles triangle, consider one of the equal sidesas base, and make on this basa an isosceles triangle equal in area

to the former isosceles triangle . Is the perimeter increased or

di

ginishe

fila

?

b th e th f h 1ow t t y continuing in way e perimeter o t e triang e

is in general continually reduced, the area remaining unchaIs there any case in which the rimeter cannot be reduced ?Wi

‘

rst is the shape of a triang e of minimum perimeter for givenarea

35. Take a 100 of string and stretch it with three pins into a

triangle. Try oving a in to increase the area of the triangle .

pin, an so on. Establish bygeneral reasoninghow each pin must be moved to increase the area. In what caseis it impossible to increase the area ?

What is the shape of a triangle of maximum area for a given

perimeter

CHAPTER XVII

SOLID GEOMETRY

1 . WE found experimentally (chap. VII, art. 14) that aline standing on a plane might be 1 to the plane, that is, J.

to every line in the plane that passed through the foot of theline. Must this remain simply the result of experiment, oris it capable of proof ?Draw a pretty large square ABCD,

and on each side ofit as base draw an isosceles triangle, all four triangles beingcongruent Cut out the whole figure and fold it

FIG . 1 .

along the sides of the square . Turn the triangles about thecreases till their vertices meet and you have the pyramidOABCD (Fig. Does the height of the isosceles triangles

300 SOLID GEOMETRY

matter — Thevertices ofthe triangleswill not come togetherif the height is too little the height must be greater thanhalf the side of the square.

2. Now draw the diagonals AC and BD of the squarebase of the pyramid, meeting in M. How does the line

with regard to the base ABCD —It appears to

Fm. 2.

be .L to the base. Can you prove it .L to any lineson the base —OMwill be shown .L to AMO if we showAs OMA and OMC congruent. 0A is the line in whichO,A and O,A of Fig. 1 unite, 00 is the line in whichand unite, and we know all the lines O,A , O,A,

are equal ; so that 0A 00. Again, the diagonals of a

square bisect one another,so thatMA MC. Lastly, MO

is a side of each triangle. Therefore the triangles OMA and

OMC are congruent, and the angles OMA and OMC are

equal, and OM is .L to AMC. A similar proof shows OMto be J. to DMB .

Draw across the base, through M,any line PQ. Can you

show this line to be .L to OM‘P Show As CMQ and AMP

congruent and CQ AP , then show As COQ and AOP

con ent and OQ OP ; and then proceed by the comparisonof s OMP and OMQ to show that OM is .L to PMQ.

It has been proved that GM is .L to every line throughM in the plane ABCD ; so that the existence of a per

pendicular to a plane is established.

302 SOLID GEOMETRY

K that containsMO and SMR . It cuts the plane ABCD in

some line TM17. So we have the planeK containing the lineMO, the line SME which is J. to MO, and the line TMU .

Further,M0 being J. to every line in the planeABCD, is .L

to TMH. Since in the plane K there is only one line throughM .L toMO, SMR and TMUmust be the same line . Thatis, any line SME .L to M0 must lie in the plane ABCD.

5. SupposeAMOand OM two rods soldered together atMat right angles. And suppose that the rod AC, while keepingits ends at A and C, turns round and carries OM withWhat kind of surface doesM0 (produced indefinitely) sweepout —It looks like a plane . Can you prove thata line turning about a fixed line and always .L to it tracesout a plane — SupposeMO andMO’

in Fig. 4 two positionsof a line turning about AM and a ways .L to it. Think of

the plane P that passesthrough OM and O

'M.

Since MA is .L to OMand to O

’M it is J. to

every line in P thatc passes through M ; in

other words, AM is J.

to the plane P. Andwe know that any line

FIG ~ 4..L at M to MA lies inthis plane which is J.

to MA . Therefore any third position 0”M of the rotatingline lies in the plane P determined by OM and O

'.M. That

is, OM sweeps out the plane P . More shortly put, twopositions of OM determine the plane P .L to AM and anythird position of OM, being 1. to A111 , also lies in the plane P .

So that OM sweeps out the plane J. to AM.

6. How does the end 0 of the rod MO move as MOrotates round AM0 — O is all the time in the plane P

,

and is all the time at the same distance from M.

We first met the circle as the path of a point at a con

stant distance from the foot of a tree (chap. It appearedlater that for the path to be a circle the point had to remainon the ground (chap. VII

,art. the complete locus of a

point moving at a constant distance from a fixed point being

CHAP . XVII, ARTS . 5-7 303

a sphere . Give now a complete definition of a circle.—It

is the path of a point that moves in a plane and keeps at

a constant distance from a fixed point in the plane .

And what is the path of 0 — By the definition the pathof O is a circle.

7. Fig. 5shows a triangular pyramid, in which the edges

AB, BC, CD have the same length, say 10 cm . also theangle BCD is right, andAB is .L to the plane BCD. Suppose

FIG . 6.

FIG . 5.

the pyramid has a covering of paper, which is slit alongBA,

AC, CD, and spread out . Draw carefully this spreadout covering (Fig. arranging the size of the angle A'

O’D

so that the covering will fold up again and form a pyramid.

Calculate the lengths of the various edges, use these lengthsto draw the figure carefully on stifi

' paper,and make a

pyramid— Since AB is .L to the plane BCD,LABC

andAC’ AB2 BOZ 200 sq. cm . ,so thatAC cm.

The angle BCD being right, we get in the same way BDThat AB is J. to BCD gives also LABD so

that AD2 AB’ + BD2 300, and AD

Do these calculations tell you anything of LACD

geygive usAD” AC2 CD2

, so that the angleACDmustrig t.

304 SOLID GEOMETRY

8 . Let us now take our paper pyramid, set it on BCD as

base, and at the point C put in a pin CK parallel to AB .

What do parallel lines mean — When all the lines dealtwith were in one plane (chap. II, art. 2) we saw that whentwo lines were ll, a J. to one was always .L to the other,and the .L distance between the l] lines was the same all

along. Late r, in dealing with figures not all in one plane

(chap. VII, art. 13) we found that for lines to be H, they mustin addition lie in one plane.

We hadAB J. to the plane BCD, and have drawn CK II toAB. Is CK also J. to BCD — If we can show CK to be J.

to two lines in BCD, we shall have shown it to be .L to the

plane. Since CB falls across the parallels BA and CK,the

angle KCB is 180°LABC or We have still to Show

CK to be J. to CD . CK being Hto AB is in the plane ABC.

NowDCwas made J. to BC, andwe have seen that it is also.L to CA ; it is therefore .L to the plane ABC that containsthese lines, and therefore to CK which lies in this plane .

Thus CK has been shown A to two lines in the plane BCDand so .L to the plane. So we have proved that i f one l ineis perpendicu lar to a p lane all l ines paral le l to itare al so pe rpendicu lar to the p lane .

9 . State and prove a converse to this theorem .—A ll

perpendicu lars to a plane are para l le l to one

anothe r. Take our paper pyramid again, and now supposethe pin CK set .L to the plane BCD (Fig. OK and

CD are therefore at right angles. We know also that CDis J. to the plane ABC. Therefore CK lies in this plane .

W e now have CK and BA in the same plane and both at

right angles to BC ; they are therefore parallel.Show how, given any two .La to a plane, to arrange a

pyramid about them so as to prove them parallel.

10 . The top of your desk is a plane. Set up .L to it a

pencil or penholder by means of a pin in the end of it .

Taking the cover of a book for a second plane, try in whatvariety of ways you can place a plane through the top of the

pencil. Then set up a second equal pencil J. to the desk.

In what variety of ways can you place a plane through thetops of the two pencils -The plane may rotate round the

line joining the two t0ps of the pencils, and any position it

306 SOLID GEOMETRY

found that atG from those at C and A . So that the distanceapart is everywhere the same.

Such planes, whose perpendicular distance apart is everywhere the same, are said to be paral le l p lanes.

12. Just as the points B, D, F determined a plane , so doB,A, C, or nearly any three points. When are three pointsnot enough to determine a plane —A plane is not determined by passing throughA , C, G, for it may turn about theline AGG. Generally three points in a line do not fix a

plane. Name some other sets of points in Fig. 7

that do fix planes.

How would you measure the angle between two planes,say between ABC and ABE -On the analogy with an

angle contained by lines we might suppose the plane ABCto turn about the pencil AB till it lies flat on ABE , and

call the angle turned through the angle between the

planes. How would you measure this angle with a

protractor Your proposal needs to be made more definite.

—The turning brings AC to lie along AE we could takethe angle CAE turned through by the line AC as the

measure of the angle between the two planes. Is

the result the same as if you took the angle turned

through by BD —Yes,for As ACE and BDF are con

gruent, so that LDEF ==LCAE . Would it do to

take the angle between the two positions of BC -Byexperiment that would be a smaller angle than CAE . If

AC AE , this angle made by BC would be CBE , and we

can show it to be smaller than CAE . For CAE'

Is the verticalangle of an isosceles triangle on base CE and with sides

of length CA . The ACBE has the same base,but longer

sides of length CB .

13. Then give a measure of the angle between two planesthat is the same for all ways of M g the measurement.— Suppose any third plane drawn at right angles to the

intersection of the given planes. The angle between thelines in which the third plane cuts them is the anglebetween the given planes. This measure is always the

same ; for let us call AB the intersection of the given planes,and call C and E two points on their intersections by thethird plane. Draw CD and EF parallel and equal to AB .

CHAP . XVII , ARTS. 12-14 307

Then as before we find that BDF is a plane .L to AB, the

intersection of the given planes, and that the angle DBF isequal to LCAE . Thus we get the same measure of theangle between the two planes whether we take the thirdplane through A or through B.

Position of electric lantp .

14 . An electric lamp L hanging from the centre C of a

ceiling by awirew is drawn aside and held so by two stringss and t, that go fromthe lamp to two cornersA and B of the ceiling(Fig. Make amodelby attaching threestrings to three pointsat the same level, 9 . g.

three corners of an Openbox. For differentlengths of the twostrings, what is the

locus of all the pointsthe lamp may occupy-A Sphere whose

radius is co and whosecentre is the point C where the wire is attached to the

What is thelocuswhen the lengths s andware given -The locus isthe points common to the

sphere of centre Cand radius10 and the sphere of centreA and radius 3, and it lookslike a circle.

