Secondary Geometry - e-Book

Prescribed by the National Curriculum and Textbook Board as aTextbook for classes IX-X from the academic year 2011.

Secondary Geometry[ For classes IX–X ]

Writien by Harunur Rashid

Translated by Prof. Harunur Rashid

Edited by Prof. Saleh Motin

Md. Sanaur Rahman

National Curriculum and Textbook Board, Dhaka.

Published by National Curriculum and Textbook Board 69-70, Motijheel Commercial Area. Dhaka

[All rights reserved by the Publisher]

First Print : 1997 Second Edition : 2005

Re-print :

Cover Design Biren Shome

IllustrationMd. Quarnruzzarnan

Kazi Saifuddin Abbas

DesignNational Curriculum and Textbook Board, Dhaka

For free distribution from academic year 2010 by the Government of Bangladesh.

Printed by :

Preface

Education is the key to development. A progressively improved education system largely determines the pace and the quality of national development. To reflect the hopes and aspirations of the people and the socio-economic and cultural reality in the context of the post independent Bangladesh, new textbooks were introduced on the beginning of the 1980s following the recommendations of the National Curriculum and Textbook Committee. In 1994, in accordance with the need for change and development, the textbooks of lower secondary, secondary and higher secondary were revised and modified. The textbooks from classes VI to IX were written in 1995. in 2000, almost all the textbooks were rationally evaluated and necessary revision were made. In 2008, the Ministry of Education formed a Task Force for Education. According to the advice and guidance of the Task Force, the cover, spelling and information in the textbooks were updated and corrected.To make assessment more meaningful and in accordance with the need of the curriculum, Creative Questions and Multiple Choice Questions are given at the end of each chapter. It is hoped that this will reduce the dependency of students on rote memorization. The students will be able to apply the knowledge they have gained to judge, analyses and evaluate real life situation. Geometry plays an important role in the application of mathematic logic the history of Geometry, the define ions axioms and proofs as given by Euclid has been presented in a modern context. Moreover to expand the use of Mathematics in applied and practical fields, Trigonometry and Measurement has been included in the Geometry textbook. This book of Geometry for class IX & X is the English Version of the original textbook entitled ÔMaydhamic JyamitiÕ written in Bangla. We know that curriculum development is a continuous process on which textbooks are written. Any logical and formative suggestions for improvement will be considered with care. On the event of the golden jubilee of the Independence of Bangladesh in 2021, we want to be a part of the ceaseless effort to build a prosperous Bangladesh.In spite of sincere efforts in translation, editing and printing some inadvertent errors and omissions may be found in the book. However, our efforts to make it more refined and impeccable will continue.I thank those who have assisted us with their intellect and effort in the writing, editing and rational evaluation of this book. We hope that the book will be useful for the students for whom it is written.

Prof. Md. Mostafa Kamaluddin Chairman

National Curriculum and Textbook Board

Contents

Chapter Subject Pages

One : Elementary conception and definition 1

Two : Theorems related to line, angle, triangle

and quadrilateral 31

Three : Some problems related to triangles and quadrilaterals 37

Four : Some theorems related to area 54

Five : Theorems of pythagoras and its application 60

Six : Some problems on the application of

the theorem of pythagoras 66

Seven : Geometrical ratio and similarity 71

Eight : Problems related to area and ratio 80

Nine : Theorems on locus 91

Ten : Theorems on circle 96

Eleven : Problem related to circle 133

Twelve : Trigonometry 146

Thirteen : Mensuration 172

: Answers 203

Chapter One

Elementary Conception and Definition

1.1. Historical BackgroundGeometry is an old branch of mathematics. Literally geometry means measurement of land. Geometry originated in the age of agriculture-based civilization, from the necessity of solving problems of measurement of land. But now-a-days geometry is not only used for measuring lands, rather knowledge of geometry is now indispensable for the solution and explanation of many complicated mathematical problems.

The culture of geometry is seen in the relics of ancient civilization. In the opinion of the historians, concepts and ideas of geometry were applied for the survey of lands about four thousand years ago. Signs of application of geometry are seen in various practical works in ancient Babylon, India and China. But geometry in a systematic form was observed in the age of ancient Greek civilization. Approximately in 300 years B, C, the Greek scholar Euclid arranged the scattered propositions in a systematic manner and composed his famous book "Elements". The book "Elements" completed in 13 volumes is the basis of modern geometry.

1.2. Concept of space, plane, line and point The space around us is infinite. Its different portions are occupied by different things, large and small. By small and large things are meant all of grains of sand, cushion pin, pencil, paper, book, chair, table, brick, box, house, mountain, the earth, planets and stars. The concept of geometry originated from the size, shape, position and specially of the portion of the space occupied by different things.The space occupied by a solid is spread in three directions. The width in these three directions denotes the three dimensions (length, breadth and height or thickness) of the solid. Hence every solid is three-dimensional. For example, a piece of brick or a box has three dimensions (length. breadth and height or thickness). A sphere has also three dimensions. Though the distinctness of its three dimensions are not clearly visible, it can be divided into parts having length, breadth and thickness.

2 Secondary Geometry

Fig-1.1The upper face of a solid denotes a surface, that is, every solid is bounded by one or more surfaces. For example, the six faces of a box are parts of six surfaces. The upper face of a sphere is also a surface. But the surface of a box and the surface of a sphere are different. The first is a plane surface while the second is a curved surface.

A two-dimensional surface has only length and breadth and no thickness. Keeping the two dimensions of a box fixed, if the third dimension is gradually reduced to zero, then we are left with a certain face of the box. In this way the conception of a surface can be derived from a solid.

Fig-1.2If two surfaces intersect, then a line is formed at their intersection. For example,two faces of a box meet at one side in a line. This line is a straight line. If a lemon is cut by a thin knife, a curved line is formed where the plane of the knife intersects the curved surface of the lemon.

A line is one-dimensional it has length only and no breadth or thickness. If the width of the face of a box is gradually reduced to zero, we are left with only one line of the surface. In this way the concept of line can be derived from of a surface.

Fig-1.3

Secondary Geometry 3

If two straight lines intersect, a point is formed that is, the intersection of two lines is denoted by a point. The two edge-lines of a box meet in a point. A point has no length, breadth and thickness, it has position only. If the length of a line gradually diminishes to zero, a point only is left. A point is regarded as an entity of zero dimension.

Fig-1.4

The idea given above about surface, line and point are not their definitions-but description only. The idea of length, breadth, thickness, dimension etc., used in the description are not defined. The definition of point, line and surface which Euclid mentioned in the first volume of his 'Elements' is also incomplete according to modern outlook. The description given by Euclid is as follows:

(1) A point is that which has no portion.

(2) The end of a line is a point.

(3) A line is that which has only length but no breadth.

(4) A straight line is that whose points lie on it in the same direction.

(5) A plane is that which has only length and breadth.

(6) The terminal of a surface is a line.

(7) A plane is that surface such that all straight lines on it lie wholly on it.

It is observed that the words bold above have been accepted without definition. As a matter of fact, in any mathematical discussion one or more elementary ideas have to be accepted. By considering point, straight line and plane as fundamental concept in modern geometry, some of their properties are also admitted to be true. These admitted properties are called geometric postulates. These postulates are determined by keeping consistency with real conception.

1.3. Some fundamental postulates about point, straight line and plane It has been mentioned earlier that point, straight line and plane are three fundamental concepts of geometry. Although it is not possible to define them properly, we have ideas about them, based on real experience.


As a concrete geometrical conception space is regarded as a set of point and straight line and plane are considered sub-sets of this universal set. That is,

Postulate 1. Space is a set of all points and plane and straight line are sub-set of this set.

From this postulate we observe that each plane and each straight line are each a set whose elements are points. Generally the use of the notation of, set is excluded in geometrical description. For example, a point included in a straight line (or a plane) is expressed by "the point lies on that straight line (or plane)" or "the straight line (or plane) passes through the point" or by sentences having similar meaning.

In the same way, if a straight line is a sub-set of a plane, then it can be expressed by sentences like "the straight line lies in the plane" or "the plane passes through the straight line." It is accepted as properties of straight line and plane that

Postulate 2. For two different points, there is one and only one straight, line on which both the points lie.

Postulate 3. For three points not lying in the some straight line, there is one and only one plane on which the three points lie.

Postulate 4. A straight line passing through two different points on a plane lie on that plane.

Postulate 5. (A) Space contains more than one plane. (B) More than one straight line lie in each plane. (C) The real numbers can be related to points on each straight line such that

every point of the line corresponds to a unique real number and every real number corresponds to a unique point of the line.

Note. The postulates from I to 5 are called incidence postulates. Diagram is used to make geometric description clear. The model of a point is drawn by very thin drop of a pencil or a pen on paper. The model of straight line is constructed by drawing a line along a straight ruler. For example,


few points a straight line

In the diagram of the straight line the arrows both sides indicate that the line is indefinitely extended. According to postulate 2, two different points A and B define a unique straight line on which the two points lie. This line is called ABor BA and is denoted by the symbol AB or BA. Describing to postulate 5(C),every such straight line contains infinite number of points.

1.4. Plane Geometry According to postulate 5 (A) more than one plane exist. There are infinite number of straight lines in every such plane. The branch of geometry that deals with points, lines and different geometrical entities related to them are discussed, is known as plane Geometry. In this book plane geometry is

the matter of our discussion. Hence, if some thing in particular is not mentioned we are to understand that all points, lines etc. lie in the .same plane. Such a definite plane is an universal set in our discussion.

1.5. Distance and number line The concept of distance in geometry is also an less elementary conception. For this, it is taken as axiom that,

Postulate 6 (A) : Every pair of points (P, Q) determine a unique real number which is called the distance between P and Q and is represented by PQ.

(B) If P and Q are different points, the number PQ is positive, otherwise. PQ = O.

(C) The distance between P and Q and that between Q and P are the same. That is PQ = QP.


Remark : being PQ = QP, this distance is generally called the distance between P and Q. In practice, this distance is measured with the predetermined unit.

Remark : Postulate 6 is called the Distance Postulate.

According to postulate 5 (C) one to one correspondence can be established between the set of points in every straight line and the set of real numbers. In this connection, it is admitted that, Postulate 7 : One-to-one correspondence can be established between the set of points in a straight line and the set of real numbers such that, for any point P and Q, PQ = |a b|, where, the one-to-one correspondence associates points P and Q to real numbers a and b respectively. The correspondence mentioned in this proposition is said to have reduced the line into a number line, if P corresponds to 'a' in the number line, then P is called the graph point of P and P, the coordinates of a. To convert a straight line into a number line the co-ordinates of one point is taken as 0 and another point 1. Thus a unit distance and positive direction is fixed in the straight line. For this, it is also admitted that, Postulate 8 : Any straight line AB can be converted into a number line such that the co-ordinates of A is 0 and that of B is positive.

Remarks : Postulate 7 is known as ruler postulate and Postulate 8 is known as ruler placement postulate.

1.6 Geometrical proof In any geometrical theory different statements related to that theory are logically established on the bases of some elementary conceptions, definition and postulates. Such statements are generally called proposition. In geometry, special importance is attached to some propositions which are taken, as theorems and in establishing other proposition, these theorems are used in serial order. In geometrical proof different data a described with the help of diagram. But the proof must be logical.

In describing geometrical propositions general enunciation or particular enunciation is used. The general enunciation is the description independent of the diagram and the particular enunciation is the description based on the diagram. If the general enunciation of a proposition is given, subject matter of


the proposition is specified through particular enunciation for this, necessary diagram is to be drawn.

Generally the discussion of geometrical theorem contains the following parts, (1) General Enunciation. (2) Diagram and particular Enunciation. (3) Description of the necessary construction and (4) Description of the logical steps of the proof.

If a proposition is proved directly from the conclusion of a theorem, then it is called a corollary of that theorem.

Besides, proof of various propositions. Proposals for construction of different diagrams are considered. These are called problems. By drawing diagrams related to problems, it is necessary to mention the description of construction and its logical truth.

1.7 Intersecting and parallel lines

Definition : Two different lines are said to be intersecting, if there is a common point lying on both the lines.

Definition : Two different lines lying on a plane are said to be parallel, if they have no common point

It may be noted that, (1) Two different lines may at best have one common point. Because, according to postulate 2, two different points may lie on one straight line only.

(2) Two different straight lines lying on a plane are either parallel, or they meet at a point only. Out of three or more straight lines if no two are intersecting, then they are said to be parallel to each other.


If three or more straight lines have a common point, then they are said to be concurrent at that point and that point is called Point of concurrence.

