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BIBLIOGRAPHIC R
Graduate
University of
Preservatio
ACV2059
UL FMT B RT a BL m T/C DT R/DT
a 03021435
035 1 : a (RLIN)MIUG86-B37774O3S 2 : a (CaOTULAS)160647553O40 : CMiU
050/ 1 :O : | a Q A46S | h .I—I19
a Halsted, George Bruce, d 1853-1922 .
245 :OO: a Metrical geom etry . lb An elementary t
George Bruce Halsted .
260 : a Boston, b Ginn , Heath co . , IC 1881 .
a xiii, 1 232 p . b diagrs . c 19 cm .
650 1 : O: a Mensuration
650 2 : O: a Metrical geom etry .
998 : c JMH s 9124
JME TR I OAL GEOJ I E TR Y.
-fl — o__ — n_
TREATI SE
M'
E N SU R A T I ON
GEORGE BRUCE HALSTED
A.B .
,A .M.
, AND EX —FELLOW OF PR I NCETON COLLEGE ; PH D. AND
EX -FELLOW OF J OHNS HOPKI NS UNI VERSI TY ; INSTRUCTOR I N
POST-GRADUATE MATHEMAT I CS , PR INCETON COLLEGE.
B OS TON :
PUB L I S H ED B Y G I NN, HEATH ,
CO .
1 8 8 1 .
Entered according to Act Of Congress , in the year 1881 , by
GEORGE BRUCE HALSTED
in the Office Of the Librarian of Congress, at IVaSh ington .
GINN HEATH , 8: Co
J . S . CUSHING,PR INTER
,1 6 HAW LEY ST REET ,
BOSTON.
INSCR I BED TO
J J . SYLVESTER,
Cam . F R S L E Cor . I ll em . I nstitute of F rance ;Sciences in B er lm
,Gettingen,
Nap les , flfilan, St. P eters
burg ,etc . L L B
,Univ. of Dublin
,U. of E. DC L
Oxford Hon. F ellow of St. J ohn’
s COL,0am .
I N TOKEN OF THE INEST IMABLE BENEFI TS DER IVED FROMTWO YEAR S ’ WORK W I TH H IM
,
BY THE AUTHOR .
PREFACE.
I‘HI S book is primarily th e outcome of work on the subject Wh ile
teach ing it to large classes .
The competent critic may recognize signs Of a B erlin residence ;
but a considerable part, it is believed,is entirely new .
Special m ention must be made Of the book’
s indebtedness to Dr .
J . W . DAV I S,a classmate with me at Columbia Sch ool ofMines ; also
to Prof. Gr. A . WENTWORTH, who has kindly looked over the proofs .
But if th e book be found especially accurate,this is due to th e pains
taking care of my friend H . B . FI NE,Fellow of Princeton .
Any corrections or suggestions relating to th e work will be thank
fully received.
GEORGE BRUCE HALSTED.
PR I NCETON ,NEW J ERSEY,
May 12, 1881.
CON TEN T S
MENSURATION .
I NTRODUCT I ONTHE METR I C SYSTE MNOTAT I ON AND ABBREV IAT IONS
CHAPTER I .
THE MEASUREMENT OF L INES .
(A ) .—STRA IGHT L INES .
I LLUSTRAT IVE PROBLE M S
(a) To measure a line th e ends Of wh ich only are accessible
(B) TO find th e distance between two Objects, one of whichis inaccessib le
(7 ) TO measure a line when both ends of it are inaccessible(8) TO measure a line wholly inaccessible
Q (B ) .— S TRA IGHT L INES IN TR IANGLES .
Article.
I . R I GHT -ANGLED TR IANGLES1 . To find hypothenuse
2 . TO find sideI I . OBL I Q UE TRIANGLESOb tuse
4 . Acute
5 . Given perpendicular6 . To find medials
I I I . STRAI GHT L INES I N SI M I LAR F I GURESTo find corresponding lineIV . CHORDS OF A CI R CLE
8 TO find diameter
Vni CONTENTS .
A rticle.
‘
9 . To find heigh t of10 . Given chord and h eigh t11 . Given chord and radius
,to find chord of half the are
12. TO find circumscribed polygon
—METHOD OF L IM ITS .
V . DEF I NI T I ON OF A L I M I T13 . Principle of L imits14 . Th e length Of the curve
(D ) .— THE RECTI F ICAT ION THE C IR CLE.
Length of semicircumferenceCircumferences are as their radiiVI . L INES I N ANY CI RCLE
Value Of 7r
To find CircumferenceValue Of diameter
CHAPTER II .
THE MEASUREMENT OF ANGLES .
(E) .—T I—IE NATURAL UN IT OF ANGLE.
V I I . ANGLES ARE AS AR CSNumerical measure of angle
VI I I . ANGLE MEASURED BY ARC
CIRCUL AR MEA SURE OF AN ANGLE.
Angle in radiansTO find length of are
To find number Of degrees in arc
IX . ARCS WH I CH CORRESPOND TO ANGLESAngle at center
I nscribed angle
Angle formed by tangent and chord
Angle formed by two chords
Secants and tangents
CONTENTS .
Article.
Given degrees to find circular measure
30 . Given circular measure to find degrees
31 . Given angle and are to find radiusX . ABBREVIAT IONS FOR AREAS
CHAPTER III .
THE MEASUREMENT OF PLANE AREAS .
— PLANE RECT IL INEAR F IGURES .
XI . MEASURING AREA OF SURFACE32. Area Of rectangle
33 . Area Of square
34. Area Of parallelogram
35 . Area Of triangle (given altitude)Area Of triangle (g iven Sides)
37. Radius Of inscribed circle38 . Radius of circumscribed circle39 . Radius Of escribed Circle
XI I . TRAPEZ OI D AND TR IANGLE As TRAPEZ OI D40 . Area of trapezoid
X I I I . COORDINATES OF A POI NT .
XIV . POLYGON As SUM OF TRAPEZ OI DS41 . TO find sum Of trapezoids42. Area Of any polygon
43 . Area of quadrilateral44 . Area Of a similar figure
XV . CONGRUENT AND E Q U I VALENTXV I . PROPERT IES OF REGULAR POLYGONArea Of regular polygon
Table Of regular polygons
—AREA S OF PLANE CURV IL INEAR F IGURES .
47. Area Of circle48 . Area of sector
49 . Area of segment
50 . Circular zone
51 . Crescent
ix
X CONTENTS .
A rticle.
Area Of annulus
Area Of sector Of annulus
XV I I . CONI CS54. Area Of parabola
Area Of ellipse
CHAPTER IV .
MEASUREMENT OF SUR FACES .
XVI I I . DEF INI T IONS RELAT I NG To POLYHEDRONS56 . Faces plus summits exceed edges by two
MIL — PR I SM AND CYL INDER .
XIX . PARALLELEP I PED AND NOR M ALMantel of prismXX . CYL I NDR I C AND TRUNCATED
58 . Mantel of Cylinder
( J ) . PYRAM ID AND CONE.
XXI . CONI CAL AND FRUSTU MMantel of pyramidMantel of cone
Mantel of frustum of pyramidMantel of frustum of cone
Frustum of cone Of revolution
i (K) .
— THE S PHERE.
XX I I . SPHERE A SUR FACE,GLOBE A SOL I D
Area of a sph ere
XX I I I . SPHER I CAL SEG M ENT AND Z ONE
Area Of a zone
Theorem of Pappus
Surface Of solid ring
CONTENTS .
—~ S PHER I CS AND SOL ID ANGLES .
Article.
XXIV . STEREGON AND STERAD IANArea of a luneXXV . SOL I D ANGLE MADE BY TWO
,THREE
,OR M ORE
PLANES
XXVI . SPHER I CAL PYRA M I D69 . Solid angles are as sph erical polygons
XXV I I . SPHER I CAL EX CESS70 . Area of sph erical triangle71 . Area of sph erical polygon
TABLE OF ABBREV IAT I ONS
CHAPTER V
THE MEASUREMENT OF VOLUMES .
—PR I SM AND CYL INDER .
XXVI I I . SY M M ETR I CAL AND Q UADERXX IX . VOLU M E AND UNI T OF VOLU M EXXX . LENGTHS
,AREAS
,AND VOLU M ES ARE RAT IOS
Volume of quaderXXX I . MASS
,DENSI TY
,WEI GHT
To find densityVolume Of parallelepipedVolume of prismVolume Of cylinderVolume of cylindric shell
— PYRAM ID AND c oNE.
XXX I I . ALT I TUDE OF PYRA M I DParallel sections Of py ram idEquivalent tetrah edraVolume of pyram idVolume Of cone
XI
xii CONTENTS .
55 (O ) . PR I SMATO ID .
Article .
XXX I I I .
—XLI . DEFINI T I ONS RESPECT I NG PR I S MATOI D,
82. Volume Of prismatoidVolume of frustum .Of pyramid
84 . Volume Of frustum Of cone
85 . Volume of ruled surface86 . Volume of w edge
87. Volume of tetrahedron
(P ) . SPHERE.
Volume Of sph ere
Volume Of sph erical segment
XLI I . GENERAT I ON OF SPHER I CAL SECTORVolume Of spherical sectorXLI I I . DEF INI T I ON OF SPHER I CAL UNGULAVolume Of spherical ungulaVolume of spherical pyramid
— THEOREM OF PAP PU S .
93 . Theorem of Pappus
94 . Volume of ring
é (R ) .— S IMILAR SOL ID S .
XL IV . SI M I LAR POLYI IEDRONS
95 . Volume of Similar solids
(S ) .— I RREGULAR SOL ID S .
96 . By covering with liquidVolume by weigh ing
98 . Volume Of irregular polyh edron
CHAPTER VI .
THE APPL I CAB I L I TY OF THE PR I SMOI DAL FORMULA .
99 . Test for applicab ility
CONTENT S . xii i
- P R I SMOIDAL SOL ID S OF REV OLUTION .
Article .
XLV . EXA M INAT I ON OF THE DI FFERENT CASES
— PR I SMOIDAL SOL ID S NOT OF REV OLUT ION .
- EL IM INATION OF ONE B A SE.
100 . Prismatoid determined from one base
CHAPTER VI I .
APPROX IMAT ION TO AL L SURFACES AND SOL IDS .
(W ) .
— W EDDLE ’S METHOD .
101 . Seven equidistant sections
CHAPTER V III .
MASS—CENTER .
102 104
105— 131 . SY M M ETRY182 . THE MASS—CENTER OF AN OCTAHEDRON
EXERCISES AND PROBLEMS IN MENSURATION.
[Th ese are arranged and classed in accordance w ith th e above
132 Articles]
LOGARI THMS
AN ELEMENTARY
TREATI SE ON MENSURATION .
INT RODUCT ION.
MENSURATION is that branch of mathematics Which hasfor its obj ect the measurem ent Of geom etrical magnitudes.
It has been called,that branch Of applied geometry which
gives rules for finding the length Of lines,the area Of sur~
faces,and the volume Of solids
,from certain data Of lines
and angles .
A Magnitude is anything Which can be conceived of as
added to itself,or of Which we can form multiples .
The measurement Of a magnitude consists in finding howmany times it contains another magnitude Of the sam e kind ,taken as a unit of measure . Measurement
,then
,is the pro
cess of ascertaining the ratio which one magnitude bears tosom e other chosen as the standard ; and the measure of amagnitude is this ratio expressed in numbers . Hence
,We
must refer to some concrete standard,some actual Obj ect
,to
give our m easures their absolute m eaning .
The concrete standard is arbitrary in point of theory ,and
its selection a question Of practical convenience .
A discrete aggregate,such as a pile of cannon-balls
,or a
number , has a natural unit,“one of them .
But in the continuous quantity ,space
,With Which We
chiefly have to deal , the fundam ental unit,a length
,is
2 MENSURAT ION .
defined by fixing upon a physical Obj ect, such as a bar Ofplatinum ,
and agreeing to refer to its l ength as our standard . That is
,We assume some arbitrary length in terms
Of which all space measurements are to be expressed .
The one actually adopted is the Meter,which is the lengthOf a special bar deposited in the French archives . This wechoose because of the advantages Of the metric system ,
whichapplies only a decimal arithm etic
,and has a uni
form and significant term inology to indicate themultipl es and submultiples Of a unit .
THE METR I C SYSTEM
convenes to designate multiples by the Greek num erals
,and submultiples by the corr esponding
Latin Words ; as follows :
NAME. DER I VATI ON . MEANI NG .
SO a millimeter (mm) is one-thousandth Of a meter .
Thus we are givena number Of subsidiary units .
For any particular class Of measurem ents, the most
convenient Of these may be chosen.
The kilom eter (km) is used as the unit of distance ;
and along roads and railways are placed kilometric poles
or stones .
INTRODUCT ION . 3
The centim eter ( ) is the most common submultiple Of
the m eter .
A chief advantage of this decimal system Of m easures is,
that in it reduction involves m erely a Shifting Of the deci~
mal point .
NOTATION AND AB B REV IATIONS
TO BE USED I N TH I S BOOK .
Large letters Indicate points ; thus, A ,B
,and 0 denote
the three angular points Of a triangle , or 0 may denote thecenter of a circl e
,While A and B are on the circum ference ;
then AB will denote the chord j oining A to B ; and, gen
erally ,AD means a straight line terminated at A and D .
In the formulae , small letters are used to denote the num erical m easures Of lines ; so that ab, as in common algebra
,
denotes the product of two numbers .
The followmg choice of letters is made for writing aformula
[TABLE FOR
a,b,
and c are the sides Of
any triangle, respectively ,op
posite the angular pom ts A,
B,0 .
I f th e triangle is right—angled,
a z altitude,b base
,0 hy
poth enuse .
I n regard to a circle,c Circum
ference,( l diameter .
e spherical excess .
f : fiat angle .
9 number Of degrees in an an
I
I
I
I
I
I
I
I
H
H
H
gle or are.
9° means expressed in degrees
only , gt minutes
, g”
seconds .
h 2 h eigh t . inates of a point.i medial line.
REFERENCE ]
j projection.
is chord.
l length .
m meter .
a any number.
6 perigon.
p perimeter .
g square .
r radius .
s (a b c) .t tangent .
14 Circular measure.
I)
w
a discrete variable.
width .
CHAPTER I .
THE MEASUREMENT OF LINES :
(8 (A ) . STRA IGHT L INES .
An accessible straight line is practically measured by thedirect application of a standard length suitably divided .
If the straight line contain the standard unit 72 times,
then a is its num erical measure .
But,properly speaking, any description Of a length by
counting Of standard lengths is imperfect and merely ap
proximate . Few physical m easurements Of any kind areexact to more than six figures
,and that degree Of accuracy
is very seldom obtainable , even by the most delicate instruments . Thus
,in comparing a particular meter With the
standard bar,a difference of a thousandth Of a millim eter
can be detected .
In four measurements of a base line at Cape Comorin,it
is said the greatest error was inch in mile,or one
part in and this is called an almost incredibl edegree Of accuracy .
When we only desi re rough results , we may readilyshift the place Of the line to be measured
,SO as to avoid
natural Obstacles . Still,under the most favorable circum
stances,all actual measurem ents Of continuous quantity
are only approximately true . But such imperfections,w ith the devised methods of correction,
have reference tothe physical measurement Of things ; to the data , then,
which in book-questions we suppose accurately given.
THE MEASUREMENT OF L INES . 5
I LLUSTRATIVE PROBLEMS .
( a ) To measure a line the ends of which only are acces
sibte.
Suppose AB the line .
Choose a point C from whichA and B are both visible .
Measure AO,and prolong it
until CD 2 AC. MeasureB C
,and prolong it until OE
B O. Then ED z AB . E D
WW . 49 106 ; (Eu . I . 15 & 4 ; CV . I . 23
NOTE Ww . refers to Wentworth’
s Geometry ,third edition
,1879 .
Eu . refers to Todhunter’
s Euclid,new edition,
1879 . refers to
Chauvenet’
s Geometry . These parallel references are inserted in thetext for th e convenience Of students having eith er one Of these geom
etries at hand. References to preceding parts of th is Mensurationwill give simply the number Of the article.
(6) Tofind the distance between two objects, one ofwhichis inaccessible.
Let A and B be the two O bj ects, separated by someObstacle , as a river .
From A measure any
straight line AO. Fix
any point D in thedirection A B . Pro
duce AC to F,making
CF 2 AO; and produce DC to E
,making
C'E z z OD . Then findE“
F
the point G at which the directions of B 0 and FE intersect ; that is, find the point from which 0 and B appear in
6 MENSURATION .
one straight line,and E
’
and F appear in another straightline . Then the triangles A013 and OF F are congruent
,
and therefore AB 0 and CFC ; whence FG=AB .
WW . 106 107 ; (Eu . I . 4 26 ; CV . I . 76
Hence,we find the length Of AB by measuring FG.
(y) To measure a line when both ends of it are inaccessible.
At a point C,in the
accessible part Of AB,
erect a perpendicularCD
,and take DE : CD .
At E make FG perpendicular to B E. Find inFG the point F which
G falls in the line B D,
and the point G in the line AD . F G z AB .
WW . 107 ; (Eu . I . 26 ; CV . I .
(8) To measure a line wholly inaccessible.
If AB is the line,choose a convenient point 0 from which
A and B are both visible , and measure AOand B 0 by (B) ;then AB may be measured by (a ) .
— STRA IGHT L INES IN TRIANGLES .
I . RIGHT-ANGLED TR IANGLES .
1 . Given the base and perpendicular , to find the hypothCHUSC .
Rule : Square thesides,add together , and extract the
square root.
Formula a2 b2 2 : c
2.
Proof : WW . 331 ; (Eu. I . 47 ; CV . IV .
8 MENSURATION .
3. If the angle contained by the given sides be obtuse.
Rule : To the sum of the squares of the given sides
add twice the product of the projection and the side on
which (when prolonged ) it fa lls ; then extract the square
root
I‘
ormula aZ —l b
2+ 2 bj c
2
Proof : WW . 336 ; (Eu . I I . 12 ; CV . I I I .
c Ans.
4. If the angle contained by the given sides is acute.
Rule : F rom the sum of the squares of the gwen sides
subtra ct tw ice the product qf the projection and the side on
which it fa lls the square root of the remainder gives the
third side.
Formula a2 b2 2 bj 2 c
2.
Proof : l v. 335 ; (Eu . I I . 13 ; CV. I I I .
EXAM . 3 . Given thesides a zz 5
,b r : 6
,con
taining an Obtuse an
gle,and given j z 4
,
the proj ection Of a on
b ; find the third Side .
THE MEASUREMENT OF L INES . 9
EXAM . 4 . Given the sides a b 2 6,containing an
acute angle,and given j : 4
,the
proj ection of a on b find the thirdSide .
a2 52 25 36 61
C2 13
C A728 .
5. If two sides and the perpendicu lar let fall on one
from the end Of the other,are given
,the proj ection can
be found by 2,and then the third side by 3 or 4 .
If three sides are given,a proj ection can be found by 3
or 4, and then the perpendicular by 2 .
6. Given three sides Of a triangle to find its three me
dials ; i .e.,the distances from the vertices to the m idpoints
of the Opposite sides .
Rule : F rom the sum of the squares of any two sides
subtract twice the square of ha lf the base ; the square root
of ha lf the remainder is the corresponding m edia l .
Formula : a2+ c
2 2 2 i 2.
Proof : WW . 338 ; (Eu . Appen. 1 ; CV . I I I .
Corollary : Dividing the difference of the squares of twosides by twice the third side , gives the proj ection on it ofI ts medial .
2 b
EXAM . 5. Given two sides,a = 7
,c :
b 4 . Find the medial .
Here a2
02 49 81 130
b2 42, é— b
‘z 8
122
61
i 2“ Ans .
Formula j
10 MENSURAT I ON .
I I I . STRAI GHT L I NES I N SI M I LAR FI GURES .
7. Given two straight lines in one figure,and a line
corresponding to one Of them in a Similar figure,to find
the line corresponding to the other .
Rule : The like sides ofsim ila r figures are p ropor
tiona l .
Formula b2
30 m eters . Ans .
8. Given the height of an arc and the chord Of half thearc
,to find the diam eter Of the circle .
WW . 278 ; (Eu . VI .
,Def. I . ;
CV . I I I .
EXAM . 6 . The height ofan Uprigh t stick is 2 m eters
,
and it casts a shadow 3
neters’
long ; the shadow Of
a flag staff is 45 m eters .
Find the height Of the staff.
3 2 : 45 : b2 .
90
3
IV . CHORDS OF A CI RCLE.
Suppose AB any chord in acircle . Through the center 0a diam eter perpendicular to ABm eets it a t its m iddle point D ,
and bisects the are at ff . DH
is the height Of the arc,and
AH the chord Of half the arc .
THE MEASUREMENT OF L I NES . 11
Rule : D ivide the square of the chord of ha lf the are bythe height of the arc .
7512
Formula d —2
h
HAF is a right angle .
WW . 204 ; (Eu . I I I . 31 ; CV . I I .
Ww . 289 ; (Eu . VI . 8, Cor . ; CV . I I I .
EXAM. 7. The height Of an arc is 2 centim eters,the
chord of half the arc is 6 centim eters . Find the diameter .
218 . Ans .
9. Given the chord Of an arc and the radius Of th e
circle “
,to find the height of the are .
Rule : From the radius subtra ct the square root of the
dif erence of the squares of the radius and ha lf the chord.
Formula : h r V7?
zi‘ h2 .
I ID Z h,H0 : r
,and D 0 2 : A0
'2-AD2 —
: r2
EXAM . 8 . The chord Of an arc is 240 m illimeters,the
radius 125 m illim eters . Find the height of the are .
D 0 2 2 T2(i t )
?(r at ) (r at ) z (125 120) (125
h z r —D 0 5: 125 35 z : 90 millimeters9 centimeters . Ans .
12 MENSURAT ION .
10. Given the chord and height Of an arc,to find the
chord Of half the arc .
Rule : Take the square root of the sum of the squares qfthe height and ha lf the chord.
Formula h;2 21-16
2.
P roof AH 2 HD2 AD 2.
WW . 331 ; (Eu . I . 47 ; Cv . IV .
EXAM . 9 . Given the chord z : 48,the height Z 10 . Find
the chord Of half the are.
100 . (sic)2
7 576 .
h 26 . Ans .
If,instead of the height, the radius is given
,substitute in
10 for h its value in terms Of r and h from 9,and we have
F rom this follows
11 . Given the chord of an arc and the radius Of thecircle , to find the chord of half the arc .
EXAM . 10 . Calculate the length Of the side of a regulardodeeagon inscribed in a circle Whose radius is 1 meter ;that is
,find by when r and h are each 1 m eter long
,for h is
here the side Of a regular inscribed hexagon, which always
equals the radius.
WW . 391 ; (Eu. IV. 15,Cor . ; CV . V .
THE MEASUREMENT OF L INES . 13
Thus r and it being unity ,our formula becomes
7_2 ME v4. I 0 51763809 . Ans .
EXAM . 11 . With unit radius , find the length of one Sideof a regular inscribed polygon Of 24 sides .
And so on with regular polygons of 48 , 96, 192, etc .,sides .
12. Given the radius Of a circl e of a regularinscribed polygon,
to find the
side of the similar circumscribedpolygon .
Formula t
P roof : Suppose AB the given
side It . Draw th e tangent atthe m iddle point H of the arc
AB,and produce it both ways
to the points E and G,where it
meets the radii GA and CB produced ;required ,
t.
In the similar triangles GEE CAD,
hrt
CD
C'D2 GA2 AD2
72
B2.
EXAM . 12 . When r==1,find one side Of a regular Ci r
cumscribed dodeeagon.
14 MENSURAT ION .
With radius taken as unity ,t
Vet
From Exam . 10,13 3 05 1763809
,and substituting this
value,tn : Ans.
In the same way ,by substituting 7534 - 0 26105238 from
Exam . 11, we find {24 : 0 263305
,and from 7543 — 0 13080626
we get 1133 : 0 131087,and SO on for etc .
,Sides .
g (C ) . METHOD OF L IM I TS .
V . A variable I S a quantity which may have successively an indefinite number Of different values .
DEF I NI TI ON OF A LI M I T .
When a quantity can be made to vary in such a m annerthat it approaches as near as we please and continually nearer to a definite constant quantity ,
but cannot be conceived toreach theconstant by any continuation Of the process
,then
the constant is called the lim it Of the variable quantity .
Thus the lim it of a variable is the constant quantitywhich it indefinitely approaches, but never
-reaches,though
the difference between the variable and its limit may be
come and remain l ess than any assignable magnitude .
EXAM . 13. The lim it of the sum of the series,
is z
EXAM . 14. The variable may be likened to a convenientferry-boat
,which w ill bear us just as close as we choose to
the dock,
—the constant lim it , —but which cannot actua lly reach or touch it ; for , if they touch ,
both explode intothe unknown infinite .
The bridge,the m ethod for passing
,in the order Of our
knowledge,from variables to their lim its
,is the
16 MENSURATION .
Hence the supposition that Ab Ac AB AG,
’ or to anyquantity less than AG, is absurd .
Suppose,then
,in the second place
,thatAb Ac : AB AG
,
”
or to some term greater thanAG. Now,there is some line ,
as AB ; less than AB ,which is to AO as AB is to ACf’
If, then, we conceive this ratio to be substituted for thatof AB to AO,
”we have
which,by a process Of reasoning similar to the above
,may
be shown to be absurd . Hence,if the fourth term Of the
proportion can be neither greater nor less than AG, it mustbe equal to AO; or we must have
Ab zAc z zAB zAC. Q . E. D .
C’or . If two variables are always equal , their limits are
equaL
14. When their sides tend indefinitely toward zero,the
p erim eter of the polygon inscribed increases, circumscribed
decreases,toward the same limit
,the length of the curve.
P roof : Inscribe in a circl e any convenient polygon,say ,
the rega l“ hexagonWW . 391 ; (Eu . Iv. 15 ; CV . V .
J oin the extremities Of each side,as AB ,
tO the point ofthe curve equally distant from them ,
as E ; that is, thepoint of intersection of the arc and a perpendicular at them iddle point Of the chord . Thus we get Sides of a regulardodecagon. Repeat the process with the sides of the dedeeagon
,and we have a regular polygon Of twenty-four sides .
SO continuing , the number of sides, a lways doubling , willincrease indefinitely ,
While the length Of a Side will tend
toward zero .