AC joined andLMdrawn .L

from L to AC (Fig. The

lengthsw and 3 being given,does the position of M varyfor difi

'

erent positions of thelamp -The triangle ACL FIG 9 .

is fixed in size and shape byAC, 3, 10 ; thus LCAL is fixed. The ALAM isfixed in size

x 2

308 SOLID GEOMETRY

and shape by s, LLAM, LLMA ; thus the length AM is

fixed. So that M 15 at the same distance from A for all

positions of L,and ML remains in the same plane .L to

AMC Since the constancy of AALM shows the lengthLM also constant, the locus of L IS a circle.

When the length t is given also, what possibilities are

there for L — L must be at an intersection of the circlejust discussed and the sphere with centre B and radius t.

It may happen that there are two intersections, or one, or

none. When there is no intersection the strings are not of

suitable length to support the lamp.

15. When there are two intersections, are they both suitable positions for the lamp One position is in the room.

The other is above the ceiling so that it would not light theroom further the strings would have to be stifi'

to hold thelamp there, they would have to be rods and not strings.

When you see the image of a lamp in a mirror, how doesthe image lie with respect to the lamp —As far as we can

judge the image lies as far behind the m irror as the lamp isin front, and the line from lamp to image is J. to the

mirror. Let us adopt this as a definition of image

even when the surface 13 not a reflecting one. Are the twointersections LI and L 2 of the sphere and circle the imagesone of the other In the ceiling that is, is the line L 1L2

perpendicular to the ceiling and bisected by the ceiling ?Take first the question ofperpendicularity of the line L1L2

(Fig. Can you name two planes of which L 1L2 isthe intersection —We haveseen that when the distancesw and s are givenML rotatesin a plane J. to AMG ; let uscall this plane P . By con

sidering the locus of L when3 and t are given but not 10 ,we find the path of L to lie

in a plane l to AB, say the

FIG . 10. plane Q. The points L I and

L2 lie In each of these planesP and Q, and therefore lie on the intersection of these planesP and Q so that the line L 1L2 is the intersection.

310 SOLID GEOMETRY

is bisected by the ceiling -We know that when w and s

are given the path of the lamp L is a circle in a plane P J.

FIG. 13.

to AMO, and we have just seen thatthe line L 1L2 is J. to the line OM in

which P cuts the ceiling. So we

have (Fig. 13) the chord L 1L2 of a

circle with centre M, and MO J. to

the chord. Therefore MO bisects thechord, L 1L2 is bisected by the ceiling,and L I and L2 are images of one

another.

18 . Suppose that we wanted to

calculate the position of the electriclamp, say to calculate its distance OL from the ceiling(Fig. the distance ON of 0 from AB, and the distanceAN of N from A . Suppose the relative positions of A B

C given by their distances a b c apart. How would youproceed What quantities would you calculate in succes

sion To answer this enumerate at every stage all the

quantities you are in a position to calculate — From a b cwecan calculate the angles A B C of the triangle ABC. Fromthe length of AC, of the wire LC, and of the string LA we

can calculate all about the triangle LAC(Fig. including the .L LM from L on

CA and the distance AM. In the same

way the lengths AB AL BL give us ANand NL (Fig.

You have now enough to draw the

quadrilateral ANOM you have AN and

AM, the angle NAM,and the angles at

N and M which are right. Enough todraw is presumably enough to calculate .

How will you do it — We might produceMO to meet AN in T (Fig. Theright-angled triangle AMT, in which AMandLA are known

,givesMT,AT, andLT.

Then NT is known because it isAT—AN .

Then the right-angled triangle TN0, inwhich NT and LTare known, gives NOand TO. TM TO

CHAP . XVII, ARTS. 18-21 311

You have now AN and N0 . Only OL remains— Thetriangle LOA (Fig. 10) has LO right

, OA has just beencalculated

,and AL is the given

length a ; so that OL is given

by OL ’ a-LAz OA3

19 . It is seen that we can cal

culate all we want about the

figure. The methods could perhaps be immoved upon. Forinstance

, what do you know of

the four points A N OM —Theylie in a circle on AO as dia

meter (Fig. And if FIG. 15.

KM is a diameter of this circle,What do you know of AKMN ? -KM 0A andLMKN LMAN. So you can calculate MN from

MN2a MA2 NA2 2MA .NA cosMAN,

and then AO or KM from

MN KM sinMKN.

If you happen to want to find LOonly, this is a considerable saving.

Ex. 1 . TakingAB 17, BC= 14, CA 12, LA = 13, LB 11 ,LO 9, all in centimetres, find by drawing the distance of L fromthe plane ABC.

20 . Assuming that a triangular pyramid, whose height ish cm . and whose base has an area of 8 sq. cm. ,

has a volumeof MS c. cm.

,show how you could calculate the volume of

the pyramid from the lengthsof the six edges— This is theproblem of the electric lamp over again. Suppose L the

vertex of the pyramid and ABC the base (Fig. W e

have seen how to calculate LO, the height of the pyramid.

And we know before how to calculate the area of a trianglefrom the lengths of the sides

21 . Some time ago (chap. II) we discussed the conditionsnecessary to fix the position of a chair on the floor. Howmany were needed Give the conditions in two or three

312 SOLID GEOMETRY

diflerent ways. If now we remove the restriction that thechair must stand on the floor, how many conditions are

necessary Howmany conditions fix the position of one

point, say one leg, of the chair in space —We could fix it,

as we fix the M p, by the intersection of three spheres, thatis by the distances from three points. Orwe could fix it bythree cc-ordinates, such as the distances from the floor, thewest wall, and the south wall. Or in other ways ; in everycase three conditions.

22. Howmany conditions will the second leg take —Itwill take a total of three, like the first leg. But one condition,its distance from the first leg, is already known ; so that itneeds only two more. And for the third leg howmany

— The first two legs being placed, the third hasalready two loci fixed for it ; it lies on a sphere of knownradius, of which the first leg is centre, and on another spherewith the second leg as centre . One condition more fixesit. How many more conditions for the fourth leg-The fourth leg, and every other point of the chair, is

already fixed by its distances from the three legs alreadyplaced.

Summarize and generalize your results— The posit ionin space of any sol id body is fixed by six condi tions.

It is fixed as soon as th ree points are fixed, thefirst of wh ich take s thre e cond itions, the second

two, and th e third one. The posi tion of any fou rthpo in t may be specified by its distances from thesethree.

Conditions thatfix aframework.

23. Suppose we are dealing with a body whose size and

shape have to be fixed as well as its position. For instance,return to the electric lamp, and suppose ABC (Fig. 8) atriangular framework of rods from which the lamp hangs,and suppose we are required to determine all the lengths

as well as the position of the whole framework. Howmany conditions will fix the dimensions of theframework, and how many its position —For the dimen

sions six, one for each length for the position of the framework six, as we found for the chair. And if we

814 SOLID GEOMETRY

If DC AB and AD are removed so that AC CB and

BD are left, what freedom is there —AC being held still,

CB may point in any direction ; that is, CB has two degreesof freedom. The position of CB being settled, BD may pointin any direction, that is, BD has two degrees of freedom a

total of four degrees for the system . Otherwise, the rigidtetrahedron with AC fixed has one degree of freedom , and

the removal of the three rods adds three more degrees of

freedom .

26. Take a system of five points joined together by rods.How many conditions fix itssize and shape?— Beginwiththe pointA (Fig. it maybe put anywhere. Next takeB ; it may be put anywhereat the proper distance fromA ,

one condition. C must be atthe proper distances from Aand B, two conditions. D

’

s

position is fixed by the distance from A B C, three con

ditions. E’

s position is also

FIG . 17. fixed by the distances fromA B 0, three conditions.

The total number of conditions is nine.

Check this by counting the conditions to fix the systemin size, shape, and position, and the number to fix it inposition only.

27. If the five points are joined in every possible wayhowmany rods are there If you were given ten rods fromwhich to make the frame

,could you in general fit the frame

together -The frame is fixed by nine rods. If we followthe same order of building up as before , the distance apartof D and E is settled before we try to add the tenth rod.

The tenth rod will fit in only if it happens to have theright length.

Suppose a frame made of five points joined by ten rods.

What freedom is given bythe removal of one rod -None,since the remaining nine rods settle the form . And

by the removal of a second rod -This gives one degree of

CHAP . XVII, ARTS . 26—28 315

freedom, and the removal of every further rod adds a degree

of freedom up to a certain point. Beyond that point it isnot constraints but moving parts that we are removing.

Ex. 2. If n points are joined, two and two, in all possible waysby rods, how many rods are there , and how many are wanted tofix the form of the frame ? Howmany conditions does it take tofix the

position of the frame, and how many to fix form and

position

Ex. 3. Remove the rods in any order from a tetrahedral frame(four points joined by six rods) and discusa at what point you beginto remove moving parts and not constraints.

Ex. 4. A doll is jointed with ball and socket joints at the

muldfrs, elbows, hips, and knees. Howmanydegrees of freedom

it

28 . Consider the triangular pyramid or tetrahedronOABC (Fig. If it has a thin cover which is slitalong OA OB and 00, and spread out, what will it look

FIG. 18.

like If such a piece of paper 19 shows is cut out

and folded along the dotted lines AB, BC, CA, in whatcircumstances can G H K be brought together ? From

such a piece of paper make a model of a tetrahedron and

use it for the following discussion.

Let a plane parallel to the base ABC cut the pyramidin the triangle DEF. Compare the pyramids OABC and

ODEF as far as you cam— Consider the lines AB and DE .

316 SOLID GEOMETRY

They never meet, however far produced, for they are in

parallel planes ABC and DEE At the same time they are

in the same plane OAB ; so they are parallel lines. ThereOD

fore As OAB and ODE are s1m11ar and52 OB AB

Similarlyeach of these ratIosmequal to 370:

BC CA.

Whence also DEF and ABC are similar triangles. Thusthe two pyramids are bounded each by four triangles, thetriangles being similar in pairs.

29 . Is the angle between two faces of one pyramid equal tothe corresponding angle of the other pyramid For shortness denote the faces of OABC by 0, a, b, 3, each small letterdenoting the face opposite the corresponding capital. The

faces a, b, c are faces of the small pyramid also ; denote itsbase by 11.— The angles between a and b, between b and e,

between c and a are the same for the two pyramids. We

have to see if Loc Lps‘ Suppose OP drawn J. from O

on the base 0 . In the plane 0 draw PR J. to AB, and

consider the plane through 0 , P , R, which we will call m.