If AB is a straight line on a plane and C is a point not lying on it, then existence of a straightline passing through C and parallel to AB is admitted. A

C

B

Postulate 9. There is one and only one straight line parallel to a given straight line and passing through a point not lying on the given straight line. This postulate is called the postulate of parallel lines. It is also known as Playfare Postulate (John Playfare: 1748 1819).

1.8. Betweeness In the adjoining figure, the point C is regarded as lying between A and B. A C BDefinition : The point C is called to lie between A and B if A, C and B are different points on the same straight line and AC + CB = AB. If the point C lies between A and B, often it is expressed as A C BDefinition : A number of points are said to be collinear if all of them lie on the same straight line. Some useful characteristics of betweenness are stated below : (1) If A C B (C lies between A and B) is true, then (A) A, C and B are different collinear points. (B) B C A (the point C lies between B and A) is true (2) If on a number line the co-ordinate of A, C, B are a, c and b respectively,

then C will he between A and B, that is, A C B is true if and only if a <c <b or a > c > b.

(3) If A and B are different points. then there are points D, E, F such that, D-A-B. A-E-B and A-B-F are true,

D A E B F (4) If A, B and C be three different collinear points, then one and only one of

them lies between the other two.


1.9 Line segment Definition : If A and B are two different points, then A and B the set of all points A B lying between them is called the line segment Joining A and B, in short line segment AB.

The line segment AB is sometimes denoted by AB . A and B are called terminal points of AB and the line AB is called the line of support of AB . All points lying between A and B, are called points of AB . It may be noted that, the line segment AB is a subset of the line AB which contains A. B and all points lying between A and B, that is AB = {P : P is either A or B or any point lying between A and B}. Definition : If A and B be two different points the distance between A and B is

called the length of the line segment AB , that is, the length of AB = AB.Definition : The point M is called the

middle point of the line segment ABif A M B is true, that is, the point M lies between A and AM = MB.

A M B

It may be mentioned that, every line segment line a unique mid-point. The line segment is said to have been bisected at its mid point. Generally, if A and B are. different points and m and n are any natural numbers,

A P B Q A B A B Q

then, (1) there is an unique point P such that A-P-B true and AP t PB = m t n, and (2) there is an unique point Q such that either Q-A-B is true, or A-B-Q is true and AQ t QB = m t n. In the first case. the point P is said to divide the line segment AB internally in the ratio m t n and in the second case, the point Q divides the line segment AB externally in the ratio m t n.

1.10 Ray If A and B are two different points, then the ray from A to B or the ray AB is the set of A, B and those P such that A-P-B. or A-B-P. Such a ray is denoted by AB .

A P B

A P B


The point A is called the terminal point of AB and the line AB is called the line of support of the ray AB . Each of the points included in AB except A is called the interior point of AB . It may be noted that the ray AB is a subset of the line AB and e line seg AB is a subset of the ray AB . th ment Definition : AB and CD are called collinear rays if their line of support is the same, that is, if AB and CD are the same line.Definition : If three points A, B, C be suchthat C A B is true, then AC is called the opposite ray of AB.

A D

A B

C A B

It may be noted that, every ray AB has one and only opposite ray AC and ABand AC are different rays but collinear.

1.11 Section of surface/ Resolution of a plane

BQA

P

R

In the adjoining figure, the three points P, Q, R lying in the plane but not lying on AB are such that P and Q lie on one side of the line and R on its opposite side. In this connection it is admitted that,

Postulate : (A) Every straight line in a plane has two only two sides of the plane. Each

such side is a non-zero sub-set of the plane in which no point of the line is included.

(B) Every point in the plane that does not lie on the straight line on either side of that line. If two different points lie on one side of the straight line, then all points between these two points lie on the same side of the line. If two different points lie on two sides of the line, then a point between the two points lies on the line.


Remark : Every straight line in a plane divides it in three disjoint sets : (1) the line itself (2) one side of the line (3) the other side of the line By fixing a point P on one side of the line, that side of the line is called P-side and the other side is called its opposite side.

1.12. Angle

Each of the above figures is the model of an angle. Definition : If two rays in a plane have the same terminal point but different line of support, then their union is called an angle produced at the common end.

A

B

A

B

C

The angle produced at A, the terminal point of the rays AB and AC is

by BAC or CAB or A in short AB and AC are arms of this angle and their common terminal point A is called vertex of this angle.

Interior and Exterior regions of an angle

Exterior R

P Interior

Exterior Q

B

AS Exterior

C

denoted


In the adjoining figure, the point P is in the interior region and points Q, S and R are on the exterior region of the angle BAC.Definition : The set of all points lying in the plane on the C side AB and B side of AC is the interior region of the BAC.The set of all points not lying in the interior region or on any arm of the angle is called exterior region of the angle. Every point that in the interior region is called an interior point of the angle. Every point that in the interior region is called an interior point of the angle and every point and that in the exterior region is called an exterior point of the angle.

Adjacent angle :

In both the above figures the angles AOB and BOC are adjacent to each other.Definition : If two angles in a plane (1) the same vertex, (2) a common arm and (3) no common point in the two interior regions, then each of the two angles is said to be an adjacent angle of the other and the two arms other than common arms are called its exterior arms. In the figure the adjacent angles AOB and BOC have the same vertex O, a common arm OB and no common point inside the angle. OA and OC are two exterior arms the two angles.Remark : If a ray meet at its terminal point with a straight line, then the two angle thus formed are also adjacent angles (Ref, Fig 2.) Here both the exterior arms OA and OC are parts of the same straight line AC .

Pair of linear angles If two exterior arms of two adjacent angles are two opposite rays, that is, parts of the same straight line, then the two angles are called pair of linear angles. In Fig-2, AOB and COB are pair of linear angles.

Fiq-1 Fiq-2

A

O B

C COA

B


Opposite angles : In the adjoining figure (1) and (3) are a pair of opposite angles which are formed by the intersection of AB and CD . Similarly, in the figure, (2) and (4) are another pair of oppositeangles formed by the same pair of lines.

Definition : Two angles are said to be opposite angles if the arms of one be the opposite rays of the other,

Measurement of angle : The unit of degree is used for the measurement of angle. In this connection it is admitted that.

Aa

Postulate : (A) (Measurement of angle) For every angle

A in a plane there is an uniquely assigned number a such that 0 a 180 . This is a called the measurement of A in degrees and is expressed as A = a or, A = a .

(B) (construction of angles) If 0 a 180 and

AB is a ray in a plane, then there is an unique

AC in any side of AB such that BAC = a .

(C) (addition of angles) If D is a point inside BAC and if BAD = x and DAC = y the BAC = (x + y) .

C

C

BAaa

yx

B

C

DA

Remark : Generally the magnitude of an angle is determined with the help of protactor.Note : Through BAC and DEF are different angle but their angles in degree is expressed as BAC = DEF (meaning equality). Similarly BAC > HGK means, the measure of BAC in degree is greater than HGK in degrees.

40

F

DE40

C

BA

1

2

3

4 D

BC

A


By the method of addition of angles. QPS and SPR measure in degrees is

equal to QPR in degree. It means QPS + SPR = QPR is written.

Right angle Definition : If an angle is equal to the angle produced by one of its side and the ray opposite to its other side then the angle is said to be right angle.

If the figure; OA and OC are opposite ray and by measure in degrees, AOB and BOC are equal. That is, AOB = BOC. Both AOB and

BOC are right angles. It may be noted that AOB and BOC are pair of linear angles. If the

measure in degree of such pair of linear angles is equal, each of them is right angle.

Four angles formed at the intersecting point by two intersecting straight line if one of the angle is right angle then each of the other three will also be right angle. We admit that,

Postulate : The measure of right angle in degree is 90 i.e. if AOB is right, then

AOB = 90 i.e., AOB = 90°

In this case AOB= 1 right angle is written. As a unit of measurement of angle 1 right angle = 90°

1° = 190 right angle.

B

O AC

C

D

OD B

O

B

A

K

G H30

R

S

QP


Perpendicular Definition : If two straight lines intersect at a point formed four angle of which one of them is right angle, then the two lines are called perpendicular to each other at intersecting point.

In the figure, line AB and line CD is intersect at O and BOC = 1 right angle (each of the other three angles is also right angle).

As a result, the lines AB and CD are perpendicular to each other at the point 0.

In this case AB CD is written. If AB CD and A-O-B and C-O-D, the ray

OC and the line AB are perpendicular to each other at the common point 0.

(Symbol : OC AB )

Ray OC and OB are perpendicular to each other at the common point O.

(Symbol : OC OB ), The line segment CO is perpendicular on the line AB or

line AB at the point O, etc is said.

Straight angle In the definition of angle the lines of support of two arms have been taken to be different. For practical purpose, sometimes we consider straight angle AOB whose arms are two opposite rays as fig-A.If in a line AB, the ray OC is perpendicular at O and A O Bstraight angle AOB = AOC + COB = 1 right angle + 1 right angle. = 2 right angle let the measurement of the straight angle AOB is equal to 2 right angle or 2 × 90° = 180°

C

D

OA B

C

O BA

A BOFig-A


Complementary and supplementary angle D

Definition : If the sum of two angles measured in degrees is 90°, then the two angles are called complementary angles.

In the figure, ACB and DEF are complementary angles.

Definition : If the sum of two angles measured in degree is 180°, then two angles are called supplementary angles.

In the figure, ACB and DEF are supplementary angles.

FB(90-x )

E x AC

Remark : If two supplementary angles are adjacent, they are called pair of linear angles. Each of such two supplement ray angles is said to be linearly supplementary to the other.

Acute and obtuse angle Definition : An angle which is less than a right angle is called an acute angle and an angle greater than one right angle but less than two right angles is an obtuse angle.

In the figure, AOC is an acute angle and AOD is an obtuse angle.

It may be noted that, by AOC < AOBand AOD > AOB, are meant comparison of the angles measured in degrees.

B

OA

C

B

O A

D

F

C

B

(180-x )

Ex

AD


Reflex angle We notice that, two rays with the sameterminal point but different lines of support from an angle at their common and whose measure in degrees is less than 180°. In the above figure, such an angle AOB with

Sometimes, for the convenience of description, it is imagined that a reflex angle is associated with such an angle. The measure of the reflex angle is the sum of a linear supplementary angle of the given angle and a straight angle. For example, in the above figure, the measure of the reflex angle AOB of magnitude x° is

Reflex angle AOB = BOA' + straight angle AOA'

= (180° x°) + 180°

= (360° x°).

Alternate and similar angles

In the figure, the lines AB and CD are intersected by EF at points G and H

respectively in the same plane. EF is called the secant of AB and CD .

In such a case, AGH and BGH are called alternate angles of GHD and

GHC respectively and AGE, AGH,BGE and BGH are called

corresponding angles of GHC, CHF,GHD and DHF respectively. Among

them, AGE, BGE. CHF and DHF

are called exterior angles and AGH, BGH, GHC and GHD are called interior angles.

B

A O

x

A

A

CH

G

E

FD

B

magnitude x° has been shown.


1.13 The Triangle

Definition : If three points in a plane are not collinear, then the union of the three line segments obtained by joining any two points at a time is called a triangle. If the points A, B, C lying in a plane are not collinear, then the triangle formed by the three line segments obtained by joining A and B, B and C, C and A is called triangle ABC and is denoted by the symbol ABC. The points A, B, C are the three vertices and the three line segments AB, BC, CA are the three sides of the triangle. ABC, BCA and BAC are the angles of the triangle

ABC and sometimes they are expressed, in short, respectively as B, C,A.

The side BC is the opposite of A and A is the opposite angle of side BC are called. The sides AB and AC are the adjacent side of A and A is called the included angle of the side AB and AC.

Interior and Exterior of a Triangle Definition : The set of all points lying on the plane inside the three angles of a triangle is called the interior of the triangle.

The set of points in the plane which do not lie inside the plane or on the sides is called exterior of a triangle. Every point inside a triangle is called its interior point and every point outside a triangle is called its exterior point.

AA

B CB C

A

B C

A

B C

Exterior Exterior

Exterior

Interior


Isosceles Triangle Definition : If two sides of a triangle are equal then the triangle is called an isosceles triangle. The side opposite to the point of intersection of the equal sides (in the sense that the length are equal)called the base and the angle at that point of intersection is called the vertex of the triangle. In the ABC of the adjoining above figure, AB and AC are equal sides. BC is the base and A is the vertex.

Equilateral triangle

Definition : The triangle of which the three sides are of equal length is called an equilateral triangle.

Scalene triangle

Definition : The triangle of which the three sides are of different length is called a scalene triangle.

Acute angled triangle Definition : The triangle of which each angle is an acute angle is called an acute angled triangle.

Obtuse angled triangle Definition : The triangle of which are of the angles is an obtuse angle is called an obtuse angled triangle.