THE MEASUREMENT OF L INES . 17
The length Of the inscribed perimeter augments with the
number Of sides, since we continually replace a side by two
which form with it a triangle , and so are together greater .
WW . 96 ; (Eu . I . 20 ; Cv.
Thus AB Of the hexagon is replaced by AH and
the dodecagon.
But this increasing perimeter can never becom e as longas the circumference , since it is always made up of chordseach of which is shorter than the corresponding are
,by
the axiom,
“A straight line is the shortest distance between two points .
"Therefore
,this perimeter increases
toward a lim it which cannot be longer than the circum
ference .
In doubling the number Of sides Of a circumscribed polygon,
by drawing tangents at the m iddle points Of the arcs ,we continually substitute a straight for a broken line ; as,
TU for TN N U. SO this perimeter decreases .
18 MENSURAT ION .
But two tangents from an external point cannot be together shorter than the included are .
E.g .
,PI T TB are I IB .
Th erefore this perim eter decreases toward a limit whichcannot be shorter than the circumfer ence .
But the lim it toward which the circumscribed perimeter decreases is identical with that toward which the corre
spending inscribed perimeter increases .
For,in a regular circumscribed polygon Of any number
of sides,n
,the perimeter is n times one Of the sides.
pn ntn .
THE MEASUREMENT OF L INES . 19
But,from 12
,
But the perimeter of the corresponding inscribedpolygon,
is2 r
the ratio Of the perimeters .
Cutting the circumference into n equal parts makes eachpart as small as we please by taking n sufficiently great .
But chords are shorter than their arcs ; therefore hntends toward the lim it zero as n increases .
2 7a1 . The vari
ables and are always equal . Therefore , byPu
13,Con
,their limits are equal
,and lim it Of p 1 .
lim . p ,,2 lim .
2?
But we have Shown that lim . cannot be longer than c,
and lim . p ,, cannot be shorter than c . Therefore,the com
mon lim it is c,the length Of the curve .
THE RECTIF ICATION OF THE C IRCLE.
15. I n a circle whose radius is unity ,to find the length
of the sem icircumference.
From 14,an approximate Value Of the sem icircumference
in this circle is given by the sem iperimeter Of every poly
gon inscr ibed or circumscribed,the latter being in excess,
and the former in defect of the true value .
In examples 10,11
,and 12
,we have already calculated
20 MENSURAT ION .
the length Of a side in the regular inscribed and circum
scribed polygons of and 48 sides . Continuing the
sam e process , and in each case multiply ing the length Of
one side by half the number Of sides, we get the following
table of semiperimeters :
3-1410319
Since the semicircumference , t o, is'
always longer than
t nhn and shorter than i nt", therefore its va lue , correct toseven places of decimals
,is 3 1415926 .
16. The circumferences of any two circles are to each
other in the same ra tio as their radii .
P roof : The perimeters of any two regular polygons ofthe same number Of sides have the sam e ratio as the radii
of their circumscribed circles .
WW . 374 ; (Eu . XI I . 1 , V . 12 ,Cv. V.
The inscribed regular polygons remaining similar toeach other when the number Of Sides is doubled , their
THE MEASUREMENT OF L INES . 21
perimeters continue to have the same ratio . Hence,by 13,
the limits,the circumferences
,have the same ratio as their
radii .
Cor . 1 . Circumferences are to each other as their diame
ters .
Cor . 2 . Since
That is,the ratio Of any circumference to its diameter is
a constant quantity .
This constant,identical with the ratio Of any semicir
cumferenee to its radius, is denoted by the Greek letter 7r .
But,in circle with radius 1
,semicircumference we
have found Therefore,the constant ratio
7r
VI . L INES I N ANY CI RCLE.
1
17. j,Z 3 1415926+T
Multiplying both sides Of this equation by d,gives
181C (577
The diameter Of a circl e being given,to find the
eumference .
Rule : Multip ly the diam eter by r .
NOTE. I n practice, for 7: the approximation or 27-2 is generally
found sufficiently close. A much more accurate value is easilyremembered by Observing that the denominator and numerator
w ritten consecutively , thus, 11 3 I 3 55 , present th e first th ree Odd
numbers each written twice . The value most used is 3 1416 .
19. Dividing 18 by 7: gives
1d 2r c X
3,c X
CHAPTER II .
THE MEASUREMENT OF ANGLES .
THE NATURAL UN IT OF ANGLE.
To say all right angles are equal,assumes th at the
amount of turning necessary to take a straight line or direction all around into its first position is the sam e for all
points .
Thus the natura l unit Of reference for angular magnitude is one whole revolution
,
called a p erigon,and equal to four
right angles .
VI I . A revolving radius describes equally an angle
,a sur
face,and a curve . Moreover
,the
perigon,circl e
,and circumference
are each built up of congruentparts ; and any pair Of angles or
sectors have the same ratio as the corresponding arcs .
WW . 201 ; (Eu . VI . 33 ; Cv . I I .
an an le its interce ted arcTherefore
,
y.
g pperi gon Ci rcumference
That is,if we adopt the whole circumference as the unit
Of are ;
20 . The num erica l measure of an ang le a t the center of a
circle is the sam e as the numerica l m easure of its intercepted
arc .
24 MENSURATION .
Of every circle by an are equal in length to its radius,and
hence named a radian,then ,
by 20 ,
21. The number which expresses any angle in radians
a lso expresses its intercepted are in terms of the radius .
SO,in terms Of whatever arbitrary unit of length the are
and radius may be expressed , if u denote the number Of
radians in an angle,then
,for every X ,
lu :
7'
Thus the same angle will be denoted by the same number
,Whatever be the unit Of l ength employ ed .
u,or the fraction are divided by radius
,is called the
circular measure Of an X .
EXAM . 16 . Find the circular measure of f.
Here,the are being a semicircumference, its l ength l : r7r .
7'
7rI A728 .
r
This is Obviously correct,since dividing a flat X by 1:
first gave us our radian.
22. Given the number Of degrees in an angle,to find
the l ength of the arc intercepted by it from a given cir
cumference .
Rule : Multip ly the length of the circumference by the
number of degrees in the ang le, and divide the product by
360 .
O
lcgFormula360
°
P roof : From VI I . we have g°
z c l .7
NOTE. I f theXbe given inminutes, the formula becomes l :cg
21600’
THE MEASUREMENT OF ANGLES . 25
EXAM . 17. HOW long is the are Of one degree in a cir
cumference Of miles ?Ans .
23. Given the length Of an arc of a given circumference,
to find the number of degrees it subtends at the center .
Rule : Mu ltip ly the length of the are by 360, and divide
the p roduct by the length of the circumference.
l 360°
Formula g°
c
NOTE. TO find the number of minutes or seconds7 77
Formulae : g'
” 1
300
and g”
l 129
i000
EXAM . 18 . Find the number of degrees subtended in
any circl e by an are equal to the radius .
Here g° becomes
7
552I
f?Ans.
Hence a radian p 57°
IX . The arcs used throughout as corresponding to theangles are those intercepted from circles whose center isthe angular vertex .
These arcs are said to measure the angles at the center
which include them ,because these arcs contain their ra
dins as Often as the including angle contains the radian .
Using m easured in this sense , wt: may state the following
Theorems :
24. An angle at the center is measured by the are inter
cepted between Its Sides .
WW . 202 ; (Cv. 11 ,
26 MENSURATION .
25. An inscribed angle is measured by . half its intercepted 8m
WW . 203 ; (Eu . I I I . 20 ; CV .
26. An angle formed by a tangent and a chord is measured by half the intercepted arc .
WVW . 209 ; (Eu . I I I . 32 ; CV . I I .
27. An angle formed by two chords,intersecting within
a circle,is measured by half the sum Of the arcs vertically
mtercepted WW . 208 ; (Eu . Appen. 2 ; CV . 11 .
28. If two secants,two tangents, or a tangent and a se
cant intersect without the circle , the angle formed is meas
ured by half the difference Of the intercepted arcs .
WW . 210 ; (Eu . Appen . 3 ; CV . I I .
29. Given the m easure of an angle in degrees, to find its
circular measure .
Rule : Mu ltip ly the number qf degrees by r,and divide
by 180 .
779°
r e'
779
180°
10800 ' 648000”
P roof : A flat X is 180? and its circular measure is 7 .
Hence
Formula u
since each fraction expré sses the ratio of any given X to aflat X . Therefore
,
and also
THE MEASUREMENT OF ANGLES . 27
This recalls to m ind again that the circu lar measure ofany X is independent of the length of the radius of the circle.
EXAM . 19 . Find the circular measure of X of one
degree .
Here1g0 Ans .
EXAM . 20 . Find the circular measure Of X of one
minute .
Dividing the last answer by 60 gives
Ans.
Of course,this number equally expresses the length of
an arc Of one m inute in parts Of the radius,and in the
same way we Obtain
Arc 1” r X
30. Given the Circular measure of an angle , to find its
m easure in degrees.
Rule : Multip ly the circular measure by 180, and divide
by r .
Formula : g° “180
EXAM . 21 . Find the number Of degrees in X whose cir
cular m easure is 10 .
180Here g
°: 10 X 10 X
7’
Ans.
31 . Given the angle in degrees and the length of the
are which subtends it,to find the radius .
Rule D ivide 180 times the length by W times the number
of degrees .
Formula r
28 MENSURAT I ON .
P roof :1g;
EXAM . 22 . An arc Of 6 meters subtends X of 10?findradius .
Here180
X1
6
006 X 572 957795+
34 03775 m eters. Ans .
X . One X is called the comp lement Of another,when
th eir sum equals a rt . X ; the supp lement, when their sum
A
f ; the exp lement,when their sum o,
REFERENCE TABLE OF ABBREV IAT IONS FOR AREAS .
When used alone,as abbreviations
,capita l letters denote
the area Of the figures ; to denote volume , a V is prefixed .
A annulus. T trapezoid.
B base. A triangle.
(j cylinder . U volume Of quader .
O circle .V : volume.
D volume of prismatoid. W: volume Of wedge .
E ellipse. X volume of tetrahedron.
F frustum . Y :
pyramid.
G 2 segment . Z zone .
H : sphere.
I volume Of an irregular poly ETU 2 TI Jr TQ 73 T”
hedron.
A'
sum of angles .
J :
parabola . A Spherical A .
K : cone . IV : spherical JV.
L lune . Y volume of spherical Y.
M midsection . 52 steregon .
N :
polygon of 7b sides . means equivalent ; i .e .,equal
0 2 solid ring . in size .
P prism . H parallel .
D :
parallelogram . _L perpendicular .
Q quadrilateral . similar .
q square . therefore.
R rectangle. X angle.
S sector .
CHAPTER III .
TH E MEA S U R EM EN T OF P L ANE A R EA S .
PLANE RECTI L INEAR F IGURES .
XI . The area Of a surface is its numerica l m easure.
Measuring the area Of a surface, whether plane or
curved,is determining its ra tio to a chosen surface called
the u nit Of area .
The chosen unit Of area is a square whose side is a unitOf length .
EXAM . 28 . If the unit Of length be a meter,the
unit Of area will be called a square m eter ( gm
) .
If the unit of length be a centimeter,the unit
of area will be a square centimeter (qcm) .
32. To find the area Of a rectangle .
Rule Mu ltip ly the base by the a ltitude.
Formula R r at) .
P roof : SPEC IAL CASE. When the base and altitude,or
length and breadth of the rectangle are comm ensurable.
In this case there is always a line which w ill divide bothbase and altitude exactly .
If this line be assumed as linear unit,a and b are inte
gral numbers .
80 MENSURATION .
In the rectangle AB OD divide AD into a ,and AB into
6 equal parts . Through the points of division draw linesparallel to the sides of the rectangle . These lines divide therectangle into a number of
of area . In the bottom row
there are 6 such squares ; and,
since there are a rows, we have
6 squares repeated a times,
which gives,in all
,at) squares .
NOTE. The composition of ratios includes numerical multiplication as a particular case. Ordinary multlplication is a growth fromaddition . The multiplier indicates the number of additions or repetitions . The multiplicand indicates the thing added or repeated .
Therefore,it is not a mutual Operation,
and the product is always
in terms of the unit of the multiplicand. The multiplicand may beany aggregate ; the multiplier is an aggregate of repetitions . To
repeat a thing does not change it in kind,so the result is an aggre
gate of the same sort exactly as the multiplicand. When the rule
says , Multiply the base by the altitude,it means
,Multiply the nu
merica l measure of the base by the number measuring the altitude in
terms of the same linear unit . The product is a number wh ich wehave shown to be the area of th e rectangle ; that is, its numericalmeasure in terms of the superficial unit.Th is is the meaning to be assigned whenever we speak of the
product of one line by another .
GENERAL PROOF .
Rectangles,being equiangular parallelograms
,have to
one another the ratio which is compounded Of the ratiosof their sides . WW . 315 ; (Eu . VI . 23 ; Cv. IV .
Let R and R ' represent the surfaces or areas of two rec
tangles . Let a and a' represent their altitudes ; b and 6 ’
their bases .
32 MENSURAT ION .
33. To find the area of a square .
Rule : Take the second power of the number denoting thelength of its side.
Formula : gz 62.
P roof : A square is a rectangle having its length and
breadth equal .
NOTE. This is the reason Why the product of a number into itselfis called the square of that number .
Cor . Given the area of a square , to find the length of aside .
Rule Extract the square root of thenumber denoting thearea .
EXAM . 1 square dekameter,usually called an
Ar G) , contains 100 square m eters . Everyunit of surface is equivalent to 100 of thenext lower denomination,
because everyunit of length is 10 of the next lowerorder . Thus a square hektometer is ahektar (
ha) .
EXAM . 26 . How many square centim eters in 10 squaremillimeters
100 square millimeters is 1 square centimeter .
10 square m illimeters is T10of 1 square centimeter . Ans .
Or,10 square millimeters make a rectangle 1 centimeter
long and 1 millimeter wide .
EXAM . 27. How many square centimeters in 10 mill imeters s uare ?q One. Ans .
THE MEASUREMENT OF PLANE AREAS . 88
REMARK . D istinguish carefu lly between square meters
and m eters square.
We say 10 square kilometers m eaning a surface
which would contain 10 others,each a square kilometer ;
while the expression 5 kilometers square means a square
whose sides are each 5 kilometers long , so that the figure
contains 25 square kilometers .
EXAM . 28 . The area of a square is 1000 square meters .
Fmd “S Slde '
vi ooo 81-628 meters . Ans .
34. To find the area of any parallelogram .
Rule J lfu ltiply the base by the a ltitude.
Formula D 3 ab.
P roof : Any parallelogram is equivalent to a rectangl eof the same base and altitude .
WW . 321 (Eu . I . 85 ; Cv. IV .
Cor . The area of a parall elogram,divided by the base ,
gives the altitude ; and the area , divided by the altitude ,gives the base .
EXAM . 29 . Find the area of a parallelogram whose base
is 1 kilometer , and altitude 1 centimeter .
b 1000 meters . a Ta,meter .
ab 10 square meters . Ans .
35. Given one side and the perpendicular upon it from
the opposite vertex ,to find the area of a triangle .
Rule : T ahe ha lf the product of the base into the a ltitude.
Formula A ab.
84 MENSURATION .
P roof : A triangle is equivalent to half a parallelogramhaving the same base and altitude .
WW . 324 ; (Eu . I . 41 ; Cv. IV .
Cor . 1 . If twice the number expressing the area of a triangle be divided by the number expressing the base , thequotient is the altitude and vice versa .
Cor . 2 . Two A’
s or D’
s,having an equal X ,
are as theproducts of the sides containing it .
EXAM . 30 . One side Of a triangle is meters, and
the perpendicular on it is 63 m eters . Find the area .
5b 2 17 87 meters .
% ab : 113 ' 581 square meters . Ans .
36. Given the three sides of a triangle , to find the area .
Rule : From ha lf the sum of the three sides subtra ct each
side separa tely ; multip ly the ha lf sum and the three re~
m ainders together : the square root of theproduct w ill be the
area .
Formula : A : \/s (s — a ) ( s - b) (s
a‘z
z b2 eg
~ 2 bj,b2 — c
2 — a
2 b
THE MEASUREMENT OF PLANE AREAS . 85
whence,
4 b21’
t2 4 6202 (b
?C2
+ c —e) (a — b + c) .
But,by 35, 2 bh equals four tim es the area of the triangle .
Cor . To find the area of an equilateral triangle , multiplythe square of a side by
EXAM . 31 . Find the area of an isosceles triangle Whose
base is 60 meters and each of the equal sides 50 meters.
Here,from last formula in Proof,
2 bh c=2) 6) (2 a b) 00 x/ 160 40 : so\/ 16OO 4.
2 6h : 60 x 40 x 2.
Area 2 1200 square m eters . Ans .
37. To find the radius of the circle inscribed in a tri~
angle .
Rule : Divide the area Of the triang le by ha lf the sum
of its sides .
Formula r
86 MENSURAT ION .
area of B OO
area of 0 0A
area of AOB
by addition,A sr .
Cor . The area of any circumscribed polygon is half theproduct of its perim eter by the radius of the inscribedcircle .
EXAM . 32 . Find radius of circle inscribed in the triangle whose Sides are 7
,15
,20 .
Here
° A : 3 x 7 x 2 : 42
r z 2 . Ans .
38. To find the radius Of the circle circumscribing atriangle .
Rule : D ivide the product of the three sides byfour times
the area of the tr i angle.
abcFormula SR
4A'
P roof : In any triangle the rectangle of two sides isequivalent to the rectangle of the diameter of the circumscribed circle by the perpendicular to the base from thevertex . WW , 300 ; (Eu. VI . O. ; Cv. I I I .
ac 2 9th .
THE MEASUREMENT OF PLANE AREAS . 37
Cor . The side of an equilateral A,b 2 ERV3 2 r V3.
EXAM . 33 . Find radius of circl e circumscribing triangle7 I 5
,20 .
Here abc 2100,A 42.
210012—1
1682 Am
39. To find the radius of an escribed circle .
Rule : D ivide the area of the triang le by the difference
between the sum of its sides and the tangent side.
P roof : Let r l denote the radius of the escribed circlewhich touches the side a . The quadrilateral OIPAO may
be divided into the two triangles, OIAB and 0 1A0
by 35 , c bi ts area —
2r1 2
73
.
But the sam e quadrilateral is composed of the trianglesand A3 0 ;
its area r1A.
38 MENSURAT ION .
c b a-
2
—r
2r1=
é—r
lA .
c b
)
_,
a71
2 A
In the same way ,
T2
ACor . 1 . Since
,by 37, r therefore ,
A4
s (s — a ) (s — b) (s - e)T T
’
I’)
EXAM . 34 . Find r 1 , r2, r3 , when a z : 7,b 15
,e z 20 .
42
17:
71
3
6
XII . A trapezoid 1S a quadrilateral with two Sides
parallel .
Cor . A triangle is a trapezoid one of whose parallel sideshas become a point .
40 . To find the area of a trapezoid .
Rule : Multip ly the sum of the para llel sides by ha lftheir distance apar t.
T : gig/1
“ l" 92Formula
2
40 MENSURAT ION .
The coordina tes of any point are its abscissa a; and its
ordina te y .
XIV . If to any convenient axis ordinates be droppedfrom the angular points of any polygon,
the polygon is
exhibited as an algebraic sum of trapezoids, each havingone side perpendicular to the two parallel sides, and hencecalled. right trapezoids .
If triangles occur,as 1D 2
,GE 5
,they considered
trapezoids,
and y5 being zero .
41 . To find the sum of any series Of right trapezoids .
Rule : Mu ltip ly the distance Qf EACH interm edia te ordi
na te from the first by the difi’
erence between its two adja centordinates
,a lways subtra cting the one following from the one
p receding in order a long the brohen line. A lso mu ltip ly
distance of last ordina te from first by the sum of last two
ordina tes . Ha lve the sum of these products .
— fv1) (yiv 1
T Q M (ye - 1 (yn’ l' yn-l-Dl '
P roof : With 0 as origin,the area of the first trapezoid ,
by 40. is (932 551)W12 and of the second is at.)
I
;yi’
THE MEASUREMENT OF PLANE AREAS . 41
Adding the two,we have
Tl
Tz 2 [can 53
1) (3h yz) m2) (fl/2
Performing the indicated multiplications,xgyz is cancelled
byfl xgyg, and
T1 + T2
= % — x1y1
— a yz + ww2 + s a— w
z?/slTI + T
=ZKr . T.) + 61
“
(AH/an
Since here my , is balanced by a lt/3 .
Thus we have proved our rule for a pair of trapezoids.
Taking three , we get,by 40 ,
(Pc.
2
71+ T24 T.= at [Ctr (ar y.) (five: it
.) (y2+all i [(w. w.) (v.+
AS before,replacing the balancing terms agy3
— x3y3 by
{tag/4 x1y4, this becomes
TiTe
“ l“ Zia : 7} s 531) l/4ll°
This proves the rul e for threetrapezoids ; and a generalizationof this process proves that if therul e is true of a series of n trape~
zoids,it is true of n+ 1 .
For,by 40
,the area of the
(n 1) th trapezoid
Tn+ 1 (fin wn+ 1) yn '“gym”
;
and,adding this to the first n trapezoids, as given by
formula,therefore
v=n+l
E(xn+ 1 x1) (yn 3/n+ 1)l "t ELIOT“
“ t 3/n + 2)l
42 MENSURAT ION .
Replacing by the balancing terms
xl yn+2, this becom es
v=n+1
T”2 TNT }. $
1) (y1 ye) "l'
(9377 4 1 931) (yn 3/n+ 2) (wn+ 2 331) (Z/n+ 1 yn+ 2ll
The sam e method proves that if the formula applies to ntrapezoids
,it must apply to n 1 . Therefore
,the rule is
true for any and every series whatsoever of right trape~
zoide.
EXAM . 36 . Find the right portion Of a railroad crosssection whose surface linebreaks twice , at the points
(x2, y2) , (x3, y3) , to the rightOf center line ; ( origin beingon grade in midpoint of road~
bed) .
This asks us to find thesum of right trapezoids cor
responding to the five points (0 , c) , (5172, yg) , (x3, y3) , (b r ; r ) ,
(b,by formula ,
0 x2
2 fl, ( 0 y3) 933 (yz r ) (b r’
) y3 brl. Ans .
In the same way ,the portion to the left of center line ,
Whether without breaks,or with any number of breaks ,
is given by our formula,Which thus enables us at once to
calculate all railroad cross-sections,Whether regular or
irregular .
EXAM . 37. To find th e area of a triangle in terms of thecoordinates Of its angular points .
THE MEASUREMENT OF PLANE AREAS . 48
Here are three trapezoids,and consequently ,
four points ;but A is both 1 and 4
,so
and the“
formula becomes
A z
v
glfl Z Ts 33
1) (yl ya) (333 m1) (ye yl ll'
A 5“ $291‘ i‘ 37392
‘ i‘ mgr/1 aLo/3] , A773
Notice the symmetry of this answer .
For practical computation this is written
2A wl (y3 ya) Cr
2 (y1 Jr fi
sh/
72 yd,
2A 3/1 (5U2 T3) "17 y2 (£t
'
3ai
l ) T (T1 " 7
To insure accuracy ,reckon the area by each .
42. To find the area of any polygon .
Rule : Ta lce ha lf the sum of the products of the abscissa
of ea ch ver tex by the difi’
erence between the ordinates of the
two adja cent vertices ; a lways m aking the subtra ction in
the same direction around the polygon .
Formula for a polygon of n sides :
N :
21— [331 (yn T wz Q/ i “
yel T xs (y2_3h) ” l a /n—l 191”
P roof : This is only that Special case of 41,where the
broken line,being the perimeter of a polygon,
ends whereit began .
44 MENSURAT ION .
J oin the vertex 1 of the polygon 1 , 2, 3 ,n
,1, with
the origin 0 . Then the area enclosed by the perim eter isthe same
,whether we consider it as starting and stopping
at 1 or at 0 . But,under the latter supposition,
though
xot/o x 1
have n+ 3 points, the coordinates (x0, yo) of the first and
last are zero,and the second (xl , y1) is identical with
point next to last ; so that formula 41 becomes
HMO w.(y. r.) a
te. y.) y.)
NOTE. No mention need be made of minus trapezoids, since th erule automatically gives to those formed by the broken line wh ile
going forward,the opposite sign to those formed Wh ile going back
ward.
Our expression for the area of any rectilinear figure isthe difference between a set Of positive and an equal number of negative terms. If this expression is negative whenthe angular points are taken in the order followed by thehands of a watch ,
then it is necessarily positive when theyare taken in the contrary sense
,for this changes the order
in every pair Of ordinates in the formula .
Observe that each term is of the form xy ,and that there
is a pair of these terms, with the m inus Sign between them ,
for each vertex Of the figure . Thus,for the vertex (xmym) ,
THE MEASUREMENT OF PLANE AREAS . 45
we have the pair or,pairing those ’ terms
which have the sam e pair of suffixes,for every vertex m ,
we have Hence,for twice the area
write down the pair xy— xy for each vertex,
and addsymmetrically the suffixes
,
n,1 1
,n .
Thus,for every quadrilateral ,
2 Q aw2511
271 T may?)7 " w
aye 31
374 $41] 3 m
4yr a°
ly4 .
But,if any point of perimeter be to the left of origin,
orif
,to shorten the ordinates, the axis
be drawn across the figure,then
one or more of the coordinates willbe essentially negative . Thus
,if in
a quadrilateral,we take for axis a
diagonal,then
al= 0
, yl , y3= O
,
and 2 Q = (cg/2 xay4 .
Here, y4 being essentially negative ,
the two terms have the same sign,and give the ordinary
rul e
43. TO find the area of any quadrilateral .
Rule : Mu ltip ly ha lf the diagona l by the sum of the per
pendicu lars upon itfrom the opposite ang les .
EXAM . 38 . The two diagonals of a quadrilateral m easure 1-492 and 375 3 meters respectively ,
and are _L to one
another . Find the area .
B 43
Q 14 92 375 3 55 99476F8 84
9 2
27 99738 square meters. Ans .
46 MENSURAT ION .
EXAM . 39 . Find the area of the polygon 1234567891 ,
the coordinates of whose angular points are (0 ,
(30 , ( 110 , (80 , (84, ( 130 , (90,
(40 , (35,By 42,
area 2 : 9375. Ans .
REMARK. The result of any calculation by coordinatesmay be verified by a simple change of origin. If the ori
gin is moved to the right through a unit of distance , thenthe numerical values of all positive abscissae will b e dim inished by one
,and all negative abscissae increased by one .