The J. at R to o is IIto the J. P0, and therefore in the same

plane on The plane m is therefore J. to the intersectionAR of the faces c and o, and the angle PRO is the angle

between 3 and 0 . Further, let m cut p in SQ ; then DSE'

,

being Hto AB, is also J. to m, so that OSQ is the angle Q ) .And RP and SQ, being in the same plane m, and in l] planeso and p , are parallel lines. Therefore the angles OSQ and

ORP are equal ; and the angle between two faces of one

pyram id is equal to the corresponding angle of the other.

30 . Suppose the area of the base ABC to be S sq. cm .,the

height OP to be H cm . and the distance OQ of the sectionfrom the vertex to be 11 cm . What is the area of the

section —Suppose the section has an area of 3 sq. cm . Then3 DE 3 DO 2 h 3 h2

a (I f?) (1 3) (17)“

ii i

Now let OP be divided into 71. equal parts, and throughevery point of division let a horizontal plane be drawn, that

L cc means angle between planes o and e.

318 SOLID GEOMETRY

the first term from each and add them . What do you get— The sum of these terms is con

Denote

this number by 82 so that the sum of the

first terms is 382 . Take the second term from the rightside of each equation, and find their sum.

—This sum is

Call this 381 . What is thesum of the third terms -Each is 1 , and there are n of

them, so the sum is n.

32. You now have the equation

and if you had a compact expression for S, this would giveyou one for S2. If you repeat the work just done, butusing (r + instead of (r + r

“,

at what resultdo you arrive —We have

- 12 2x 1 1,

(2+ - 23 2x2 1,

—32 2x3 1,

(73+ 2x” 1 .

The sum of the left sides is (n + 12. The sum of the

first terms of the right side is 281, where S, means

1 2 3 nas before. The sum of the second terms is n. So that wehave the result

(n+ —1 or 8,

Put this value for S, in(93+ — 1 382 + 3Sl + n,

and find an expression for S2 . The equation becomes

na+ 3n3 3n 382+3742+ 1)

3 (n + 1)2

or 38 - 1}

2n3+ 3n°+ nso that

6

CHAP . XVII, ARTS . 319

Ex. 5. Solve the equation 2712+ 3n 1 0. Use your results to

express 2112 3n 1 as the product of two factors, and to express S,

as the product of three factors.

33. We had the expression (1 22 32 n”) for the

volume of the pyramid together with certain small addi

tions. And now we have an expression for 1 22 ”2.

By means of this give a simpler expression for the volume,

and by taking a 10, 100, see how the volume

depends on the number of slabs the pyramid is cut into.

22”

232W”

HS. When n 10 the

multiplier of HS is or 0385; when n 100 it is

i030]

or 033835 When n the multiplier is

2x 1012+3x 10°+ 1 1 1 1

3x 1012 3+2x 106

+ex 1012

‘

the second term affects only the sixth or seventh decimalplace, so that the value 4} is very approximate.

By putting the expression for the volume in the form

1 1 1

(3 2” 6a”)it is made obvious that the greater the number of slices wecut the pyramid into the more nearly i HS gives the

volume including the additions. How does the volume of

the additions, considered by themselves, vary with the

number of slices Imagine all these additions loweredvertically till they rest on the base of the pyramid.

When resting on the base of the pyramid all the additionsfit inside one another and occupy part of the volume of a

— The expremion is

slab of the thickness of a slice of the pyramid, that is -H

and whose area is the base of the pyramid. So their total

volume is less than iHS, and it diminishes as it increases.

So that by increasing the number of slices we can show

320 SOLID GEOMETRY

that to any degree of accuracy we choose QHS is the

volume of the pyramid

Ex. 7. Assuming that pyramids on

the same base and of the same heighthave equal volumes, show how to out upthe prism of Fig. 20 (bounded by two llplanes, and three other planes .L to them)mto three pyramids equal in volume .

Work with paper models. (See Fig . 21 ,FIG 20°

showing the dissected prism, and note

that a pyramid CABE would have thesame 138 89 and 119 1t 8 8 FgAn Es. )

34. Can two tetrahedra ABCD and that havethe six edges equal in pairs be brought to have each pairof edges coinciding Actual material tetrahedra would, of

FIG . 21 .

course, get in one another’

sway ; assume that they penetrateone another as freely as ghosts— Place A ’

on A and A’B

’

along AB ; the lengths being equal B’ lies on B . The pos

Ex. 6. Show that if two pyramids of

equal base-area and equal height stand

on the same plane, they are cut by any

plane I] to the bases in sections of equalarea.

Deduce that pyramids of equal heightand base-area have the same volume

322 SOLID GEOMETRY

portionally reduced from the corresponding edge of theoriginal tetrahedron, and having every angle equal to the

ing angle. If two tetrahedra ABCD and E FGH22) have all their edges in proportion, that is, ifB a s FG/BC

'can you place them so that E lies

FIG . 22.

on A, while EF, EG, and EH lie on AB, AG and AD ? Is

the base FGH then Hto BCD‘

P Denote the face opposite toA by a, the face Opposite to G byg, and so ou.

-Place E on

A , and EF along AB as AF'

. Turn the face It or EFG aboutAB to lie flat on the face (2or ABC, so that EG lies to thesame side of AB as A0 does. We know that EF/ABFG/EG

' GE /CA so that the As are similar, andLFEG LBAO, and therefore EG lies along A0, say as

AG’

.

This fixes the position of the tetrahedronEFG-H andH liesat one of the intersections of certain three spheres. Doeseither of these inte rsections lie onAD -Through F ’ drawF

’K I I BD, meeting AD in K. Since AB/EF BD/FHDA/HE , As 0 and y are similar ; .and since E

’

X I I BD ,

As 0 and AF’K are similar. Hence As 9 and AF

’K are

similar ; and having EF AF’

, they are equal. ThusKA = EH and KF

’HF, so that K lies on two of the

spheres that fix the position of H . Again, since As b and

f have their sides proportional, they are similar, and LOADSo that As f and AG

’E have AG

’E G ,

AK EH,LG

’AK LGEH thus the As are congruent

and G’K GH. So that K lies on the third sphere, with

centre G’and radius GH.

We shall discuss later what happenswhen H does not fall

CHAP . XVII, ARTS . 37, 38 323

WhenH does fall atK, are the bases 6and a parallel?—Draw a plane through F’

Hto the base a . It cuts face atin a lino to BC', that is in F

'

G’

; and it cuts face c ina line I I to BD, that is in E

’H. So that the base 6 takes up

a position I I to base a.

37. Two points A and a being images of one another in a

plane, and two other points B and b images of one anotherin the same plane, comparethe distances AB and ab.—Let Aa meet the planein C, Bb in D (Fig.

By the meaning of imC'A = C’

a and DB = Db ;and Au and Bb being .L tothe plane, they are .L to CDand are in the same plane.

Suppose the plane figureAO'BB rotated about CD tillit lies flat on aC'Db. Since

both LACD and LuC'D are

right, and CA Ca, C'Amust lie along Ca and coincides with it. So does DB withDb. Thus AB coincides with ab and is equal to it.Is the straight line ab the image of the straight line AB,

or is the image of AB perhaps curved — Take a point P on

AB . To find its image we suppose a .L PE drawn to the

plane, and produce it its own length to p . This .L, being II

to AC, is in the same plane with AC, namely, in the planecontaining AC and APB. This plane contains OD and ab, so

that PE cuts these lines, say in E andp’. Also the rotation

of ACDB about CD brings P to the point p’

, and so Ep’EP .

Thereforep’isp , the image of P . And the straight lines ab

and AB are images of one another.

88 . Take any triangle ABQ and its image abq (Fig. 23

Are the angles AQB and aqb equal —The As havesides equal in pairs and so are congruent, and LAQBLaqb. Is the angle contained by two lines equal tothe angle contained by their images —Join two points on

The figure shows the vertices of the As, but not all the sides.at 2

the legs of the

SOLID GEOMETRY

reduce the problan to that

just d'

mcussed, a A and its

If we take the image of

do

plane— Suppose we fix

plane by three points in it,Then we

know that the images of

joining them are threepointsa b c and the lines joiningthem. Take D, any otherpoint in the plane ABC

D E F am thme points d c f in a straight finq and since

abc, so does d. Thus everyits image in the plane abc.

89 . Is the angle between two planes equal to the angle

FIG . 25.

is right. In the

of the planes P and Q.

The As ABC and aimare congruent, so thatLcab is LCAB andis right, so that Load is

—Let AB (Fig. 25) bethe intersection of the

planes P and Q, AC’and

AD lines in the twolanes 1 to AB , so that

CAD is the angle between the planes. Takethe images a b c d ofthese four points. Thenthe planes abc and abd

326 SOLID GEOMETRY

and specified by the sides of the AABO, the distances o f

D from the points A B C, and the distances of E

from the points A B C. Suppose a second figure abode

specified by corresponding lengths, every pair of corresponding lengths being equal. Are the two figures either con

e,

nt or images one of the other — We apply Aabo toE EC. Then d may lie at D or at its Image D

’, and 6 may

lie at E or at its Image If . If d coincideswith D and cwithE

,the figures are congruent. If d and c coincide with D ’

and E'

, one figure is the image of the other. But if d

and c with E ’

,one figure is neither con

gruent to, nor the image of, the other.

42. Give the corresponding properties for two such5-point figures that have the nine lengths in questionproportional instead of equal. —One figure may be similarto the other, or similar to its image, or it may be thatneither relation holds.Can you suggest amethod of specification that will exclude

the third possibility —If we choose the AABO so thatD and E lie on the same side of it, and 1m the conditionthat d and e are to lie on the same side of a ,bc abode mustbe similar either to ABCDE or to its

.

Can you suggest a specification that willmake the figuressimilar -Take an ordinary right-handed corkscrew or

screw-nail. Suppose our figure solid, so that it will holda corkscrew or screw nail

, and bounded by the plane ABC.

Screw the instrument into the plane ABCat right angles

,naming the points A B C

in such an order that the direction of

rotation is from A to B, from B to C, and

from C to A (Fig. The screw is at the

same time going through the plane towardsD and E . Make it a condition that a screw

FIG . 27 screwed through the plane abc in this wayshould also go through the plane towards

d and 6. Then when we come to fit the figures together,there is no ohmes of position for d and e, and the figuresmust be similar.

CHAP . XVII, ARTS. 42—44 327

48. Consider in particular two figures of frequent occur

rence, the circle and the sphere, and let us try to find

expressions for certain lengths, areas, and volumes.