A

B C

A

B C

A B A B

A

B

CCC

A

B C

A B

C


Right angled triangle Definition : The triangle of which one of the angle is a right angle is called a right angled triangle. The side opposite to the right angle of a right angled triangle is called its hypotenuse and one of the side adjacent to the, right angle is called the base and the other side is called the altitude.

A

90

B C

Interior and exterior angles of a triangle If any point D outside ABC lie on BC, then D is called a point on BC produced. If B-C-D (i.e. C lies between B and D) then BD is the line segment obtained by producing BC to D. the angle ACD formed at C by CD

D

A

CB

and is called an exterior or external angle of the triangle ABC In this respect, the angles of the triangle are called its interior or internal angles.

ACB

ACD

is the adjacent internal angle and CBA and CAB are the opposite internal angles of the external ACD.

1.14. Quadrilateral Definition : If four points A, B. C, D in a plane are such that (a) any three or them are not collinear (b) any two of the line segments AB BC CD DA line do not have common point other than their terminal points and (c)C and D are on the same side of AB and A, B are on the same side of CD, then the union of AB, BC. CD, DA is called the quadrilateral ABCD. The quadrilateral ABCD issometimes denoted by ABCD. The four points A, B, C, D are the vertices of the quadrilateral. A, C and B, D are vertices one opposite to the other, the line segments AB, BC, CD, DA are the sides, AB, CD and BC, DA are one opposite to the other side and the line segments ACand BD are called diagonals of ABCD.

CD

BA


ADCB, BCDA etc. denote the same quadrilateral. This cannot be denoted by

ADBC, because the common point of DB and CA is their interior point.

DAB, ABC, BCD and CDA are the four angles of the quadrilateral. Sometimes they are expressed, in short, as A, B, C, D respectively. AB is the common side of A and B and they are adjacent angle to each other.

Similarly, B and C, C and D, D and A are adjacent to each other. A

and C, B and D are opposite angles to each other.

If E is the point on AB produced of a quadrilateral ABCD is such that A-B-E,

then CBE is an external angle of the quadrilateral. CBE is an adjacent internal angle of this external angle and ADC is opposite internal angle.

Different kinds of quadrilaterals

The above figures 1, 2, 3, 4, 5 are diagrams of parallelogram, rectangle. square, rhombus and trapezium respectively.

Parallelogram : The quadrilateral whose opposite sides are parallel is called a parallelogram.

Rectangle : The parallelogram one of whose angles is a right angle is called a rectangle.

It is noted that if one angle is right angle then all the angles are right angle of a parallelogram.

A (Fig.-l) B A (Fig.-2) B A (Fig.-3) B A (Fig.-4) B A (Fig.-5) B

90 90

D C D C D C D C D C

A B E

CD


Square : If two sides of a rectangle passing through any vertex are equal then it is called a square. It is to be noted that if two sides passing through any vertex of a rectangle are equal, then all its side are equal.

Rhombus : The parallelogram of which two sides passing through any vertex of a parallelogram are equal and one angle is not a right angle then it is called a rhombus. It is noted that if in a parallelogram two sides passing through a vertex of a parallelogram are equal then all its sides are equal and if one angle is not a right angle, then none of the angles is a right angle.

Trapezium : The quadrilateral only two of whose side of a quadrilateral are parallel then it is called a trapezium.

One of the parallel sides of a trapezium is called the base and one of the nonparallel sides is called the oblique side. It two non-parallel sides of a trapezium are equal it is called an isosceles trapezium. It is noted that the parallel sides of a trapezium cannot equal.

1.15. Polygon To denote some points in order in a plane sometimes the points are termed by P1, P2, P3 etc.

P1

P2 P3

P1

P2 P3

P4

P1

P2 P3

P4

P5

P1

P2 P3

P4

P5P6

P1

P2 P3

P4

P5P6

P7

P1

P2

P3

P4 P5

P6

P7

P8

Triangle Quadrilateral Pentagon

Hexagon Heptagon Octagon


Each of the above figures denotes polygons. They are named according to the number of sides. A polygon having n-sides where n > 3 is called n-gon in short.

Definition : Let P1, P2................. Pn be n points in a plane, where n 3. Then the union of the line segments P1, P2, P3 .................. Pn-1 is called a polygon of n-sides if—(a) the common point of any two line segments is one of their terminal points.

(b) those two line segments which have a common terminal point have different lines of support ( ) all points on each line segment other than the terminal point lies in one side of the line of support.

This Polygon is denoted as P1, P2, P2, P3, ........., Pn and the points P1, P2, ........., Pn are its vertices and the line segment P1 P2, P2 P3 , .......... Pn P1 are its sides.

The angle included between the two sides at each vertex is called an angle of the polygon.. Corresponding to n = 3, 4, 5, 6, 7, 8. the polygons are respectively called triangle, quadrilateral, pentagon, hexagon, heptagon, octagon.

Congruence of two polygons If in two polygons having equal number of sides, the vertices of one correspond respectively in a certain order to the vertices of other, then the similar angles and sides of the polygon are specified. For example, between the quadrilaterals ABCD and EFHG, if the vertex A corresponds to the vertex E, i.e, A E, B

F, C G. D H, then A and E, B and F. C and G, D and H are

similar angles, also AB and EF , BC and

FG , CD and GH , DA and HE are similar sides.

Note : The correspondence of the above two quadrilaterals in expressed in short as ABCD EFHG.

G H

A B

D C

E F


Definition : If two polygons having equal number of sides correspond to each other such that their similar angles are equal in magnitude and their similar sides are equal in length, then the two polygons are said to be congruent.

O xxBD

Ax

CE

F

In the above figure, the triangles. are congruent and ABC FED such that A= F, B = E, C = D and AB = FE, BC = ED, AC = FD. Note : The method of correspondence explained above is sometimes called method of superposition in which one figure is supposed, to be placed on the other.

1.16. Rectilinear regions F E

A

DA

B C B C B CFigure-1 Figure-2 Figure-3

AB

C

D

E

F

G

H

Figure-4

A

D


Each of the above diagram, represents the rectilinear regions in a plane, Fig. 1 is a triangle, Fig. 2 is a rectangle. Fig. 3 is a hexagon whose boundaries are respectively ABC, ABCD, and hexagon ABCDEF. Each of them and Fig. 4 is a rectilinear region. The boundary of such a region is formed by the union of some line segments.

Definition : The subset of the plane formed by the union of a triangle and its interior in a plane is called a triangular region. The triangle is the boundary of this triangular region which is included in the triangular region The. triangle, which is induced in the triangular region. The triangular region whose boundary is the ABC is called -region ABC.

Definition : The subset of the plane formed by the union of any polygon and its interior in a plane is called a polygon region and the polygon is called the boundary of the region.

According to the nature of the polygon, the region may be triangular region, quadrilateral region, rectangular region, square region, pentagon region. It is to be noted that, each polygon region can be divided into some triangular regions in which two triangular regions are either disjoint or their intersection is a line segment or point.

The Fig. 4 given above is a rectilinear region whose boundary formed by the

union of the line segments AB, BC, DE, E F , FG, GH and HA is a closed polygon line in the plane. The boundary is also included in the rectilinear region.

Definition : If P1, P2, P3 ,............, Pn-1, Pn, are n different points (n > 2) in a plane, then the subset of the, plane formed by the union of the line segments P1P2 , P2P3........... Pn-1Pn and PnP1 is called a closed polygon line, if no two straight lines have common point other than their terminal points.

AreaAt this stage, we admit that Postulate : (a) For every rectilinear region in a plane, a unique real number, can be specified this number is called the area of the region. (b) If two polygons are congruent, then the areas of two polygon regions bounded by them are equal.

CD,


(c) The area of the rectangular region ABCD = AB × BC D C

A B

(d) If two rectilinear regions R1 and R2 be such that they intersect at best in a finite number of line segments or points and rectilinear region is formed at their union, then the area of region R = area of region R1 + area of region R2

Exercise 1

1. State the concepts of space, plane, line, and point. 2. Fill in the gaps :

(i) Space is the ........ of all points and plane is also ........ are subsets of his set.

(ii) For two different........ there is one and only one straight line on which both the ........ lie.

(iii) For ........ different points not lying on the same straight line, there is one and only one plane on which the three ........ lie.

(iv) The ........ passing through two points on a plane lie on the plane. (v) There exist more than one plane, more than one ........ lie on each

plane.3. State the incidence postulate.


4. What is plane geometry? 5. Fill in. the blanks : (i) If A and B are two different points, then the number AB is otherwise

AB = O. (ii) The distance between A and B is —— to the distance between B and

A.6. State the distance postulate. 7. State the ruler postulate. 8. Explain number line. Also explain the graph of the point P on a number

line and the coordinates of a point. 9. State the ruler placement postulate. 10. Define intersecting straight lines and parallel straight lines. 11. Fill in the blanks. (i) Two different straight lines can have at best one ........ point. (ii) According to postulate ....... two different ...... may lie in one straight

line. (iii) Two different straight lines in the same plane are either or they

intersect only in one. (iv) Three or more straight lines are said to be ........ if no two of them are

intersecting. (v) If three or more straight lines have a common point, then they are

said to be ........ at that point. 12. State the condition for which the point K will lie between the points M

and N. 13. State the condition for which P. Q, R. S, T may be collinear. 14. State four characteristics of betweenness. 15. Define a ray, terminal point of a ray, line of support of a ray and interior

point.16. State collinear rays and opposite rays. 17. Fill in the blanks : (i) If the three points A, B, C line on the same straight line and if AB = 20,

BC = 7, AC = 13, then C is an point of A and B.(ii) If the points A, B, C are not collinear, then the three straight lines defined

by them are AB . ............ ............ the three line segments defined are

........, BC , ......... and the six rays are ......... BA . ........... .......... , AC

............


18. Explain which of the following are true or false :

(i) PQ = QP (ii) PQ = QP (iii) PQ = QP (iv) PQ = QP. 19. If A, B, C are three different points, then which of the following is true or

false for the condition AB + BC = AC: (i) A-B-C (ii) A-C-B (iii) B-C-A (iv) B-A-C (v) C-A-B (vi) C-B-A 20. State the postulate of resolution of a

plane.21. Define angle, arm of an angle and

vertex of an angle. 22. If D, B, E are three points in the same

straight line, then state the names of the angles formed in the adjoining figure.

23. Define interior and exterior of an angles.

24. In the adjoining figure, which are the interior and exterior points of the given four angles.

25. In the above figure determine which of the points are neither interior nor exterior points.

26. Define an adjacent angle. Explain its exterior side with the help of a diagram.

27. Define a pair of linear angles and explain with a diagram. 28. Define the following and explain with diagram. Vertically opposite

angles, straight angle, complementary angle, supplementary angle, right angle, perpendicular, acute angle and obtuse angle.

29. Explain alternate angles, similar angles and reflex angle with the help of those diagrams.

30. Define triangle, vertices, sides and angles of a triangle.31. State the names of the triangles shown in the following figure.

A

B C

D

E

A

C B

DN P

TL

QE

G

R

O

M


32. State the total numbers of angles and sides of the triangles shown in the above figure.

33. Define the interior and exterior of a triangle. 34. Define an isosceles triangle, its base and vertex. 35. Define : Equilateral triangle, scalene triangle, acute angled triangle,

obtuse angled triangle. 36. Define and explain with diagram : Right angled triangle, its hypotenuse

base and altitude. 37. Explain the internal and external angles of a triangle with diagram. 38. (i) Define a quadrilateral and different kinds of quadrilaterals. (ii) State the conditions for which a quadrilateral and a parallelogram

may be a square. (iii) State the conditions for which a quadrilateral may be a rectangle

and a rhombus. 39. Fill in the blanks by using the necessary words from quadrilateral, parallelogram, rectangle, rhombus, square and trapezium. (i) Square is always -———, parallelogram and———— (ii) Parallelogram is always ————— and ————— (iii) Rhombus is always ——————and——— (iv) Rectangle is always —-——— and ————— (v) Trapezium is a ———— 40. Fill in the blanks : (i) If three angles of a triangle are acute angles, then if it is called

......... triangle. (ii) The triangle of which one angle is ....... called a right angled

triagle.

A

B D C

F

E


(iii) The triangle of which three sides are unequal is called triangle. (iv) The parallel sides if a trapezium can never be........... 41. Define polygon. 42. Fill in the blanks of the following :

43. Explain when two polygons are congruent. 44. Explain what does it mean by ABC DEF 45. If ABC DEF, then state the corresponding angles and sides of

ABC

and DEF.46. Define a triangular region. 47. Define a polygon region. 48. Define a rectilinear region. 49. Define it closed polygonal line. 50. Fill in the blank:

is called a ............... is called a ...............