Thus,to verify our last answer
,move the origin thirty
units to the right,and the question becomes
EXAM . 40 . To calculate the area of polygon whose coordinates are
H
- 50 and 5 ><
area 2 9375,as before .
EXAM . 41 . From the data of Exam . 40 construct thefigure .
Choose a convenient axis and origin,noticing that the
By 42,twice the area equals
sum of30 70 2100
50 4000
50 2500
54 x 50 2700
60 z 6000
BOX 40 2400
19700
950
18750
48 MENSURATION .
44. Given the area and one side of a figure,and the cor
responding side of a sim ilar figure , to find its area .
Rule : J lfulttp lg the green a rea hg the sguarecl ra tio ofthe sides .
A a2
Formula : AI2 1
P roof : The areas of similar figures are to one another as
the squares of their like sides .
343 ; (Eu . VI . 20 ; 0v . IV .
Cor . The entire surfaces of two similar solids are proportional to the squares of any two homologous lines .
EXAM . 42 . The side of a triangle containing 480 squaremeters is 8 meters long .
Find area of a similar triangle whose homologous sideis 40
A 480 X 25 12000 square meters . Ans .
XV . Magnitudes which can be made to coincide arecongruent.
Magnitudes which agre e in size,but not in shape
,are
equiva lent.
THE MEASUREMENT or RLANE AREAS . 49
XVI . A regular polygon is both equilateral and equi~
angular .
The bisectors of any two angles of a regular polygonintersect in a point equidistant from al l the angular pointsof the polygon,
and hence alsoequidistant from all the sides
,
and at once the center of an
inscribed and a circumscribedcircle .
J oining this center to everyangle of the polygon cuts it upinto congruent isosceles triangles .
Hence the area of the regularpolygon is the area of any one
of these triangles multiplied by the number of sides of the
polygon .
45. To find th e area of a regular polygon.
Rule J lfultip lg together one
side,the perpendicu la r from the
center, anel ha lf the number of
Or,in other words
T ahe ha lf the p roeluet of per ini eter 6g apothem .
Formula N
EXAM . 43 . The Side of a regular hexagon is 98 centimeters
,and its apothem centimeters ; find area .
Area 2 3 X 98 X 848 7
249517 8 square centimeters . Ans.
50 MENSURAT ION .
46. By the aid of a table of polygons, to find the area ofany regular polygon.
Rule Mu ltip ly the square of one of the sides of thepolygon by the area of a similar polygon whose side is unity .
Formula IV, l,,
2XVI .
P roof : This follows from 44,all regular polygons of the
same number of sides being similar .
TAB LE or REGULAR POLYGONS .
Area in Terms ofSquare on S ide.
Name.
EXAM . 44 . The Side of a regular hexagon is 98 cent-imeters ; find its area .
Arear : 98 X 98 X 2 5980762
249517 8 square centim eters . Ans .
EXAM . 45 . If the side of a regular decagon is 0 6 meters,
its area is
0 -6 x 0 6 X 76 942088 2 76991524 square meters . Ans .
THE MEASUREMENT OF PLANE AREAS . 51
AREA S OF PLANE CURV IL INEAR F IGURES .
47. To find the area of a circle .
Rule Mu ltip ly its squared radius by
Formula 6 73W .
P roof : If a regular polygon be circumscribed aboutcircle
,its area
,by 45,
is
N : % 7’
pn
and,by 14, as n increases , p ,,
decreases toward 0 as l imit,
and N toward (9 . But the variables N and are always
in the constant ratio % r ; therefore , by 13, their limits arein the sam e ratio
,and we have
EXAM . 46 . Find the area of a circle whose diameter is75 m eters .
Here T2
2 14 0625 .
square meters . Ans.
48. To find the area of a sector .
the length of the
are by ha lf the radius .
Formula : 8 2fl
.lr z :
“
2116?
W . 382 ; (cv . v.
52 MENSURATI ON .
follows
EXAM. 47. Find the area of a sector whose are 99 58
meters long , and radius 863 4 m eters .
995 8 X 43-17 2 4298-8686 square m eters . Ans .
EXAM. 48 . Find the area of a sector whose radius is 28
centimeters , and which contains an angle of 50° 36f
Here,by 29 ,
u 08 83
r2
z: 282 z : 784
3532
7064
6181
8 : 3461 36 square centimeters . Ans.
49. To find the area of
is the difference betweenangle AB C.
a segment less than a semicircle .
Rule : F rom the sector hav
ing the sam e are a s the seg
m ent,
subtra ct the tr iang le
form ed by the chord and the
two radii from its extr em ities .
Formula
0h2( l h) if h
2
( l h‘
)T
4 h
P roof : The segm ent AHBthe sector AHB O and th e tri
THE MEASUREMENT or PLANE AREAS . 53
AB C : é—AB X OD :1 h (r — h) .
G = S ~ A z % lr — % k(r — h) .
2G= lr — hr + hh .
HD DF AD?
w . 307 ; (Eu . VI . 13 ; UV . I I I .
h (2r h) i kh
i—kM—h?
2h
Substituting this value of r in the expression for 2G,we
obtain
s a— k)
7} h2
( l h) h2( l h)4 h
71‘ 762
(b h) h‘zl h2/c 2h2h
2 h
Cor . The area of a segment of a circle is equal to halfthe product of its radius and the excess of its are over halfthe chord of double that arc . For
sector AHB O é lr ,A AB C : 15,- r B L .
segment AHB 371i B L ) .
Approximate Rule for Segment T ahe two- thirds theprod
uct of its chord and height.
Approximate Formula G 2 hh.
EXAM . 49 . If the chord of a segment is r X 09 59851
and its height is r X 0 0 12241743 then an approximation toits area is
473 0 959851+ 0 0 122417+ z 347
12 0 r2X
54 MENSURAT I ON .
But if,also
,we can measure the arc
,and here find it
equal to radius,then
G4 r ><
Gz i -
1—TI
-239919++T1 -30216+
Gz i rz
(0
G 2 ,3 Ans .
Proceeding directly by Rule 49 , instead of Formula 49,
we h ere getS = %fi,
A so (r r 0 1224174 9A r x 0 4799255+ r x 0 877583
A = 2X 0
-421174
G S A r2>< 0
-07883 .
Since,in this example
,arc z r
, G is,in any
segment whose hf is p.
50. A circu lar zone is that part of a circle included between two parallel chords
,and may be found by taking
the segment on the shorter chord out of that on the longer .
51 . A crescent is the figure included between the cor
responding arcs of two intersecting circles,and is the dif
fer ence between two segments having a common chord,
and on the same side of it .
52. To find the area of an annulus ; that is, the figureincluded between two concentric circumferences .
Rule Mu ltip ly the sum of the two radii by their difif
er
ence,and the product by 77 .
Formula A 2 (71 “ l (71 w
56 MENSURAT I ON .
sectorial area AB ED is the difference between the sector AB C and thesector ODE .
by 48 ,
S . A. 33 (r h) l1 % r l211,— ( ldl l
lr l
zr ) .
Now,since ll and Z2 are arcs
subtending the sam e angle at O,
by IX ,
1,
z,
r h r
blz
l,r l
zr .
Substituting,
S . A. we,
z 3Mt,
Cor . By comparison with 40 ,we see an annular sector
equivalent to a trapezoid whose parallel sides equal thearcs
,and are at the sam e distance from one another .
EXAM . 51 . The upper arc of a circular arch is 35 25
meters ; the lower , 24 75 meters ; the distance betweenthe two is 3 5 meters . How many square m eters are therein the face of the archHere
S . A .2 17 5 X 60 105 square meters . Ans .
XVI I . Comes .
If a straight line and a point be given in position in aplane
,and if a point move in the plane in such a m anner
that its distance from the given point always bears thesame ratio to its distance from the given line
,the curve
traced out by the moving point is called a conic .
The fixed point is called the focus , and the fixed linethe directrix of the conic .
THE MEASUREMENT OF PLANE AREAS . 57
When the ratio is one of equality ,the curve is called a
parabola .
54. To find the area of a parabolic segment ; that is, thearea between any chord of a parabola and the part of thecurve intercepted .
Rule : Tahe two-thirds the product of the chord by the
height of the segment.
Formula J : 3?hh .
P roof : A parabolic segm ent is two-thirds of the triangle
made by the chord and the tangents at its extrem ities .
If AB,AO
,be two tangents to a parabola
,to prove that
the area between the curve and the chord B 0 is two-thirdsof th e triangle AB 0 .
Parallel to B 0 draw a tangent P P B . J oin A to thepoint of contact P
,and produce AP to cut the chord B O
at JV.
By a property of the parabola deducible from its defini
tion,
AP PN.
Ww . 276 279 ; (Eu . VI . 2 (St 4 ; Cv. I I I . 15
by 85, A B PO= 2 -ADE.
This leaves for consideration the two small trianglesPDB
,PEG
, each made by a chord and two tangent-s .
With each proceed exactly as w ith the original triangle '
e. g .
,draw the tangent F Q G parallel to PB j oin D Q ,
and
produce it to AI ; then
D Q Q M.
PB = 2 -FG.
A P Q B = 2 -FDG.
58 MENSURATroN .
This leaves four little tangential triangles, like P F Q .
In each of these draw a tangent parallel to the chord , etc .
,
and let this process be continued indefinitely .
Then the sum of the triangles taken away within theparabola is double the of the triangles cut off without
it . But the of the interior triangles approaches,
its limit,the parabolic segment . For the triangle B P O
,
since it is half of AB C,is greater than half the parabolic
area B Q P C, and so successively w ith the smaller inter iortriangles . Therefore
,the difference between the parabolic
segment and the sum of these tr iangles can be made lessthan any assignable quantity .
WW . 198 ; (Eu . XI I,Lemma ; CV . V .
Ther efore,the constant segm ent is
,by definition V .
,the
lim it of this variabl e sum .
THE MEASUREMENT or PLANE AREAS . 59
Again,each outer triangle cut off is greater than half
the area between the curve and the two tangents ; e. g .
,
ADE,being half the quadrilateral AB P C
,is more than
half the area AB Q P O. Therefore,the limit of the sum of
the outer triangles is the area between the curve and the twotangentsAB
,AO. But these two variable sums are always
to each other in the constant ratio of 2 to 1 . Therefore,
by 13,their lim its are to each other in the sam e ratio
,
and the parabolic segment is two-thirds its tangentialtriangle .
But the altitude of this triangle is twice the height ofthe segment .
A hlc,
J : 3M.
55. To find the area of an ellipse .
Rule M i ltip ly the product of the semi-axes by 77 .
Formula : E abr .
P roof : Let ADA ’D ’ be a circle of which AO,CD are
radii at right angles to one another .
In 01) let any point B be taken ; then,if this point
move so as to cut off from all ordinates of the circl e thesame part that B O is of D C
,the curve traced is called an
ellipse .
In one quadrant of the circl e take a series of equidistantordinates
,as Q 1P IM , Q 2P 2AI L) , Q 3P 3M
'
s, etc . Draw P I R I ,
Q I B 1 , etc .
, parallel to the axis AAf Then,by 32 ,
area area R ’lrlf
lFIJ I
,B 0 AC;
and each corresponding pair being in this constant ratio,
the sum of the rectangles R 211 is to the sum of E'M as
B 0 AU. But the sum of B flf differs from one-quarter
60 MENSURAT ION .
of the ellipse by l ess thanthe area BM ,which can be
made less than any assignable quantity by taking 01111, thecommon distance between the ordinates
,sufficiently small .
Hence,A'B Ois the limit of the sum of the rectangles RM ;
and,in the same way ,
the quadrant of the circle is the lim itof the sum of E'flf . Therefore
,by 13,
E B O b
(Db a‘lff
Cor . The area of any segment of an ellipse,cut off by a
line parallel to the m inor axis,will be to the corresponding
segm ent of the circle upon the maj or axis in the ratio ofb to a .
EXAM . 52 . Find the area of an ellipse whose maj or axisis 616 m eters
,and minor axis 44 4 m eters .
E 308 X 22 2 X 3 14159
2148-094 square m eters . Ans .
CHAPTER IV .
THE MEASUREMENT on THE AREAS or B ROKEN AND
CURVED SUREAcEs .
XVIII . A polyhedron is a solid bounded by polygons .
A polyhedron bounded by four polygons is called a tetra hedron ; by six
,a hexahedron ; by eight , an octahedron ;
by twelve , a dodecahedron ; by twenty ,an icosahedron.
The faces of a polyhedron are the bounding polygons .
If the faces are all congruent and regular,the polyhedron
is regu lar .
The edges of a polyhedron are the lines in which itsfaces meet .
The summits of a polyhedron are the points in which itsedges m eet .
A section of a polyhedron is a polygon formed by theintersection of a plane w ith three or more faces .
A convex figure is such that a straight line cannot meetits boundary in more than two points .
56. The number offaces and summ its in any polyhedron
taken together exceeds by two the number of its edges .
Formula :
P roof : Let 6 be any edge j oining the summits and
the faces AB,and let 6 vanish by the approach of ,
8 to a .
If A and B are neither of them triangles,they both re
62 MENSURAT I ON .
main,though reduced in rank and no longer collateral
,
and the figure has lost one edge e and one summ it,8 .
If B is a triangle and A no triangle,B vanishes with 6
into an edge through a,but A remains . The figure has
lost two edges of B ,one face B
,and one summ it ,8 . If B
and A are both triangles, B and A both vanish with e,
five edges forming those triangles are reduced to two
through a ; and the figure has lost three edges, two faces,and the summ it
,8 .
In any one of these cases,whether one edge and one
summit vanish ,or two edges disappear with a face and a
summ it,or three edges with a summit and two faces, the
truth or falsehood of the equation
s + e = e + 2
remains unaltered .
By.
causing all the edges which do not m eet any face tovanish
,we reduce the figure to a pyram id upon that face .
Now ,the relation is true of the pyram id ; therefore it is
true of the undiminished polyhedron.
i a) . PR I SM AND CYL INDER .
XIX . A prism is a polyhedron two of whose faces arecongruent
,parallel polygons, and the other faces are paral
lelograms .
64 MENSURATroN .
By 46 ,the area of the pentagonal base is
9 X 17 204774 15 4842966 .
Doubling this for the base and top together,and
the lateral area of the prism, which , by 56 , is
12 14 z 168,
the total surface
168 309 685932 2 198 9686 square meters . Ans .
XX . A cy lindr ic surface is generated by a straight lineso moving that every two of its positions are parallel .The generatrix in any position is called an element of
the surface .
A cy linder is a solid bounded by a cylindric surface andtwo parallel planes .
The axis of a cylinder is the straight line j oining thecenters of its bases .
A trunca ted cylinder is the portion between the baseand a non-parallel section .
58. To find the curved surface or mantel of a right cironlar cylinder .
Rule Mu ltip ly its length by the cir
cumference of its base.
Formula 0 cl 2 7rr l .
F irst P roof : Imagine the curvedsurface slit along an elem ent and thenspread out flat . It thus becomes a
rectangle having for one side the circumference and forthe adj acent Side the length of the cylinder .
Inscribe in the right cylinder a rightprism having a regular polygon as its base . B isect the
AREAS or B RoKEN AND CURVED SUR FACES . 65
arcs subtended by the sides of this polygon ,thus
scribe a regular polygon of double th e
number of sides,and construct on it
,as
base,an inscribed prism .
Proceeding in this way continually to
double the number of its sides,the base of
the inscribed prism,by 14,
approaches thebase of the cylinder as its lim it
,and the
prism itself approaches the cylinder as its
lim i t . But,by 56 ,
P :
and always the variable P bears to the variable p theconstant ratio l . Therefore
,by 13 ,
their lim its are in thesame ratio
,and
0 z d .
Cor . 1 . The curved surface of a truncated circular cylinder is the product of the circum ference of thecylinder by the intercepted axis . For
,by sym
m etry ,substituting an oblique for a right section
through the same point of the axis alters neitherthe curved surface nor the volum e
,since the
solid between the two sections will be the same above andbelow the right section .
Cor . 2 . The curved surface of any cylinder on any curveequals the length of the cylinder multiplied by the perim ete r of a right section .
EXAM . 54 . Find the mantel of a right cylinder whosediam eter is 18 m eters and length 30 m eters .
0 : 30 X 18 X 3 14159 2 : 1696 4586 square m eters . Ans .
66 MENSURAT ION .
b (J ) . PYRAM ID AND CONE.
XXI . A regu lar pyram id is contained by congruentisosceles triangles whose bases form a regular polygon .
A conica l surface is generated by a straight line movingso as always to pass through a fixed point called the vertex .
A cone is a solid bounded by a conical surface and a
plane .
The frustum of a pyram id or cone is the portion in
cluded between its base and a cutting plane parallel tothe base .
EXAM . 55 . Find the lateral area of a r egular heptagonalpyram id whose slant height is 13 56224 m eters , and basaledges each I }; m eters .
One quarter of 13 56224 is 3 39056 . Adding these and
dividing their sum by 2 gives 8 4764 for the area of one
triangular face . The lateral area is 7 times this, or
59 3348 square m eters . Ans .
59. To find the area of the lateralsurface or mantel of a regular pyramid .
Rule : Mu ltip ly the p erim eter ofthe base by ha lf the slant height.
Formula : YZ t hp .
P roof : The altitude of each of theequal isosceles triangles is the slantheight of the pyram id , and the sum
perim eter of its base .
AREAS OF B ROKEN AND CURVED SUR FACES . 67
60 . To find the area of the curved surface or mantel ofa right circular cone .
Rule : Mu ltip ly the circum
ference of its base by ha lf the
slant height.
Formula I f 2 é—eh 7rrh .
First P roof : The distancefrom the vertex of a cone of
revolution to each point on the circumference of its base is the slant height ofthe cone . Therefore
,if the surface of the
cone be slit along a slant height and
spread out fiat,it becomes the sector of a
c ircle, w ith the Slant height as radius and
the circumference of cone’
s base as are.
by 48 ,its area is % eh .
Second P roof : About the base of the cone circumscribea regular polygon,
and j oin its vertices and points of con
tact to the vertex of the cone .
Thus is circumscribed aboutthe cone a regular pyram idwhose slant height equals theslant height of the cone .
By draw ing tangents,
cir
cum scribe a regular polygon ofdouble the number of sides
,
and construct on it,as before
,
a circumscribed regular pyram id . Thus proceeding continually to double the numberof sides
,the base of the circumscribed pyram id , by 14,
ap
proaches the base of the cone as its lim it , and the pyramiditself approaches the cone as its limit .
68 MENSURATION .
But,by 58 ,
Y : 3769 .
and always the variable Y has to the variable p the con
stant ratio fi h . by 13,their limits are in the same
rat1o,and
K é- ch .
Cor . 1 . In the Proof of 47 we find
0 7 —91
the slant height of a right circular cone has the sameratio to the radius of the base that the curved surface hasto the base
,or
K s z h z r .
Cor . 2. Calling A the sector angle of the cone , we have
A: 360 z r : h .
EXAM . 56 . Given the two sides of a right-angled triangle . Find the area of the surface described when the
triangle revolves about its hypothenuse .
Calling a and b the given altitude and base
,and x the length
of the perpendicular"from the
right angle to the hypothenuse , by59
,the area described by a is 7rxa
,and described by b is
7rxb. Thus the whole surface of revolution is vr (a b) x .
But
Eu . I . 47 & VI . 8 .
( 32
61 . To find the lateral surface or mantel of the frustumOf a regular pyramid .
AREAS OF BROKEN AND CURVED SURFACES . 69
Rule : Af ultip ly the slant height of the frustum by ha lfthe sum of the perim eters of its bases .
Formula : F % Zi (p1
P roof : The base and top being similar regular polygons,the inclined faces are congruent trapezoids, the height ofeach being the Slant height of the frustum . If n be thenumber of faces
,by 40 ,
n n
area of lateral surface F % h (p1
area of each face % h —1
EXAM . 57. Find the lateral area of a regular pentagonalfrustum whose slant height is 110 382 meters
,each side of
its base being 23m eters, and of its top meters .
The sum of a pair of parallel sides is 1
73
110 382 X 13 143 4966 .
143 4966 6 23 9161,
the area of one trapezoidal face . The lateral area is fivetimes this
,or
119 5805 square meters . Ans.
62. To find th e curved surface or mantel of the frustumof a right circular cone .
70 MENSURAT ION .
Rule : Mu ltip ly the slant height of the frustum by ha lfthe sum of the ci rcumferences of its bases .
Formula z 1 h ( cl 02) r: 7rh (r 1 rz) .
P roof : Completing the cone and
slitting it along a slant height,the
curved surface of the frustum developsinto the difference of two sim ilar sec
tors having a common angle,the arcs
of the sectors being the circumferencesof the bases of the frustum . By 53
,
th e area of th is annular sector F 1 h (c1
EXAM . 58 . Find the mantel area of the frustum of aright cone whose basal diameter is 18 m eters ; top diam eter
,9 m eters ; and slant height , 171-0592 meters .
3 1416 X 9 282 744 circumference of top.
Twice top circumference 56 5488 circumference of base.
Half their sum is 42 4116,and this multiplied by the
slant height 1710 592,gives
,for the curved surface
,
7254-89 square meters . Ans .
63. To find the curved surface of a frustum of a cone ofrevolution.
Rule : Mu ltiply the projection of the frustum’
s slant
height on the axis by tw ice 7: times a perpendicular erected
a t the midpoint of this slant height and terminated by the
Formula : F 2 7raj .
P roof : By 62,the curved surface of the frustum whose
slant height is P B and axis MO is
72 MENSURAT ION .
Now ,if the whole figure revolve round B D as axis
,the
sem icircl e will generate a sphere,while each side of the
inscribed polygon,as PR
, will generate the curved surface of the frustum of a cone . By 63, this
F z 2 7rllfN >< OQ ;
and the sum of all the frustums thatis
,the surface of the solid generated
by the revolving semi-polygon,equals
2 71-0 62 into the sum of the proj ections .
PETE, 2 7r 0 Q >< BD r 2Tra 2r = 4ar7r .
As we double n the number ofsides of the inscribed polygon,
by 14,
its semiperimeter approaches thesem icircumference as limit
,and its solid of 1 evolution ap
proaches the sphere as limit,while OQ or a ,
its apothem,
approaches r the radius of the sphere as lim it . But thevariable sum bears to the variable a the constant ratio 4 r7r .
Therefore,by 13, their lim its have the same ratio, and
H : 4 r27r .
Cor . 1 . A sphere equals four times a circle with sameradius .
Ger . 2 . A sphere equals the curved surface of its circumscribing cyl inder .
EXAM . 59 . Considering the earth as a sphere whoseradius is 63 709 X 108 centimeters
,find its area .
H X 1016
H 4 X 40 58836681 31 4159265 1016 .
H 2 square centimeters . Ans .
Cr,about 510 million square kilometers.
AREAS OF BROKEN AND CURVED SUR FACES . 78
XXIII . A spherica l segm ent is the portion of a globecut off by a plane , or included between two parallelplanes .
A zone is the curved surface of a Spherical segment .
The Proof of 64 gives also the following rul e for the area
of a zone :
65. To find the area of a zone .
Rule : bfultip ly the a ltitude of the segment by twice 77
tim es the radius of the sphere.
Formula Z 2 7rra .
Cor . 1 . Any zone is to the sphere as the altitude of itssegm ent is to the diam eter of the sphere .
Cor . 2 . Let the arc B P generate a calotor zone of a single base . By 65,
its area
WW . 289 ; (Eu . VI . 8,Cor . ; Cv. I I I .
Hence,a calot or zone of one base is
equivalent to a circle whose radius is thechord of the generating arc .
EXAM . 60 . Find the area of a zone of one base,the di
ameter of this base being 60 meters,and the height of the
segm ent 18 m eters .
Using Cor . 2, the square of the chord of the generating
arc isz 1224
which , multiplied by gives,for the area of the calot
,
square meters . Ans.
74 MENSURATI ON .
66.THEORE M OF PAPPUs .
If a p lane curve lies wholly on one side of a line in its
own p lane, and revolving about that line as axis genera tes
thereby a surfa ce of r evolution,the area of the surf ace is
egua l to the product of the length of the revolving line into
the pa th described by its center of mass .
Scholium . The demonstration given under the next rul e,
though fixing the attention on a single representative case,
applies equally to all cases where the generatrix is a closedfigure
,has an axis of symmetry parallel to the axis of revo
lution,and so turns as to be always in a plane with the
axis of revolution,while its points describe circles perpen
dicular to both axes .
EXAM . 61 . Use the Theorem of Pappus to find the dis
tance of th e center of mass of a semicircum ference from
the center of the circl e,by reference to our formula for the
surface of a sphere .
By 64, H 4 r27r.
By 66 , I I : M"X 2cm.
Equating, we get 2 r = 2 7.Ans .
67. To find the area of the surface of a solid ring .
Rule Mu ltip ly the genera ting circumference by the pa th
of its center .
Formula 0 z 4 71-2 r 1 r 2 .
P roof : Conceive any plane to revofve about any straightline in it . Any circle w ithin the plane , but without theaxis
, will generate a solid ring .
AREAS OF BROKEN AND CURVED SURFACES .
Draw the diam eter B OD parallel to the axis AO. Di
vide the sem icircumference B PD into n equal arcs , and
call their equal chords each h . From the points of divis
ion drop perpendiculars to the axis, thus dividing the
other sem icircumference B Q D into n corresponding parts .
Let B P,B Q ,
be a pair of arcs . If we draw their chords,
we have a pair of right-angled trapezoids, AB PN and
AB Q JV, which ,during the revolution
,describe frustums
whose curved surfaces,by 62,
are
F1z 7th (BA PN ) ,
Fa
: nh (BA Q N ) .
J II L GF is the m edial line , then,by Proof to 40 ,
F1= 2 7rk >< FM
F2= 2 7Th >< LM
F,F,warn/ 4 LM ) 2 7 um; 4 GM + GM GL ) .
But the diameter B CD is an axis of symm etry ,and
Gill : CO= r
76 MENSURATION .
the radius Of the path of 0 .
F1 +F2 2 7 76
This is the expression for each pair ; and,as we have n
pairs,therefore
,the whole surface generated by a symmet
rical polygon of 2n sides equals
2nlc .27rr2 p 2 7rr
since 2nh is p the whole perim eter .