Suppose two circles of different radii drawn, and radii0A OB ou ob drawn in each at intervals of a degree(sketch in Fig. Join AB ab The figureABC a plane figure bounded by a number of straightlines, is called a polygon, and the distance it measuresround, is called its perime ter . Compare

the As OAB and cab, and compare the perimeters of the

two polygons ABC and abc —The angles A0B and

aob are equal as each is 1 degre e, and AO/BO ao/bo as

each ratio is unity ; so that the triangles AOB and aob are

similar. And calling the two radii B and r,AB/R ab/r.

Also there are 360 triangles, OAB all congruent,so that the perimeter P of the polygon is 360 xAB ; the

perimeter p of the second polygon is 360 xab. Hence

P/B p/r

44 . The perimeter of the polygon ABC 1s for mostpurposes the same as the perimeter or circumference of the

circle. But if We want a polygon that will fit the circlecloser we may suppose the angle AOB a hundredth or a

millionth of a degree, or even less. What relation have wethen — Just as before P/B p/r, and the smallerwe make

SOLID GEOMETRY

the anglesAOB and aob the more exactlydo P and p become

the ministers of the circles.

The ratio of the circumference of a circle to its radius,which has been shown to be the same for all circles, may befound in various ways. You found it some time

(chap. IV) experimentally. Calculation by methods youdo not yet know gives it as accurately as we like ; it isabout The ratio of the circumference to the diameter,which is half this ratio, has a special symbol Its valueto greater accuracy than we are likely to want is 31 4159.

Ex. 9. Assuming thatwhen a: is the number of degrees ina small3

angle sin s:1156

6show that in article 43 AB is

2B sin and find by how much the ratio P/B differs from 211when the angle AOB is

45. Find expressions for the areas of the A OAB and the

polygon ABC and deduce one for the area of a circle.

Draw OM .L to AB, and the area of AAOB is fiOMxAB.

The polygon contains 360 congruent triangles, so that itsarea is i OMxAB x360 or QOMXP . The smaller the

angle AOB the more nearly does OM become equal to R ,

the more nearly does P become the perimeter of the circle,and the more nearly does the polygon coincide with thecircle ; so that the area of the circle is QR x or

Ex. 10. Using the expression Q0A 0B sinAOB for the area of

A AOB, find by how much the area divided by R3 differs from a

when the angle AOB is (See Exercise

Ex. 11 . Draw radii of a circle at intervals of there beingthus 20 radii or 10 diameters. Cut out the circle and cut it in twoalong one of the diameters. Cut each semicircle along the nine

radii to the circumference, but not through it . Spread out each halfcircumference as straight as you can, and in this form fit the two

pieces together so as to give an area roughly rectangular. Whatare its length and breadth ?Ex. 12. Draw a large circle or a quarter of a circle on squared

paper. By counting squares calculate

of circle square of radius.

48 . What is the nature of the curve in which a sphere iscut by a plane through its centre —All the points of the

330 SOLID GEOMETRY

MNP and the sphere in QRS. The planes X and Y cut

a strip from the cylinder and a strip from the Sphere. Thesestrips we will now compare. Call the radius of the sphere r,and the distance between the planes X and Y 11, and giveexpressions for

,the length and breadth of the strip cut from

the cylinder.— Suppose a plane through OD to cut the

Sphere in the line AQH (line not drawn in Fig. 30) and thecylinder in the line AME . The breadth EM of the strip is

between length is

FIG . 30.

the ci1

2

-cumference of the circle EFG or of the circle ABC

it is 11 r.

Now join QH,and from O draw OT .L to QH and there

fore bisecting it. Denote the angle TOD by w. What isthe relation between EM and HQ -LEHQ 10 , so thatEM HQsin 10 or HQ h/sinw. Suppose a planethrough Tand .L to OD to cut OD in U. Give an expressionfor the length TU.

-LTUO being a right angle, TU TO

x sin w s sin w“. Now suppose AOUT to turn

about OD, carrying QH with it. What sort of surface doesQH trace out -H moves along the circle EHL and Qalong the circle QRS, so thatHQ traces out a strip not unlike

Denoting TObys ; s r approximatelywhen the strip is narrow.

CHAP . XVII, ART. 43 331

the strip cut by the planes X and Y from the sphere. Thestrip cut by these planes from the sphere would be traced bythe rotation of the arc QH of the circle AQH,

and not the

48 . Consider the strip traced by the chord QH , and takingits area as given approximately by its breadth QH,

multipliedby the length of the path traced by T, give an expression forthe area and compare with the strip cut from the cylinder.

—TU being an s sin 10 , the length of the path of T is211 ssin 10 . We have seen that QH h/sinw. So that the

area is 21rs sin 30 xsilica or 211 310, or approximate ly 21m».

We have seen that the strip of cylinder has a length 2Mand a breadth 71

, so that its area is 21rrh.Inwhat circumstances do the areas of the strips traced by

the arc QH and the chord QH differ greatly, andwhen little ?—When Q and H are far apart, the arc and the chord

separate widely about the middle, and a comparison of the

areas traced by them is not easy. But when Q and H are

close together the curvature of the arc is slight, and the stripspretty nearly coincide. For instance

,if the sphere

you were considering was the Earth, you could take thestrips a mile wide, that is, take Q and H a mile apart,without finding a sensible difference between the arc and

the chord and between the corresponding strips.

Let us consider more closely the area of the strip tracedby the chord QH . Suppose QH pro

duced to meet the axis OD in V.

How do you know QH produced Q

will meet OD Consider the sur

face traced out by Q7 by rotationround the axis OD ; this surface

is called a cone . Suppose a sheetof paper of the form of this cone slitalong VHQ and Spread out. Whatform h as it then (Fig. Sucha figure bounded by an arc of a circleand two radii is called a se ctor of FIG . 31 .

the circle . Give an Sxpression forthe area of the sector in terms of the radius VQ a: and

332 SOLID GEOMETRY

the length l of the arc.—We divide up the area by a great

number of radii into pieces approximately triangular, andso find the area to be i s t.Consider the cone traced out by VH, and the sector into

which it spreads out. Express the area of the sector in termsof the radius VH y and the length m of the arc. Hence

give the area of the strip into which the surface traced byHQ spreads out— The area of the strip is the differencebetween the sectors, namely .3cl ym. Show thatthe two sectors are Similar, and express a: and y and the area

of the strip in terms of l m and HQ t. —The SimilarityQ

7 5, and we have x—y t. These two equations

give x and y in terms of l m t ; namely, 3;M t

l—m° We have then for the area of the strip “

é

,”

Prove now that the expression 21rsh used for the area

traced by the chord QH was not merely an approximuion,but was accurate.

e radius of the circle QRS of Fig. 30 to beV0 to be 78 cm. Calculate the angle of

e cone s reads out.Draw the sector, cut it out an make the cone from it.

49 . It appears then that two planes .L to OD and prettyclose together cut from the Sphere and the enclosing cylinderstrips of approximately equal area. Canyou deduce amethodof finding the area of the whole spherical surface —We con

ceive drawn a series of planes .L to OD and cutting the

whole Sphore into pretty narrow strips. Each strip is equalin area to the corresponding strip of the cylinder. So thewhole area of the sphere is equal to the part of the cylinderenclosed between the two planes that bound the Sphere.

These planes enclose a length 2r of the cylinder. Slit thecylinder along a line [I to the axis OD, and the areawe wantspreads into a rectangle, measuring by 2r. Its area is

therefore 4m“.

We have shown that, approximately, a Sphere of radius

r cm. has a surface of area 4m“sq. cm . the form of the

334 SOLID GEOMETRY

i 1 38 and rzS. Now suppose the divisions of the spherical

surface to become smaller and smaller. Our polyhedron fits

the sphere more and more exactly as the divisions become

smaller. The plans As, whose total area is S, fit the surface

more and more accurately, so that 8 becomes more and

more nearly 411 13 . And the distance of each plane triangle from the centre of the sphere (including f l and r?)becomes more and more nearly equal to r. Therefore theexpression 5rx411 1

"I for the volume of the sphere is

52. We have already (chap. VII) had co-ordinates as a

method of giving the positions of points. Consider againa lamp L hung from threepoints A B C of the ceiling.

Let us take as cc-ordinateplanes the floor, the north walland the west wall, and as co

ordinate axesOx running south,03] running east, and 05 upwards (Fig. The co-ordi

nates of a point will always begiven in the same order so thatto get from the origin 0 to thepoint 20) a point

FIG 32 must travel units in the direction 0x, 12 in the direction

and 20 in the direction 03.

The co-ordinates of the lamp being and

the co-ordinates of the three points of suspension (0, 0,(0, and the unit a metre, calculatethe distances apart of the points of suspension, and the

distance of the lw p from each.

58 . Another method of Specification is by projection on

two planes at right angles. The p roj ect ion of a point ona plane is the foot of the .Llet fall from the point on theplane. Suppose the lamp L, the points of suspensionA B C, and every point of the suspending strings projected

CHAP . XVII, ARTS. 52-54 335

on the north wall and also on to the floor . Now supposethe wall turned about its base XY into the horizontal plane.

TakingFig. 33 to represent floor andwall and the projections

FIG. 33.

on a scale of find apart of the pointsA B C, and the length of each suspending string.

A projection on a horizontal plane is called a p lan, a

projection on a vertical plane an e levat ion.

54. Volumes of similar solids. We have seen that a tetrahedron of height H and base-area S has a volume V equal to

5HS ; the units of length , area, and volume being the linear,square, and cubic centimetre, or any other corresponding setof units. Suppose a second tetrahedron to be similar to thefirst, the ratio of similarity being 11, so that every dimension

of the second tetrahedron is 70 times the correspondingdimension of the first. What are its height, base-area, andvolume -The height is kH,

the base-area 7928 (chap. IX,

zsrt. and in consequence the volume 3

1, xkH x7028 or

V

When two tetrahedra are similar and the height of one istwice, or three times, the height of the other, Show how thegreater maybe cut up into smaller tetrahedra each congruentwith the smaller given one.

336 SOLID GEOMETRY

Show how to divide two Similar pyramids on polygonalbases into similar tetrahedra, and compare their volumes.

Show how to divide two similar solids, bounded by planefaces, into similar pyramids, and compare their volumes.

Show how to extend the result to similar solids boundedby curved surfaces. You thus arrive in another way at the

result of article 35of chapter IX,that the volumes of similar

solids are as the cubes of their linear dimensions.

EXERCISES .