Chapter Two

Theorems related to Line, Angle, Triangle and Quadrilateral

Theorems from 1 to 19 related to line, angle, triangle and quadrilateral learnt earlier are stated again for ready reference. Theorem 1. If a line segment meet a ray at its terminal point, the sum of the adjacent angles thus produced is equal to two right angles. Theorem 2. If the sum of two adjacent angles is two right angles, then the exterior arms of the angles lie on one straight line. Theorem 3. If two straight lines intersect, the vertically opposite angles are equal.

Theorem 4. If a straight line intersects pair of parallel straight lines, then (a) the alternate angles are equal, (b) the corresponding angles are equal, and (c) the sum of the interior angles on the same side of the intersecting

line is equal to two right angles.

Theorem 5. If two straight lines are intersected by another straight line if, (a) the alternate angles are equal, or, (b) the corresponding angles are equal, or, (c) the sum of the interior angles on the same side of the intersecting

line is equal to two right angles then, in each case, the two straight lines are parallel.

Theorem 6. All straight lines parallel to a given straight line are parallel to one another.

Theorem 7. If two sides and the included angle of a triangle are respectively equal to the corresponding sides and the included angle then two triangles are congruent.

Theorem 8. The angles opposite to two equal sides of a triangle are equal.


Theorem 9. If two angles of a triangle are equal, then their opposite angles are also equal.

Theorem 10. If three sides of one triangle are equal to the corresponding three sides of the other triangle then two triangles are congruent.

Theorem 11. If two sides of a triangle are unequal, the angle opposite to the greater side is greater than the angle opposite to the less.

Theorem 12. If two angles of a triangle are unequal, the side opposite to the greater angles is greater than the side opposite to the shorter angle.

Theorem 13. The sum of any two sides of a triangle is greater than the third.

Theorem 14. Of all line segments that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest.

Theorem 15. The sum of three angles of a triangle is equal to two right angles.

Theorem 16. If any two angles and a side of a triangle are respectively equal to the two angles and the corresponding side of another triangles then two triangles are congruent.

Theorem 17. If the hypotenuse of two right angled triangles are equal and one side of one triangle is equal to the corresponding side of another triangle, then two triangles are congruent.

Corollary 2.1 The exterior angle so formed by producing any side of a triangle is equal to the sum of the interior opposite angles.

Corollary 2.2. The exterior angle formed by producing any side of a triangle is greater than of the two opposite interior angles;

Theorem 18. If two opposite sides of a quadrilateral are equal and parallel, then its other two sides are also equal and parallel.

Theorem 19. The opposite sides and angles of a parallelogram are equal to one another and each diagonal divides the parallelogram into two congruent triangles.

Corollary 2.3. The diagonals of a parallelogram bisect each other.

Corollary 2.4. The diagonals of a rhombus bisect each other at right angles.


The diagonals AC and BD of the rhombus ABCD intersect at a point O. It is to be proved that, 1) AO = CO and BO = DO 2) AOB = AOD = COB = COD = 1 right angle. Proof : AB and DC are two parallel lines and AC and BD are their intersectors.

BAC = ACD [alternate angles] and BDC = ABD [alternate angles] In AOB and COD,

OAB = ODC and AB = DC AOB COD

AO = CO and BO =DO Now in AOB and AODAB = AD, BO = DO and AO is their common side. Hence, AOB = AOD

AOB = AODAgain AOB + AOD = 1 straight angle.

AOB = AOD = 1 right angle Again, COD = Opposite AOB = 1 right angle. and COB = Opposite AOD = 1 right angle Hence, AOB = AOD = COB = COD = 1 right angle.

Colollary 2.5. The line segment joining the middle points of any two side of a triangle is parallel to and half the third side.

Example 2.1 Prove that if an obtuse angle of an obtuse angled triangle, the side opposite to the obtuse angle and another side are respectively equal to the obtuse angle of another obtuse angled angle and the corresponding side, then two obtuse angled triangle are equal.

A B

D C

O

×

×

A

B C E F

D

G


In the triangles ABC and DEF, let DEF be obtuse angles. In ABC and DEF, ABC = DEF. AC = DF and AB = DE.

It is to be proved that, ABC DEF.Proof : Now BAC and EDF may or may not be equal. i) Let BAC = EDF. Now in ABC and DEF AB = DE, AC = DF and the included BAC = included EDF

ABC DEF.ii) Let BAC EDF, let us suppose, BAC > EDF. or, BAC < EDF. Construction : (i) Let BAC > EDF, EDG = BAC is constructed. Let the ray DG meet EG at G. As BAC > EDF, G will be exterior point of line segment EF. Now, in ABC and DEG,AB = DE, ABC = DEG and BAC = EDG

ABC DEG AC=DG.But AC = DF [given]

DF = DG, Hence, DGF = DFG and each of them will be an acute angle [Because in any triangle two angles cannot be obtuse or right angles].

DFE will be an obtuse angle, [ DFE is supplementary to DFG]Now, in DEF, DEF and DFE are both obtuse angles, which is impossible. Hence, BAC cannot be > EDF

BAC = EDG

(ii) Let BAC < EDF in this case as in (i) BAC = EDG is constructs G is a point on the line EF and in this case in DEG, DEG and DGE both will be be obtuse angle which is not possible.Hence, BAC cannot be > EDP

BAC = EDG.

(ii) Let BAC < EDF in the case as in (i) BAC = EDG' is constructed, G' is a point on the line EF and in this case in DEG', DEG" and DGE' both will be obtuse angle which is not possible.

BAC can not be < EDFBAC = EDF

Hence, from (1) ABC DEF [proved]


Exercise 2

1. Prove that, the bisector of the vertical of an isosceles triangle bisects the base and perpendicular to it.

2. Prove that, the triangle formed by joining the middle points of the sides of an equilateral triangle equilateral.

3. Prove that, the medians of an equilateral triangle are equal to one another.

4. Prove that, the sum of any two exterior angles of a triangle is greater than two right angles.

5. D is a point inside a triangle ABC. Prove that, AB + AC > BD + DC,

6. If D is the middle point of the side BC of ABC then prove that, AB + AC > 2AD.

7. Prove that, the sum of the medians of a triangle is less its perimeter.

8. A is the vertex of an isosceles triangle ABC, and the side BA is produced to D such that BA = AD; prove that. BCD = 1 right angle.

9. The bisector of the angles B and C of a triangle. ABC intersect at O.

Prove that BOC = 90° + 12 A.

10. If the sides AB and AC of a triangle ABC are produced then the bisector of the exterior angles formed at B and C meet at O, prove that,

BOC = 90° 12 A.

11. In the adjoining figure, C = 1 right angle and B = 2 A Prove that, AB = 2BC.

12. Prove that, the line segment joining the middle points of any two sides of a triangle is parallel to and half the third side.

13. Prove that, the exterior angle so formed by producing any side of a triangle is equal to the sum of the interior opposite angles.

14. In ABC, the perpendiculars drawn from vertices B and C to opposite sides are BE and CF. If. BE = CF, prove that AB = AC.

A

C B


15. Prove that, the difference between any two sides of a triangle is less than the third.

16. In the adjoining figure, B = 1 right angle and D is the middle point of the hypotenuse AC of the triangle ABC.

prove that. BD = 12 AC.

17. If one side of a triangle be produced, then show that, the exterior angle so formed is equal to the sum of the two interior opposite angles.

18. In the ABC, AB > AC and the bisector AD of A intersects BC at D. Prove that ADB is an obtuse angle.

19. Show that, any point on the perpendicular bisector of a straight line is equidistant from the terminal points of that line.

20. Show that, any point equidistant from two particular points lies on the perpendicular bisector of the line joining that particular points.

21. Prove that, the sum of the diagonals of a quadrilateral is greater than half the perimeter.

22. Prove that, the quadrilateral formed by joining the middle points of the adjacent sides of quadrilateral is a parallelogram.

23. Show that, a rectangle is formed by joining the middle points of the adjacent sides of a rhombus.

24. If the diagonals of a parallelogram are equal, prove that, it is a rectangle.

25. Prove that, the diagonals of a parallelogram bisect each other.

A

B C

D

Chapter Three Some Problems Related to Triangles and Quadrilaterals

3.1. Construction of triangles Every triangle has six parts such as three sides and three angles. But description of all these six parts is not necessary to specify the shape and area of any triangle. It is found from the theorems regarding the congruency of triangles stated in the previous chapter that, out of the six parts of a triangle if only the following three parts are equal to the corresponding three parts of another triangle, then the two-triangles are congruent and in that case a triangle of definite shape and area can be constructed: 1) Two sides and the angle included between them (Theorem 7.); 2) Three sides (Theorem 10); 3) Two angles and a side (Theorem 16); 4) Two sides and an angle opposite to one of them, where the angle is not an

acute one (Theorem 17 and Example 2.1).

For example

Figure : 1

Figure : 2

DA

BaCx

b

In the case of 1,

x

ab

In the case of 2, A

BCa

cb

D

a

bc


In the case of 3.

In the case of 4.

P

x

yD B C E F

A

c

B C Dxx

ab

(Right angle) Figure : 4(a)

(obtuse)

x

B DC

c

x

b

A

GAH

x2

x2

y2

y2


It is observed that, in each of the above cases, three parts of a triangle have been specified. But if any three are specified, then the triangle is not specified. As for example (5) If three angles are given, an infinite number of triangles of different

dimension can be drawn (which are called similar triangles).

Figure : 5

(6) If two sides and an angle opposite to one of them are given and if the given angle is acute, then two triangle can be constructed.

Figure : 6

Some times, for the construction of triangle such three conditions are given which are applied in different drawings for specifying the triangle. A few such problems are stated below.

x

c

cA

B C DC ©

b

b

bx

A

x

y

y

z

x

x

z

y

z

C B

A B

C


Problem 3.1 The base of a triangle and angle adjacent to the base and the sum of other two sides are given the triangle is to be constructed.

Let a be the base of a triangle, x be an angle adjacent to the base and S be the sum of the other two sides are given the triangle is to be drawn.

Construction : From any ray BE we cut off the line segment BC equal to a. At B of the line segment BC we draw CBF = x. Then we cut off line segment BD = S from the ray BF, We join C, D. At C draw DCA = BDC, at that side of DC in which B lies. Let the ray CA meet the line segment BD at A. Then

ABC is the required triangle.

Proof : In the ACDADG = ACD [by construction] AC =AD

Now in ABC,ABC = x, BC = a [by construction]

and BA + AC = BA + AD = BD = S [ AC = AD] ABC is the required triangle.

Problem 3.2 The base of triangle an acute angle adjacent to the base and difference of the other two sides are given. The triangle is to be constructed.

a

x

s

B C Ex

a

sA

DF


Let 'a' be the base of any triangle, x be an acute angle adjacent to the base and 'd' be the difference between the other two sides are given. It is required to draw the triangle.

Construction : From any ray BF, the line segment BC = a is cut off. At the point B of the line segment BC we draw CBE = x. We cut off BD = d from the ray BE. We join C, D. On that side of the line segment DC in which the point E lies, we draw DCA = EDC. The ray CA cuts the ray BE at A. Then


Proof : By construction, in ACDADC = ACD. AC = AD

Hence difference of the two sides AB AC = AB AD = BD = d.

Now, in ABC

BC = a, AB AC = d and ABC = x.


N. B. If the given angle is not acute, the construction of the triangle in the above method is not possible.

Problem 3.3 To construct a triangle when two angles adjacent to the base and the perimeter are given.

x b

aB

D

C F

AEa

b


Let P be the perimeter and x and y be two angles adjacent to the base of the triangle are given. The triangle is to be constructed.

Construction : From any of ray DF we cut off DE equal to the perimeter p. At

D and E draw EDG = 12 x and DEH =

12 y on the same side of line

segment of DE. Let the rays DG and EH intersect each other at A. At the point A draw DAB = ADE, and EAC = AED. Two rays AB and AC intersectthe line segment DE at B and C respectively. Then ABC is the required triangle.

Proof : In ADB,ADB = DAB. [ by construction] AB = DB.

Again, in ACE, AEC = EAC. CA = CE.

Hence in ABC, AB + BC + CA = DB + BC + CE = DE = p.

ABC = ADB + DAB = 12 x +

12 x = x,

and ACB = AEC + EAC = 12 y +

12 y = y.

Hence, ABC is the required triangle.

D B C E F

GAHx

x2

x2

y

y2

y2

P


Problem 3.4 To construct a triangle when a side adjacent to the right angle and the difference between the hypotenuse and the other side of a right angled triangle are given.