But,as we increase 2n the number of sides of the in
scribed polygon,by 14 , p approaches c as its limit
,and the
sum of frustral surfaces approaches the surface of the ringas limit . But the variable sum bears to the variable pe
rimeter the constant ratio 2 7rr2 . Therefore,by 13
,their
lim its have the same ratio,and
O : c 2 a'
r2z 2 7rr
12 7rr
where r l is the radius of the generating circle,and r2 the
radius of the path of its center .
EXAM . 62 . Find the surface of a solid ring,of which the
thickness is 3 meters,and the inner diameter 8 meters .
Here r l is 13 m eters , and r2 is meters .
4 427-
1 74
22 81 416 x 8 8-1416 11
9 4248 X 34 5576
325 698 square meters . Ans .
EXAM . 63 . Find the area of the surface of a square ringdescribed by a square m eter revolving round an axisparallel to one of its sides
,and 3 meters distant .
Here the length of the generating perim eter is 4 meters .
The path of its center is 77r,since r2 is 33 m eters .
0 r : 28 7: 879 648 square meters . Ans .
AREAS OF BROKEN AND CURVED SUR FACES . 77
EXAM . 64. A circle of 13 5 m eters radius,with an in
scribed hexagon,revolves about an axis 62 5 meters from
its center and parallel to a side of the hexagon. Find thedifference in area of the generated surfaces .
H eerr1
13 5 and r2
62 5 .
Therefore,area of circular ring is
asx 2-7 x 12 5 9 3693 33 75 333 099 .
For the hexagonal ring
IVW . 391 ; (Eu . IV .
the length of the generating perim eter is
6 X 1 35 8 1 .
The path of its center is
n'
X 12 5 392 7
Therefore,its area is
392 7 X 8 1 318-087.
Thus the difference in area is15 012 square meters. Ans .
SPHER ICS AND SOL ID ANGLES .
XXIV . A great circle is a section of a globe made bya plane passing through the center .
A lune is that portion of a Spherecomprised between two great sem icircles .
The ang le of two curves passingthrough the same point is the angleformed by the two tangents to thecurves at that point .
A spherica l ang le is the angle included between two
arcs of great circles .
78 MENSURAT ION .
Ap lane angle is the amount of divergence between twostraight lines which m eet in a point .A solid ang le is the amount of spread between two or
more planes which meet at a point .
Two polyhedral angles, having alltheir parts congruent
,but arranged
in reverse order,are symm etrica l .
A steregon ,the natural unit of solid
angle,is the whole amount of solid
angle round about a point in space .
As a perigon corresponds to acircle and its circum ference
,so a
steregon corresponds to a globe and
its sphere .
The steregon is divided into 360 equal parts,called
spherical degrees of angle,and these divide the whole
sphere into 360 equal parts,each called a degree of spheri
cal surface .
A steradian is .the angl e subtended at the center by thatpart of every sphere equal to the square of its radius .
68. To find the area of a lune .
Rule Mu ltip ly its angle in radians by twice its squared
radius .
Formula L 2 : 2 r2u .
P roof : Let PA Q B P ,PB Q OP be two lunes having
equal angles at P ; then one of these lunes may be placedon the oth er so as to coincide exactly with it : thus luneshaving egua l ang les are congruent. Then
,by the process
of Eu . VI . 33 ; (Ww . 766 ; CV . VIII . it follows that alune is to the sphere as its ang le is to a perigon ;
L u
4 r27r 2 a; L 2 rgu.
80 MENSURAT I ON .
From any property of polyhedral angles we may inferan analogous property of spherical polygons . Reciprocally ,
from any property of spherical polygons we may infer acorresponding property of polyhedral angl es .
XXVI . A spherica l pyram id is a portion of a globe
bounded by a spherical polygon and the planes of the Sidesof the polygon.
The center of the sphere is the vertex of th e pyramid ;the spherical polygon is its base.
69. J ust a s plane angles at the center of a circle areproportional to their intercepted arcs
,and also sectors ;
so solid ang les a t the center of a sphere are proportiona l to
their intercepted spherica l polygons , and a lso spherica l
pyram ids .
XXVII . The spherica l excess of a spherical triangle isthe excess of the sum of its angles over a flat angle . Thespherical excess of a spherical polygon is the excess of thesum of its angles above as many flat angles as it has sides
less two.
70 . To find the area of a spherical triangle .
Rule : Mu ltip ly its spherica l excess in radians by its
sguared radius .
Formula A er2
.
P roof : Let AB O be a spheri cal triangle . Produce thearcs which form its sides until they m eet again,
two and
two . The A AB O now forms a part of three lunes,name
ly ,AB DOA
,B OFAB
,and OAFB O.
S ince the A’
s ODE and FAB subtend vertical solid
AREAS OF BROKEN AND CURVED SUR FACES . 81
angles at 0,they are equivalent , by Therefore
,the
lune OAFB O equals the sum of the triangles AB Oand ODE . Thus the luneswhose angles are A
,B
,and O
,
are together equal to a hemisphere plus twice A AB O.
Subtracting the hem isphere,
which equals a lune whoseangle is a flat angle
, we have
2A AB O :
lune whose 4 is (A B O— f ) .
A lune whose AZ
by 68 , A er2
.
Oor . 1 . A A contains half as many degrees of sphericalsurface as its e contains degrees .
Oor . 2 . A A m easures as many steradians as its e containsradians .
Oor . 3 . Every X of a A is é- e.
EXAM . 66 . Find the area of a tri—rectangular A .
Here6 rt 4 Z
‘
A z—é— rrr ;
or,a tri—rectangular triangle is one—eighth of its sphere .
By Cor . 1,a tri-rectangular A contains forty -five de
grees of spherical surface .
71 . To find the area of a spherical polygon .
Rule : M i ltip ly its spherica l excess in radians by its
squared radius .
Formula : 17 2 [ é (n 2) wjrz
82 MENSURAT I ON .
P roof : From any angular point divide the polygon into
(n 2) A’
s . by 70 ,
A
This expression is true even when the polygon has reentrant angles
,provided it can be divided into A ’
s with each
X less than f .
Oor . 1 . On the sam e or equal spheres,n-gons Of equal
angle-sum are equivalent ; or ,
Oor . 2 . To construct a dihedral solid X equal to any
polyhedral X ; that is, to transform into a lune any sphe
rical polygon ; add its angles, subtract (n 2)f, and halvethe remainder .
EXAM . 67. Find the ratio of the vertical solid anglesof two right cones of altitude a l and a2, but having the
same slant height h .
These solid angles are as the corresponding calots on thesphere of radius h .
Therefore,from 65
,the required ratio is
2 7rh (h a,) h a
the ratio of the calot -altitudes . For the equilateral and
right—angled cones this becom es
CHAPTER V
THE MEASUREMENT OF VOLUMES .
3(M) . PR ISM AND CYL INDER .
XXVIII . Two polyhedrons are symmetr ica l whose facesare respectively congruent
,and whose polyhedral angles
are respectively symmetrical ; e. g .
,a polyhe
dron is symmetrical to its image in a m irror .
A quader is a parallelepiped whose six facesare rectangles .
A cube is a quader whosesix faces are squares.
XXIX . The volume of asolid is its ratio to an assumed
unit .
The unit for measurement of volume is a cube whoseedge is the unit of length .
Thus,if the linear unit be a meter
,the unit of volume
,
contained by three square meters at right angles to eachother
,is called a cubic m eter (
Chm) .
XXX . Using length of a line to mean its numericalvalue
,l engths
,areas
,and volumes
,are all three quantities
of the same kind,namely ra tios . All ratios
,whether ex
pressible as numbers or not,combine according to the sam e
simple laws as ordinary numbers and fractions .
WW . Bk . I I I . ; ( Eu . Bk ‘ V ‘ ; CV ' Bk . I L ) .
THE MEASUREMENT OF VOLUMES . 85
Therefore, we may multiply lengths, areas, and volum es
together prom iscuously ,or divide one by the other in any
order .
If we ever speak of multiply ing a line , a surface,or a
solid, we mean always the length of the line , the area of
the surface,or the volum e Of the solid .
72. To find the volum e Of a quader .
Rule : Afu ltip ly together the length ,breadth
,and height
of the guader .
Or,in other Words
,
Mu ltip ly together the lengths of three adja cent
Formula U r : abl .
P roof : By 32,th e number of square units in the base of
a quader is the product of two adj acent edges , bl .If on each of these square units we place a unit cube
,
for every unit of altitude we have a layer of bl cubic
units ; so that , if the altitude is a,the quader contains abl
cubic units .
Oor . 1 . The volume of any cube is the third power ofthe length of an edge ; and this is Why the third pow erof a number is called its cube .
Oor . 2 . Every unit of volum e is equivalent to a thousandof the next lower order .
Oor . 3 . The arithm etical or algebraic extraction of cuberoot makes familiar the use of the equation
(a b)3
a —1 3 u‘zb 3 a b2 b".
86 MENSURAT ION .
I ts geometric meaning and proof follow from inspectionof the figure of a cube on the edge a + b, cut by threeplanes into eight quaders.
The cube as of the longer rod a,taken out
,had faces a2
in common w ith three quaders of altitude b had edges ain common with three quaders of base b2, and one cornerthe corner point of the smaller cube b3 .
XXXI . MASS,DENSI TY
,WEIGHT .
The unit of capa city is a cubic decimeter called theliter
The quantity of matter in a body is termed its m ass .
The unit of mass is called a gram (g) . Pure water at tem
perature of maximum density is gram per cubiccentim eter (
ccm) . So
,in phy sics , the centimeter is chosen
as the unit of length ,because of the advantage of making
the unit of mass practically identical with the mass ofunit-volum e of water in other words
,of making the value
of the density of wa ter practically equal to unity ; density
being defined as m ass per unit—volume.
88 MENSURAT ION .
EXAM . 69 . How many cubic centim eters in one
hektoliter (m) ?
Since 1 liter 1000 cubic centimeters,
1 hektoliter cubic centimeters . Ans .
EXAM . 70 . If the density of iron is 77 88 ,find the mass
of a rectangular iron beam 7 m eters long,25 centim eters
broad,and 55 m illimeters high .
The volum e of the beam in cubic centimeters is
700 X 25 5 5 cub ic centimeters .
Therefore,its mass is
X 1 : grams . Ans .
74 . To find the volum e of any parallelepiped .
Rule : J lf ultip ly its a ltitude by the area of its base.
Formula : V . P abl .
P roof : Any parallelepiped is equivalent to a quader ofequal base and altitude . For
,supposing AB an Oblique
parallelepiped on an Oblique base,prolong the four edges
parallel to AB and cut them normally by two parallelplanes whose distance apart
,OD
,is equal to AB . This
gives us the parallelepiped ODE, which is still oblique,
but on a rectangular base . Prolong the four edges parallelto DE
,and cut them normally by two planes whose dis
tance apart FO is equal to DE . This gives us the qua~
Now,th e solids AO and B DE are congruent , having all
their angles and edges respectively equal . Subtractingeach in turn from the whole solid ADE leaves ODE
eq uivalent to AB .
THE MEASUREMENT OF VOLUMES . 89
Again,the solids ODE and E9 are congruent . Taking
from each the common part EF l eaves ODE equivalent
to FG. Therefore,the parallel epiped AB is equivalent
to the quader FO of equal base and altitude .
EXAM . 71 . The square of the altitude of a parallelepipedI s to the area of its base as 121 to 63
,and it contains
cubic centimeters . Find its altitude .
HereaB and 63 a2 121 B .
63 3 3 121
53
4 63 2
a r : 154 cubic centimeters . Ans .
75. TO find the volume of any prism .
Rule : Al u ltip ly the a ltitude of the prism by the area ofits ba se.
Formula : V . P aB .
MENSURAT I ON .
P roof : For a three-sided prism this rul e follows from 74,
since any three—sided prism is half a parallelepiped of thesame altitude
,the base of the prism being half the base of
this parallelepiped . TO showthis
,let AB OaB
‘
y be any
three-sided prism . Extending the planes of its bases,and through the edges Aa
,
Oy ,drawing planes parallel to
the sides,B y ,
AB,we have the
parallelepiped AB OD aByS,whose base AB OD is doubl ethe base AB O of the prism .
WW . 123 ; (Eu . I . 34 ; CV . I .
Also,this parallelepiped it
self is twice the prism . For,
its two halves,the prisms
,are congruent if its sides are all
rectangles . If not,the prisms are symmetrical and equi
valent . For,draw planes perpendicular to A0. at the
points A and a . Then the pr ism AB OaBy i s equivalentto the right prism AEMamu,
because the pyram idAEB OM is congruent to the pyramid In thesame way ,
ADOaSy equals ALMaAu.
But AEManpt and ALflfaAu are congruent . Therefore,
ABOaBy and ADOaSy are equivalent, and the parallelepiped AB OD aByS is double the prism AB Oafly .
Thus,the rule is proved true for triangular
prisms,and consequently for all prisms ; since
,
by passing planes through any one lateral edge,
and all the other lateral edges,excepting the
two adj acent, we can divide any prism into a
number of triangular prisms of the sam e altitude
,whose triangular bases together make the given
polygonal base .
92 MENSURATION .
In 58,Second Proof
,we saw the cylinder to be
of an inscribed prism when the number of sidesof the prism is increased indefinitely ,
and
the breadth of each side indefinitely dim inished
,the base of the cylinder being conse
quently the lim it Of the base of the prism .
But,by 75 ,
alw ays V . P is to B in the constant ratio a ; hence , by 13 , their limits willbe to one another in the same ratio ; and
Scholium . This applies to.
all solids whosecross—section does not vary ,
whatever be the shape of thecross—section .
Oor . 1 . B etween any two parallel planes, the volum e ofany cyl inder equals the product of its axis by thecross-section normal to it .
Oor . 2 . By 58, Cor . 1,the volume of any trun~
cated circular cylinder equals the product of itsax is by the circl e normal to it .
EXAM . 73 . A gram of mercury ,density 13 6, fills a cylin
der 12 centim eters long ; find the diam eter of the cylinder .
Volume of cylinder cubic centimeters .
Area of base of cy linder 2 7r 0 73529 1"-00612741% square centimeters .
Therefore T2 0 06127414 31 416 0 01950 .
Thusr
°044 centim eters Z 4 4 millimeters,
and d 2 2 r 8 8 m illimeters . Ans .
77. To find the volum e of a cylindric shell .
Rule : Afu ltip ly the sum of the inner and outer radii bytheir difierence
,and this product by 77 tim es the a ltitude of
the shell ,
THE MEASUREMENT OF VOLUMES . 98
Formula : V . C1 V . C2 a 7r (r 1 rz) (rl r2) .
P roof : Since a cylindric shell is the difference between
two circular cylinders of the sam e altitude,its volum e
e ualsqar1
27r a ri a
"an (r? r
2
2
) .
EXAM . 74 . The thickness of the lead in a pipe weighingkilograms is 6 millim eters
,the diam eter of the open
ing is 4 8 centimeters ; taking 7r 2 272,and density 11, find
the length of the pipe .
Herer2
2 4 centimeters and r,
3 centimeters.
11 ln‘
(ri (r1 T2)
11 l -272—54 X 6
1 43 l 3°24
717- l 784’ 08
7840 8 l.
l : 784-08 840 centim eters8 4 meters .
PYRAMID AND CONE.
XXXII . The altitude of a pyram id is the normal distance from its vertex to the plane of its base .
78. Parall el plane sections of a pyramid are similar figures
,and are to each other as the squares of their distances
from the vertex .
P roof : The figures are sim ilar,since their angles are
reSPt VelY equal , WW . 462 ; (Eu . XI . 10 ; UV . VI .
and their sides proportional .
WW . 279 ; (Eu . VI . 4 ; CV . I I I .
94 MENSURAT I ON .
By 44, they are to each other as the squares of homologous sides
,and hence as the squares
of the normals from the vertex .
WW . 469 ; (Eu . xi . 17 ; ov. VI .
Scholium . This is the reason Whythe strength of gravity ,
light,heat ,
magnetism,electricity ,
and sound,de
creases as the square Of the distancefrom the source .
Image part of the beams from aluminous point as a pyramid of light . If a cutting planeis moved away parallel to itself, the number of units ofarea illum inated increases as the square of the distance .
But the number of rays remains unchanged . Therefore,
the number of beams striking a unit of area must decreaseas the square of the distance .
79. Tetrahedra ( triangular pyramids) having equiva
lent bases and equal altitudes are equivalent .
P roof : Divide the equal altitudes a into n equal parts,and through each point of division pass a plane parallel toth e base . By 78 ,
all the sections in the first tetrahedronare triangles equivalent to the corresponding sections inthe second .
B eginning with the base of the first tetrahedron,con
a,struct on each secti on as lower base a pri sm high Wi th
n
lateral edges parallel to one of the edges of the tetrahedron.
In the second,similarly construct prisms on each section
as upper base .
Since the first prism- sum is greater than the first tetrah edron
,and the second prism-sum l ess than the second
96 MENSURAT I ON .
P roof : Any triangular prism ,as AB O—EDE
,can be
divided into three tetrahedra,two (B —DEF and
D —AB O) having the sam e altitude as the prism,and its
top and bottom respectively as bases,while the third
(B ODE ) is seen to have an altitude and base equal toeach of the others in turn by resting the prism first on itsside OE and next on its side AF . Hence
,by 79 ,
thesethree tetrahedra are equivalent
,and therefore
,by 75, the
volume of each is AaB .
The rule thus proved for triangular pyram ids is true forall pyram ids, since , by passing planes throughany one lateral edge
,and all the other lateral
edges excepting the two adj acent to this one,
we can exhibit any pyram id as a sum of tetrahedra having the same altitude Whose basestogether make the given polygonal base .
EXAM . 75 . If the altitude of the highest Egyptian pyram id is 138 meters
,and a side of its square base 228 meters,
find its volume .
Here V . Y r"
ri 138
46 X
cubic m eters . Ans .
81 . To find th e volume of any cone .
THE MEASUREMENT OF VOLUMES . 97
Rule : Mu ltip ly one- third its a ltitude by the area of itsbase.
Formula. when Base is a Circle : V . K 3 a ;
P rom”: In 60
,Second Proof
, we saw thatcone was the limit of the base of the circumscribed Or inscribed pyram id , and thereforethe cone itself the lim it of the pyram id.
But,by 80 ,
always the variable pyramid isto its variable base in the constant ratio
3a .
Therefore,by 13 ,
their limits are to one
another in the sam e ratio and
V . K §~aB .
Scholium . This applies to all solids determined by an
elastic line stretching from a fixed point to a point describing any closed plane figure .
Oor . The volum e of the solid generated by th e revolu
tion of any triangle about one of its sides as axis is one
third the product of the triangle ’
s area into the circumference described by its Vertex .
V : % 7rrA .
EXAM . 76 . Find the Volume of a conical solid whose altitude is 15 meters and base a parabolic segment 3 metershigh from a chord 11 m eters long .
By 54, hereV . K z
l 11
5 x 2 x 11
110 cubic meters . Ans .
98 MENSURAT ION .
EXAM . 77. Required the volume of an elliptic cone,
the maj or axis of its base being 152 meters ; the minoraxis
,10 m eters ; and the altitude
,22 meters .
By 55 ,here
V . K Z —zg
- 7°6 7T 5
38 3 -73
7 2783
875-45+ cubic meters . Ans .
triangle , eachof the cone .
Herex/ i 22 62 x/ i os.
é arQn 2 : fix/ 108 7 36
391-78 cubic m eters . Ans .
3 PR I SMATOID .
XXXIII . If,in each of two parallel planes is construct
ed a polygon,in the one an m -
gon,e. g .
,AB OD ; in the
other,an n-
gon,e. g . ,
A 'B ’O' then, through each side of
one and each vertex of th e other polygon a plane may be
passed .
Thus,starting from the Side AB ,
we get the n planes
AB A: AB B f AB O'; again ,
W ith the side B O, the n
planes B OA'
,B OB ,
’
B OO,
’etc . Using thus all on sides
of the polygon AB OD ,we get mn planes . Also , combin
ingeach Of the n sides A ’B !B
'OfiO’D '
with each of the
m points A ,B
,O
,D
,gives nm planes A 'B 'A ,
A ’B 'B .
EXAM . 78 . The section of aright circular cone by a planethrough its vertex
, perpendicu
lar to the base is an equilateralwhich is 12 meters find the volume
100 MENSURATION .
From the prismatoids thus pertaining to the sam e two
bases, XXXIII . chooses the greatest .
XXXV .The a ltitude of a prismatoid is the normal dis
tance between the planes of its bases . Passing throughthe middle point of the altitude a plane parallel to thebases gives the midcross—section. I ts vertices halve thelateral edges of the prismatoid . Hence , its perimeter is
half the sum Of the basal perimeters . But, if one base
reduces to a straight line , this line must be considered a
digon,i . e. ,
counted twice .
XXXVI . In stereometry the prism , pyramid , and pris
matoid correspond respectively to the parallelogram , tri
angle,and trapezoid in planimetry .
XXXVII .
Though ,in generalf the lateral faces of a
prismatoid are tr iangles, yet if two basal edges which form ,
THE MEASUREMENT OF VOLUMES . 101
w ith the sam e lateral edge,two sides of two adj oining
faces are parallel , then these twotriangular faces fall in the sameplane
,and together form a trape
zoid .
XXXVIII . A prismoid is a prismatoid whose baseshave the same number of sides
,and
every corresponding pair parallel .
XXXIX . A frustum of a pyramid is aprismoid whose two bases are sim ilar .
Oor . Every three-sided prismoid is the frustum of apyramid.
XL . If both bases of a prismatoid become lines, it is a
tetrahedron.
XLI . A wedge is a prismatoid whose lower base is arectangle
,and upper base a line parallel to a basal edge .
82. To find the volum e of any prismatoid .
Rule : Add the areas of the two bases and four tim es the
m idcross- section ; mu ltip ly this sum by one-sixth the a lti
tude.
Prismoidal Formula : D z t a (B ; 4 111 B 2) .
P roof In the midcross-section of the prismatoid takea point J V
,which j oin to the corners Of the prismatoid .
These lines determ ine for each edge of the prismatoid a
102 MENSURAT ION .
plane triangle,and these triangles divide the prismatoid
into the following parts
1 . A pyram id whose vertex is N and whose base is B 2,
the top of the prismatoid . Since the altitude of this pyramid is half that of the prismatoid , therefore , by 80 , its volume is t aB z.
2 . A pyramid whose vertex is N and whose base is B 1,
the bottom of the prismatoid . Since the altitude of thispyramid also is f a ,
therefore its volum e is t aB l .
104 MENSURAT ION .
midcross-section we get
TS 80 meters,
LB 87 5 meters,
NP 90 meters,
TL 2 75 meters,
L I 32 5 meters,
IN 75 meters .
Thus the areas are
B1
6500 square meters,
B2
Z 2325 square meters,
ill : square meters,
and for the Whole volum e we get
D -5 cubic m eters . Ans .
NOTE. In a prismoid the midcross- section has always the same
angles and th e same number of sides as each base,every Side being
half the sum of the two corresponding basal edges . The rectangular
prismoid has its top and bottom rectangles ; hence, by 32, its volume
D 4 15 +
—g1(5
-
a (2w1b1
wlb2
wzb1
2w2b2) .
Oor . If a prism has trapezoids forbases
,its volume equals half the sum
of its two parallel side-faces multiplied by their normaldistance apart .
83. To find the volum e Of a frustum of a pyram id .
Rule To the areas of the two ends of thefrustum add the
square root of their product ; multip ly this sum by one- third
the a ltitude.
Formula V . F 2 3; a (B 1—l~ \/B 1B 2
Jr B 2) .
THE MEASUREMENT OF VOLUMES . 105
P roof : If w ; and w ; are two corresponding Sides of theW I
‘ 1‘ 70 2C
bases B 1 and B 2, then a side of the m idsection is
Since in a frustum B 1, B 2, and M are similar,by 44, we
have”
KL —L
;w,”
A
—/Ll
4M = B,
2x/ BIB,B,
.
Substituting thi s in 82 gives
V . F %=Ci (2B 12\/B
1B2
2 B2) .
Oor . By 44,
Substituting this for B 2 gives
WE F —T -
jlgCtB l
w1
MENSURAT I ON .
EXAM . 80 .
i
The area Of the top of a frustum is 160
square meters ; of the bottom ,250 square meters ; and its
altitude is 24 meters . Find its volume .
HereV . F s (250 200 160)
4880 cubic meters . Ans .
If,instead of the top
, we are given w ; w2 : 5 4,then,
by our Corollary ,
V . F x 2000 (1 é—g4880 cubic m eters . Ans .
84. To find the volume of a frustum of any cone .
Rule : To the areas (pfthe two ends add
the square root of their product ;
mu ltip ly this sum by one-third the
Formula for Circular ConeV . F 2 6671
"
(71
12
‘ j" 7”
n
P roof : As in 81, so the frustum of a cone is the limit ofthe frustum of a pyramid .
EXAM . 81 . The radius of one end is 5 meters ; Of theother , 3 meters ; the altitude , 8 m eters . Find the volum e .
Herev. F 15 9) 3 1416
410-5024 cubic meters. Ans .
85. To find the volume of any solid bounded terminallyby two parallel planes, and laterally by a surface generatedby the motion of a straight line always intersecting the
planes, and returning finally to its initial position.
108 MENSURAT ION .
13,their lim its w ill be to one another in the sam e ratio ;
and
for the prismoidal solid .
EXAM . 82 . The radius of the m inimum circle in a hyperboloid is 1 meter . Find the volume contained betweenthis circle of the gorge and a circle 3 meters below it whoseradius is 2 meters.
Solution : The hyperboloid of revolution of one nappeis a ruled surface generated by the rotation of a straight
line about an axis not in the sameplane with it . All points of thegeneratrix describe parallel circleswhose centers are in the axis . The
shortest radius, a perpendicular both to axis and generatrix
,describes the circl e of the gorge
, which is a plane ofsymmetry . Hence , taking this circl e as midsection
,and
for altitude twice the distance to the base below it,the
Prismoidal Formula gives twice the volume sought in
Exam . 82.
2V=D % 6 [22rr 4 (1)
’27r 2277 ] 12 7:
Therefore, V : 6 7r cubic meters . Ans .
86. To find the volume of any w edge .
Rule : To twice the length of the base add the opposite
edge ; m ultiply the sum by the width of the base,and this
product by one-sixth the a ltitude of the wedge.
Formula W t aw (2 b1 bg) .
P roof : Since the upper base of a wedge is a line, so,by
the Prismoidal Formula ,
4M) .