14. In a pyramid CABOB the base ABCD is a square a inchesThe centre of the base is Mand the line OM is per

the base and is bBy considering in

AMO, ANO, findthe face ABO, in

terms of a and b.

15. If a parallel be drawn to one side of a

cut the other two sides, show that it will divide

The volume of a cone of he'

ht h and radius of base a being11 0

211,find the e ression for t e volume of a truncated cone of

eight k, radius of a, and radius of top b.A chimney of solid brickwork is 120 feet high ; the radii of its

base are 12feet external, 9 feet internal, and of its top 6and 5. Findthe volume in cubic feet.

16. A 45°

set square resting on a table is turned about itshypotenuse as a hinge throug

h an angle of 56°

set

square, full size, in lan an elevation in the rngas d

)

in length and parallel to the plane of

17. Determine the height of a mountain C, the following databeinggiven

— a base line AB 3000 ft. is measured in the horizontal

ge and the angles CAB and CBA are observed to be 55

°

20'

and

8°18

’

respectively, and the angle of elevation of C from B

is 10°

18. The two faces of a burning glass are parts of spheres of radiia and b, and the diameter of the glass is 0 . Find an expression for

the thickness of the glass at the centre .

19. A tent is supported by two upright posts with a bar fastenedacross them at the top. Discuss how many ropes should be used

338 SOLID GEOMETRY

28. A straight bar 2feet long is suspended horizontall by two2 feet long, attached to its ends . T 0 bar rs

tight, and the bark

Through what angle is

29 . The following statements contain obvious errors ; Show howto detect them with the least possible work. You are not to

rectify them but to indicate why you consider them wrong. Thelast figure in each result is to be regarded as

(0 ) (0°

a : 1 °153937.

(b) The side of a regular pentagon‘ inscribed within a. circle of

diameter 4'

7in. is 2°34 in. lolo .ng

(c) The area of a triangle with sides 10cm . , 14 cm. , 15cm. rsis98 sq. cm.

273x 7x23 _(d)

8 x22x277

(e) If one root of the equation x2 101 1+ l7= 0 is the othermust be 7

30. A garden l ,000 square yards In area is watered with s hoesthat delivers 5 one a minute. How long will it take to deliverwater that wo d cover the garden to a de th of a quarter of aninch if it did not run away ? A gallon rs cubic inches.

31. The amount of water which can be carried ofi'

by a circular

drain pipe, whose diameter is D feet, is0

4

+if)

; cubic feet perminute, where A m 4543 and C 1 Find, with as littleworkas possible, whether a drain pipe 21 inches In diameter will sumacto off 2,000 cubic feet per minute.

Fin inmiles an hour the average speed of the water in the p ipe ,the area of the cross-section of the pipe being 0

°785xD2 squar e

A pentagon is a rectilinear plane figure with 5 sides, a

po lygon is a rectilinear planec

l

ifiure with any number of sides .

A pen n, or any polygon, 1s ed regular when all its sid e sare equa and all its angles are equal.

CHAP . XVII . EXERCISES 339

32. Figure 34 shows a road that Is to bemade . The breadth ofeach part rsmarked on thefigure, and the length CE rs also given; all

these The bearings of various lines given, by the

FIG. 34.

the lines make with the north measured round to the east.these data draw the figure to scale and see if there are data

enoughd

and not more than enough. Then calculate the lengthsAB an CD.

ADDITIONAL EXERCISES

Length, Area and Volume.

1 . Compute to two significant figures, the diameter and cross

sectional area of a s an of copper wire, one mile of whichweighs 4 lb . One on is inch of copper weighs lb .

2. A reservoir, with vertical walls, has an area of square

yards . It is supplied by a pipe, 3 square feet in section, throughwhich water flows at the rate of 1 mile an hour. Find how muchthe water will rise in the reservoir in one hour .

3. Calculate the weight of a cast-iron pipe from thediameter 15 inchea thickness

length 8 ft . Density of metal 465lb . per cu . ft . It is assumedthat there are no flanges or other enlargements at the ends.

It is required to coat a mile of such pipe inside and outside withs

ame protective material at the rate of 20 sq. ft . for 7d . Calculate

t e cost.

4. It is desired to dig a trench 4 feet wide with vertical sides

and horizontal base across a piece of ground, the contour of whichis given bythe followingmeasurements. The bottom of the trench

is the datum line . Calculate the volume of earth excavated in cubic

eet .

Horizontal measurements

along datum line in feet .

datum line in feet.

5. A river basin of a square miles receives a rainfall of b inchesper 24 hours, half of which drains into the river ; whose section,

where it leaves the district, is 0 square yards. Find a formula

for the velocity of the river at this point, in feet per second,supposing the whole process to be steady, and the speed of the

water in the river to be the same at all depths.

Give anumericalvalue to youranswerwhen a 180, b c

ADDITIONAL EXERCISES

For

draw s m ph depths

(scalesShow how m ph n ables you to tell the dq rth of the watar at

any time when the bath is being filled, if the rate of infiow of wateris p

‘

ven ; and, the rate of infiow as fiflfill l litn s an hour ,find the depth after

l l . A man ph ying5holes of a golf coume walks first 260 yda duethen l40 yda m

°south of east, thm M yda due south ,

“ yards “ wewd m rmwenm m 30°

1mst of south,

plan of the conrse to a

scale d lmyda to the inch, and find fmm 1t how far the fifth hole

equal parts. Call the points of divisionC, Draw the ect ot chords D,Cl

13. A a housq the bottomand a man 14

a scale of l/ 00 the

14. Draw a horizontal line ABC from left to right, BC re

sen 50 feet to a convenient scale. Above A 1s a moun on

which Is situated a tower whose foot we will call D and top E .

It is observed that the e ABD the angle ACD 15

and the angle ACE = 25°. w this figure to scale as accurately

as you can, and measure from your drawing the height of themound AD and the height of the tower DE .

Check your results trigonometrically, that is,15. The wheel in Fig. 2 is 5inches in diameter and makes one

revolution per second. On the knob A, which is fixed to the

wheel, rests a door 15 inches high, free to slide between vertical

guides Show in a gra h variations during2 seconds In the heightof the centre 0 of thedoor above the bottom of thewheel repre

sent the height full size, and a second by 4inches.

Give an expression for the height of C rn terms of the posrtron

of A.

ADDITIONAL EXERCISES 343

Show what change would be necessary in the gra h if the doorrested on the hub B all the time the knob A was be ow B.

FIG. 2.

FIG. 3.

16. Fig. 3 shows a quarter of a desi which is symmetricallines 0B and OE .

°

ck off the figure and

17. ABCD (Fig. 4) is a frame of four orinted rods, stand

a vertical plane on a table. The rod A is fixed to the ta l

i

e

n

.

Drawthe frame from the given dimensions, and show by thick linesthe paths that C and D are at liberty to trace out, and

FIG. 4.

brieilfiy the reasoning by which you determine the of eachtpa

B the help oftracing-paper, or otherwise, draw the path which

P , e middle point of CD, is at liberty to trace out.

344 ADDITIONAL EXERCISES

the path the positions inwhich P is nearest toand furthest from O,

the middle pornt of AB.

18. Within every parallelogram, no shapethere is a point such that all lines drawn through it from boundary

,

to boundary bisected at that point. ’

Find in anywayyou likethe position of this point, and give a geometrical proof of the above

parallelogram on the same base and betweenthe same parallels are equal in area.

How could this be verified by means of a pack of cards or a

pile of slates ?

From this illustration deduce the corresponding

parallelepipeds. A parallelepiped is the solid figure

three parrs of parallel planes.

20. Make a quadri lateral ABCD, given that AB 4 inchesB0 5inches ; CD 6 inches LABO and LBCD

Find the area of the figure.

If the figure represents an estate on a scale of 5 inches to themile, what is the area of the estate in acres ?

21 . Draw a rectangle ABCD having AB 5 inches and BC2 inches, and bisect its angles by

lines meeting at P Q R S (seeroug sketch in Fig. Bymment and by general reasomng, determine what kind of a quadrilateralPQRS is.

If ADwere to remain constant whileAB changed in value from 0 to AD,

Show that the area of the figure PQRS(formed as before) would change invalue from }AD

' to 0.

22. Draw a fair-sized figure of fivesides, and find its area in squareinches, by reducing it to a triangle of

equal area, and then finding the area

of the triangle by construction.

23. Fig. 6 is a sketch of a framework made of six rods each 8 cm. long, the rods AB and CD beingjoined to the middle points of the other four rods. All the jointsare hinged so that P and Q may be brought together or pulled

a art .pDraw a straight line KL, and suppose the point P of the frame

work held at the int K while Q is moved to and fro along the

line KL. Draw til)

: path that E would trace out.

FIG . 5.


26. Take two points A and B 12cm. apart to represent two

positions of a

(l) SAB SBA (2) SAB SBA 30°

(3) SAB SBA 33°

(4) SAB SBA

of the ship. (b) Assuming that the shi’

s

two straight lines, find how near 9

(c) Assuming that she sails equally near

tack, draw a line showing the direction of the

27. AOB is an acute angle ; from a point P , either within or

without the angle, the line PQ is drawn perpendicular to GA andthe line PB perpendicular to OB . Show thateither equal or supplementary to the angle AOB .

State and prove any one proposition you have made use of.

28. ABCD is a rectangular piece of paper, 9 cm . byi

rsa line drawn W

rit (see

ig. 8 FD = 2cm. an angleEX ahd EL are

creases formed by folding E Cand EB against EF. Draw a

value.

Prove thatwhatever angle EF

FIG. 8. are supple

mentary, so long as the creasedlines cut the sides CD and AD.

29. AB and AC (Fig. 9) areon them and reflected

in the directions EF angle AED the angleBEF and the angle AHG the angle CHK. Prove that (1) theangle between themirrors the sum of the anglesAED and AHG

(2) the e between the two reflected rays EF and HK is twicethe angle tween the mirrors.

Two angles that together make up 180°

are called eupple

m entary.


30. InFig. 10 OADB is a square, andof two lines eachmoving uniformly andboth together from

the one,MN, moving parallel to OA , and reachin its

finalm

rfosition BD, in the same time that the other, OP, take?! torevolve about 0 to its final position OB. Their point of inter

section C traces out a curve. Obtain this curve for a quadrant ofradius 9 cm . Give the value of the angle AOC when MN hasmoved (a) cm. , (b) 6 4 cm . , and compare these angles. Copy

the angle XYZ, use the curve for dividing it into two parts in

the ratio of 2 to 3, and state your method ; the angles may betransferred by means of tracing

-paper.