Let 'a' be a side adjacent to the right angle of a right angled triangle and 'd' be the difference of the hypotenuse and the other side. The triangle is to be constructed.

Construction : Let from any ray BE, BC = a is cut off, at the point C a perpendicular FG is drawn on BE, CD = d is cut off from ray FG. B, D is joined. At B of line segment BD, DBA = CDB is drawn. The ray BA intersects CF at A Then ABC is the required right angled triangle.

Proof : In ABD. ABD = ADB [construction] AD = AB. AB AC = AD AC = CD = d

Now in ABD. AB AC = d. BC = a and ACB = 1 right angle.

ABC is the required right angled triangle.

a

d

B a C E

d

DG

A

F


3.2 Construction of quadrilateral We have seen that, in many cases if three parts of a triangle are given, it is possible to construct the triangle definitely. But in general it is not possible to construct a definite quadrilateral when four part of it are given. For example, a quadrilateral is not Specified when its four sides are given. In order to construct a definite quadrilateral, five independent data are required. For example,

(1) Four sides and an angle,

Figure : 7

(2) Four sides and a diagonal,

Figure : 8

a

b

cd

ab

cdk

x A a B E

cb

DF

d C

D b C

A B

d

c

ka


(3) The intercepted parts of two diagonals and a interior angle of two diagonals,

Figure : 9

(4) Three sides and two diagonals,

Figure : 10

In many cases for the construction of particular types of quadrilaterals, the given data are such that five independent conditions can be obtained from them to construct. Then a define quadrilateral can also be constructed with the help of the data. For example, a square can be constructed when a side is given.

Problem 3.5 To construct the parallelogram when two diagonals and an angle between them are given.

a

a

cd

b

x

EA

BF

CG

D C

A B

b n

a

m c

D

Oda

x bc H

bcnm


Figure : 11 Let 'a' and 'b' be two diagonals of the parallelogram and an angle x are given between them. The parallelogram is to be constructed.

Construction : From any ray AM take line segment AC = a. We fix up the middle point 0 of AC. At O we draw AOP = x. We draw the line segment

OQ opposite to OP . Then from two rays OP and OQ , we take two line

segments OB and OD respectively equal to 12 b. We join A, B; A, D; C, B and

C, D.Then ABCD is the required parallelogram. Proof: In AOB and COD,

OA = OC = 4 a, OB = OD = 12 b [by construction]

and AOB = COD [ opposite angles] AOB COD

Hence, AB = CD and ABO = CDO, but they ore alternate angles. AB and CD are equal and parallel. Similarly, AD and BC are equal and parallel.

Hence, ABCD is a parallelogram whose diagonals AC = AO + OC =12 a +

12 a

= a and BD = BO + OD =12 b +

12 b = b and the angle included between them is

AOB = x, ABCD is the required parallelogram.

a

b

P

B

A

x DQ

C MOx


Problem 3.6 To construct a parallelogram when two diagonals and a side are given.

Figure : 12 Let 'a' and 'b' be the diagonals and 'c' be a side of a parallelogram. Parallelogram is to be constructed.

Construction : We bisect the diagonals 'a' and 'b ' Then from any ray AX we

take AB = c. Taking A and B as centres a 2 and

b 2 radii respectively, we draw

two arcs on the same side of AB. These two arcs intersect at O, we join A. O

and O, B. AO is produced to AE and BO to BF. We take OC = 1 2 a from OE

and OD = 1 2 b from OF. We join A, D; D,C and B,C.

Then ABCD is the required parallelogram. Proof : In AOB an d COD

OA = OC = a 2 , OB = OD =

b 2 [by construction]

and AOB= COD [opposite angles] AOB COD

AB = CD and ABO = ODC but they are alternate angles. AB and CD are equal and parallel.

Similarly, it can be proved that BC and AD are equal and parallel. Then ABCD is the required parallelogram.

a

bA

FD C

E

B

O

X

c

b2

b2

b2

a2


Problem 3.7 To construct a rhombus when its perimeter and an angle are given.

Figure : 13

Let perimeter a of the rhombus x ( 90°) are given. We are to construct the rhombus.

Construction : From any ray AE take AP = a. We bisect AP at Q where

AQ =12 a. Again we bisect AQ at B where AB =

14 a. At the point A of AB

we draw BAX = x. AD =12 a is taken from the ray AX. Taking B and D as

centres we draw two arcs equal to radius14 a interior to the angle BAD, Let,

they intersect each other at C. B, C and D, C are joined. Then ABCD is the required rhombus.

Proof : In ABCD, AB = BC = CD = DA =14 a and BAD = x

ABCD is the required rhombus.

Problem 3.8 To construct a trapezium when two parallel sides of trapezium and the angles adjacent to the greater side are given.

a

x D C

Q P EBAx


Figure : 14

Let 'a' and 'b' (a > b) be two parallel sides of a trapezium where a > b. Let xand y be the angles adjacent to the greater side. The trapezium is to be constructed.

Construction : We take AB = a from any ray AX . Of the line segment AB, we

draw BAY = x at the point A and ABE = y at the point B. Let two rays

AY and BE intersect at E. At E we draw EZ || AB and take EF = b from EZ. At

F we draw FC || EA and let FC intersect BE at C. At C we draw CD || FE and let

CD intersect AE at D. Then AB CD is the required trapezium.

Proof : As ED lies on EA, so ED || FC [ FC || EA. by construction]

and CD || FE [ by construction]

DEFC is a parallelogram and DC = EF = b.

In the quadrilateral ABCD, AB = a, CD = b,

AB || CD [by construction]

and BAD = x, ABC = y [ by construction]

ABCD is the required trapezium.

ab

x yE

D

Ax

b

ay

B X

C

F Z

Y


Exercise 3

1. Construct a triangle when two angles adjacent to the base and the length of the perpendicular from the vertex to the base are given.

2. Construct a square when its perimeter is given.

3. Construct a right angled triangle when the hypotenuse and the sum of the other two sides are given.

4. Construct a triangle when an angle adjacent to the base, the atitude and the sum of the other two sides are given.

5. Construct the quadrilateral ABCD when the sides AB and BC and the angles x, y and z, are given.

6. Construct a quadrilateral when two sides and three angles are given.

7. Construct an equilateral triangle whose perimeter are given.


About Triangle

Multiple Choice Questions (MCQ) : 1. If the difference between the acute angles of a right angle triangle is 6°. then

the smaller angle is A. 38° B. 41° C. 42° D. 49°

2. Which of the following is the characteristic of a parallelogram? A. The diagonals bisect each other B. Opposite sides are not parallel C. Opposite sides are not equal D. Each diagonal divides it into two unequal triangle

3. If PM is median on the side R of a triangle PQR then A. Area of PQR = Area of PRM B. QPM = RPM C. Area of PQM = Area of PRM D. PMQ = PMR

4. Look at the following information: i. The area of the triangular regions on the same base and between the

same parallel lines are equal ii. The area of the parallelogram regions on the same base and between

the same parallel lines are equal iii. The diagonals of a trapezium divide it by four equal triangular

regions.

Which of the following is correct? A. ii and iii B. i and iii C. i and ii D. i, ii and iii


The medians AD, BE and CF of ABC intersected at point G.The following information refers to questions 5 to 7 : 5. What is the name of the point G? A. Perpendicular point B. Incentre C. Circumcentre D. Centroid

6. Which one is correct? A. AD + BE + CF > AB + BC + AC B. AD + BE + C < AB + BC + AC C. 2(AD + BE + CF) > 3(AB + BC + AC)D. AD + BE + CF AB + BC + AC

7. Three medians are divided by point G in a ratio ofA. 2 t 1 B. 2 t 3 C. 1 t 3 D. 1 t 4

Creative Questions (CQ) : 1. Mr. Joki and Mr. Jhafrul's houses are in the same boundary line and their

areas are also same. Mr. Joki's house is rectangle shaped and Mr. Jhafrul's house is parallelogram shaped.A. Draw the boundary line of their houses taking base/length as 10

units and breadth as 8 units. B. Show that the perimeter of Mr. Joki's house is smaller than that of

Mr. Jhafrul's house. C. If the ratio of length and breadth of Mr. Joki's house is 4 t 3 and the

area is 300 sqr. unit then determine the ratio of the area of their houses.

A

F E

DB C

G


2. Lengths of the three sides of ABC triangle are x cm. (x 2) cm and (x + 2) cm respectively.

A. Draw the triangle with a short description. B. Give your opinion about the nature of the triangle and justify your

opinion if x = 8. C. Prove that. the areas and height are proportional to each other of two

triangles which are produced by a line segment joining the middle point of the longest side and the opposite vertex of that side.

3. Let, you three friends are standing in a shade like an equilateral ABC triangle. After some times you three were moving at the opposite direction of a watch and came across the middle point of the three sides of the triangle.

A. Draw the picture with a short description visualizing the distance between each others were 10 metre.

B. Prove that, in the second position you three were standing like an equilateral triangle.

C. Based on the above statement, show that the medians of an equilateral triangle are equal to each other.

4. ABC is an isosceles triangle. BC is produced to E such that BC = CE and join A, E. AD BC.

A. Present the above information in a geometric figure. B. Prove that, 4BD2 = BC2. C. Show that, AE2 + CD2 = AC2 + DE2.

5. Perimeter of a triangular area is 'a' unit. A. Divide perimeter 'a' in three equal parts with a short description. B. Draw a triangle of perimeter 'a' with a short description.

C. Draw a rectangle of above same perimeter whose ratio of length and breadth is 2 : 1.

Chapter Four Some Theorems Related to Area

4.1. Area of plane surface Every closed plane region has a definite area. In order to measure such area, usually the area of square having side of unit length is taken as the square unit. For example, the area of the square whose side is one metre in length is 1 square metre. We know,

E

C

B

D

A

a C

B

D

A

a

C

B

D

A

(A) If the rectangular region ABCD of length AB = a unit (say, metre) breadth BC = b unit (say, metre) then the area of the region ABCD = ab square unit (say, square metre).

(B) If the length of the side of a square region ABCD = a unit (say metre) then the area of the square region ABCD = a2 square unit (say, square meter).

When the areas of two regions are equal, the sign '=' is used between them. For example, Area of rectangle ABCD = Area of triangle AED. It may be mentioned that if ABC and

DEF are congruent, then it is denoted by ABC DEF. In that case, certainly the area of ABC = area of DEF.

D

E FCB

A


But two triangles are not necessarily congruent when they have equal areas. For example, in the diagram, area of ABC = area of DBC but ABC DBC

DA

CB

4.2 Some theorems related to area The explanation of five theorems with statements and figures of the areas related parallelogram region and triangular region are given below. These can be proved with the help of the theorems mentioned in second chapter. But at present these theorems can be excepted without proof.

Theorem 20 If a triangular region and a parallelogram region stand on the same base and lie between the same parallel lines then the area of the triangular region is half that of the parallelogram region.

CB

DE A

The triangular region ABC and the parallelogram region EBCD stand on the same base BC and lie between the same parallel lines BC and ED.

Then triangular region ABC =12 (parallelogram region EBCD.)


Theorem 20

The areas of the triangular regions on the same base and between the same parallel lines are equal.

CB

DA

The triangles ABC and DBC stand on the same base BC and lie between the same parallel lines BC and AD, Hence, the triangular region ABC = triangular region DBC. or region ABC = region BDC.

Theorem 22

All triangular regions of equal areas standing on the same base and on the same side of it are between the same parallel lines.

CB

DA

In the figure, the triangular regions ABC and DBC of equal areas lie on the same base BC and on the same side of it. Hence, AD || BC


Theorem 22 (A)

The areas of parallelogram regions on the same base and between the same parallel lines are equal.

D E CF

A B

In figure the parallelogram region ABCD and ABEF are on the same base AB and between the same parallels AB and FC. Hence the parallelogram region ABCD = the parallelogram region ABEF.

Theorem 22 (B)

The areas of parallelogram region on equal bases and between the same parallel lines are equal.

GHCD

EBA F

In figure the parallelogram regions ABCD and EFGH stand on equal bases AB and EF respectively and lie between the same parallel lines AF and DG. Hence parallelogram region ABCD = parallelogram region EFGH.


4.3 Area of a parallelogram regionsDistance between parallel lines : If two lines are parallel then the length of the perpendicular distance of any point on one of them to the other is constant. This distance is the distance between two parallel lines. The base and the height of a parallelogram region : Any of the sides of a parallelogram region may be taken as its base. In that case, the perpendicular distance between the base and its opposite side is the height of the region. That is the perpendicular from any point of the side opposite to the base and the base line is the height of the region. In figure, the opposite side CD of the base AB of the parallelogram region ABCD is produced to F and perpendiculars AF and BE are drawn to it from the end point of AB. Then, the parallelogram region ABCD and the rectangular region ABEF stand on the same base AB and lie between the same parallels AB and CF.

h

CEDF

BA a

Hence, their areas are equal. But rectangular region ABEF = a. h square unit, where AB = a unit and BE = h unit. Then, the parallelogram region ABCD = a. h square units. In this case, length of the base of the region = a unit and its height = h unit, i.e. the area of a parallelogram region = (base × height) square unit.