THE MEASUREMENT OF VOLUMES . 109
Therefore,by 32,
W 2 50. wbl + 4
é l gb2 x %
Oor . 1 . If the length of edge equals the length of base ;
b b then Hf : %~ awb,1
the simplest form of wedge .
Oor . 2 . The volum e of anytruncated triangular prismis equal to the product Of itsright section by one-thirdthe sum of its lateral edges .
EXAM . 83 . Find the volum e Of a wedge , of which thelength of the base is 70 meters ; the width , 30
'
meters ; thelength of the edge
,110 meters ; and the altitude
,24 8
m eters .
Here IV: ( 140 2248
( 140 110) 10 12 4
2500 x
cubic meters. Ans .
87. To find the volume of any tetrahedron.
Rule Mu ltip ly double the area of a para llelogram whose
vertices bisect any four edges by one—third
theperp endicu lar to both the other edges .
Formula : X : 3aM
P roof : When,the bases being lines
,
B1= B
2= 0
,then D : —a 4M :
~ —aM
110 MENSURATION .
Since M bisects the line perpendicular to the two basaledges
,it bisects the four lateral edges ; and is a parallelo
gram .
EXAM . 84 . The line perpendicular to both basal edgesOf a tetrah edron i s 2 r long the length of the top edge is2 r
,and of the bottom edge , 2 7rr . The m idsection is a rec
tangle . Find the volume of the tetrahedron .
Herea : 2 r
,and M : r
27r
Therefore,
X 37.-r3 Ans .
6(P ) . SPHERE.
88. To find the volume of a sphere .
Rule : Mu ltip ly the cube of its radius by
Formula : V . H z -
gr r
3.
P roof : Any sphere is equal in Volum e to a tetrahedron
whose midsection is equivalent to a great circle of thesphere
,and whose altitude equals a diam eter .
Let the diam eter DO be normal to the great circl e AB
at O. Let Q be the point in which the m idsection LNbisects the altitude J K at right angles . In both solids
112 MENSURAT ION .
B ut OH : 2L U : e /d,
and EF
For any one of the equal lateral edges,
+ D T2
+ L V
SO it is not a regular tetrahedron .
89. To find the volum e of any spherical segm ent .
Rule : To three times the sum of the squared radii of the
two ends add the sguared a ltitude ; mu ltip ly this sum by
the a ltitude,and the product by 5 236
Formula V . G 3 an [8 a2
] .
P rocy’: In 88 , we proved any spherical segment equal in
volume to a prismoid ofequivalent bases and altitude . Therefore
,by 82,
V . G 6 a (r127r 4 7°
3
27r r
2
27r) .
To eliminate rs call x thedistance from center ofsphere to bottom of seg
ment,and r the radius Of
sphere ; then,by Eu. II . 10
,
az=z
Doubling and subtracting both m embers from
a2 r2 — 2 r2 — 2x2 = 4 r2 — 4
§+ x
2
or,by 2, 2 r
,
?a2 = 4 r
3
2.
Substituting , V . G Jr
THE MEASUREMENT OF VOLUMES . 113
Oor . In a segment Of one base,since r2 0
,we have
V . G if; an (3 i f a
2
) .
But now,by WW . 289 ; (Eu. VI . 8
,Cor . CV . III .
a (2 r — a ) .
Substituting ,
v. G ra s t er — a ) e ) W e )“ 9 ) em 3a )
EXAM . 86 . If the axis of a cylinder passes through the
center of a sphere,the sphere-ring SO formed is equal in
volume to a sphere of the same altitude .
For,since the bases of a middle segm ent are equidistant
from the center , V . G 2 4; m. (6 732
a2
)( 57 7 1
23” 66
3
But,by 76 , the volume of the cylinder cut out of the
segment is ar 127r
,and the remaining ring “Aw-at"is, by 88 ,
the volum e of a Sphere of diam eter a .
XLII . When a semicircl e revolves about its diameter,
the solid generated by any sector of the semicircl e is calleda spherica l sector .
90 . TO find the volume of any Spherical sector .
Rule M i ltip ly its zone by one—third
the radius .
Formula : V . S 2 Fe ar?
P roof : If one radius of the generating sector coincides with the axisof revolution
,the spherical sector is
the sum of a spherical segment of onebase and a cone on same base
, with vertex at centersphere .
114 MENSURAT ION .
v. e g7r (r gm) .
V . K i (r a ) r12 '
rr ;
or,substituting for r l
z,its value used in 89
,Oor .
,
V . K i (r a) a (2 r a) 7r §~7r (2r2a 3 ra2 a
3) .
Adding, we have
Any other spherical sector is the difference of two suchsectors .
V . S V . SIV. S
2 § r27ra
1 gem ,
«gems,
az) .
a z a,
the altitude of 8’
s zone, whose area , by 65, is Z a ra . Thus,
for every spherical sector the volume is zone byJ-g r .
Cor . If r1 and r2 are the radii of the bases of the zone , itsaltitude
,
a \/r2
EXAM . 87. Find the diameter of a sphere of which asector contains r
7°854 cubic m eters when its zone is O°6
meters high .
V . S ‘2—6 7W‘2
z : 7 854.
Dividing by 77 31 416,
0-4r 2 5 . 25 .
2r 5 m eters . Ans .
XLIII . A spherica l ungu la is a part of a globe boundedby a lune and two great sem icircles .
91 . To find the volum e of a spherical ungula .
MENSURATION .
EXAM . 88 . Find the distance of the center masssemicircle from the center of the circle .
By 88 ’V I—I g r
3
BY937 V . E m
ir277 2x713
Equating , we get tw3 = T
27T” M293
gi r z rrzc.
94. To find the volume of a ring .
Rule : Multiply the genera ting area by the path of its
center .
Formula for Ellipse : V . O Z rzabr .
P roof : Conceive any ellipse to revolve about an exterioraxis parallel to one of its axes . Divide the axis of symme
try AE into n equal parts, as
A V= VT : z,
and from these points of division drop perpendiculars on
the axis of revolution PO. J oin the points where theseperpendiculars cut the ellipse by chords FD ,
DA,AG,
GI T, etc .
The volume generated by one of the trapezoids thusformed
,as DGHF ,
is the difference between the frustums
generated by the right-angled trapezoids Q EF W and
Q G’H W Therefore
,by 84
,
V . by DGHF
D Q +D Q2
) GQ +GQ2
)
(co + F T ) (co +D V) (00 D V)2
-(co— FT ) (00 -DV) ( 00 — Dw]gmeco,
F T coo,D V)
2 7,CO
,z (F T +D V)
FH + DG
THE MEASUREMENT OF VOLUMES . 117
Thus,by 40
,the volume generated by the polygon
HGADF ,etc .
,equals its area multiplied by the path of
the center . But,as we increase n,
and thus decrease z
indefinitely ,as shown in 55, the area of the polygon ap
proaches the area of the ellipse as its lim it . But alwaysthe variable volum e is to the variable area in the constantratio 2 7rr ; therefore , by 18 , their lim its will be to one an
other in the same ratio and
V . O 2 7rrab7r.
EXAM . 89 . Find the volum e of the ring swept out by an
ellipse whose axes are 8 and 16 m eters,revolving round an
axis in its own plane , and 10 meters from its center .
Here
640 772
6316 5 cubic meters . Ans .
118 MENSURAT ION .
S IM I LAR SOL ID S .
XLIV . Sim ilar polyhedrons are those bounded by thesame number of faces respectively sim ilar and Sim ilarlyplaced
,and which have their soli d angles congruent .
Given the volume and one line in a solid,and the
homologous line in a sim ilar solid,to find its volum e .
Rule : Mu ltip ly the givenr
volume by the cubecl ra tio ofhomologous lines .
Formula VI
P roof : The volum es of two sim ilar solids are as thecubes of any two corresponding dim ensions .
w . 590 ; (Eu . XI . 33 ; CV . VI I . 73
Thus,for th e sphere
,by 88 ,
H. if
H.
NOTE. I f a tetrahedron is cut by a plane parallel to one of its
faces,th e tetrah edron cut off is sim ilar to th e first . I f a cone b e out
by a plane parallel to its base, th e whole cone and th e cone cut Off
are similar .
120 MENSURATION .
97. If the solid is homogeneous, weigh it . Also weigh acubic centimeter of the sam e substance . Divide the weightof the solid by the weight of the cubic centimeter . Thequotient will be the number of cubic centimeters in thesolid .
From 78 , we have the
g( 1)
Formula Vccm
8
EXAM . 92. A ball 5 centim eters in diameter weighs481-97 grams . An irregular solid of the same substance
weighs 13 2 grams ; find its volume .The volume of the ball is
53 X 0 5236 654 5 .
431 97 65 45 2 6 6 grams,
the weight of a cubic centimeter .
6-6 2 2 cubic centimeters . Ans .
98. To find the volume of any irregular polyhedron .
Rule : Out the polyhedron into prismatoids by passing
para llel p lanes through a ll its summits.
Formula for n consecutive prismatoids
[ 2l
[$2<B l B g) j L' $ 3 (B g B 4) “
1L etc"
(Eu—1 B?l+ 1) "l“ xn+1 ( -Eu F l" B 7t+ l>]
“ t“ [3321111 t ( 973 372) J l[ 2
“ l" (554 373) jl[ 3 *l‘ etc .
(mu xn)fljiz]
NOTE. T2is the distance of B
2from B 1 , and x
3is the distance of
B3from E
l ,etc .
P roof : This formula is obtained directly by the methodof 41 .
CHAPTER VI .
THE APPL I CAB I L I TY OF THE PR I SMOIDAL FORMULA .
99. To find whether the volume of any solid is determined by the Prismoidal Formula .
Rule : The P rism oida l Formu la app lies exa ctly to ALL
SOL I DS conta ined between two para llel p lanes , OF WH I CH
the area of any section para llel to these planes can be etc
pressed by a ra tiona l integra l a lgebra icfunction ,of a degree
not higher than the third,of its distance
from either of these
bounding p lanes or bases .
Test : Acc g mx nx2+fa3
5.
NOTE. A, is the area of any section of the solid at the distance as
from one of its ends . The coefficients g, m ,n
,f, are constant for th e
same solid,but may be either positive or negative ; or any one
,two,
or three of th em may be zero .
P roof : Measuring x on a line normal to which the sec
tions are made,let q!) (a ) be the area of the section at the
distance x from the origin .
The problem then is,What function <1) will fulfil the con
ditions of the Prismoidal Formula ?
For any linear unit , the segment between (I) (0) and cs(4)is the sum of the segm ents between (I) (0) and cs(2) and between qb (2) and gb (4) . Therefore
,if c]; is such a function
122 MENSURAT ION .
as to fulfil the requirements of the Prismoidal Formula, we
have identically
— 4¢ (1) M4) 0 .
But for 9500) 9 mm 71332+115
3
9684.
$500) 64502) M4)
becomes 9— 4g 4m 4n ~ ~ 4f 49
48f + 96 g— 4 g
- 12m — 36n — 108f —~ 324g
9 + 4m + 16n + 64f + 256g
0 0 0
So the conditions are satisfied only by functions whichhave no fourth and higher pow ers . Hence (My ) must bean algebraic expression of positive integral powers not exceeding the third degree .
Thus,in general
,the cubic equation
expresses the law of variation in magnitude of the planegeneratrix of prismoidal spaces ; i . e.
,solids to which the
Prismoidal Formula universally applies .
Cor . 1 . for prismoidal solids
Mm) no
nix n
zzz:2
therefore,
‘P(0) 44) (t a) <1) (9 )
no
2 l 3+ 4n0 + 2 an1 + a n
2 + 2 a n3
no
ani
agn2
a3n3
2 3 32 a n2 + 7 a n3
1 74H MENSURAT ION.
XLV . EXAMINATI ON or THE DI FFERENT CASES .
Let my22
g ; y is constant,and the solid is a circu
lar cylinder .
(2) Let 7ry2
r -mx, y
20c 33
,and
the solid is a paraboloid ofrevolution ; for in a para~
bola,the square of the ordi
nate varies as the abscissa .
(3) Let r yz
z naf"; y oc x
,and
the solid is a rightcircular cone .
Let 7ry2r s fr
3; y
2oc and the solid is a
semicubic paraboloid of revolution.
Let wy2
(1+ m ic ; y2at (h a ) where h
is constant , and the solid is a frustum of a paraboloid of revolution,
h being the height of the seg
ment cut off.
Let r y2 z
g+ naQ
; supposing g and n positive,this is
th e equation to a hyperbola , the conjugate axis beingthe axis of a“
,and the center the origin. Hence , the
solid is a hyperboloid of one nappe .
Let wyg
g +fa3
. In this case,the solid is generated
by the revolution of a curve , somewhat sim ilar
in form to the semicubic pa
rabola,round a line parallel
to the axis of cc,and at a
constant distance from it .Let 7ry
2=mx+ nafi In this case ,the solid may be a Sphere , aprolate spheroid
,an oblate sphe
roid,a hyperboloid of revolution,
or its conjugatehyperboloid .
THE APPL I CAB I L I TY -OF THE PR I SMOIDAL FORMULA . 125
(9 ) Let 7rg2
g mx nrz
. In this case,the solid w ill
be afrustum of a circular cone
,or of the sphere ,
spheroids,or hyperbo
loids of revolution,made
by planes normal to theaxis . In the frustum ofthe cone g, m ,
and n areall positive . The othersolids in (8) and (9) aredistinguished by the Values and signs of the constants m and n .
( 10 ) Let g mx nx2+fzc
3. In this case
,the solid
is a frustum of a semicubic paraboloid of revolu
tion. For,if a: be the distance of the section from
the smaller end of the frustum ,and h the height
of the segment cut off, y2oc (h x)
3fly
2 is
of the form 9 ma: nag+fx
3
( 11) Let fl y22 : ma: +f.v
3.
( 12) Lot nx2+fx
3.
( 13) Let g ma: +fx3
.
( 14) Let 7ry2
g nx2+fx
3.
( 15) Let 7z-y2 mcc nx
‘"+fx
3.
EXAM . 98 . Since , for an oblate spheroid ,
32
: 0,4 2lf = 4 rra2
,and h 2 b
,
therefore its volum e
0 :
Sim ilarly ,the volume of a
prolate spheroid
0 : gr ab?
Thus each is,like the sphere
,two-thirds circum
scribed cylinder .
126 MENSURATION .
EXAM . 94. The volume of the solid generated by therevolution round the conjugate axis of an arc of a hyperbola,
cut off by a chord and II to the conjugate axis,is
twice the spheroid generated by the revolution of theellipse which has the same axes .
Making the conjugate the axis of a",
B1is the value of fly
? when x s b ;
B,is th e value of fly
? when w b
JVI is th e value of when w 0 ;
Since the conjugate axis is the height of the solid,
h z : 2 b.
Hence its volume X gwa2b. Ans.
EXAM .
.
95 . To find the volume of a paraboloid of revolution. Let h be its height
,that is
,the length of the axis
,
r the radius of its base , and p the parameter of the gene~
rating parabola , y2z
px .
Thenit
B10 , 3
2rrph ,
4M = 4 7rp-
é27rph .
C % 7rph2 Ans .
Cor . Sincenphh % 7T9
‘2h,i
on
r = ph,
or a paraboloid is half its circum scribing cylinder .
EXAM . 96 . To find the Volume of any frustum of aparaboloid contained between two planes normal to theaxis .
F . C: % 7Tph257170 711
2 —
21 h
1
2
) h1) (h2 h
l ) .
128 MENSURAT ION .
common vertex,the plane of the ellipse being always
normal to this axis ; for , in this solid,the area of the
section at distance x from the vertex wil l be 7ryy ,
'
where y and y' are the ordinates of the two parabo
las,and since both y
2and vary as x
,wyy
'
varies as 37.
Let A,na
g. This is a property of all pyram ids and
cones,whatever may be their
bases .
Let A,
z fr3
. The solid will be
an elliptic semicubic paraboloid .
Substitute semicubic for common
parabolas inLet Ax
:
g+ mx . This is a prop
erty of a frustum of an elliptic
paraboloid .
Let A,2
9 + nag
. This is a property of a groin,a
simple case of which is the square groin seen in thevaults of large buildings .
This solid may be generated by a variable square,
which moves parallel to itself, w ith the midpoints of
two Opposite sides always in a semicircumference ,the plane of which is perpendicular to that of the
square .
If y is a side of the square when at a distance xfrom the centre ; y
2: 4 (r
2 — x2
) .
Let s mx —l— naz
. This is a property of the ellip
soid,and of the elliptic hyperboloid .
Let Am:
9 mp + nrz
. This is a property of a pris
moid, and of any frustum of a pyram id or cone ,
whatever may be the base ; also , of any frustum of
an ellipsoid , or elliptic hyperboloid made by planesperpendicular to the axis of the hyperboloid , or to
any one of the three axes of the ellipsoid .
THE APPL I CAB I L I TY OF THE PR I SMOI DAL FORMULA . 129
EXAM . 97 To find the volume of an ellipsoid . Let a,b,
c be the three semi-axes,a the greatest
h = 2a,B,= 0
,B,= 0
,4M = 4 7rbc.
7;z : fi
gvrabc . Ans.
EL IM INAT ION OF ONE BASE.
For all solids whose section is a function of degree nothigher than the second
,or
g, m ,n
,and consequently Ax,
for all values of x,are deter
mined if the value of A,for three values of a: is known.
Measuring x from B 1 we have
Supposing we know the section at 1 the height of thez
solid above B 1, we have for determining m and n the twoequations
,
B,
B,ma na
2,
a a2
Agz B
,m,
”5"
Hence,
2Ae (z2 1)B ,
B,
(z 1) a
B,
z (z 1)B ,22A‘1
(2: 1) a2
the volume of the solid wehave
V B,a 4ma
2
gnai’,
a3)B , (z — 1) (z 3)B ,
180 MENSURATI ON .
For 7: z 8,this gives
Again,for z 2 I t ,
V ”: BAgal
These give the following theorem
100 . To find the volum e of a prismatoid,or of any solid
whose section gives a quadI 'ELt ’
Rule Mu ltip ly onefour th its a ltitude by the sum of one
base and three tim es a section distant from tha t ba se two
thirds the a ltitude.
Cor If B 2 reduces to an edge or a point,
_ BV —4aAg
132 MENSURAT ION .
Therefore the whole content of the figure is
A + 320n2
12?
04En4
h
n_1 m 7712
7714
which equals6 h (A 3 -8 -Eh4 l i
e. an) .
Now Weddle’
s Rul e gives for the same figure
5 (y2 y,) 1O(A 16Eh4 64Gh6)6y, 6A
3h 2 (A 90 h2 81Eh4 729Gh6)
y, y 2 (A 0 h2 + Eh4 + GM)
5 : 35 h [20A + 6OC
’h2 + 824Eh4 + 2100Gh6]
5 :7 3 5 Gh6) .
Therefore the value given by our formula is in excess
the quantity3
753‘ t
Therefore the error in the last term is
_3 6 Gh7
& %2 9 Gh7
0
or the error is less than 09
0 0of the last term .
Hence the error in the whole quantity w ill be very much
sma ller than this,since the most important terms in the
expression for y are the earlier terms.
Change of origin does not change the degree of an equation ; hence , we have demonstrated that by means ofWeddle
’
s Rule we can find exactly the contents of any sur
face Or solid in which the sections may be expressed by arational integral algebraic function
,of a degree not higher
than the fifth ,of its distance from either end ; and very
approximately when the expression is of the sixth or seventhdegree .
Cor . Suppose there are (6n 1) equidistant sections
y1 ’ y2 ’ ya) 94. a y6n+L
APPROXIMATION To ALL SURFACES AND SOL I DS . 133
yl and being the extreme or bounding sections ; andlet h be their comm on distance ; then,
5 T36h fzyodd ‘ t‘ 5 2yeven Xyevery th ird] ,
observing not to take either of the two extreme sectionstwice ; that is, begin Ey every third with y4, and end itwith yen_2.
EXAM . 98 . B etween 91 and 919
5 2
T3sh fy1+ “ t‘ Z/7
‘ tj ?/s‘ t‘ “ t“ ‘ t 910+ 913+
EXAM . 99 . Find the volume of the middle frustum of aparabolic Spindle .
A spindle is a solid generated by the revolution of anarc of a curve round its chord
,if the curve is symmetric
about the perpendicular bisector of the chord . Hence aparabolic spindle is generated by the revolution of an arc ofa parabola round a chord perpendicular to the axis .
Let the altitude of the middle frustum be divided intosix parts each equal to h
,and let AI , A2, AS, A4 , A5 , A6, A7
be the areas of the sections .
Taking origin at center and r the longest radius,or
radius of mid-section,by the equation to a parabola , we
have for any point on the curve
w2 = a (r —y)
£132
ny2 = 7r r
2 — 2 r
will be the area of the section at the distance r from O;
and this being a rational integral function of a: of the
134 MENSURAT I ON .
fourth degree , Weddle’
s Rule w ill determine the volumeexactly .
Now,if r 1 is the shortest radius
,or radius Of the two
bases, we have
” 2 2 74 h)
? 16 h4
a
c h2 h4A
3Z A
5Z
2
CL a
Therefore,by 101
2 h? 2 9 2 23 r —+ a r -t u r q r — 2 ra
9
1
z o ffihn [18 r2+— 2 r 2
i (r — 2 (r —~ rJ fl
h [32 r27~ 16 rr — l 2 r
,
5”2 "if ” (8 7
2 "f 8 r12
) . Ans .
Making r l 0,gives for the volume of the entire spindle
,
1 6 hm
But 6 ha ) “2 is the volume of the circumscribing cylinder ; hence Volum e of spindle is
T83 circum scribing cylinder . The
middle frustum of a parabolici
spindle isa very close approximation to the general
form of a cask,and hence is used in cask-gauging .
EXAM . 100 . The interior length of a cask is 30 decim eters ; the bung diam eter , 24 decim eters ; and the head diameters
,18 decimeters . Find the capacity of the cask .
I 7. (2304. 864 486) z 7. 3654 . Ans .
CHAPTER VIII .
MASS-CENTER .
102. The point whose distances from three planes atright angles to one another are respectively equal to them ean distances of any group of points from these planes,is at a distance from any plane whatever equal to themean distance of the group from the sam e plane .
103. The mass—center of a system of equal materialpoints is the point whose distance is equal to their averagedistancc from any plane whatever .
104. A solid is homogeneous when any two parts ofequal volum e are exactly of the same mass . The determination of the mass-center of a homogeneous body is, therefore
,a purely geometrical question. Again
,in a very thin
sheet of uniform thickness,the masses of any two portions
are proportional to the areas . In a Very thin wire of uniform thickness
,the masses of different portions will be
proportional to their lengths . Hence we may find themass-center of a surface or of a line .
SYMMETRY.
105. Two points are symmetric when at equal distanceson opposite sides of a fixedpoint, line , or plane .
106. If a body have a plane of symmetry , the mass-center ”O lies in that plane . Every particl e on one side cor
MASS—CENTER . 137
responds to an equal particle on the other . Hence the ”O
of every pair is in the plane, and therefore also the ”0 ofthe whole .
107. If a body have two planes of symm etry ,the ”0 l ies
in their line of intersection ; and if it have three planes ofsymm etry intersecting in two lines
,the ”O is at the point
where the lines cut one another .
108. If a body have an axis of symmetry ,the ”O is in
that line
109. If a body have a center of symm etry ,it is the ”0.
110. The ”Oof a straight line is its midpoint .
111 . The ”O of the circumference or area of a circl e isthe center .
112. The ”0 of the perimeter or area of a parallelogram is the intersection of the diagonals .
113. The ”0 of the volume or surface of a sphere is thecenter .
114. The ftC’ of a right circular cylinder is the m idpointof its axis .
115. The ”(I of a parallelepiped is the intersection oftwo diagonals .
116. The ”0 of a regular figure coincides with the ”0 ofits perimeter
,and the ”C of its angular points .
117. The ”C of a trapezoid lies on the line j oining them idpoints of the parallel sides .
138 MENSURAT I ON .
118. Since in any triangle each m edial bisects every linedrawn parallel to its own base
,therefore the ”O of any tri
angle is the intersection of its m edials . By sim ilar triangles,
this point lies on each m edial two- thirds its length from its
vertex,and so coincides w ith the I
“Oof the three vertices .
119. The ”O of the perimeter Of any triangle is the cen
ter of the circle inscribed in the triangle whose vertices arethe m idpoints of the sides .
P roof The mass of each side is proportional to its length,
and its ”0 is its m idpoint . So the ”O of M}, and 1113 is ata point D
,such that
2M.2 tAB JAM
BM,
b 52 0 MM,
Hence,the f‘C of the whole perim eter is in the line DM ;
and since D i ll , divides the base ..M2M, into parts proportional to the sides , it bisects 1111 . Similarly the ”C is in
the line bisecting hfM} ,
120. The ”O of the surface of any tetrahedron is thecenter of the sphere inscribed in the tetrahedron whosesummits are the l“(7
’
s of the faces .
121. If the vertex of a triangular pyramid be j oinedwith the of the base
,the ”0 of the pyramid is in this
line at three—fourths of its length from the vertex . (Proofby sim ilar triangles . )
122. The ”O of a tetrahedron is also the ”0 Of its foursumm its .
123. The ”C of any pyramid or cone is in the line j oin~
ing the ”0 of the base w ith the apex at three - fourths of itsl ength from the apex .
140 MENSURAT ION .
center,and at a distance from it equal to one-third of the
distance betw een the two centers .
That is,L OG will be a straight line , and
GL = §OG.
128. In any body between parallel planes, we can reckonthe distance of its ”0 above its base
,if every cross-section
is a given function (Mr ) of its distance x from thebase .
The prism whose base is the section (Mr ) , and whoseh eight is the nth part of the altitude a of the body ,
has for
volum e ; (Ma ) . I ts ”G i s a +z
a
” from the base of the body ,
and has the coefficient Now suppose the ”G of then
body distant 1 from its base , and form the product of 1 with
the sum D of the values whichCf. (Maj) takes when equalsn
0,l a
,ga
,
72 .— la ; also form the sum 8 of the values
n n
which ( 5c) takes for the sam e worths of an The
product 7 D equals the sum S when the arbitrary number n
is taken indefinitely great .
rD z S,for n J—z oo .