31 . Make as many of the following triangles as you can. In

cases where you cannot, state a property of tri angles to explain

the impossibility.

Sides, 5cm 5cm. , 9 0m.

(b) Sides, 5cm. , 4 cm. ,10cm.


(0) A3 81“ , 90°

09 A0 81“ , 1”

3 . ABCD l l ) represents a rectangular sheet of paper thecornerDofwhifh

l

rgs to be turned down so that the crease lineMN shall

make an angle of 30°with AD .

FIG . 11 .

FIG . 12.

of

f

MDN when turned down and indicate how (I, the

o

and so as to cut

34. In Fig. 12 M is a rough drewin of a Sheet of paper,inches by 7 inches . The corner b is fol ed down to meet the

Side ad in k in such a way that the creased line ef passes throughc, which is inches from a . Give a careful drawing half-size ;show the exact position of the part turned down, and find bymeasurement the length of of and the size of the angle kcf .

35. ABC and A ’B

’ C’are two difl

'

erent positions of a paper

triangle in the same plane and with the same side of the paper

11 wards. Find a point 0 which is equidistant from A and A’and

from B and B’. Join O te A A’ B B

’C C

'

,and then

prove that the angles AOA’ BOB'

and 000 are all equal, and

that O is equidistant from C and C'

.


40. Prove two tu ang’

les ual inall respects when the three sidesknown to be

eqequal to the three sides of the other .

In Fig. 15, assuming that the lines are all unequal, and that ABis parallel to DC, and AE

'

parallel to BD, state what is the least

measurements that must be made in order that thefrom these measurements alone

of measurements, either of

41 . A surveyor gives the following dimensions of a four-sidedfield ABCD z—AB 1130 links, BC 1370, CD 1440, DA1040, BD= 1960, CA 1470, I. CAD 67

°Drawthefield

using the values given for BD and CA, and state the stepsdrawing. Check yourwork by measuring BD and CA .

42. Two sides of a triangle are 3 inches and 4 inches long, andone angle is Sketch figures, with dimensions marked, showingthat four difl'

erent triangles answer to this description. State howwould determine which of these four has the greatest area,

with help of your instruments find approximately what thisarea is.

43. A surveyor fixes the°

tion of a rectangular buildingABCD, meu uri 320 linksmmlinks, as follows —When hehas gone 1,248 along a straight road to P (Fig. he is in


line with the shorter side AB and A is 142 links away. When hehas gone links

, A is 75links awa andwhenhe hasgonelinks A is SS links away and he is in e with the long side AD of

the building. Take a line with a point P on it

road and the point P and draw the building on 1 inchto lOOIinks, sta how you do so .

Do you need the data given ? How many could you dowithout ? Use these extra data to check the accuracy of your work.

44 . Two quadrilateral figures ABCDDC QB, GD = ES, and DA =

IfP .

80 9

45. Criticize the propositionsufficient and necemary to determine a

and size

46. You are to of the circle and regular hexagon in17 and 18 the least possible number

FIG . 18.

of points that will enable you to the figure with yourmethod in each case, and give the

geometrical principles you make use of for the completion of your

Algebraic Symbols.

47. 0 men are each 3: feet high, b are 3] feet, and c are 2 feet .

Give their average height .

48 . A certain sum of money contains 0 pounds b shillings and c


121. 183. 11d. Verify

49 . A and B travel alo the same road in the same direction,

A at a miles and B at b°

es an hour. When A is at

B is m miles ahead. If B is overtaken H hours later

Also give the values of -9 .

50. A r

gclist can travel qmiles an hour on a still day, but his

speed is eel-eased m miles an hour against a certain wind, and

increased n miles an hour with this wind. How longwould it takehim to go 20 miles and to retum , going against the wind andreturni with it ?Give e time to the nearest hour when q 9, m an 3, n 2.

Takingq 9, drawagraph to show the time hewould take to ride

20 miles against the wind for values of 1» between 0 and 8 . For

the time represent an hour by 1 cm . and for the speed representone mile per hour by 2cm .

A tax of

formula for

52 The exposure required to take a photogra h is portional

to the square of the stop-number and inverse ygroporg

l

oo

nal to the

speed-number of the plate. It is found that 1} secs. is the correct

exposure with an ordinary plate (speed 80) using a st0p numbered16. What would be the e ure with a fastplate (speed 350) whenthe stop

-number is 68 ? race your answer in the form l/n of a

second, gi n to the nearest whole number .

Whatwoul be the exposurewith a plate of speed numberN and

a stop numbered N ?

53. On a long voyage the speed of a ship gradually decreases asthe ship’

s bottom gets fouled. If u and c are the speeds in miles

per day at the beginning and end of the journey, T the number of


60. A signals l45words in23minutes. B 94 words in 15and C 133words in 21 minutes. Represent the times taken byhorizontal distances, 1 cm. representing 2minutes and the numberofwords signalled byvertical distances, 1cm . representingfivewords.

Mark the three points representing the above results.

Who is the fastest signaller ? Show on the diagram the graphcorresponding to a rate of six words a minute .

61. A mine shaft is feet in length. It slo downwards atan angle of 45

°

to the horizon for a certain part oi’e

i

s

ts totalsay 1 : feet, and at an angle of 35

°

for the rest of its length. If thetotal (i3th reached is 1,000 feet, obtain an equation for

hence culate 3 . Draw a rough figure of the shaft .

62. A carriage is moving on a level road without any slipping of

the wheels. One of the wheels of diameter 3 feet picks up a speckof mud which adheres to the rim of the wheel. On squared

out to scale the curve described in the air by theud, and carefully mark the points at which the kbe when the wheel has turned through

and Also in the case of check your result

63. Draw up a table showing in three columns the value of

10 sin 6, l000s 6, and 8 sin 6+6cos 6 for each 30° from 0° t0 360°.

From the table draw the graphs of y 10 sin 6 and y= 8 sin 66 cos 6, and from the curves determine approximate ly a value of

6 for which tan 6 3.

64. On a line OB of indefinite length set ofi’

OP 10 cm. long.

Now imagine OP to revolve about.

0 through an angle of butto vary in length in such a wa that its length at any moment isequal to its original length mu tiplied by the cosine of the angle ithas revolved through ; thus at 60

°its lo h will be 5cm. Use

the tables to determine the path of P , an justify any statement

you can make about the locus of P .

65. Four rods are made into a jointed parallelogram ABCD

FIG . 19 .

(Fig. Two of the rods, BA andBC, project beyond the corners


of the parallelogram. A fixed point E on the roduction of BA is

taken, and F is taken on BC produced so that or the sha of the

in the fitiure F is in line with and D.

e parallelogram this point F is in

This framework is laid on a map, F pinned down, and D is madeto trace out the boundary of a country. Say how E will move .

If the frame used is represented to scale by Fig. 19, what is ther

atio of

?

any line on the newmap to the corresponding line on the

0 one

66. The base BC of a lo is 7cm . long. Find the locus of

the vertex when AB/AC 2/ byplotting a number of points anddrawing a freehand curve through them. In the same way findthe locus of the vertexwhen AB AC 10 cm.

One of these loci looks like a circle . Prove that it is a circle.

67. Drawa circle 10cm. in diameter and take aits centre . Sup B a point on the circle andof PB. Draw t a locus of S as B moves round the circle .

proof, the nature of the locus of S, whatever thesize of the given circle and the distance of P from it .

68 . Find the diameter of the saucer, of which a broken fragmentis shown in Fig. 20. Prick the figure 03.

69 . Three ships, A, B , and C, at anchoroccupy the following positions with regard to

a buoy P z—A is 350 yards north of P ; B is

170 yards north-east of P ; C is 420 yards east

of P . Draw a plan showing the positions ofthe ships and the buoy, on a scale of 100 yardsto an inch .

A fourth ship, D, anchors in a position equidistant from A, B, and 0 . Determine its posi

tion on your plan, stating its distance from the

uoy.

State your method of finding the position ofD, and prove that the method is correct.70. Prick off the curves 0 and

and find whether they are arcs

plain your method of testing the curves

FIG. 20.

(0 ) inside thetria ngle, (b) outside the triangle, (c) on one of the sides ?

A a 2


Youfpart 0

not be given.

72. Two cog-wheels gear together ; one of them has a cogs, the

other b cogs, and the are set c inches a on each wheel.the first wheel es one revolution, ow far does a int

in its rim travel ? And a point in the rim of the second w eel ?And howmany revolutions does the second wheel make ?

Fic . 21.

If the cog-wheels were connected by a chain instead, howmany

ml

ftions would the second wheel make for one revolution of the

I wish to settle the number of teethdriven bicycle with a rear wheel 30 into travel 6 yards for each revolution

showing the number of turns of the

1, 2, 3, 4, &c . , turns of the pedals. Deduce from your pha convenient number of teeth for each of the two cog-whes

l

s

‘

on

73. Two circles have radii and 21 inches, and their centresare inches apart. Draw a circle of radius 1 1 inch touchingthem both. Can a circle of radius inch be drawn to touch thegiven circles ?

74. A int P moves round the circumference of a circle whosehorizon diameter is AOB, 0 being the centre . Two circles are

drawn on AO and OB as diameters, and OP always meets one of

these circles. Showthat the piece cut off OP by the circle it meetsis equal to the projection of OP on the diameter AOB .

Give agraph showing (for half a revolution of P ) the projectionof OP as a function of the angle AOP ; that 18 , mark of! distances on


2cm. from the centre . Show, both bydrawing and bycalculation,

that AB is trisected at the points where it cuts the circumference

Prove that chords

79. A stream 20 feet wideis to

radius of the arch and the angleit subtends at its centre. Checkyour results by a drawing to

Fro . 23. scale .

80. Prove that angles in thesame segment of a circle equal.State (without proof) what property of a circle may be deduced

at once from this theorem by conceiving the angular point to movealong the circumference into coincidence with either extremity ofthe segment.

81 . The distance of the horizon d, the radius of the Earth R, andone

’

s height above the sea It (all in miles) are connected by theequation

d2 h?

the Earth ’

s radius as miles, find at what height theis 34 miles away, and express this height in feet to the

nearest foot.

82. Prove that a straight line drawn from the centre of a circleto the middle point of a chord cuts the chord perpendicularly.

P is a point external to a°

ven circle whose centre is C. With C ascentre describe a circle wit radius twice that of the given circle .