4.4 The area of triangular region The base and the height of a triangular region Any of the sides of a triangular region may be taken as its base. The length of the perpendicular on the base from the vertex opposite to the base is the height of the region. In the adjoining figure, BC is the base of the triangular region ABC and its height is AQ. From the vertex opposite to the base a line parallel to the base is

Q CB

E


drawn and perpendiculars BE and CD are drawn from the end points of the base to its parallel line. Then region ABC and rectangular region BCDE stand on the same base BC and lie between the same parallels BC and ED.

Hence, region ABC = 12 (rectangular region BCDE)

= 12 b. h square units,

where, BC = b unit and CD = h unit. Here the length of the base BC of the triangular region b unit and altitude AQ = DC = h unit.

Hence, the area of the triangular region = 12 (base × height) square unit.

Exercise 4 (* marked propositions can be used to prove other propositions.) *1. Prove that, any median of a triangle divides the triangular region into two

triangular regions of equal areas. 2. Prove that, the diagonals of a parallelogram divide the parallelogram

region into four equal triangular regions. 3. A parallelogram region and a rectangular region of equal area stand on

the same base and on the same side of it. Show that, the perimeter of the parallelogram region is greater than that of the rectangular region.

4. X and Y are the middle points respectively of the sides AB and AC of the

triangle ABC. Prove that, -region AXY =12 ( -region ABC.)

5. In the figure, ABCD is a trapezium. The sides AB and CD of it are parallel. Find the area of the trapezium region ABCD.

Hint : Draw perpendiculars AL and CM from A and C respectively on the line CD and AB. A, C be joined. Then trapezium region ABCD = ( -region ABC) + ( -region ACD).

6. P is a point interior to the parallelogram ABCD. Prove that,

-region PAB + -region PCD = 12

(Parallelogram region ABCD)

L CD

A BM

Chapter Five Theorem of Pythagoras and its Application

5.1. Theorem of Pythagoras The theorem stated below is well-known as "The Theorem of Pythagoras". Pythagoras, the Greek scholar enunciated this theorem about 600 years before the birth of Christ. But the Egyptian land surveyors had an idea about the theorem even about 1000 years before Pythagoras. This theorem is proved in different methods. Two methods of proof are presented here. The first one was presented by Euclid.

Theorem 23 In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides.

The proof given by Euclid

Let ABC be a right-angled triangle, in which A = one right angle.

It is to be proved that, the square on the hypotenuse BC = the square on the side AB + the square on the side AC.

i.e. BC2 = AB2 + AC2.

Construction :

Let us draw three square BCDE, ACFG and ABLM in the external side of the triangle ABC. Through A. let us draw AN parallel to BE which intersects .the line segments BC at P and ED at N. Join A, E and C, L.

L

B

M

A F

G

C

DNE

P


Proof : BAC = 1 right angle (given) and BAM = 1 right angle [ ABLM is a square ]

BAC + BAM = 2 right angle CA and AM are in the same straight line.

Again, CBE = ABL = 1 right angle (by construction) CBE + ABC = ABL + ABC (Adding ABC both side) ABE = CBL

Now in ABE and CBLAB = BL, BE = BC [by construction]

ABE = CBLABE= CBL-region ABE = -region CBL

Now ABE and rectangular region BPNE stand on the same base BE and lie between the same parallels BE and AN.

Rectangular region BPNE = 2( -region ABE.)Again, CBL and square region ABLM stand on the same base BL and lie between the same parallels BL and CM.

Square region ABLM = 2( -region CBL) Rectangular region BPNE = Square region ABLM.

Similarly by joining A, D and B, F it can be proved that rectangular region CDNP = Square region ACFG. Rectangular region BPNE + Rectangular region CDNP = Square region ABLM + Square region ACFG.

Square region BCDE = Square region ABLM + Square region ACFG. Square on BC = Square on AB + Square on AC.

i.e. BC2 = AB2 + AC2 [proved]

Alternative proof Let ABC be a right angled triangle in which A = 1 right angle, BC = a, AB = c and AC= b. It is required to prove that, BC2 = AC2 + AB2

i.e. a2 = b2 + c2bc

b

E

a

C

DBA

Construction : We produce AB to D such that BD = AC = b. At D draw DE perpendicular to AD such that DE = AB = c. Join C, B and B, E.


Proof : In ABC and DEB

AB = DE = c, AC = DB = b [by construction] And the included angle BAC = included angle EDB [each being I right angle].

ABC = DEB BC = EB = a and BCA = EBD.

Now since CA AD and ED AD, hence CA || ED CADE is a trapezium.

Again, ABC + BCA = 1 right angle. ABC + EBD = 1 right angle.

But. ABC + CBE + EBD = 2 right angles. CBE = 1 right angle.

Now, trapezium region CADE = -region CAB + -region CBE + -regionEBD.

12 AD (AC + DE) =

12 bc +

12 a + 2 1

2 bc

or,1212

(c + b)(b + c) = bc + 12 a2

or, (b + c)2 = bc + 12 a2

or,12 (b + 2bc + c ) = bc + 2 2 1

2 a2

or,12 b + bc + 2 1

2 c = bc + 2 12 a2

or,12 b + 2 1

2 c =2 12 a2

b2 + c2 = a2 [Proved]

Remark : In the alternative proof given above, the formula of the area of a

trapezium region = 12 (distance between the parallel sides) (sum of the lengths

of the parallel sides) has been used.

Corollary : If in the ABC. A = 1 right angle and. AD BC at D, then AB2 = BC.BD and AC2 = BC.DC.


Converse of the theorem of pythagorasTheorem 24 If the square on one side of a triangle is equal to the sum of the squares on the other two sides, the angle contained by these two sides is a right angle.

Let ABC be a triangle in which AC2 = AB2 + BC2 It is required to prove that, B = 1 right angle.

Construction : Let us draw the triangle DEF such that E = 1 right angle DE = AB and EF = BC.

DA

B EC F

Proof : Since E =1 right angle, hence by the theorem of pythagoras, DF = DE22 + EF2

= AB2 + BC2 [ by construction] = AC2 [by hypothesis]

DF = AC. Now in the ABC and DEF. AB = DE, BC = EF [by construction] and AC = DF

ABC = DEF B = E

But E = 1 right angle. B = 1 right angle [proved].

Example and practical application Example 1. In the right angled triangle ABC, A = 1 right angle and BE and CF are medians. Prove that. 4(BE2 + CF2) = 5BC2


Proof : C

BE2 = AB2 + AE2

and CF2 = AC2+ AF2

4(BE2 + CF2)= 4(AB2 + AE2 + AC2 + AF2) E= 4(AB2 +AC2) + 4 AE2 + 4AF2

= 4BC2 + (2AE)2 + (2AF)2

= 4BC2 + AC2 + AB2BFA

= 4BC2 + BC2

= 5BC2

Example 2. A man starting from a fixed point A travels a distance of 12 km. right to the north and from there covers a distance of 5 km. right to the east. How far from A will be remain at the end of the journey? Solution : Let the man reaches B after travelling 12 km. to the north from A and from B travelling a distance of 5 km. right to the east, reaches C. Then AB =12 and BC = 5. It is required to find the distance AC. In the right triangle ABC,AC2 = AB2 + BC2

= 122 + 52

= 144 + 25 = 169 = 132

AC = 13 At the end of his journey the man is 13 km. away from A.

5B C

12

A

Exercise 5 1. One end of a ladder reaches the roof of house 15 metre of height above the

ground and the other end remains on the ground at a distance of 8 metre from the house. Find the length of the ladder.

2. Two posts of height 25 metres and 32 metres stand 24 metres a part. Find the distance between their tops.


3. A man starting from a fixed place travels 27 km right to the north. from then 24 km. right to the east and finally 20 km right to the south. How far is be from the starting point?

4. The base and one side of an isosceles triangle are 4.2 cm and 7.5 cm. respectively, find its height and area.

5. Find the height and area of an equilateral triangle if one side of an equilateral triangle is a cm., find its height and area.

6. In the triangle ABC, A = 1 right angle and D is a point on AC. Prove that, BC2 + AD2 = BD2 + AC2

7. ABC is an equilateral triangle and AD is perpendicular to BC. Prove that. 4AD2 = 3AB2

CB

A

D

DCB

A

DCB

A8. ABC is a right-angled isosceles triangle. P is a point on BC, its hypotenuse. Prove that, PB2 + PC2 = 2PA2

9. In ABC, C is an obtuse angle and AD is perpendicular to BC. Prove that, AB2 = AC2 + BC2 + 2BC. CD.

10. In ABC, C is an acute angle and AD is perpendicular to BC. Prove that, AB2 = AC2 + BC2

2BC.CD.

11. In figure, AD is a median of ABC. Prove that, AB2 + AC2 =

2(BD2 + AD2)

12. O is a point inside a rectangular region ABCD. Prove that, OA2

+ OC2 = OB2 + OD2.

Chapter Six Some Problems on the Application of the Theorem

of Pythagoras

Problem 6.1 To draw a square such that the square region is a given multiple of the area of a given square. Let OA = a unit be the side of a square. Its area is = a2 sq. unit. It is required to draw a square region whose area is a given multiple of a2.Let, the area of the required square region be = k.a2 where k is a positive integer.Construction : Let OX and OY be two rays perpendicular to each other at O. Now OA and OP each equal to a be cut off from OX and OY respectively. Let us join P and A.

Hence, PA2 = OA2 + OP2 = a2 + a2 = 2a2

Le., the square region on PA is twice the area of the given square region and PA = 2 .aAgain, we cut off OB equal to PA from OX and join P, B.Then, PB2 = OB2 + OP2 = 2a2 + a2 = 3a2

i.e., the square region on PB is thrice the given square region and PB = 3 .a

X

Y

P

NDCBAaO

a


Proceeding similarly we cut off PN from OX. K times and Join P and N at line PN2 = k. a2 i.e. at the (k-1) th step, the square region on PN is k times the given square region and PN = k .a

Problem 6.2 To divide a line segment into two parts such that the sum of the areas of two square region on the two parts is equal to the area of a given square region.

Let AB be a given line segment an CD = 'a' be the side of a square region. It is required to find a point in AB such that the sum of areas of two square regions on two parts is equal to the area of the square region described on CD. Construction : Let us draw ABP = 45°. Taking A as the centre and CD = a as radius we draw an arc. Let it intersects BP at E and F. Draw perpendicular EM and FN on AB from F and F respectively. Then M or N is the required point. Proof : ABP = 45° and EMB = 90° [ by construction].

BEM=45°Again, EM = BM [ EBM = BEM]Hence. AM2 + MB2 = EM2 + AE2 = CD2 = a2

Similarly. AN2 + NB2 = AN2 + FN2 = AF2 = CD2 = a2

Hence, M or N is the required point.

PF

a

aA

N MBA

45O


N.B. The arc drawn by taking A as centre and any radius CD = a may not meet BP in two points. They are shown in the following two figures [Fig. l and Fig. 2] Construction : ABP = 45° is drawn at B and AL is drawn perpendicular to BP from A.

Fig.- 1 Fig.-2

1) If CD = (a) is less than AL, then the arc drawn with A as centre and CD as radius will not intersect BP. (Fig. 1) and in that case the required point on the line segment AB cannot be obtained. 2) If CD (=a) is equal to AL, then the arc drawn with A as centre and CD as radius will meet BP only at L. (Fig-2) and in that case only one point on the line segment AB can be obtained.

Problem 6.3 To divide a straight line two parts such that the area of the square region on one part is double the area of the square region on the other part.

C

E

BDA

22o1

2

22o1

2

a

A B45o

L

P

a

A B45o

L

P


Let, AB be a given straight line. It is required to find a point on AB such that the area of the region on one part of it is twice the area of the square region on other part. Construction : At A on AB, let us draw BAE = 45° and at B draw

ABC = 22 10

2

Let, BC intersect AE at C. We draw BCD = B at C. Let, CD intersect AB at D. Then AD and BD are the required parts. Proof : BCD = CBD

CD = BD Again, CDA = BCD + CBD = 45°

AC = CD [ CAD=45°] Hence, AC = CD = BD. Again, in ACD, ACD = 90° and AD is its hypotenuse AD2 = AC2 + CD2

= BD2 + BD2 = 2BD2.i.e. the area of the square region on the part AD of AB is twice the area of the square region on the part BD.Hence, AD and BD are the required parts of AB. Note : Angle equal to 45° can be obtained by bisecting a right angle. Bisecting

the angle of 45° thus obtained an angle of 22 10

2 can be obtained.