But for n 00,the sum D expresses the volume of the
body .
a
The sum 8 consists of the sum 0 of terms fromgp cfifii ) ,
2 7,But the sum
E has the value5
— D,and vanishes for n
L n
Therefore,for the determ ination of 7
,we have the equa
tion
and Of the sum E of terms from
MASS-CENTER . 141
If qS( £c) is of degree not higher than the second,then
2 93(x) , which we will call f (v) , is not higher than the third .
Therefore,by 99,
+ 4232 )0 were) wea ) + f (a )l
But
TTherefore
,
a NO) M i en+ 202 )
Therefore,
7 = % a +
129. For the applicability of the Mass-Center Formula
12D
the bCSb ISAx q ma,
For an examination of the possible varieties, see 99 , (T) ,and
Of course it applies also to the corresponding planefigures .
EXAM . 102. For trapezoid
ag(b, — b
,) a (2b + b )120 2 + bag
—12T :
EXAM. 103 . For pyramid or cone
a2B
T = % a + — J
mecca ):
from B 1, or t a from B 2.
142 MENSURAT ION .
EXAM . 104. The ”0 of a pyrimidal frustum is in the linej oining the ”G’
s of the parallel faces,and
a (B ,B,)
4 (B ,x/B
,B,B,)
r = § a +
So,if any tvio homologous basal edges are as l to A
the distance of the frustral ”G from one base w ill be
and from the other
EXAM . 105 . From Exam . 95,for paraboloid
T = % a +
For frustum of this we obtain an expression similar tothat for trapezoid . It is
5a 12 71" 3
EXAM . 106 . For ”0 of half-globe,from center ,
130. The average haul of a piece of excavation is the distance betw een the ”0 of the material as found and its ”0 as
deposited .
131 . The ”0 of a series of consecutive equally-long quadratio shapes may be found by assum ing the ”G of eachShape to be in its mid- section
,then compounding
,and to the
“2
(B_2— ~B 1)distance of the point thus found from B , adding
12D
144 MENSURATION .
ratio 1 2. The same is true of the oth er three tetrah edraand the mid-points of HB
,DO
,GG. Hence
,the mass
centers of the four tetrahedra are in one plane passingthrough the point S found by the above construction
,and
therefore the /‘G of the whole solid is in this plane . SO,
also,it is in the other two planes determ ined by dividing
the solid into tetrahedra having the common edge B G and
the common edge OH respectively . Therefore it coincideswith the point S .
Cor . By making the pairs of faces ABE AHG,
‘ AUG,
GFG; CDF ,B HF to be respectively copular
,we pass to
a truncated triangular pyramid . If we j oin its cross- centerK with its mid- center M and produce KM to S
,making
MS I 3KM then S w ill be the (“Gof
'
the trunc .
NOTE. This corollary was Sylvester’
s extension of the geometricmethod of centering the plane quadrilateral , and suggested to Cliffordthe above.
EXERCISES AND PROBLEMS INMENSURATION.
PROBLEMS AND EXERCI SES ON CHAPTER I .
1. a2+ b
2: e
2.
EXERCI SE 1. Find the diagonal of a square whose sideis unity . 2 2 1 Answer .
2. Find the diagonal of a cube whose edge is unity .
z : 1 Ans .
3. To draw a perpendicular to a line at the point 0 .
Measure CA 2 3 m eters,and fix at A and 0 the ends of
a line 9 m eters long ; which stretch by the point B taken
4 meters from C and 5 meters from A .
B C is 1 to A0 because 32 42 52 .
4 The whole numbers which express the l engths of thesides of a right-angled triangle
, when reduced to the lowestnumbers possible by dividing them by their common divisors
,cannot be all even numbers . Nor can they be a ll odd.
For,if a and b are odd
,a2and b2 are also odd ,
eachbeing an odd number taken an odd number of times .
a2 b2 is even
,and c is even.
2.
5. To obtain three whole numbers which shall representthe sides of a right-angled triangle .
146 MENSURATION .
Ru le of Pythagoras . Take 77. any odd number , then2
n + 1 the third numberthe second number,and
P la to’
s Ru le. Take any even number m,then also
2
and 21;
1 .
Euc lid’
s Ru le. Take a"and y ,
either both odd or both
even,such that xy z 62
,a perfect square ; then a
x —y
and c
R ule of Maseres . Of any two numbers take twice theirproduct
,the difference of their squares
,and the sum of
their squares .
P roof : Let the whole numbers which express the lengthsof the two sides of a rt . A be a and m . If 0 be a wholenumber
,it must 2 a n
,where n is a whole number .
By 1 ,a2+ m
2c
m2 = 2 an + n2
2an = m2 ~ n2
.
”fizz — n
?
a + n = c =
m — n2 2o7m
Z n Z n
Magnifying the rt . A 2 71 times,its sides are ex~
Therefore,the three sides
,a,m
,c,are and
m2+ n
2
2 77,
pressed by the whole numbers m2
712,2 77m
,and m2
148 MENSURAT ION .
TAB LE I . Continued.
Area .
EXERCI SES AND PROBLEMS . 149
TAB LE I .
—Concluded.
Area .
159 1405 19 225 0586 81
150 MENSURAT ION .
3 .
6. Two sides,a 6 708
,l) 2 5
,contain an obtuse angle .
If j : 3, find 0 . 10 . Ans .
7. If a : 13,l) 11 , c : 20
,find j. 5. Ans .
4 . a2 62 2 6
97.
62
.
8. The three sides of a triangle are 2 5 meters,4 8
m eters,3 2 meters ; find the proj ections of the other two
sides on meters.
1 985 m eters and 28 15 meters. Ans .
9. TWO sides of a triangle,3 m eters and 8 meters long
,
enclose an angle of 60?
Find the third side .
7 meters . Ans .
HINT . J oining the midpoint of 3 with the vertex of the rt . 4'
made in projecting 3 on 8 gives two isosceles triangles, and
j = 15
10 . Two Sides of a triangle are 13 meters and 15 meters .
Opposite the first is an angle of 60? Find the third .
7 or 8 meters. Ans .
HINT . Drop 1 to 15-meter side. The segment adjacent the th irdside is half the third side.
11 . Two sides of a triangle are 96 meters and 12 8
meters ; the perpendicular from their vertex on the thirdside is 76 8 meters ; find that side . 4 . Ans .
6 .
2 2a —l—e —
91 l92= 2 l 2
12. Given 21 , the medial from A to a ; B to b ;and 'i3 , from 0 to 0 . Find a
,l),and 0 .
From 6,
a2+ c
2
a2+ b
2
2 02 = 2 i
3
2,
b2 + c2 —
2 a
152 MENSURATION .
22. On the three sides of a triangle squares are describedoutward . Prove that the three lines j oining the ends oftheir outer sides are twice the m edials of the triangle and
perpendicular to them .
23. Prove that the m edials of a triangle cut each otherinto parts which are as 1 to 2 .
24. The intersection-point of medials is the center ofgravity of the triangle .
25. Prove an Ain a A ,acute
,rt .
,or Obtuse
,according as
the m edial through the vertex of that 2Cis or halfthe opposite side .
26. The three sides of a triangle are 174 meters, 23 4
m eters,318 meters . The smallest side of a sim ilar tri
angle is 5 8 meters . Find the other sides.
78 meters and 106 m eters . Ans .
27. Find the height of an obj ect whose shadow is 378
meters,when a rod of 2 75 meters casts 14 m eters shadow .
742 5 m eters . Ans .
28. The perpendicular from any point on a circumference to the diameter is a mean proportional between thetwo segments of the diameter .
29. Every chord of a circl e is a m ean proportional between the diam eter drawn from one of its extrem ities
,and
its proj ection on that diameter .
30. From a hypothenuse of 72 9 m eters , a perpendicularfrom the right angle cuts a part equal to 64 m eters ; find
the sides . a z : 216 meters,6 2 696 2 m eters . Ans .
EXERCI SES AND PROBLEMS . 153
31. The hypothenuse is 325 meters,and the perpendicu
lar on it 15 6 meters ; find the segments .
117 and 20 8 . Ans .
32. I II a rt . A ,0 2 36 5 meters ; a + b z 511 meters ;
find a and l) . a 219 meters,6 z 292 meters . Ans .
33. The sides of a triangle are 45 5 ,6-3
,and 44 45 ; the
perim eter of a similar triangle is 4 37. Find its sides .
13,1-8
,and 12 7. Ans .
34. A lamp-post 3 meters high is 5 meters from a man 2
meters tall find the length of his Shadow .
35. The sides of a pentagon are 12,20
,11
,15
,and 22 ;
the perimeter of a similar pentagon is 16 meters . Find itssides . 24
,4
,2-2
,3,and 44 Ans .
36. Through the point of intersection of the diagonalsof a. trapezoid a line is drawn parallel to the parallel Sides .
Prove that the parallel sides have the sam e ratio as the
parts into which this line cuts the non-
parallel sides.
1612—
k
wea, le.
37. Find the Side of a regular decagon.
If a radius is divided in extrem e and mean ratio,the
greater segment is equal to a side of the inscribed regulardecagon.
W . 394 ; (Eu . Iv. 10 ; Cv. v.
kw) 72
10
21
r2
r lc
L2 r le r2,
L2 rk 111-
72
244 2
.
7510 Z
154 MENSURATION .
38. Prove that the Sides of the regular pentagon,h exa
gon,and decagon w ill form a right-angled triangle ; or
73102
‘ l‘ 1562
'I052
39. Show
40. Show that k, r \/2 .
41 . If a,,is the apothegm of a regular inscribed n-
gon,
prove a ?”
HINT . Use 7 and 2.
42. If also be perimeter of inscribed polygon, in
”of
o
Ci rcumscribed,prove P
'zn
2
43. pn zp'n z zr z an.
44. Indicate how to reckon 77' by 40 ,
41,and 42
,or by
132.+ 19.and p
'gn Vp
'npgn .the following pg”
13 .
45. When a quantity increases continually without becoming greater than a fixed quantity ,
it has a limit equalor inferior to this constant .
156 MENSURAT ION .
55. The hypothenuse is 10 ,and one side is 8 ; sem icircles
are described on the three Sides . Find the radius of thesemicircle whose circumference equals the circumferencesof the three semicircles so described . 12 . Ans .
56. Find the radius of a Sphere in which a section 30
centimeters from the center has a circumference of 251-2centimeters . 2512 2
Ans
EXERCI SES AND PROBLEMS ON CHAPTER II .
20 . Are measures 25 a t center .
57. Find the third angl e Of a triangle Whose first angleis 12
°
56"and second angle Ans .
58. The angles of a triangle are as 1 : 2 3 . Find each .
Ans .
59. Of the three angles of a triangle Xa is 12°
20 ' smaller than 2; fl,
and 2; y is smaller than >7CB. Find
each . a z 53°
41', B z 66
°
1 ’, y
z 60°
18 ’. Ans .
60 . The sum of two angles of a triangle is 174°
the difference of the same two is Find allthree . Ans .
61. Three angles of a quadrilateral are 125°
127°
find the fourth . Ans.
62. The angles of a quadrilateral are as 2 3 4 7 find
each . 67°
157°
Ans.
63. In what regular polygon is every angle 168°
A 32—gon. Ans .
64. The vertical angle of an isosceles triangle isfind the others . 15
°
Ans .
EXERCI SES AND PROBLEMS .157
65. The sum Of two angles adj acent to one side of an
isosceles triangle is 35° find the three angles .
and Ans .
66 . Find the circular measure of the angle subtended bya circular watch-spring 3 millimeters long and radius 13millimeters .
67. If a perigon be divided into n equal parts,how many
of them would a radian contain ? nAns .
2 71'
2 2, zz fif/i360
°
68. Find the arc pertaining to a central angle of 78°
when r : 15 meters . l : 2 ' 042 m eters . Ans .
69. Find the arc intercepted by a central angle of 36°
25 ’
when r z 8 5 meters . Z : 5-39 meters . Ans .
70. Find an arc of 112° which is 4 meters longer than itsradius . l : 8 189 meters . Ans .
71. When cl : find are 4 6 meters long .
2; r Ans .
72. Calling 71 -
7find r when 64° measure 704 meters .
r z : 63 meters. Ans .
HINT . 2 7rr : 360 : 64.
24, 25, 26 27, and 28 0
73 . Find the compl ement , supplement , and explement of
158 MENSURAT ION .
74 . Find the angle between the bisectors of any two ad
j acent angles.
75. If a m edial equals half its base i ts angle is right .
76. If one angle in a r ight-angled triangle is one
side is half the hypothenuse .
77. In every isosceles right triangle half the hypothenuseequals the altitude upon it .
78. One angle in a right triangle is into what partsdoes the altitude divide the right angle ?
79. How large is the angl e between perpendiculars on
two Sides of an equilateral triangle ?
80. Find the inscribed angle standing on an arc ofAns .
81 . Find the inscribed angle cutting out one-tenth cir
cumference . Ans .
82. Find the angle of intersection of two secants whichinclude arcs of 100° 48 ' and 54° 23
°
Ans .
83 . An angle made by two tangents is measured by thedifference between 180
°
and the smaller intercepted arc .
84. From the same point in a circumference two chordsare drawn which cut off respectively arcs of 120° and
find the included angle .
85. The four angle-bisectors of any quadrilateral froma quadrilateral whose opposite angles are supplemental .
29 . u j ig
180
86. Find the circular measure of “73303 . Ans.
87. Find the circular measure of “785398 . Ans .
160 MENSURATION .
102. Find the number of degrees,minutes
,and seconds
in an angle whose n z fi. Ans .
103. The circular m easure of the sum of two angles is
T5—2- 7r
,and their difference is find the angles.
27°
15' and Ans .
104. Express in degrees the angle whose n z : rim :
Ans .
105. How many degrees, m inutes, and seconds are therein the angle whose circular measure is at?
47°
nearly . Ans .
106. Express in degrees and circular measure the verti~
cal angle of an isosceles triangle which is half of each of theangles at the base .
107. How many times is the angle between two consecu~
tive sides of a regular hexagon contained ( 1) in a right
angle ? (2) in a radian ? ( 1) 2, (2) 23
Am ,
7?
108 . Two wheels with fixed centers roll upon each other,
and the circular m easure of the angle through which one
turns gives the number of degrees through which the otherturns in the same time ; find the ratio of the radii of the
wheels . 180Ans .
7T
9‘
[ 180°
7rg°
109. The length of an arc of 60° is 363 ; find the radius .
7" 35 . Ans .
110. Find circumference where Z 30°
is subtended byare 4 meters.
111. If ll be the length of an arc of 45° to radius 73 , and
Z2 the length of are 60° to prove 3 7 3 72 r : 47 2 ll .
EXERCI SES AND PROBLEMS .161
EXERCI SES AND PROBLEMS ON CHAPTER III .
32. R z 035.
112. The area of a rectangle is square meters ; thediagonal measures 775 m eters . Find the sides .
5 2 465 meters,0; z 62 m eters . Ans .
HI NT . a2 52 I .
ab I I .
Substitute in I . the value of a, from I I . Th is gives a biquadraticsoluble as a quadratic . Or
,to avoid the solution of a quadratic ,
multiply I I . by 2,th en add and subtract from I . This gives
a2 52 —25
,
a” 52 240-25 .
a b 1085,
a b 155 .
113 . Find the perimeter and area of a rectangle whosealtitude a -02 m eters and base 6 8456 meters .
p meters,
R -712 square meters . Ans .
114. Find the number of boards 4 m eters long,0 5 me
ters broad,necessary to floor a rectangular room 16 m eters
long and 8 m eters broad . 64 . Ans .
115. Of two equivalent rectangles,one is 48 7 m eters
long and 2 84 m eters broad,the other is 4 26 m eters long .
How broad is it ? 3 246 m eters . Ans .
116. The perimeter of a rectangle is 245 4 m eters ; thebase is double the altitude . Find the area .
R 2 33 44562 square meters . Ans.
117. The difference of two Sides of a rectangle is 1 4 me
ters ; their sum ,8 2 meters . Find its area .
R 2 16 32 square m eters . Ans .
162 MENSURATION .
118. The base of a rectangle of 46 44 square m eters is3 2 meters longer than the altitude find these dim ensions .
5 8 6 m eters,a Z 54 m eters . Ans .
HINT . b a 3 2 meters .
d (a 3 2) 46 44,etc .
119. The perim eter of a rectangle is 13 m eters longerthan the base ; the area is 20 88 square m eters . Find thesides . 58 and 36 m eters
,or 72 and 2 9 meters . Ans .
120. The perim eter of a rectangle is 3 78 m eters ; its diagonal
,13 5 . Find the area .
R square meters . Ans .
121. A rectangular field is 60 m eters long by 40 m eters
w ide . It is surrounded by a road of uniform w idth , thewhole area of which is equal to the area of the field . Findthe w idth of the road . 10 m eters . Ans .
122. A rectangular court is 20 m eters longer than broad ,and its area is square meters ; find its length and
breadth. 78 and 58 m eters . Ans .
33 . gz : 52 .
123. Find the area of a square whose Side is 15 4 m eters .
237-16 square m eters . Ans .
124 . The area of a square is 8 square m eter ; find its side .
m eters . Ans .
125. The side of a square is a ; find the Side when thear ea is n tim es as great . a l
2 n . Ans .
126. The sum of two squares is 900 square m eters, the
difference 252 square m eters ; find the sides .
24 and 18 meters . Ans .
HINT . “1
032
2z : 909
,
a2
2 252.
576 .
164 MENSURATION .
134. From any point in an equilateral triangle the threeperpendiculars on the Sides together equal the altitude .
135. Find the area of a right triangle whose two Sidesare 248 2 and 160 5 meters .
square meters. Ans .
136 . If 18 4 meters is the altitude of a A 125-36 squarem eters
,find 6. 13 626 meters . Ans .
137. The two diagonals of a rhombus are 8 52 and 6 38
meters ; find the area . square meters . Ans .
138. The altitude of a triangle is 8 meters longer than itsbase
,and area I S 44 02 square m eters ; find b.
6 2 meters . Ans .
HINT .
139. The altitude Of a right triangle cuts the hypothe
nuse into two parts, 7 2 and 16 2 m eters long ; find thearea . 126 36 square meters . Ans .
140. 2 logA= log s+ log (s
— a ) log (s - b) log (s
141. If Z) is the base of an equilateral triangle , find the
area .
Here
The altitude of an2eq
u
ilateraltriangle I S 8 5 meters ;
find the area A 41 71 square m eters . Ans .
143 . The area of an equilateral triangle is 50 0548 square
meters ; find the Side . 34 . Ans.
EXER CI SES AND PROBLEMS . 165
144 . The side of an equilateral triangle is 4 meterslonger than the altitude ; find both .
2 2 258 56,5 2 298 56 . Ans.
2I‘I INT .
b2 (bI
;
b” 32 93 64.
145. The three Sides of a triangle are 10 2,136
,and 17
meters ; find the area . 69 36 square meters . Ans .
Here
8 204 . A x/2o4 (2O-4 102 ) 2 136 ) (204
146 . By measurem ent , 6 2 meters,6 2 48 72 me
ters,c 2 56 46 meters ; find A .
H°1 °
log s 1 85236
log (3 a ) 15 3148
log (s b) 13 5141
log (s c) 1 16791
logA2 59 0316
logA 2 2 95158
A 2 8945 0 square meters . Ans .
147. If three arcs , whose radu are 3, 2,1,at their centers
subtend angles of and intersect each otherat their extrem ities, prove that the Sides of the triangleformed by their chords are 3, 2
, V3 ; and its area
HI NT . The perpendicular from vertical 2C 1200 Of isosceles Aequals half a side, since joining its foot with midpoint of side makesan equilateral A .
148. The area of a triangle is 1012 ; the length of theSide a is to that of b as 4 to 3
,and c is to 5 as 3 to 2. Re
quired the length of the Sides .
2 2 52 470,b 2 393 53
,e 2 59 029 . Ans .
166 MENSURAT ION .
Here
s - b
s — a = 2— 3f§1—O2
T7Q- b
,
8 4 2 446,
3 C 2 .3113s 2
I°2 b
VSS55 52 1012 x 144
941 01 b?
149. The area of a triangle is 144,and one of two equal
sides is 24 ; find the third Side,or base.
Here
s = 24 + and s — a = s — c = 21—b. s — b = 24 - } b.
144 x/ (24280
2
) :102
.
150. Show that , in terms of its three m edials,
A 2 11V2 {127722 202
2232 223
2212
4023 if 3 (a2 b
‘2
16 (,i 2 1, 2 i
1
2 i3
2 43
2
) 9 (a2b2 (11
202
16 ( i14 0
2
“ 03
4
) 2 9 (a4 54
But,by multiply ing out 36 , we have
A l\/2 (a
2b2 C202 bee?) (a
4 64 645.
151 . Prove that the triangle whose Sides equal the me
dials of a given triangle is three-fourths of the latter .
168 MENSURAT ION .
TABLE I I .
-C’ontinued.
Area . Area .
136 183
231
EXERCI SES AND PROBLEMS .
TAB LE I I .
—Continued.
257
169
zi rea .
170 MENSURAT I ON .
TAB LE I I .
— Ooneluded.
Sides . Area .
152. Any right triangle equals the rectangle of the segments of the hypothenuse made by a perpendicular fromcenter of inscribed circle .
153. In any triangle AB O,let M be the m id point of the
base AO,I the point of contact of the inscribed circl e
,H
and I f the points where the perpendicular from the vertexB
,and the bisector of the angle B meet AO; prove the
relation Al f . I f ] : Mff . If ] .
172 MENSURAT I ON .
164. Find the radius circumscribing the equilateral triangle whose base equals 86 6 m eters . 5 meters . Ans .
165. In every triangle , the sum of the perpendiculars fromthe center of the circumscribed circl e on the three sides isequal to the sum of the radii of the inscribed and circumscribed circles .
166. If from the vertices of an equilateral triangle per
pendiculars be drawn to any diam eter of the circle circumscribing it
,the perpendicular which falls on one Side of this
diam eter will be equal to the sum of the two which fall onthe other side .
167. If the altitudeof an isosceles triangle is equal to itsbase
,I) 2 ER.
168. If A '
,B '
,0 ’ be the feet of the perpendiculars from
the angles of a triangle upon the sides,prove that the
radius circumscribing AB C is twice the radius circumscribing169. If a
, ,8
, y be the perpendiculars from the center ofthe circumscribing circl e upon a
,b,c,the sides of a triangle
,
Prove abc b
aflr [3
A39 . r l
z8 a s b s c
170. If the sides of a triangle be in arithmetical progression
,the perpendicular on the m ean side from the opposite
angle,and the radius of the circle which touches the m ean
side and the two other Sides produced , are each three timesthe radius of the inscribed circl e .
171. Each of the common outer tangents to two circlesequals the part of the common inner tangent interceptedbetween them .
EXERCI SES AND PROBLEMS . 173
172. Each tangent from A to the c ircle escribed to a
equals 3 ; from B to circle escribed to a equals 8 c ; from0 to circle escribed to a equals 3 — 6. Sim ilar theorems
hold for the escribed circles which touch 6 and c .
173. The area of a triangle of which the centers of the
escribed circles are the angular points is c
gbc
174. If a ; b, 0 denote the Sides of a tri angle ; kl , 712, kg thethree altitudes ; 91 , 92, Q 3 the Sides of the three inscribedsquares
, prove the relations
91 hi a 92 hz b 93 ha 0
175. Prove
176 . Prove
173 7
4
3177. Prove r ial
r l r r3
178. If 7 ,7 7, 7 8, 7 9 are the distances from the center of the
circumscribed circle to the centers of the inscribed and
escribed circles, prove the relations
HINT .— 2 r51t ; T
2z flt2 + 2 r flt
174 MENSURAT ION .
abc
2 (a b c)’
abc
4 (s — b);
179. Prove
180. Prove rl r2 73,z : r 4 SR.
181. In any triangle prove 82 81
2z : rrl ; etc .
40 . T : xy1+ y2
2
182. The base of a triangle is 20 meters,and its altitude
18 m eters . It is required to draw a line parall el to th ebase so as to cut off a trapezoid containing 80 squaremeters . What is the length of the line of section,
and its
distance from the base of the triangle ?Calling 192 the line of section,
and x its distance from bl ,
T : 80 :
— x : 18 .
l g(18 w) = (18 — x) = 20 —1—Q :v.
8
6
0 :
Z ( 9O+ 2O —19°
720 = 18Oa3 - 5 cv2 .
x2 ~ 36 3: z 144.
3: 18
as 18 - 13 4 584,
5 2 20 — 1
183. In a perpendicular section of a ditch , the breadth atthe top is 26 feet , the slopes of the Sides are each and
the area 140 square feet . Required the breadth at bottom
and the depth of the ditch .
176 MENSURAT ION .
187. T : square meters,a z : 124 m eters
,51 r : 186
meters . Find the other parallel side .
22 5 9 meters . Ans .
188. The altitude and two parallel sides of a trapezoid are
2 3 5,and T 2 1270 08 square m eters . Find the parallel
sides . 51 z 68 m eters 52 z 878 m eters . Ans .
189. The triangle form ed by j oining the mid-point of oneof the non-parallel sides of a trapezoid to the extrem ities ofthe opposite side is equivalent to half the trapezoid .
190. The area of a trapezoid is equal to half the productof one of its non-parallel sides, and the perpendicular fromthe mid-point of the other upon the first .
191. The line which j oins the mid-points of the diagonalsof a trapezoid is parallel to the bases
,and equals half their
difference .
192. Cutting each base of a trapezoid into the same num
ber of equal parts,and j oining the corresponding points
,
divides th e trapezoid int-o that,number of equivalent
parts .
193. If the m ean line of a trapezoid be divided into nequal parts
,and through these points lines , not intersecting
w ithin the trapezoid ,be extended to its bases
,they cut the
trapezoid into n equal trapezoids .
194. In every trapezoid ,the difference of the squares of
the diagonals has to the difference of the squares of the non
parallel sides the same ratio that the sum of the parallelsides has to their difference .
195. Let 51 be the longer , 52 the Shorter , of the two parallelSides in any trapezoid , 21 and 22 the other two sides
,and take
EXER CI SES AND PROBLEMS . 177
5
From the intersection point of 52 and 22 draw a line
Prove
parallel to 21 ; the base of the triangle so formed is (51 52)and its other sides are 21 and 22 .
196. The two parallel sides of a trapezoid are 184
68 meters ; the two others,84 and 72 m eters . Find the
area . 6586 squar e m eters . Ans .