With P as centre and PC as radius, describe a third circle cuttingthe second in 0. Join CQ, cutting thefirst circle in B. Prove thatPR is a tangent to the first circle.

83. I wish to fix the size of a circular enclosure which I may notenter I therefore note the following points It is surrounded bya footpath 10 feet wide, and four lamp posts stand along the outer

circle are drawn through a given

point inside a circle, the rect

angles contained by their segments are ual.If the radius of the circle

is 3 inches, and the given point21 inches from the centre , calculate the value of this rect

angle and verify the result from


boundary of the footpath at the corners of a square . Each side of

this square cuts from the enclosure a strip whose greatest width iswidth of the th.

t e enclosure calculation and also bydrawing. For our drawing take a circ e of radius 4 inches forthe outer boun of the footpath.84. Find the area of the rectangle KLMN (Fig. 24) inscribed in

a semicircle in terms of CK :c cm. and the radius CL r cm.

Taking r a z 5, find the area

for a: 0, l , 2, 3, 4, 5, and

give a general idea of the

of per of radius 5cm.

tats and prove a relation FIG . 24.

that holds between KA, E Land KB, whatever values r and a: have.

85. Draw to a scale of 10 feet to the inch the junction lines of

a small tramway for a factory from the dimensions given in the

(Fig. All lengths are in feet . The arcs AB and BC

have a common tangent at B . What is the distance between thecentres of the area AB and BC ? Howdoes the knowledge of thisdistance enable youM ad the centre of the arc BC, andthe posiof the point B ?


86. In the rough sketch in FigJ Bthe circle represems a secfionthe eenh e cf the earth by a plane in which a dar fi

'

lies.

ed are the directions in which the star S is seen frcm tw<>ii

3

8

B d ef ats tangents at a and c. Prove that thebetween the angles def and bah is equal to the angle aOc.

FIG . 26.

87. The section of a tube railway is a segment of a circle (Fig. 27The breadth of the tube on the level of the rails is 8 feet, and its

greatest height above the rails is 10 feet . Draw the section on anyconvenient scale, and write down the greatest breadth of the tubein feet to the nearest tenth.

State how you drew the section, and state and prove a propertyof the circle that you used .

88. Prove that of all straight lines drawn from a given point toa straight line the pe endicular is the lA instructs B to ma e a triangle with these measurements : two

sides are to be 7 inches and inches in length ; also the angleopposite the shorter side is to be B says it is impossible todo so . Prove that B is right .89 . Given the lengths of two sides of a triangle, and of the r

pendicular from the point where they meet on the third side, ow

ow to find the length of the third side by drawing.


IN miles at any speed fiem Oto fi miles pcr-hour ;

the mcst economieal spsed and the cxpense at that

95. The shortest side of a h-iangle is 10 cm. long, the other two10r cm. and l0r3 cm . long. Determine the greatest value r can

96. ABC is a right ~angled h im gle having AB =-8 cm . , B C

4cm. , LB -w’. O is a point on AC, OL and on

BA and BC. Show by a graph how the are a of the rectangleLO pames from A to B C,

97. Assummg'

thah if a gun is fired over a plam,

’

the range ofthe pro

'

ectile ra given in by y = 5000 3m 2z (where f isthe angfe of elevation of barrel), out a curve

ranges hetween the limits cf 5°

and for the angle of elevafion .

Statc what must be the valuc of z to give the greatest range .

9& A man in a launch at a point L is 30 miles from the nearest

point A of a straight ccast-lim and wants to reach a point B on

division of your squared paper for a mile and 20 divisions foran hour the time 0 hours need not be shown.

Tota l time

99. AB is the base (100 feet long) of a triangle AB andpoint. The ABC and

C, C m

to be 63°

14’and 80

°

Calculate the distance


100. To determine the size of a piece of ground, pegs were

driven at the four corners A B C D. AB was found to bechains, AC chains, AD chains, LEAO 26

°

L CAD 43°

Find the area of the ground by calculation.

101 . A section of corrugated iron roofing consists of circular arcs

of angle 11 alternately curved in opposite directions. Supposingthe section of the sheet to contain an exact number of thesearcs, find (expressed as a decimal) the ratio of the area covered

at to the entire superficial area of the sheet-iron of which it ise .

102. A meteor is seen from two places, A and B ; B being 20miles from A in a direction 15

°west of south . At A the meteor

appears to be 5°35

'

east of south, and at B it appears 7°

24’

northof east . The altitude of the meteor as observed from B is

From these data find the height of the meteor .

103. Taking the earth as a sphere of radius miles, find thedistance from its axis of London whose latitude is 51° 30' N . Find

also the distance London travels in an hour in consequence of the

rotation of the earth .

104. The extreme range of the guns of a fort is metres .

A ship metres distant sailing due east at 24 kilometres an

hour notices the bearings of the fort to be 20° 30' north of east.

Find, to the nearest minute, when the ship will first comewithin range of the guns. Check your result by a drawing to

scale .

105. Draw a triangle ABC such that tan A 2 and tan B

Find, by measurement, the height h (perpendicular to AB) and thearea S of the triangle you have drawn. Find the value of the

ratio S/h’as a decimal.

Find an exact formula for the area of any triangle in terms of itsheight h and the tangents of its base anglesA and B, and comparethe numerical value of S/h’ given by your formula with yourprevious result .

106. If one angle of a triangle is

greater than another, prove

that the side opposite the former ang e is greater than the side

op°

te the latter .

tate , with accompanying proof, whether the following theorem

is or is not true n every triangle, the magnitudes of the angles

proportional to the magnitudes of the opposite sides.

107. A rod BC of length 58 cm . rotates about B . Another rodCA of length cm . has one and C hinged to the first rod, whilethe other end A slides along a fixed line BO. By drawin the rods

in various positions (taking values of LB at intervals of or so),find how the length BA varies as the angle B increases ; and show


BA as a function of LB in a gra h for one revolution of BC,showing the actual length of BA an re resenting 30

°by 1 cm.

Write down an equation connecting t e angle B and the lengths

of the three sides of the triangle ABC.

Solve the equation to find the length of BA when LBTo check your result find the length by drawing.

108. Draw a circle of radius R 10 cm . and draw two radii at an

angle of 6 Join the ends of the radii , cutting off a segmentof height Inem. and base ccm. Find a formula for the area of

the segment in terms of R and 6. The area is sometimes taken as

that of a triangle on the same base as the segment of times

the altitude of the segment . The formula 2391'

E;is

to give the area. Use these three methods to determine the area

of the given segment, and tabulate your three results.

109 Fig . 28 represents a two-foot rule, which is ivoted at O

and hinged about ac and bd . Imagine it placed, as s own, flat on

table, so

Ighat the angle AOB is and calculate the distan

'

cc

rom A to

Ac and Ed are now turned about the hinges ac and0°from the plane of the table, what is the new

A to B ? Briefly give reasons for your statements .

110. Sufiicient data are given below to draw three triangles .

State, with reasons, whether any two of the triangles are similar toone another(a ) One angle hypotenuse 15cm . , one side 9 cm.

(b) Two sides in. and in. , included angle

(c) Sides 36yards, 27yards, 45yards.

Equations.

111 . The speed (s) of sound in air at a temperature of t°

Fahrenheit) is given by the expression 3 a ti . When t 32,8 ft. per sec. ; and when t 41 , s ft. per sec. Find

the value of s when t 70.


119 . A and B are two coal pits m miles apart. The price of coal

at A is a shillings a ton, and at B it is b shillings a ton. The cost

of carriage fiem each pit is c shillings per n per mile . Give,in terms of a b c, thc distance froo f a point X, on the road

from A to B, atwhich the total cost of the coal is the same, whetherit comes froo r from B . Give a numeriu l result whcn ¢ = 29,b = 27

2 5 A

xii- 3a: - 9f“ “n

values of 3 , find the value of A in terms of .r .

121. A screen is

screen by either light varieso the screen from the

find the distance, x feet, from

interpretation you place on the

122. Solve az b2 c’ 2bo cceA as a quadraticwhich b is the unknown quantity . And hence, or otherwisl ,calculate the positive value of b when a ==11 cm. c = 9 cm.,

A Check your result by an accurate drawing.

123. The formula for bodies

Find the time a body takes to rise 192 feetthe initial velocity is 128 feet per second .

Indices.

124. Find the values, giving not more than

w assumed by a + a4, when a has the20, 3, 2, l ,

Tabulate the values of a , a4, c + c4, and explain what yourtable shows as to the relative importance of the two termsaccording as a is large or small

125. When a man lends £a at b per cent . compound interest, the

value of his loan grows so that after cyears it is £a (1 + Ute

this formula to find the value after two years of a loan of £460 at

3} per cent . Give the value to the nearest shilling.

126. The formula is used by engineers. Find its

value to two decimal places when n and 2.


127. Find between whatpair of consecutive integers each of the

following lies : 2 /4,

128. The discharge Q of water through a rectangular submergedhole is given in cubic feet a second by the equation

0 go's (H3/2 [cs/2)fl } ,

all the dimensions being in feet. Find what thegallons per 24 hours, when C 060, y 32,K 1 . One cubic foot gallons.

129. The pressure p and volume 0 of a gas which expands in a

vessel impem ous to heat are connected by the relation p07 con

stant, where y A mass of gas occupies a volume of 1 cubic

foot at a pressure of 100 lb . per square inch. Find its pressure

when its volume is 2, 3, 4, 10 cubic feet respectively, and

represent the relation between pressure and volume within theselimits by a curve drawn on squared paper.

130. Aftervery careful calibration itwas found that the dischargeD gallons per minute over a certain weir was given by the equation

D 20H1 ‘48

when E was the head of water in feet above the sill of the weir.

Calculate with the aid of a table of logarithms the head necessaryto discharge gallons per minute .

131 . A workinsgoerule $011; the collapsing pressure of a boiler flue

is given by P—zD

where P the collapsing pressure in

D diameter of the flue in inches, L the

the thickness of the plate in inches . Find the

pressure when t = 3, D = 31}, L = 25.

Cc-ordinates.

132. Draw OX horizontally towards the ht

upwards. Values of :c are measured to the right along OX from O

and values of y from 0 along 0Y, unit 1 foot . A telegraphgiven in position approximately by the equation

(100 x)” 10000The centre line of the j ib of a builder

’

s crane is given in its

highest position bythe equation 3! 3 ix. The heightlower end of the

'

ih above the ground (that is, above OX) is 30 fta

lnd the length 0 the j ib is 43ft . Find whether the crane fouls

t 0 wire .