Exercise 6

1. Draw a square region whose area is equal to the sum of two given squares regions.

2. Construct a square region of area equal to the difference of two given square regions.

3. Construct a rectangle such that the square region on its diagonal is 5 times of a given square region.

4. To divide a given line segment into two parts such that the difference of the square regions on them is equal to a given square region

5. To divide a given straight line segment into two parts such that the square region on one part is thrice the square region on the other parts.

[Hint : AB is the given line segment. Draw ABC such that A = 45° and B = 60°.

Draw ACD = A. then AD = CD In BCD, BC = 2BD. AD2 = CD2 = BC2 BD2 = 3BD2 ]

C

BDA

4590 90

4530

60

Chapter Seven Geometrical Ratio and Similarity

71. Ratio and proportion The ratio of two quantities are considered for comparing the measure of two quantities measurable in the same unit. Elaborate discussions have been made about this in the Algebra portion. It may be mentioned that if A an B are two

such quantities, their ratio is expressed as A t B or AB . In the same unit u if the

measure of A and B be au and bu respectively,

then A t B = au t bu = ab

For example, if the length and breadth of a table be 2.25 m. 0.9 m respectively,

the length of the table : breadth of the table =2.250.9 =

52 from which it may be

said that, the length of the table is 52 or 2

12 times its breadth. The equality of

two ratios is called proportion. The quantities A. B, C, D are called proportional if A t B = C t D.

Internal and External section of a line segment If A and B be two different points in a plane surface and m and n be two natural numbers, we take it as true that, (1) There is a unique point P in the

line AB that P lies between the points A and B and AP t PB = m t n.

(2) There is a unique point Q in AB where the point A lies between Q and B or the point B lies between Q and A and AQ t QB = m t n

Fig-2

nm

QBA

nmQ BA

nmP

Fig-1BA

Fig-3

In the first case the line segment AB is said to be divided internally at the point P in the ratio m t n and in the second case the line segment AB is said to be divided externally at the point Q in the ratio m t n. It may be noted that in case of external section if m < n, then Q-A-B (Fig-2 and it m > n, then Q-B-A (Fig-3).


Some useful geometric proportion 1) If the height of two triangular regions are equal then their areas and bases are proportional.

h

FB

A

dE

D

h

a C

Let BC = a units and EF = d units be the bases of the triangular regions of ABC and DEF respectively and in both the cases the height is h unit. (All lengths are expressed in the same unit.)

Then, - region ABC = 12 ah square unit and

- region DEF= 12 dh square unit

- region ABC t - region DEF =

12 ah

12 dh

=ad =

BCEF = BC t EF

(2) If the bases of two triangular regions are equal, their areas and heights are proportional.

h

A

B CPX

D

k

E FQX


Let AP = h and DQ = k units be the heights of the triangular regions ABC and DEF respectively and let the base of both the regions = x unit.

Then, -region ABC = 12 h square unit

-region DEF = 12 k square unit

-region ABC t -region DEF =

12 h

12 k

= hk = AP t DQ

Two useful theorems

The description and explanation of two geometric hypothesis without proof are given below. Their proofs are given in higher mathematics.

Theorem 25 A straight line drawn parallel to one side of a triangle intersects the other two sides or those sides produced proportionally.

QP

A

QP

CB

A

In the figures, the straight line PQ is parallel to the side BC of the triangle ABC. PQ intersects AB and BC or their produced parts at P and Q.

X

X

X

X


Then AP t PB = AQ t QC. Corollary : if PQ || BC.

then, (i) ABPB =

ACQC and

(ii) APAB =

AQAC

Theorem 26 If a straight line divides two sides or their produced parts of a triangle proportionally then it is parallel to the third side.

QP

A

QP

CB

A

In the figures, the straight line PQ divides the two sides AB and AC of ABCor their produced parts in equal proportions, that is AP t PB = AQ t QC, then, PQ || BC.

7.2. Similarity of triangles The congruency of triangles has been discussed earlier. Two triangles are. congruent if and only if the vertices of one are made to coincide with those of the other such that their corresponding angles are equals and corresponding sides are also equal.

x

A

x

zyBB

zyC

A

C


In the above figures, ABC and DBF are congruent, that is, ABC DEF and A = D, B = E, C = F,

BC = EF, AC = DF, AB = DE. Such triangles have the same size and shape.

In many cases ABC and DEF of different size but of the same shape are considered where A = D, B = E, C = F, and BC t EF = AC t DF = AB t DETwo such triangles are said to be similar.

Generally,Definition : Of two polygons with equal number of sides, if the angles of one be equal to those of the other in order, then the two polygons are said to be, equi-angular.

Definition : If the vertices of two polygons with equal number of sides are made to coincide in such a way that, the two polygons

(a) have their corresponding angles equal, and

(b) ratios of the corresponding sides are equal, then the two polygons are said to be similar.

The symbol~ is used to denote similarity. For example in Fig-2, ABC and DEF are similar in order of their vertices which is expressed by ABC~ DEF.

It may be noted that, if two polygons are similar, they are necessarily equiangular. But two equiangular polygons may not be similar. For example,

A

B C

x

y z

D

E F

x

y z

5c 5b

5a2a

2b2c


x

In the above figures the rectangle ABCD and the square PQRS are equiangular but not similar. But two triangles are similar if they are equiangular, (the theorems stated below may be noted.)

Some theorems related to similarity of triangles

Some theorems relating to similar triangles are described here without proof.

The proofs of these theorems will be given in higher mathematics.

Theorem 27 If two triangles are equiangular their corresponding sides are proportional,

In the figures, in ABC and DEF,A = D, B = E, C = F

then,ABDE =

ACDF =

BCEF

A 2x

D 2x C S R

xxxx

xP QB

E

D

QP

C

A

FB


Theorem 28 If the sides of two triangles are proportional then they are equiangular and the opposite angles of their corresponding sides are equal.

E

D

QP

CB

A

F

In the figures, in ABC and DEF,BCEF =

ACDE =

ABDE

then, A = D, B = E and C = F

Theorem 29 If one of the angles of a triangle is equal to one of the angles of the other and the sides adjacent to the equal angles are proportional, the two triangles are similar.

E

D

QP

CB

A

F

In the figures, in ABC and DEF.A = D and AB t DE = AC t DF, then ABC DEF


Theorem 30 The ratio of the areas of two similar triangles is equal to the ratio of the squares on any two corresponding sides. In the figures, ABC ~ DEF. That is, A = D, B = E, C = F and

ABDE =

ACDF =

BCEF then

ABCDEF =

AB2

DE2 = AC2

DF2 =BC2

EF2

7.3. Example In the figures, ABC a DEF are two triangles F is any point on BC.EF || BA and FG || BD; Show that, FG || AD.

Solution : Since EF || BA, CEEB =

CFFA

Again, since EG || BD, CEEB =

CGGD

CFFA =

CGGD

Hence, FG || AD.

A

B C

P Q

D

E F

A

D

B E

F

G

C


Exercise 7

1. ABC is a triangle. The line DE parallel to BC intersects the other two sides at D and E respectively.

(i) AB = 3.6 cm, AC = 2.4 cm. and AD = 2.1 cm. Find the length of AE.

(ii) AB = 6 cm, AC = 4.5 cm. and AE = 2.7 cm. Find the length of BD.

2. ABC is a triangle. A line DE parallel to BC intersects the other two sides produced at the points D and E respectively. AB = 45 cm., AC = 3.5 cm. and AD =7.2 cm., find the length of AE.

3. Prove that, the line segment joining the middle points of the oblique sides of a trapezium is parallel to its parallel sides.

4. In the figure, (A) AC = 2CD, BC = 2CE; Show that, ABC ~ DEC (B) ED || AB, Prove that, ABC ~ DEC.

5. Prove that, if the perpendicular is drawn from the vertex of a right angle triangle. The two triangle formed they are equivalent to each other and similar to the original triangle.

A B

C

DE

Chapter Eight Problems related to area and ratio

Problem 8.1 To draw such a parallelogram whose one angle is equal to a given angle and the region bounded by it is equal to a triangular region.

Let ABC be the given triangular region and x be the given angle. To draw a parallelogram whose one angle is equal to x and the region bounded by it is equal to triangular region of ABC. Construction : Let BC be bisected at E, at E in line segment EC. let us draw

CEF is equal to x. We draw the ray AG parallel to the side BC through

A and let it intersect the ray EF at F. We draw the ray CG parallel to EF

through C and let it intersect the ray AG at G. Then, ECGF is the required parallelogram.Proof : Let us join A, F. Now -region ABE= -region AEC [ BE = EC and both have the same height]

-region ABC = 2 ( -region AEC.) Again, parallelogram region ECGF = 2 ( -region ABC) [ Both have the same base EC and EC || AG]

Parallelogram region ECGF = region ABC.Again, CEF = x [ EF || CG by construction].

Parallelogram ECGF is the required parallelogram.

EB

A F G

Cx

xx


Problem 8.2 To draw a triangle whose bounded region is equal to a given quadrilateral regions.

Let ABCD be the given quadrilateral region. It is required to draw a triangle whose bounded region is equal to the quadrilateral region ABCD. Construction : Join D, B. Through C, let us draw CE || DB. Let it intersect AB produced at E. Join D, E.

Then, DAE is the required triangle.

Proof : The triangle ABC and BDE are on the same base BD and DB || CE [ by construction ]

-region BDC = -region BDE. -region BDC + -region ABD = -region BDE + -region ABD.

the quadrilateral region ABCD = -region ADE. -ADE is the required triangle.

Note : An infinite number of triangles can be drawn by applying the above method.

Problem 8.3 To bisect a quadrilateral by a straight line drawn through a vertex.

D

A B

F

C

E


Let ABCD be a given quadrilateral region, A its vertex. It is required to bisect the quadrilateral region ABCD by drawing a straight line through A.

Construction : Let us join A, C. Through D, let us draw DE parallel to AC. Let it interect BC produced at E. Let us join A. E. Then quadrilateral region BCD= -region ABE. Bisect BE at F, join A, F. Then, AF bisects the quadrilateral ABCD, Proof :

Bisect BF = 12 BE [ F is the middle point of BE ].

-region ABF = 12 ( -region ABE) =

12 (quadrilateral region ABCD)

AF bisects the quadrilateral region ABCD.N. B. In the above construction the condition that -region ABC > -regionADC is necessary. Otherwise, F will fall on BC produced and then it will not be possible to bisect it by applying the above method.

Problem 8.4 To draw a triangle whose one side is equal the given line segment and the region bounded by it is equal to a given triangular region.

Let ABC be a triangle and a be a given line segment. It is required to draw a triangle such that the region bounded by it is equal to triangular region of ABC and its one side is equal to a.Construction : The side BC of ABC is produced and a portion BE = a is cut off from it. Let us join A, E and at C draw CD || AE. Let, CD intersect AB at D and join D, E. Then DBE is the required triangle.

A

D

B a

a

C E


Proof : Both the triangles ADC and DCE stand on the same base DC and they lie between the same parallels DC and AE. Hence, -region ACD = -region CDE Now, -region ABC = -region DBC + -region ADC = -region DBC + -region DCE = -region DBE. -DBE is the required triangle.

Problem 8.5 To draw a parallelogram such that its one angle is given and the region bounded by it is equal to the given quadrilateral region.

Let the given quadrilateral region be ABCD and x be the given angle. We are to draw a parallelogram whose one angle is equal to x and bounded region by it is equal to ABCD region.

Construction : Join B, D. Through C, draw CF || DB and let CF intersect the produced part of AB, at F. Bisect AF at G. Draw GAK = x at A of the line segment AG. Draw GH || AK, through D draw KDH || AG and let it intersect AK and GH at K and H respectively. Then parallelgoram AGHK is the required parallelogram. Proof : Join D, F. AGHK is a parallelogram [ by construction] Where GAK = x and -region DAF = quadrilateral region ABCD. Parallelogram region AGHK = -region DAF.

AGHK is the required parallelogram.


Problem 8.6 To bisect a triangle by a straight line drawn through a point lying on any side of it.

Let ABC be the given triangle and P be the given point on the side AB. It is required to bisect the ABC by drawing a straight line through the point P. Construction : Bisect AB at Z. Join P, C and draw ZQ || PC through Z. Let ZQ intersect BC at Q. Join P, Q. Then PQ bisects the triangle ABC. Proof : Join Z, C.A-region ZQP = -region ZQC [Both the s stand on the same base ZQ and lie between the same parallels ZQ and PC]

-region ZBQ + -region =ZQP -region ZBQ + -region ZQC.