197. The diagonal of a symmetric trapezoid is Veg 51 122 .
198. The altitude of a trapezoid is 80 m eters ; the twodiagonals 110 and 100 m eters . Find the area .
square m eters . Ans .
HINT.
— a
199 . In a trapezoid a I 140,51 z 160
,52 : 120 m eters
if the area is halved by a line parallel to the bases , find itslength and distance below the Shorter base .
Z : 1414 2 m eters,( Z 2 : m eters . Ans .
Z bzz b
l~ b
zz z cl z u
T 91
‘ l‘ 63 a z 9800
2
I . 7 l 2 cl 840
I I . 120 ( Z Zd
Substitute th e value of ( Z from I . in I I .
200 . In a trapezoid 51 z 812,52 z 89
,21
r : 850,22z 287
m eters if cut by parallels to 5 into three Sim ilar trapezoids ,
178 MENSURAT ION .
find where the two parallels cut the sides,
the areasof the three trapezoids .
I fZ2 n
,then Z
,nb
,.
192
2Z1
nl2
n 52
.
2 3bl
— nll— n l
2_ n 5
2— 89n
n z 2 .
2 10
50 meters .
41 meters,etc .
7
41”
3T 3 : 551) (y, 551) (y, ,4) I t
(xn 581) - 1 yn+1) (mu 7 ” 551)“ i“
201. Find the distance between the points 1 and 2 .
B etween two points (xl yl ) , (xzyz) the distance
202. Find the sum of the two right trapezoids determ inedby the ordinates of the three points ( 12 8 , (78 9 ,
(24,
203. If the cross section of an excavation is a trapezoid,
b breadth of top,75 depth
,w ith side slopes an and n in 1
,
which m eans that one side falls 777. m eters vertically for oneon n
meter of hori zontal di stance then show T r : 5759
752
L 7nn
2° N : E[371 t 37280 ‘ 7 t flfs fgz/z JD~i~
1” (yn—l
204. Prove th at a polygon may be constructed when all
but three adj acent parts ( 1 Side and 2 g4s , or 2 Sides and
1 26) are given. What theorem for congruence of polygons
follows from this ?
180 MENSURAT ION .
I
o
4
85
8
5
85
m
g
84
4
2mo
i
4
a
s“a
t.
an
m
gm
g
84
£
2
o
EXERCI SES AND PROBLEMS . 181
209. Find the area of a hexagon from I ts coordinates
(719 , (512, (719 , (720 ,
2 Q 2 33192“ 932 372 333?/2d
" 373 374
210. The area of a quadrilateral inscribed in a circle isa+5+ c+ d
2 P ’
211 . If through the mid-point E of the diagonal B D of aquadrilateral AB OD ,
FEG be drawn parallel to the otherdiagonal AO
,prove that the straight line AG divides the
quadrilateral into two equivalent parts .
212. Show that two quadrilaterals whose diagonals con
tain the sam e angle are as the products of their diagonals .
213 . A circle of r is inscribed in a kite,and another of r '
in the triangle formed by the axis of the kite and the twounequal Sides ; Show that , if 2 Z be the length of the otherkite -diagonal , 1 1 1
r'
r Z
214. To find the area of any quadrilateral from one side
,and the distances
from that side of the other two vertices
,and the intersection-point of the
diagonals .
Given the side AB = 5,and the
ordinates from 0 , D ,andE namely ,
Parallel to B D draw GM to intersection with AD prolonged
,and drop yG, th e ordinate of M
Then6 .
But
Q
182 MENSURAT ION .
44. A12
215. The area of a triangle equals square meters ;one side equals 1124 meters . Find the area of a sim ilartriangle whose corresponding side equals 28 1 meters .
208 725 square meters . Ans .
216. The sides of a triangle are 8892 ,4865
, and 2919
meters . The area of a sim ilar triangle is 2098 14 squaremeters . Find its sides . 748
,56 1
,985 meters . Ans .
217. The areas of two Sim ilar triangles are 248 6 and
1827 square m eters . One Side of the first is 8 5 m etersshorter than the homologue of the second . Find thesesides . and 188 8 m eters . Ans .
£02
.
218. Two triangles are 216 6 and square meters,
and have an equal angle whose including sides in the firstare 76 and 57 m eters . The corresponding sides in seconddiffer by 2 7 meters. Find them .
108 and 8 1 meters . Ans .
219 . The areas of two sim ilar polygons are 468 7 and
1854 8 square m eters . A Side of the first is 15 m eterssmaller than the corresponding side of the other . Findthese Sides . 15 and 80 meters . Ans .
HINT . 468 7 : 1854 8 : 552 : (x
45 . N2 2
220 . The sum of perpendiculars dropped from any pointwithin a regular polygon upon all the sides is constant .
221 . In area,an inscribed 2n—
gon is a m ean proportional
between the inscribed and circumscribed n-gons .
184 MENSURATION .
47. C) :
227. There are three circles whose radii are 20 ,28
,and
29 m eters respectively . Required the radius of a fourthcircle whose area is equal to the sum of the areas of theother three .
91
1- 400 7
,784 5 , O 341 5
42 2025 77 r z : V2025 45 . Ans .
228. If a circl e equals 848 6 square m eters,find its radius .
88 m eters . Ans .
229. Two Q’
s together equal 7404 282 square m eters,and
differ by 688 8744 square m eters . Find radii .r 2 15 056 m eters and r
’ 8 m eters . Ans .
2 : 7404 282 .
TF/r2 688 -8744.
230. If ( 9 be the area of the inscribed circl e Of a triangle ,G) C) Q 3 the ar eas of the three escr ibed circles , prove
1 1 1 1
V653 V6
231 . If from any point in a sem I CIrcumference a per
pendicular be dropped to the diam eter , and sem icirclesdescribed on these segm ents
,the area between the three
semicircumferences equals the circle on the perpendicular
as diam eter .
232. The perim eters of a circle,a square
,and an equi
lateral triangle are each of them 12 m eters . Find the areaof each of these figures to the nearest hundredth of a squaremeter . 114 6
,9
,square m eters . Ans .
EXERC I SES AND PROBLEMS . 185
238. Find the side of a square inscribed in a semicircle .
% 7'V5. Ans .
234. An equilateral triangle and a regular hexagon havethe same perim eter ; show that the areas of their inscribedcircles are as 4 to 9 .
235. How far must the diameter of a circle be prolonged,
in order that the tangent to the circl e from the end of theprolongation may be m long ? 4m 2
( Z Ans .
48 . S = ft lr z
236 . Find the area of a sector of 68° 86 ’ When 7“Z 72 .
810 8898 square meters . Ans.
237. When circle equals 482 square m eters,find sector of
84°
100 8 square meters . Ans .
H INT . 432 : S 360 : 84%
238 . Find the number of degrees in the arc of a sectorequivalent to the square of its radius .
239. In different circles,sectors are equivalent whose
angles have a ratio inverse to that of the squared radii .
240. Find radius when sector of 7° 12’ is 2 square centinaeters .
241 . Find sector Whose radius equals 25,and the circular
m easure of whose angle equals 8. 2848 75 . Ans .
242. The length of the arc of a sector of a given circle is16 m eters ; the angle of the sector at center is t of a rightangle . Find sector . 488 9 square meters . Ans .
186 MENSURAT ION .
243. AB is a chord of a given circle ; if on the radius
0A,which passes through one of its extremities
,taken as
diam eter,a circl e be described
,the segments cut off from
the two circles by the chord AB are in the ratio ofto 1 .
that,if a. is the angle or arc of a segment
,for
.2
a : G1
7
2( Q w
— SVS) ;
G7
(4W— SV3) ;
12
0°
G72
a : 97Top
a
a . .
245. In a segment of to how many places of decimals
is our approximation correct ?
246 . Prove that there can be no segm ent with is : 120,
Zz 156,k 12 .
247. In a circle,given two parallel chords kl and 152, and
their distance apart 7 ; find the diam eter .
(412
‘ t 7522
” l Ans .
I f a: z W 2y a; and y
2 we H3,
11 . x2
ga g
188 MENSURAT ION .
252. Find the annulus between the concentric circumferences
,0 1
2 219 8 meters and 02: 18 84 meters
,taking
71 8 14 . A square m eters . Ans .
53 . S . Az 1 it ( Z1 Z2) .
253. To trisect a sector of an annulus by concentric circles .
54 . J : {4 Me.
254. What is the area of a parabola whose base is 18meters and height 5 m eters ? 60 square meters . Ans .
255. What is the area of a parabola whose base is 525meters and height 850 m eters
square m eters . Ans .
55 . E r
256 . The area of an ellipse is to the area of the circumscribed circle as the m inor axis is to the maj or axis .
257 . The area of an ellipse is to the area of the inscribedcircl e as the maj or axis is to the m inor axis .
258 . The area of an ellipse is a mean proportional betweenthe inscr ibed and Ci rcumscribed circles .
259 . What is the area of an ellipse whose maj or axis is
70 meters,and minor axis 60 meters ?
829536 7 square m eters . Ans .
260 . What is the area of an ellipse whose axes are 840
and 810 ? -896 . Ans .
EXER CISES AND PROBLEMS .189
EXERC ISES AND PROB LEMS ON CHAPTER IV
PoLYHEDEONs .
56 . s e z e 2 .
261 . The number of plane angles in the surface ofpolyhedron is twice the number of its edges .
H INT . Each face has as many plane angles as sides . Each
pertains , as side,to two faces .
262 . The number of plane angles on the surface of apolyhedron is always an even number .
263 . If a polyhedron has for faces only polygons w ith an
odd number of sides,e .g .
,trigons
,pentagons
,heptagons , etc .
,
it must have an even number of faces .
264 . If the faces of a polyhedron are partly of an even,
partly of an odd number of sides,there must be an even
number of odd-sided faces .
265 . In every polyhedron f; 8 (E.
H INT . Th e number of plane angles on a polyh edron can never beless than thrice the number of faces .
266 . In every polyhedron fi g <
4 (S.
267 . In any polyhedron (S—l 6 3 8 6 .
268 . In any polyhedron C5 6 3 8
269 . In every polyhedron (55 8 Q 7, and
270 . In a polyhedron not all the summ its are more than
five—sided ; nor have all the faces more than five sides .
271 . There is no seven- edged polyhedron .
190 MENSURAT ION .
272 . For every polyhedron 3s : 6 (C32 that is,the sum
of the plane angles is as many perigons as the differencebetween the number of edges and faces .
273 . For every polyhedron r é ( @ just as for everypolygon : f (n
274 . How m any regular convex polyhedrons are possible ?
275 . I n no polyhedron can triangles and three - faced sum
m its both be absent ; togeth er are present'
at least eight .
276 . A polyhedron w ithout triangular and quadrangularfaces has at least twelve pentagons ; a polyhedron withoutthree—faced and four- faced summits has at least twelvefive—faced .
5a p z a.
277 . In a right prism of 9 m eters altitude,the base is a
right triangle whose legs are 8 and 4 m eters . Find themantel .
192 MENSURAT I ON .
287 . Find a cylinder equivalent to a sum of right circularcylinders of the same height .
H INT . Find a radius whose square equals squares of the n radii .
288 . How much must the altitude of a right circularcylinder be prolonged to make its mantel equal its previoussurface
289 . A plane perpendicular to the base of a right cylindercuts it in a chord whose angle at the center is a ; find theratio of the curved surfaces of the two parts of the cylinder .
59 . Y : ‘i hp .
290 . The surface of any regular tetrahedron is to that ofthe cube on its edge as 1 to 2 .
291 . Each edge of a regular tetrahedron is 2 meters .
Find mantel .
292 . Each edge of a regular square pyram id is 2 meters .
Find surface .
293 . From the altitude a and basal edge 6 of a regularhexagonal pyramid , find its surface .
8 56 Ans .
294 . In a regular square pyram id , givenp ,the perimeter
of the base,and the area A of the triangle made by a basal
diagonal and the two opposite lateral edges ; find the surface of the pyramid .
60 . I f : t ck z w rk.
295 . The convex surface of a right cone areaof the base ; find the vertical angle .
Herec X —é—h z 2 c x —é r .
EXERCISES AND PROBLEMS . 198
Thus the section containing the axis is an equilateraltriangle ; so the angle equals Ans .
296 . Find the ratio of the mantels of a cone and cylinder
whose axis-sections are equilateral .
297 . Find the locus of the point equally distant from threegiven points .
298 . In a right cone
( 1) Given a and 9' find K VCL
2+ T
2. Ans.
(2) Given a and 74 ; find K
(8) Given K and IL ; findI f
wit
299 . In an oblique circular cone,given kl , the longest
Slant height , 712, the shortest , and a,the altitude find the
radius of the base . hf) Ans .
300 . How many square meters of canvas are required to
make a conical tent which is 20 meters in diameter and 12
meters high ?Here
K : wrh 31 4159 x 10 x V144 100 .
K : 814 159 15 6205
490-78204 square meters . Ans.
301 . Given a basal edge 171 , and a top edge 62, of the frustum of a regular tetrahedron ; also a
,the altitude of the
frustum . Find la,its slant h eight
,and F
,its mantel .
302 . Same for a regular four-sided pyram id .
F (b,
194 MENSURAT ION .
62. F : “ t“ 02) I "0 (7 3+ 5 2)
303 . In the frustum of a right circular cone,given 7 3,
and a ; find it .
304 . In the frustum of a right circular cone,on each base
stands a cone with its vertex in the center of the other base ;from the basal radii and rz find the radius of the circle inwhich the two cones cut .
Ans .
fr l 7 2
305 . Given 61, 62, the basal edges, and Z, the lateral edgeof a frustum of a regular square pyram id ; the frustum of acone is so constructed that its upper base circumscribes theupper base of this pyramid-frustum ,
while its lower base isinscribed in its lower base . Find the slant height of thecone- frustum . \/Z2 l 51
2
( 1 Ans .
306 . How far from the vertex is the cross-sectionwhichhalves the mantel of a right cone ? 7} a \/2 . Ans .
307 . Reckon the mantel from the two radii when the inclination of a slant height to one base is
7TVIZ. A728 .
308 . I f in the frustum of a right cone the diam eter of theupper base equals the slant height
,reckon the mantel from
the altitude a; and the per im eter p of an axial section.
8
71
6 (292+ 12 04
3 —13 29 V292
“ 12 a2
) . Ans .
63 . F : 2 7rd] .
309 . In the frustum of a cone of revolution,given 73 , 7 5, it ;
find a .
310 . Find the altitude of the frustum of revolution fromthe mantel la and the bases B 1 and B 2 .
196 MENSURAT ION .
65 . Z : 2 717 03.
321 . Out a sphere into n equal parts by parallel circles .
322. In a calot,
( 1) Given and a ; find 7 3 .
(2) Given 7”
and 71 ; find Z 1 .
(8) Given a and TI ; find Z .
(4) Given a and H ; find Z l . a \/H7r . Ans .
323 . In a segment 6 centim eters h igh,the radii of base
and top are 9 and 8 Find area of the zone .
86 ? 10 square centim eters . Ans .
324 . In a segment of altitude a,and congruent bases
,
calling the top and base radii find the zone .
325 . How far above the surface of the earth must a person
be raised to see one- third of its surfaceHere
and,by Similar triangles,
m+ r z r = r z r ~ a
§ r (m = r .
A7ZS .
326 . A lum inous point is distant 7' from a sphere of radius
how large is the lighted surface ? 2 7 7 2 77Ans .
327 . Find a zone from the radii of its bases 75 , and theradius of the sphere fr 2m [V 7
42 Ans .
328 . How far from the center must a plane be passed todivide a hem isphere into equal zone and calot ?
EXERCISES AND PROBLEMS . 197
66 . THEOREM OF PAPPUS .
329 . An acute-angled triangle is revolved about each Sideas axis ; express the ratio of the surfaces of the three doublecones in terms of a
,5,c,the Sides of the tr iangle .
034—5 d —l— C 5+ cAns
C 5 a
330 . The Sides of a symmetric trapezoid are 51 , 52, and 2 .
Express the surface generated by rotating the trapezoidabout one of the non-parallel sides .
3512 bl z 52 z) \/z2 4 (51 a)
? Ans.
67. 0 4 772717 2 .
331 . An equilateral triangle rotates about an axis w ithout it
,parallel to
,and at a distance a from one of its Sides 5.
Find the surface thus generated . 57r (5 8 —l 6 a ) . Ans .
332 . A rectangle w ith sides ( 6 and 5 is revolved about an
axis through one of its vertices,and parallel to a diagonal .
Find the generated surface .
4 56577 (a, “ l“ 6) Ans .
8 (L ) . SPHER ICS AND SOL ID ANGLES .
5 2 2 r2u .
Find the area of a lune whose angle is
% r27r
Find lune of 86° when r 12 6156 .
69 .
335 . A conical sector is one-fourth of a globe ; find itssolid angle . Ans .
Find the vertex-angle of an ax ial section. Ans .
H INT . By 65 , Cor . 2, generating arc GOO
198 MENSURATION .
70 . A Z 6 72
.
336 . If two angles of a spherical triangle be right,its
area varies as the third angle .
337 . In a cube each solid angle is one-eighth of a stere
gon. (For eight cubes may be placed together , touching ata point . )
338 . Find the ratio of the solid angle of a regular righttriangular prism to the solid angle of a quader . 2 : 8 . Ans .
339 . Find the ratio of the trihedral angles of two r egularright prisms of 7 n and n sides. ( 7 n 2) n Ans .
(n 2) 7n340 . Find the area of a spherical triangle from the
radius and the angles a l t - 20°
,8 55
°
y 114°20
’14" 0 1818 7
2. Ans .
341 . Given and a 78°,8 85
°
771 :
82°9’ find A . 0 18598 7 2. Ans .
342 . Given and a 2 114°20 '
,8 : 80
°57’
y 90°9
’ 41”-67 ; find A . 0 9678 7 2. Ans .
343 . Spherical triangles on the sam e base are equivalentif their vertices lie in a circumference passing through the
Opposite extrem ities of Sphere-diameters from the ends of
the base .
344 . All trihedral angles having two edges common,and
their third edges prolonga tions of elem ents of a right cone
containing the two common edges,are equivalent .
P roof : On the edges of a trihedral angle take SA , SB ,
SO equal ; and pass through the three extremities a circl eAB C of center 0 . J oin SO
,and suppose three planes to
Start from SO and to pass one through each edge of thetrihedral angle . These planes form three new trihedrals
900 MENSURAT ION .
345 . Equivalent spherical triangles upon the same base ,and on the same Side of it
,are betw een the sam e parallel
and equal l esser Circles of the sphere .
346 . The locus of B ,the vertex of a spherical triangle of
given base and area,is a lesser circle equal to a parallel
lesser circle passed through A and 0,the extremities of the
given base .
347 . Find the spherical excess of a triangle from its area
and the radius .A
Ans .
7"7r
348 . Find the ratio of the spherical excesses of two equiv
alent triangles on different spheres . 61 62z 7 2
2 Ans .
349 . A spherical triangle whose
a 91°
,8 at 120
°
9 ’
72 100
°
42’
contains square meters . Find the sphere .
square meters . Ans .
71. e f’ — (n
350 . Find the ratio of the vertical solid angles of tworegular pyramids of 777 and n Sides
,having the inclinations
of two contiguous faces respectively ,0. and B.
2 7: 777. ( 71 a ) Ans .
2 71 n (71 B)351 . What is the area of a Spherical pentegon on a Sphere
of radius 5 meters, supposing the sum of the angles 640°
48-688 square meters . Ans .
EXERC ISES AND PROBLEMS . 201
EXERCISES AND PROBLEMS ON CHAPTER V
72. U : d 5l.
352 . The diagonal of a cube is n ; find its volum e .
353 . Find the volum e of a cube whose surface is 89 402square meters .
354 . The edge of a cube is n ; approximate to the edgeof a cube twice as large .
355 . Find the edge of a cube equal to three whose edgesare 03
,5,Z.
356 . Find the cube whose volume equals its superficial
area .
357 . If a cubical block of marble , of which the edge is 1
meter , costs 1 dollar , what costs a cubical block whose edge
is equal to the diagonal of the first block .
8V8 dollars . Ans .
358 . In any quader ,
(1) Given a,5,and mantel ; find U.
(2) Given a,5,U ; find Z.
(8) Given U,B
,and find Z and 5.
(4) Given U, 7)
find a. and 5.
(5) Given (d l 5Z) ; find 0. and 5.
359 . If 97 centim eters is the diagonal of a quader with
square base of 48 centim eters Side , find its volum e .
Rm .( 72
202 MENSURAT I ON .
360 . What weight will keep under water a cork quader55 centimeters long , 48 centim eters broad
,and 97 centi
meters thick,density 0 24 ?
55 0572 kilograms . Ans .
361 . The volume of a quader whose basal edges are 12
and 4 meters is equal to the superficial area . Find its
altitude .
362 . In a quader of 860 square meters superficial area
the base is a square of 6 m eters edge . Find the volum e .
363 . A quader of 864 cubic centimeters volume has a
square base equal to the area of two adj acent sides . Findits three dimensions .
364 . In a quader of 8 m eters altitude and 160 squaremeters surface the base is square . Find the volume .
365 . The volum e of a quader is 144 cubic centimeters ; itsdiagonal 18 centimeters ; the diagonal of its base 5 centim eters . Find its three dimensions .
366 . In a quader of 108 square meters surface the baseequals the mantel . Find volum e .
367 . If in the three edges of a quader, which meet in an
angle,the distances of three points A
,B
, and 0 from thatangle be a ,
5,c ; then triangle AB C : {5d
—
52—l—a2024—5
202
.
368 . How many square m eters of metal will b e requiredto construct a rectangular tank ( open at top) 12 meterslong
,10 meters broad
, and 8 m eters deep . 472. Ans .
369 . The three external edges of a box are 8,2 52
,and
15 28 m eters . It is constructed of a material 0 1 meters inthickness . Find the cubic Space inside .
8 594208 cubic m eters . Ans .
204: MENSURATION .
379 . The vertical ends of a hollow trough are parallelequilateral triangles
,with 1 meter in each Side
,the bases
of the triangles being horizontal . If the distance betweenthe triangular ends be 6 meters
,find the number of cubic
meters of water the trough will contain.
2 598 cubic meters . Ans .
V . O 2
In a right circular cylinder ,
( 1) Given a, and c ;
(2) Given a and O;
(8) Given (V . 0 ) and 0 ;
V . C ar27r
,
v. C 0“
av. C
737: 2 7 71 0
381 . If 0 : 918 4 square meters,and V . 0 = 145 cubic
meters,find a . 0: 2 4 628986 meters . Ans .
382 . A right cylinder of 50 cubic centimeters volume has
a circumference of 9 centim eters ; find mantel and volum e .
383 . In a right cylinder of 8 cubic centim eters the man
tel equals the sum of the bases ; find altitude .
384 . If,in three cylinders of the sam e height , one radius
is the sum of the other two, then one curved surface is the
sum of the others, but contains a greater volume .
385 . Find the ratio of two cyl inders when the radius of
one equals the altitude of the other .
386 . Find the ratio of two cylinders whose mantels areequivalent .
EXER C ISES AND PROBLEMS . 205
387 . If 1728 cubic meters of brass were to be drawn into
w ire of one-thirtieth of a meter in diam eter , determine thelength of the wire .
Here 1 2 an
1728 aQéB) 3600.
1728 X 3600meters . Ans .
7,
388 . What must be the ratio of the radius Of a rightcylinder to its altitude , in order that the axis- section may
equal the base ? 2 Ans .
389 . A cylindric glass of 5 meters diameter holds half aliter ; find its height .
390 . A rectangle whose sides are 8 m eters and 6 m etersis turned about the 6-meter side as axis ; find the volumeof the generated cylinder .
391 . The diagonal of the axis-section of a right cylinderis 5 centimeters ; the diam eter of its base is three-fourthsits height . Find its volum e .
392. In a right cylinder , from A ,the area of the axis
section,reckon the area of that section which halves the
basal radius normal to it . 5A\/8 . Ans .
393 . The longest Side Of a truncated circular cylinder of15 meters radius is 2 meters ; the shortest
,1 75 meters .
Find volume .
394 . If a room be 40 meters long by 20 meters broad ,wh at addition will be made to its cubic contents by throwing out a semicircular bow at one end ?
cubic m eters . Ans .
395 . The French and German liquid measures must becylinders of altitude twice diam eter . Find the altitude formeasures holding 2 liters
,1 liter
,and 8 liter .
216 7, 172 1, and 136 5 millimeters. Ans.
206 MENSURAT ION .
396 . The German dry measures must be cylinders of alti~
tude two-thirds diam eter . Find diam eter of a m easurecontaining 100 liters . 575-9 m illim eters . Ans .
397 . In the French grain measure the altitude equalsdiameter . Find for hectoliter . 508 7 m illim eters . Ans .
77. V O1 — ‘ V
398 . How many cubic m eters of iron are there in a roller
which is half a meter thick ,with an outer circumference of
61 m eters,and a width of 87 meters ? 2— 2
1858 cubicmeters . Ans .
399 . Find the amount of m etal in a pipe 8 1881 m eterslong
,w ith 12 meters and 7 2 8 m eters .
800 cubic meters . Ans .
400 . The amount of metal in a pipe is 175 9292 cubic
meters,its length is 8 5 m eters, and its greater radius is
5 meters. Find its thickness . 2 meters . Ans .
SECTIONS SIMILAR .
401 . A regular square pyram id , whose basal edge is 5,is so cut parallel to the base that the altitude is halved ;find the area of this cross-section .
402 . A section parallel to the base of a cone (base radius
its altitude in the ratio of m to 77 . Find the area of this777 7
271"
secti on . Ans .
403 . On each of the bases of a right cylinder , radius 7 ,stands a cone whose vertex is the center of the other base .Find the circumference in which the cone-mantels cut .
Ans .
208 MENSURAT ION .
410 . A cone and cylinder have equal surfaces,and their
axis-sections are equilateral find the ratio of their volumes .
2
Surface of cylinder2
cl ?3 ( i
)
71'
2 2 2
Surface of coneh 7 h 7" 3 77 M
4 2 4
h d\/2.
411 . In a triangular prism of 9 m eters altitude,whose
base has 4 square m eters area and 8 85487meters perimeter,
a cylinder is inscribed . Find the base and altitude of anequivalent cone whose axial section is equilateral .B 17 1286 square meters
,at 4 048788 m eters . Ans.
412 . Find the edge of an equilateral cone holding a liter .
16 4 centim eters . Ans .