133. Draw the quadrilateral ABCD, the corners A, B, C, Dhaving the co-ordinates (14, (12, (3,

368 ADDITIONAL EXERCISE

Discus how many conditicns will fix a polygon os ides (1 )specifyingm and

rfi)£yp

csifion. and givc two ways of

cc-ordinates of O in

also when O is as far

Solid Geometry.

134. Am ing

uare ,houae m m feet each way, has a roofI

?rsm all four wulls at §5° to the horimntah

FIG . 29 .

136. A be shown as a


dcpth of water at the

these

line OA, OB ,

water to

are in the tank, 0, . s gallons,rate of filling in terms of thee quantity

in turn ;

Two straight lines are called parallel when, hbwever far prothey never meet.

through a certain point within called the centre are

drawn .

TABLE OF WEIGHTS AND MEASURES

1 metre 100 cm . mm. km .

1 yard 3ft. 36in. O°9I44metre.

1 mile 8 fur. 80 chains 320 po. yds.

1 hectare 100ares sq. metres.1 acre 4 roods 10 sq. chains sq. yds.

1 litre 1 0 . dm . 100 cl h] . c. metre.

1 gallon 4 qt. 8 pt. c. ft. litres.

1 quarter 8 bush . 32pk. 64 gal.

1 kilogram 1 000 g. tonne .

1 lb. 16oz. grains grammes.

1 ton 20 cwt. 80qr. lb.

1 litre of water weighs 1 kg.

1 c. ft. of water weighs lb.

Ratio of circumference of circle to diameter is 31 416.

ANSWERS TO CERTAIN NUMERICALQUESTIONS

Orin-ru m

2. 6 0r 7hect’5res or 16acres. 4. 200 sq. cm .

sq15. 2°

.6 sq m. ; 18 or 19 ft. 17. acre . 18 . 2°9 cm .

21 .

Cm m a IV

5. in. 9 . kgm1000 bc/a -d .

11 . £13. 83. 4d.

13. 302 and 26880/sm sq. ft. 14. 18 pence per ton.

15. 9s. 8d.

° 0 ’000, ,009 13mmiba shi llings.

16. bole /2240 or tons ; k/bc or 28 ft.17. sq. ft. ° 180 18 . 0°13a in.

19 c. ft . or 20 . 361 .

21 .

Cm rrsii v

9 . cm. 40sq. cm. 59 sq. cm.

14. 4ac. 1 rd. 15. 73gals./min.

Cm m n VI

1 . 0°8 1b. 2. 3ih 3. 482 ; 7°

.21b

4. 106°6c. c. 6. 2O

0

E . .P to nearest

A rice

I] . 0v

20mc

gn

o

i.p

13. in.

14.8°7aa sq. ft .

16. £920. 20 . 3.20p.m.

374 ANSWERS TO NUMERICAL QUESTIONS

Art . 8 . For indices 20, . 240, e powers of are

2705,8° 27, 10

Ex. 2. At .C at 8 40 a.m 3. 1, ,600 000gal

°

0

290 gal.4. 7 ; more . 6.

7. 99 gal.° true volume 103

5

gal. 8 . miles.

CHAPTER XIII

21 . c. ft . 24. kilos ; 32 sbillmgs'

26. 75 gal. sec. 27. tons. 28 . 6ft. 9 in.31 . 8,000 mil 32. 13ft .

Gnu-ma XIV

1 . 8°6cm. 2. 4. 400 ft 2°5sec.

6. 2001b . sq 12 1, 15°

.2cm

8 . In m 40 01

5;9 . 65million miles.

12. Son’

s weight 737 of father’

6

s,

9

but he needs 81Z of father ’

s

food, i. .e

15. in. 16. 48 swings. 17. ft .

18 . 19 . 21 . 13°1 lb ./sq. in.

24. z = 1 °

,22 k = 470, p = 49.

25. 27. 210 kilos. 32. 100°8 .cm

11. 86ft. 13. Side of table in.

ves for least value of D.

19 . sine = 0°

,4 9 cosine = 0°.908 20 . 11 °3cm.

21 . 3oh . 22. ml./br.

24 . miles. ft .

CHAPTER XVI

Art. 16. Area sq. cm.

Art . 17. (1) A (2) a (3) c or

5. 68yds. 6. miles. 7.

8 . 19s. 2d. 9 . a = o°

,39 b=

10 . y 2° °55z . 14.

15. 17. km. 444m.

18 . 980 ft. 19 . 376metres.

ANSWERS TO NUMERICAL QUESTIONS 375

20 . :c —13 y 21 . y = 1° -62 :c 53.

128 at 2and 4, 144 at 3secs.

inches. 24. 205°

Fahr .

-I-B) ;o

308 0 . yds. 80 1b./yd.

27. 4°

.9 °

or 150 cm 7 .6oft 9 . ft2. 80 sq31 . l 17ft. 32. 219 links 71 °7

°S . of E .

33. 5,090 1inks.

Ciu rm XVII

1 . 6°

.9 cm 9 . 10 . 13.

Art. 52. AB == BC 2°94, CA 3°09, AL 2°

,51 BL 2°42,

c. ft. 17. 98oft. 20 . 3°69 in .

21 . 5°oft. 22. 8,748 c. ft . 24. 21 in 100.

ft. 26. 253ft.30 . 3hrs. 54min. 31 .

32. In links to nearest integer AB = 800, CD

INDEX TO EXPLANATIONS OF TERMS

AND PROPERTIES OF FIGURES

Reference-cre wman .

Angle, 4.

Amus’L aa footnm ni .

1. between two planes,L greater than 219.

Bisection ofan L$64.

Correspopdi n

zLs, 1

0

Minute of L,69 footnote.

Obtuse L, 83 footnote, 71 .

Right L, 3.

Areas, 27.

Circle, 51, 828.

Parallelogram, 188

Triangle 83 291 .

(a t e) ch i-9 1 260, 257

Average, 51.

Axes of cc-ordinates, 116.

Base of logaritbin, 200.

Base of A, 85.

Binomial series and theorem ,181 .

Bisseti of M’

t 18.

To find centre of circle, 216.

Centre of sphere, 100.

Are of a cirele, 220.Area, 51, 528.Centre, 1 .

72.

Number of points that fix a

Radius, 1 .

Ratio of circumference to d ia~

meter, 50, 328.Rectangle properties of circle ,

Sector of circle, 331.

Segment of circle, 220.

216.

Circumference of a circle, 49 .

878 INDEX

Parallel rulers, 132.

Area, 138 .

Oplpé

site sidesequal conven es ;

Pentagon regular pentagon ; 338footnote .

Perimeter, 270, 327.

129shortest distance to a line,

J. toaplane, 106, 299.

L a to a plane are ll converse ;

Theorem of Pythagoras ; conProperty of .L bisector of a line, verse 259.

11 . Root of equation, 84.

To draw a .L to a line, 13, 14,221 . Sector of circle, 331 .

Plan and elevation, 835. Segment Of circle, 220.

Plane, 100. Set-square, 131 .

Plane table, 62. Shape of A, conditions that fix it,Plumb line , 107. 153

Polygon, 140, 827, 388 footnote . Simi lar figures, 161, 167.Perimeter, 827.

Regular polygon, 388 footnote.

Polyhedron, 833Power, 152.

Prism, 820.

Projection, 103, 115.

Projection on a line, 287.

Proj ection on aplane, 834.

Plan and elevation, 335.

Proportion, 281 .

Proportional ,mean, 231.Pyramid, 299 .

Volume, 811, 820, 836.

Pythagoras, theorem of con

verse 259 .

Extension to any A, 290.

Quadrant, a quarter of a circle

bounded by two .L rad i i .

Quadratic equations, 92, 264, 291 .

Quadrilateral, a figure ofsides.

Sum of Opposite Ls of cyclic

quadrilateral is 222.

Radius of circle, 1 .

Ratio, 150.

Ratio of circumference of circle todinneter, 50, 338.

Rect

an

gle ; rectangular ; 27, 88,

Area of rectangle, 29.

-_20 c, 257.

Regular pentagon and polygon,338 footnote.

Right L, 3, 4.

Right Ld A, 31 .Calculation of,Hypotenuse of, 41 .

To draw, from various data,

Similarity of As, condi tions thatensure it

,164.

Simple equations, 86.

Sine of an L, 165.

S ine of an obtuse L, 282.

Slide rule,209 .

Slider crank pair, 78 .

Solid body, position fixed by sixdata, 312.

Congruent solids, 321 .Solution of equation, 84.

Sphere , 100.

Area , 382.

Centre, 100.

Section by a plane, 329.

Volume, 338.Square, 36, 136.

To draw square-c a given rect

angle or A,229 .

Square root, 212, 244, 248.

Straight path shortest, 188.

Supplementary Ls,Surface , 99 .

Symbols, table, 23.

and 155.

V 218 .

INDEX

Precedence of x over 4» 29 .

Right use of symbols and con

tractions, 23, 30, 40, 55.

Table of weights and measures,871

Tangent of an L, 195.

Tangent of an obtuse L, 284.

Tangent to circle,Lbetween tangent and chordL in alternate segment, 225.

To draw a tangent,T-square, 131 .

Tetrahedron, 315.

Transversal, 129 .

Trapezium ,139 .

Triangle,Acute LdA, 83Angle

-sum property, 127.

Area, 38,Base, 35.

Bisector of vertical L dividesbase in ratio of sides, 191 .

Conditions that fix A, 65.

Congruence of As, 65.

Exterior L of A, 126.

Greater Lopposite greater sideconverse ; 190.

Height of A,35.

Interior and opposite L, 126.

Isosceles A, 6.

379

Obtuse Ld A, 33 footnote .

Hto base cuts sides in same

ratio, 157.

.L bisectors of sides are con

current, 217.

.La of a Aare concurrent, 289 .

Pythagorean theorem genera

lization ;Right Ld A, 31 .Shape of A

,conditions that fix,

Similarity, 164.

A given by cc-ordinates, 292.

Two sides are greater than

third,3.

Vertex, 123.

Trisection of line, 146.

Weights and measures, table of,371

UNN . OF MlCHlOAN ,

Vertical line, 107.

Vertical plane , 107.

Vertically Opposite Ls, 24.

Volumes, 45.

Cylinder, 48.

Rectangular room, 45.

Similar solids, 167, 836.

Sphere, 833.

Pyramid, 31 1, 317, 336.

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