413 . Halve an equilateral cone by a plane parallel to thebase .
414 . Find the ratio of the volumes of the cones inscribed
and circumscribed to a regular tetrahedron whose is n .
82 . PR I SMOIDAL FORMULA : D 71;MB , 4M
415 . Find the volume of a rectangular prismoid of 12meters altitude , whose top is 5 meters long and 2 metersbroad and base 7 m eters long and 4 meters broad .
220 cubic meters . Ans .
416 . In a prismoid 15 meters tall , whose base is 86 squaremeters
,the basal edge is to the top as 8 to 2 . Find the
volume . 880 cubic meters . Ans .
EXERC I SES AND PROBLEMS . 209
417 . Every regular octahedron is a pr ismatoid whosebases and lateral faces are all congruentequilateral triangles . Find its volum e interms of an edge 5. 53 3
13 Ans .
418 . The bases of a prismatoid are con
grueri‘
t squares of side 5,whose Sides are
not parallel ; the lateral faces are eightisosceles triangles . Find the volum e .
saws vs) . Ans .
419~ If,from a regular icosahedron
,we
take off two five—sided pyram ids whosevertices are opposite summits
,there re
mains a solid bounded by two congruentregular pentagons and ten equilateral triangles . Find its volum e from an edge 5 .
4 aw . ) Ans .
420 . B oth bases of a prismatoid of altitude a are squares ;the lateral faces isosceles triangles ; the sides of the upperbase are parallel to the diagonals of the lower base , andhalf as long as these diagonals ; 5 is a side of the lower base .
Find the volum e . 3—6452
. Ans .
421 . The upper base of a prismatoid of altitude a r : 6 is
a square of Side 52 2 7 07107 ; the lower base is a square ofSide 51 I 10
,with its diagonals parallel to Sides of the upper
base the lateral faces are isosceles triangles . Find volum e .
40 (1912 J
r 61 52V2 522
) t : 500 . Ans .
422 . Every prismatoid is equivalent to three pyram ids ofthe same altitude w ith it , of which one has for base half the
sum Of the prismatoid’
s bases, and each of the others its
midcross section .
210 MENSURAT ION .
423 . Every prismoid is equivalent to a prism plus apyramid , both of the same altitude w ith it
,whose bases
have the same angles as the bases of the prismoid ; but thebasal edges of the prism are half the sum
,and of the pyram id
half the difference,of the corresponding sides of both the
prismoid ’
s bases .
424 . If the bases of a prismoid are trapezoids whose midlines are 51 and 52, and whose altitudes are and ( 72,
5 5 0 — 0 5 — 5D z a,
I g
2 2 2 2
83. V . F 23 03(B i
“ l‘ VEIB 2‘ 1‘ B 2)
425 . A side of the base of a frustum of a square pyram idis 25 m eters
,a Side of the tOp is 9 meters , and the height is
240 meters . Required the volume of the frustum .
Here V . F 2l 240 (625 225 81)80 X 981 cubic meters . Ans .
426 . The sides of the squar e bases of a frustum are 50
and 40 centimeters ; each lateral edge is 80 centim eters .
Find the volume . 59 28 liters . Ans .
427 . In the frustum of a pyramid whose base is 50 squaremeters
,and altitude 6 m eters
,the basal edge is to the cor
responding top edge as 5 to 8 . Find volume .
588 cubic meters . Ans .
428 . Near Memphis stands a frustum whose height is142 85 m eters
,and bases are squares on edges of 185 5
and 8 714 meters . Find its volum e .
429 . In the frustum of a regular pyram id , volume is 827cubic m eters
,al titude 9 meters
,and sum of basal. and tOp
edge 12 meters . Find these . 7meters and 5 m eters . Ans .
212 MENSURATION .
434 . In a frustum of 8 meters altitude and 68 cubic mo
ters volum e , 7 1 z 2 712 ; find 74
2.
8 1 Ans .
435 . In the frustum where a z 8 meters,7 1
: 10 meters,
6 m eters,the altitude is cut into four equal parts by
planes parallel to the base . Find the radii of these sections.
H INT . The altitude of the completed cone is 8 12 : 20,and of
the oth ers , 18, 16 , 14 by similar triangles,
7,8,9 meters. Ans.
436 . The frustum of an equilateral cone contains 2 hec
toliters, and is 40 centimeters in altitude . Find the radii .277 85 and 50 879 centimeters . Ans .
85 . PR I SMOIDAL FORMULA : D 6 0 (B 1 4 11./H B 2) .
437 . A solid is bounded bythe triangles AB O
,OB B
,the
parallelogram AODE,and the
Skew quadrilateral B A E B
whose elements are parallel tothe plane B OD . Find its volume. % CL . AB G Ans .
The Skew quadrilateral is part Of a warped surface called the
hyperbolic paraboloid.
438 . A tetrahedron is bisected by the hyperbolic paraboloid whose directrices are two opposite edges
,and whose
plane directer is parallel to another pair of Opposite edges .
439 . A solid is bounded by a parallelogram ,two Skew
quadrilaterals,and two parallel triangles ; find its volume .
0 (A 1 A2) . Ans .
EXERC ISES AND PROBLEMS . 218
440 . Tw ice the volume of the segm ent of a ruled surfacebetween parallel planes is equivalent to the sum of thecyl inders on its bases
,dim inished by the cone whose ver
tex is in one of the parallel planes,and whose elem ents are
respectively parallel to the lines of the ruled surface .
86 . W :6 0 70 (2 51
441 . A wedge of 10 centimeters altitude,4 centim eters
edge,has a square base of 86 centim eters perimeter . Find
volume .
442. The three parallel Sides of a truncated prism are 8,
9,and 11 meters . The section at right angles to them is a
right-angled triangle , with hypothenuse 17 m eters, and one
Side 15 meters . Find volume .
443 . The volum e of a truncated regular prism is equal tothe area of a right section multiplied by the axis or mean
length of all the lateral edges .
V=All
[2
71
444 . To find the volume of any truncated prism .
Ru le Multiply the length of each edge by the sum Of
the areas of all the triangles in the right section which have
an angular point in that edge . The sum of the products
will be three tim es the volum e .
Formu la. 8 V : EA l .
87. X :
445 . Given V,the volume of a parallelepiped ; in each
of two parallel faces draw a diagonal , so that the two diag
ouals cross .Take the ends of these as summits of a tetra
hedron,and find its volume . 2V. Ans.
214 MENSURAT I ON .
88 . V . H —3—7r7 .
446 . Find the volume of a Sphere whose superficial areais 20 m eters .
447 . Find the radius of a Sphere equal to the sum of twospheres whose radii are 8 and 6 centimeters .
9 centim eters . Ans .
448 . Find the radius of a golden globe , densityweighing a kilogram .
449 . A solid metal globe 6 meters in diameter is formedinto a tube 10 meters in external diam eter and 4 m eters inl ength . Find the thickness of the tube . 1 meter . Ans .
450 . If a cone and hemisphere of equal bases and altitudes be placed with their axes parallel , and the vertex ofthe cone in the plane of the base of the hem isphere , and becut by a plane normal to their axes, the sum of the sections
will be a constant .
ARCHIMEDES’
THEOREM .
451 . Gone , hemisphere , and cylinder , of same base and
altitude,are as 1 2 : 8 .
452 . The surfaces and the volum es of a sphere , a circum
scribed right cylinder , and a circumscribed right cone whoseaxial section is an equilateral triangle , are as 4 6 9 . Therefore
,the cylinder is a geometric mean between the sphere
and cone .
H INT . H ==4 73 77 . V . I I 47371:
O+ ZE = 6 72 . V . 0 = -g- 73
.
K + B 9 7275: V . K
4537 A quader having a square base of 5 centim etersedge
,is filled 28 centimeters high with water . Into it is
216 MENSURAT ION .
458 . Find the volum e of a segm ent of 12 centimetersaltitude
,the radius of whose single base is 24 centimeters .
7“z : 80 centimeters
V . G °014976 7r cubic meters . Ans .
459 . In terms of sphere-radius, find the altitude of acalot n times as large as its base .
2 7 . Ans .
460 . Find the ratio Of the volum e of a sphere to the volume of its segment whose calot is n times its base .
V . H : to V . G : : n3: (n Ans .
461 . Find volume of a segment whose calot is 15 085Square m eters, and base 2 m eters, from Sphere-center .
V . G 2 5 787025 cubic meters . Ans .
462 . In a Sphere of 10 centim eters radius,find the radii
7 1 and 7 2 of the base and top of a segment whose altitude is6 centim eters
,and base 2 centim eters
,from the sphere
center . 71 centim eters, 7 2r: 6 centimeters . Ans .
463 . Out of a globe of 12 centim eters radius is cut a segm ent whose volume is one- third the globe
,and whose bases
are congruent and 8 centim eters apart ; find the radius ofbases .
7 2: 4 2 centimeters . Ans .
90 . V . S T : 8 71-6772
464 . In a Spherical sector,
( 1) Given 7 1 , find V . S .
(2) Given 7 2 ; find V . S .
465 . In a sphere of radius 7,find the altitude a seg
ment which is to its sector as n to 777 .
EXERC I SES AND PROBLEM S . 217
466 . A sector is 1 of its globe,whose diameter is cl find
n7rCl 3 8n — 2
the volume of I ts segment . V . G6 3
Ans .
n
467 . A sphere of given volume V is cut into two seg
ments whose altitudes are as 777 to n ; find both calotsand 22, and the segm ents .
2. Q /ssfl-V
Z;
(7n n)3
(m n)3
91
468 . Find the volume of a spherical ungula whose radiusis and 31 18
°
92 958 . Ans .
469 . Z 7“ 18 2 . Find 17. 6984 5 . Ans .
470 . A lune of 192 square meters has radius 15 meters ;find volume of the ungula . 960 cubic meters . Ans .
471 . Given L and find v . Ans .
270 6472 . Given 1) and find 31.
473 . Given L and 4 ; find
92. Y : t 7'3e.
474 . In a spherical pyramid given the angles of its triangular base , y
i 108°
15',and
given findY. 1106 61 . Ans .
475 . Given a, B, y , and A ; findY. Ans .
218 MENSURAT ION .
476 . A 486,
a 84°
s 96°
y 112°
20'
find Y. 2548 06 . Ans .
477 . Given8 112
°
find the four—facedY. 5114 88 . Ans .
THEOREM OP PAPPUS .
478 . If an equilateral triangle whose sides are halvedby a straight line rotates about its base
,the two volumes
generated are equivalent .
479 . A trapezoId rotates first about the longer,then
about the shorter , of its parallel sides ; the volumes of thesolids generated are as 777. to n . Find the ratio of theparallel sides . 2” V 777
Ans .
2m n
94. V . O 2 7132 57:
480 . Find the volume of a solid generated by rotating aparallelogram about an axis exterior to it ; given the areaof the parallelogram D
,and the distance 7 of the inter
section-point of its diagonals from the axis . Ans .
481 . The volume of a spiral spring , whose cross—section
is a c ircle , equals the product of this generating circle bythe length of the helix along which its center moves .
The helix is the curve traced upon the surface of a cir
onlar cylinder by a point , the direction of whose motion
makes a constant angle with the generating line of the
cylinder .
482 . A regular hexagon rotates about l, one of its Sides ;find the volume generated . gl
37r . Ans .
220 MEN SURAT ION .
98 . I“
2 23' [552 (B 1 3 3) ‘ i‘ 373 (B 2 B 4) etc .
513 (B Bn-f-l ) “ h xn+ 1 <B n
‘ l‘ B n+ 1>l
4” fi‘“ i" (5133 ”
ROM ”
1L (x4 [EQ M “ i‘ etc .
mm.
489 . If the areas of six parallel planes 2 meters apartare l
,3
,5,7,9
,11 square m eters
,and of the five mid-sec
tions 2,4,6,8,10 square m eters
,find the whole volume .
99 . A, g+ mm 72332+fx
3
490 . Find an expression for the volum e of a semicubic
paraboloid generated by the revolution of a sem icubic
parabola round its axis . In this curve y?cc 513
3,the revolv
ing ordinate being y .
491 . A paraboloid and a sem icubic paraboloid have acommon base and vertex ; show that their volum es are as
492. A vessel, whose interior surface has the form of a
prolate spheroid,is placed w ith its axis vertical
,and filled
with a fluid to a depth it ; find the depth of the fluid whenthe axis is horizontal .
493 . A square-threaded screw,with double thread , is
formed upon a solid cylinder 8 m eters in diameter ; thethread proj ects from the cylinder T
5? of meter , and the
screw rises 8 meters in four turns . Find the volume,if
the screw be 9 meters in length .
494 . Find the volum e of a square groin,the base of
which is 15 meters square,and the guiding curve a sem i
circle .
495 . The prismoidal formula applies to any shape con
tained by two parallel bases , and a lateral surface generated
EXERC ISES AND PROBLEMS . 221
by the motion of a parabola or cubic parabola whose planeis always parallel to a given plane , but whose curvaturemay pass through any series of changes in amount
,direc
tion,and position.
496 . No equation of finite degree , representing a bounding
swf aee, can define the lim i ts of applicability of the prismoidal formula
,because surfaces of higher degrees enclose
prismoidal spaces .
100 . V 8fi ga) 9 <B g+
497 . Show how existing rules for the estimation of railroad excavation may be improved .
101. e: e.,-M56 2 + e4+ y6) as-l n]
498 . If a parabolic Spindle is equal in volume to One-fifth
of the sphere on its axis as diam eter,Show that its greatest
diam eter is equal to half its length .
499 . A parabolic Spindle is placed in a cylinder half—fullof water
,the greatest diam eter of the spindle being equal to
that of the interior of the cylinder ; find the height of thecylinder so that the water may just rise to the top .
500 . A vessel , laden with a cargo,floats at rest in still
water,and the line of flotation is marked . Upon the re
moval of the cargo every part of the vessel rises 8 meters ,when the line of flotation is again marked . From theknown lines of the vessel the areas of the two planes offlotation and of five interm ediate equidistant sections arecalculated and found to be as follows , the areas being ex
pressed in square meters : 8918 ,8794
,8661
,8517
,8861
,
8191,8004 . Find the weight of the cargo removed .
222 MENSURAT ION.
EXER CISES AND PROB LEMS ON CHAPTER VI I I .
501 . A square on the line I) is divided into four equaltriangles by its diagonals which intersect. in 0 ; if one triangle be removed
,find the ”0 of the figure formed by the
three remaining triangles . CL é Ans .
9
HINT . For such problems let L be the ”0 of the part left, and 0of the part out out ; then
CL area left 00 X area out out .
502 . If a heavy triangular slab be supported at its an
gles,the pressure on each propwill be one
-third the weight
of the Slab
A weight (0 is placed at any point 0 upon a triangular table AB 0 (supposed without weight) .
Show that the pressures on the three props
(viz .
,A
,B
,O) are proportional to the areas
Of the triangles B OO,AGO
,A03 respec
tively .
Draw the straight lines AOF,B OI—I
,
OOE ; and let A ’,B ’
,0 ' be the pressures at
A,B
,0 respectively . Then
O’X OE = w X OE
'
.
C"A0 80) AB C
.
Similarly ,for A '
and B '.
504 . The mid-point of one side Of a square is j oined withthe mid-points of the adj acent sides
, and the triangles thusform ed are cut Off ; find the of the remainder .
224 MENSURAT ION .
518 . Find the ”O of any polygon by dividing it into triangles .
519 . I f the sides of a triangle be 8 4 , and 5 m eters , findthe distance of ”(I from each side . 4, 1 , 4meter . Ans .
MI SCELLANEOUS .
520 . Find both Sides of a rectangle from their ratio on n,
and its area B .
a Ans .
521 . If two triangles have one angle of the one equal to
one angle of the other , and the sides about a second anglein each equal , then the third angles will be either equal or
supplemental .
522 . Two triangles are congruent , if two sides and a
medial in the one are respectively equal to two sides and a
corresponding medial in the other .
523 . Two triangles are congruent , if three m edials in one
equal those in the other .
524 . On a plane lie three tangent spheres of radius
upon these lies a fourth Of radius How h igh is its cen
ter above the plane,and how large at least is r
',since the
sphere does not fall through ?
LOGAR I T HMS .
183. The logarithm a of a number n to a given base 6
is the index of the power to which the base must be raisedto give the numberSo
,if ba z n
,then b logn
z a,or the b- logarithm of n
I S a .
1139.
b log b"
1is called the modu lus or multiplier transform
b log b'
ing the log of a number to base Z) to the log of same numberto base
140. The base of the common system of logarithms is 10 .
10 loog (n
10 log n + 2)
1010 0g
.(n10 logn p .
142. The onantz’
ssa is the decimal part of a logarithm .
The e/zm aezfmisme is the integral part of a logarithm .
1 .
]“log l .
b log/m b logn .
b logm~
b logn .
-*
p Xb logn .
b lo n .
178
b logn
226 MENSURATION .
The logs of all numbers consisting of the sam e digits inthe sam e order have the sam e mantissa .
143. The characteristic of the log Of a number is one less
than the number of digits in the integral part .
144. When the number has no integral figures,the
characteristic of its log is negative,and is one more than
the number of cyphers which precede the first significantdigit ; that is, the numb er of cyphers ( zeros) imm ediatelyafter the decimal point .
2 8 MENSURAT I ON .
L OGARITHMS . 229
9042
9085 9090 9096
9188 9149
9196 9201
9248
9294 9299 9804
9850 9855
9895 9400 9405
9445 9450 9455
9494 9499 9504
9542 9547 9552
9590 9595 9600
9688 9648
9685
9781
9777
9827 9882
9868 9872
9917 9921
9965
280 MENSURATION .
0484
0478
0512 0515 0519
0558
0588 0592 0596
0626 0680 0688
0668
0550
0774
~0810 0818 0817
0849 0858
0881
0917
0441
04 17 0481
0671
0708
0745
0885
0484
0528 0527
0561 0565
0599 0608
0687
0711 0715
0785 0788
0821 0824
0856 0860
0896
0924 0927 0981
0952 0955
0990 0998
0962 0966
1000
1021 1024 1028 1081 1085
1055 1059 1062 1065
1089
11 2 8
1156
1222
1819
1851
1888
1414
1446
1477
1508
1092 1096
1126 1129
1159 1168
1198 1196
1225 1229
1258 1261
1294
26
5
1 28
1 5 8
1
5 1‘
OJ
03
8
8
1418 1421
1449 1452
1480 1488
1511 1514
1541
1572 1575
1605
1188
1166
1199
1282 1285
1265 1268
1800
1829 1882
1864
1892 1896
1424 1427
1455
1486
1547
1578 1581
1611
28 MENSURATION .
O 1 3 4:
2558 2555 2558 2560 2562
2577 2579 2582 2584 2586
2601 2608 2605 2608 2610
2625 2627 2629 2682 2684
2648 2651 2658 2655 2658
2672 2674 2676 2679 2681
2695 2697 2700 2702 2704
2718 2721 2728 2725 2728
2742 2744 2746 2749 2751
2765 2767 2769 2772 2774
2788 2790 2792 2794 2797
2810 2818 2815 2817 2819
2888 2885 2888 2840 2842
2856 2858 2860 2862 2 865
2878 2880 2882 2885 2887
2900 2908 2905 2907 2909
2928 2925 2927 2929 2981
2945 2947 2949 2951 2958
2967 2969 2971 2978 2975
2989 2991 2998 2995 2997
2567 2570 2572 2574
2591 2594 2596 2598
2615 2617 2620 2622
2689 2641 2648 2646
2662 2665 2667 2669
2686
2709
2782
2755
2801
2824
2869
2891
2914
2986
2958
2633 2693
2711 2714 2716
2735 2739
2760 2762
2731 2733 2735
2804 2806 2808
2826 2828 2881
2849 2851 2858
2871 2874 2876
2896 2898
2916 2918 2920
2988 2940 2942
2960 2962 2964
2984 2986
8002 8004 8006 8008
6 1./WV, HEA T/J ,
(5° CO.
’S P UB LI CA TI ONS .
MATHEMATICS.
Wentwortn’
s Elements of Plane and Solid Ge
OMET [ 1’ Y. By GEORGE A . WENTWORTH, Phillips Academy, Exeter,1 2mo . Half morocco . 40 0 pages . Mailing price, Introduc
tion, Exch ange, 60 cts.
Wentworth’
s Elements of Plane Geometry.
1 2 1110 . 250 pages . Mailing Price, 85 cts . ; Introduction, 75 cts. ;
Exch ange, 40 cts .
T his work is based upon the assumption that Geometry is a
b ranch of practical logic , the Object of wh ich is to detect , 5nd state
clearly and precisely , the successive steps from premise to conclusion .
In each proposition, a concise s tatement of what is given is
printed in one kind of type , of what is required in another, and thedemonstration in still another . The reason for each step is indicated in small type , b etween that step and the one following , thus
preventing the necessity of interrupting the process of demonstration by referring to a previous proposition . T he number of the
section, however , on which the reason depends , is placed at the
side of the page ; and the pupil should be prepared , when called
upon, to give the proof , of each reason .
A limited use has been made of symbols , wherein symbols stand
for words , and not for operations .
Great pains have been taken to make the page attractive . T he
figures are large and elegant, and the propositions have been so
arranged that in no case is it necessary to turn the page in readinga demonstration.
A large experience in the class-room convinces the author that ,if the teacher will rigidly insist upon the log ica l form adopted inthis work , the pupil will avoid the discouragmg difficulties which
usually beset the beginner in geometry ; that he will rapidly developh is reasoning faculty , acquire facility in simple and accurate expres
sion, and lay a foundation of geometrical knowledge wh ich will bethe more solid and enduring from the fact that it will not rest uponan effort of the memory simply .
Strong evidence of Me meru‘
of this boo/e is found in we fact
Mat since ine oeg inning of Me senool y ear ,1877
—78 , it nas been intro
duced into Tnifl'
y-sz
'
x Colleges and near ly Four Hundred Prefiara lorySc/cools .
T eachers should not fail to examine this book before formingnew classes .
TEST IMONI AL S .
J oseph F ick l in , P ro/C of Ma i/e.,
U niv. of [Mi ssour i I have examined ,
with considerab le care ,W entworth
’
s
Geometry ,and the result is a decidedly
favorab le op inion of th e b ook . Profes
sor W entworth is evidently a practicalteacher. He has sh own in th e execu
tion of h is work that h e knows just
wh ere b eginners in Geometry encoun
ter difficulties , and,in my judgment
,
he has b een eminently successful in his
attempt to make th ose difficulties dis
appeau
S amue l Hart , P rof. of Mafia,
Tr ini ty Col leg e : T here are some
things inW entworth'
s Geometry wh ich
makes it spec ially well adapted for
use in Sch ools and Academies .
T he clear method in wh ich each
proposition is presented as a Who le ;
the manner of reproduc ing th e state
ments of former th eorems to wh ich
r eference is made,so th a t th e student
is almost forced to recall th em ; th e
careful ‘
and precise use of the method‘
J oh n R . F rench , P ro/i of Mat/z
of limits — these are among th e th ings Sy r acuse Univ . : The distinctness of
wh ich will tend to make the work es its statements,and the c learness and
pec ially serviceab le . And I am satis ccimpactness of its demonstra tions ,a re
fied that it has a suffic iently extensive adm irab le . W e purpose to test it in
view of Geometry for most students,recitation-room with our next c lass.
treated in a way wh ich will help both
the teacher and the taught.
(Nov. 2 2 ,
S e ld en J . Coffin , P rof of M ac/z
L afayefle Col l P a . W e are pleased
with th e simplic ity and c learness of th e
demonstrations in W entworth’
s Geom
etry ,and h ave adopted it as our text
book ih th at subject.
E. Ot is Kend a l l , Pr of. of Maria,
Univ. of P ennsy lvania I have no hes
ita tion in saying that it is th e b est b ookfor b eginners that I have ever seen.
T h e demonstrations are rigorous , th e
language clea r,and I have not discov
ered any defect in the reasoning .
W . C . Esty , Prof : of Ma t/e.,Am
nerst Col l . I t is a step in advance of
th e text-b ooks of its kind. I t must
m ake the course in Geometry easier for
b oth teacher and pupil.
MA THE/WA T I CS .
ness enough to enab le those who have forgotten most of theiralgeb ra and geometry to follow the reasoning .
Chapter XV . contains a short history of Astronomy
New York Tr ib une : I t is one
of the c learest,most comprehensive ,
and most inform ing expositions of th e
princ ip les and facts of Astronomy th at
are to b e found in th e language . T he
student wh o h as made h imself master
of its contents,will b e in possession of
an amount of knowledge for the atta in
ment of wh ich he migh t cheerfully paya h igh price , and for wh ich h e would
h ave b een envied by th e late most
g raceful American h istorian,who could
never c learly compreh end th e differ
ence b etween the ec liptic and the equa
tor. Compared with th e popular sci
entific treatises wh ich swa rm to such an
th e palm ,b oth for exac tness ofmethod
solidity Of information,and appropri
ateness of expression.
New York Na t ion : I t is full
w ith out b eing h eavy ,precise w ithout
b eing pedantic ,and exp lanatory with
out b eing diffuse . I t nowh ere disap
points,and least of a ll by any enthusi
a stio ascrib ing of pre-em inence to its
own subject-ma tter. As regards th e
b ody of th e work we h ave not detec tedan error of statement in any part of it ;
wh ich is a good deal to say, or would
b e of any author but an astronomer
T he most readab le treatise upon As
extent in England,th e present b ears tronomy that we have ever seen.
Q uestions and Exercises on Stewart’
s Elementa ry P/zy sics, w il l; Answers a nd occasiona l Solutions . By GEORGE
A . HILL, formerly Assistant Professor of Physics in Harvard University .
181110 . Boards . 192 pages . Mailing Price, 40 cts . ; Introduction, 35 cts .
T hese Q uestions and Exercises have been drawn up with the aim
of making Mr . S tewart’s excellent work more useful in elementary
teach ing .
Part I . consists of questions upon the text ofMr . S tewart’s b ook
which are intended to be direct and exhaus tive . T hese will befound useful for review and exam ination purposes .
Parts I I . and I I I ., which form the principal part of the work , have
been written with two Objects in view : First, to s timulate originalthough t on the part of the student, and to give the teacher the
means of testing th oroughly the student’s knowledge of principles ;
Secondly , to make certain needful additions to the felicitous butcursory sketch of Mechanics , Hydrostatics , and Pneumatics . con
tained in the first two chapters ofMr . Stewart’s book .