-region PBQ = -region ZBC = 12 ( -region ABC)

Problems 8.7 To divide a given line segment internally in two given ratios.


Let AB be the given line segment. It is required to divide AB internally in the ratio m t n. Construction : Let us draw a ray AX making any angle at A. From AX we cut off AE = m. Again from EX, we cut off EC = n. Then we join B, C. Through E, draw ED || CB. Let ED intersect AB at D. Then AB is divided internally at D in the ratio m : n. Proof : In ABC, AD t DB = AE t EC [ ED || CB] = m t n. Hence, D is the required point which divides the line segment AB internally in the ratio m t n

Problem 8.8 Given three line segments of length a, b, c. To find a line segment of length such that a t b = c t d.

a b c

Let a, b, c be three line segments. It is required to find the line segment of length d such that a t b = c t d. Construction : Let us draw two intersecting rays AB and AC. From AB we cut off AD = a and DE = b. Again we cut off AF = c from AC. Let us join F, D. Again through E we draw EH || DF. Let EH intersect AC at H. Then FH is the required line segment of length d.Proof : If AEH, FD || HE,

AD t DE = AF t FHor, a t b = c t d [AD = a, DE = b, AF = c, FH = d]

BEbDa

c

Fd

H

C

A


Remark . The third proportional of two line segment a, b can be determined in the above method. In that case, it is to be shown that a t b = b t d.

Problem 8.9 The lengths of two line segments are a and b respectively. To find the line segment of length c, such that ab = c2

Let a, b be two given line segments. It is required to find a line segment of length c such that a t c = c t b or, ab = c2

Construction : From any ray AX, let us take off the portions AB = a and BC = b. Taking AC as a diameter, let us draw a semi ADC. At B on AC, BD is drawn perpendicular to AC. BD intersects the semi cord, at D. Then BD will be the required length c.

Proof : In ADC ADC = 1 right angle and BD AC [ by construction] ABD and CBD similar.

BDAD =

BCBD

or, AB.BC = BD2

ab = c2 [ AB = a, BC = b, BD = c]

C b b

Baa

c

A X

D


Exercise 8

1. Construct parallelogram such that one of its side is equal to a given line segment and the region bounded by it is equal to given rectangular region.

2. Construct a triangle such the region bounded by it is equal to the given region bounded by a pentagon ABCDE.

Hint : Let us draw a quadrilateral region EDGA so that quadrilateral region EDGA = pentagon region ABCDE. Then draw DGH such that

-region DGH = quadrilateral region EDGA.

3. Trisect a triangle by two straight lines drawn from a point on one of its sides.

4. Construct a parallelogram such that its base is equal to given line segment and an angle is equal to a given angle and the region bounded by it is equal to the given triangular region.

E

xx aa CDB

K

A

AH

E

D

C

GB

G H


Hint : Let ABC be a given triangular region. Let x and a are given and side respectively of the parallelogram. From BC a length = 2a is cut off. On BE as base a triangle KBE is drawn such that -region KBE =

-region ABC. Now, parallelogram EDGH is drawn such that EDGH region = -region

KBE and EDG = x.5. Construct a parallelogram a such that one of its side is equal to a given

line segment and the region bounded by it is equal to a given parallelogram region.

a

Hint : ABCD is a given parallelogram and a is a side of the required parallelogram. AB is produced to E such that BE = a. The line EC is drawn and it intersects produced part of AD at F the parallelogram AEGF is completed. The lines BC and DC intersect FG and EG at K and L respectively. KGLC is the required parallelogram.

6. ABCD is a quadrilateral. X is a point on the side DC. Draw a triangle with X as a vertex and base along AB such that the triangular region is equal to quadrilateral region.


About Quadrilateral

Multiple Choice Questions (MCQ) :

1. If ABCD is a parallelogram then which of the following is correct? A. AC = BC B. AD = AC C. AO = OB D. OA = OC 2. Which of the following information is needed to draw a quadrilateral? A. Three sides and two diagonals. B. Segments of two diagonals and one side. C. Four sides and two angles D. Two sides and two angles 3. If AB || CD. AC BD and A = 90° of a quadrilateral ABCD, then what

type. of quadrilateral is ABCD? A. Rectangle B. Parallelogram C. Rhombus D. Square

The following information refers to questions 4 to 6.

In the figure AB = 10 unit, CD = 7 unit. BC = 4 unit, BC AB and AB || CD

A10

EB

4

CD 7

4. Which of the following is the correct value of AD? A. 3 B. 4 C. 5 D. 6 5. If A = 90° of AECD quadrilateral then what type of quadrilateral is it? A. Parallelogram B. Rectangle C. Trapezium D. Rhombus


6. What is the area of ADC in sqr unit? A. 28 B. 20 C. 14 D. 6

7. Look at the following information: i. The diagonals of a parallelogram is perpendicular bisector each other ii. If AB II CD and AD BC of a quadrilateral ABCD; ABCD is a

trapezium. iii. If opposite sides of a quadrilateral are equal as well as parallel and

one of its angles is 90°, then it is a rectangle. Based on the above information which is correct? A. i and ii B. ii and iii C. i and iii D. i, ii and iii

8. P is any point on side AB of ABCD rectangle. The area of ABCD rectangle is

100 sqr. cm and area of PBC is 20 sqr. cm. What is the area of APD in sqr. cm?

A. 50 B. 40 C. 30 D. 20

Creative Questions :

AD is a median of ABC. AD || CE, DC || AE and AD = AE. A. What type of quadrilateral is ADCE and why? B. Bisect quadrilateral ABCE by a straight line drawn through point B.

(Sign of drawing and description are required) C. Find the mid point O of AC and prove that O is also mid point of

DE.

A

B D C

E

Chapter Nine Theorems on Locus

9.1. Locus Suppose an ant is let move on a rectangular table. The ant keeping equal distances from two angular terminal points A and B moves along CD starting from C, the middle point of AB. Here the ant moves satisfying a condition. So the path of the moving ant is a locus. In this case the locus is a straight line.

One end of a thin thread is tied to a pin and the other to a pencil. The pin is fixed on a piece of paper and the pencil tied to the other end of the thread is moved on the paper keeping the thread always straight. The path in which the end of the pencil moves is a locus. In this case also, the end of the pencil moves under certain condition. The locus here is a circle. Generally speaking, under certain given condition, the locus of a point is a line containing points which satisfy the given condition. This locus is described as the locus of the moving point or points satisfying the given conditions. To make the locus finite, care is taken to see that : a) All the points satisfying the conditions lie in the locus; and b) Every point on the locus satisfies the given conditions.

9.2. Two Theorems Two theorems on locus are stated and proved below : Theorem 31 The locus of a point equidistant from two given points is the perpendicular bisector of the straight line joining the given points.

A BC

D


Let A and B be two given points. It is to be proved that the locus of a point equidistant from A and B is the perpendicular bisector of the line segment AB. (1) First take a point P equidistant from A and B (Fig 1) and prove that the point P lies on the perpendicular bisector of AB.

Construction : Join A, B; P. A and P, B. Let Q be the middle point of AB and join P and Q.

Proof : Note that, if P lies on the line segment AB, then P and Q are coincident points P must lie on the perpendicular bisector of AB. Otherwise, PA = PB [by hypothesis] and AQ = BQ [by construction] Now, in PAQ and PBQ,PA = PB, AQ = BQ and QP = QP [common side]

PAQ PBQPQA = PQB.

N.P. As these two angles are on adjacent sides of a line, they are equal and as such, each is equal to one right angle. PQ is the perpendicular bisector of AB, that is, P lies on the perpendicular bisector of AB.(2) Let P be a point on CD, the perpendicular bisector of the line segment AB and let us prove that P is equidistant from the points A and B.

Construction : Join P, A and P, B.

Proof : By hypothesis, CD passes through Q, the middle point of AB. AO = BO. also, AQP = BOP = 1 right angle.

P

A BQ

CP

A BQ

Figure : 1 Figure : 2


It is noted that P lies on AB, then P coincides with Q. Hence P is equidistant from A and B. Otherwise, In PAQ and PBQAQ = BQ, QP = QP [ common side ] and AQP = BQP.

PAQ = PBQ. PA = PB.

i.e.; P is equidistant from A and B.

Theorem 32 The locus of a point equidistant from two intersecting straight lines is the bisectors of the internal angles between the two given straight lines. Let AB and CD are two intersecting line intersect at a point 0. To prove that the locus of any point equidistant from AB and CD is the bisectors of the internal angles between AB and CD; That is (1). Any point equidistant from AB and CD is on any line on the bisectors of the internal angles between AB and CD and (2). any point on that bisectors of angle is equidistant from AB and CD. Now,1) First we take a point P equidistant from AB and CD [Fig-1] and prove that P lies on either of the bisectors of the internal angles between AB and CD.Construction : Perpendiculars PM and PN are drawn from the straight line PO and P on AB and CD respectively. Proof : PM and PN are the perpendiculer distances of AB and CD respectively from P.

PM = PN. In the right angled POM and PON hypotenuse PO = hypotenuse PO [common side] and PM = PN.


POM PON. POM = PON PO bisects one of the internal angles between AB and CD Or P lies on

either of the bisectors of the internal angles between the lines. (2) A point P is taken on QR or ST, the bisectors of the internal angles between AB and CD (Fig. 2) and we prove that P is equidistant from AB and CD. Construction : Let us draw PM and PN perpendiculars from P on AB and CD respectively. Proof : Here PM and PN are distances of AB and CD respectively from P. In the POM and PON

POM = PON [by hypothesis] PMO = PNO [ by construction ]

and PO = PO [ construction side] POM PON

PM = PN The point P is equidistant from

AB and CD. Hence it is proved from (1) and (2) that, the locus of the point P equidistant from AB and CD is the bisector of the internal angles between AB and CD.

Point of intersection of loci. More than one locus can be obtained when the points satisfy two or more conditions simultaneously. The point of intersection of these loci satisfies the given conditions. Example 1. Determine a point on a given line XY which is equidistant from two points A and B outside XY.

BAD

CP Y

XThe locus of points equidistant from A and B is the perpendicular bisector of the line AB.Hence, the required point lies on CD. Again, the required point lies on XY. Hence, the required point is the point of intersection of CD and XY. Again, the required point lies on XY. Hence, the required point is the point of intersection of CD and XY.


Example 2. Determine the point equidistant from three points A, B and C not lying on the same straight line. The locus of the point equidistant from A and B is the perpendicular bisector PQ, of the line AB. Again, the locus if the points equidistant from the points A and C is the perpendicular bisector RS, of the line AC.

As A, B and C do not lie on the same straight line, PQ and RS must intersect at a point. As the point of intersection O is the only point common to both the loci. it is equidistant from A, B and C. Hence, O is the required point.

Exercise 9

1. Determine the locus of a point that lies at a particular distance from a particular straight line.

2. Determine the locus of a point equidistant from two parallel lines.

3. Determine the point equidistant from the three sides AB, BC and CA of the ABC.

4. Prove that, the internal bisector of one angle of a triangle and the external bisectors of the other two angles are concurrent.

5. Prove that, the perpendicular bisectors of the sides AB, BC and CA of the ABC are concurrent.

6. Prove that, three bisectors of the angles of a triangle are concurrent.

7. AB and AC are two given lines. Determine the point 2 cm. and 3 cm. away from AB and AC respectively.

A

B CQ

P

S

R

O

Chapter Ten

Theorems on Circle

10.1. Circle Definition : If O be a given point on a plane and r a given positive real number, then the set of all points on the plane which lie at distance r from O is called a circle whose centre is O and radius is r.

Remark : An infinite number of straight lines lying on the plane pass through O. All such lines have two and only two points which are at distances r > o from O. Hence the circle and any line through its centre have two and only two common points. In the above figure, the points A, A' ; B, B'; C, C'; D, D' are the points on the circle. The circle is a line which is not a straight line. No straight line contains more than two points equidistant from a point outside the straight line.

D

BD

B

A

C C

A

Remark : The distance of a point on the circle from the centre has been called the radius of the circle. The line segment joining the centre and any point on the circle is generally called a radius of the circle. For example, in the figure, O is the centre of the circle. A, B and C are points on the circle. Each of OA , OB and OC is a radius of the circle.

C

OA B

Definition : Some co-planar points are called concyclicif a circle passes through these points or, there is a circle on which all those points lie.


O

P

Q CB

ARemark : Any two different points A and B are concyclic. Any point O on theperpendicular bisector of the line segment AB is equidistant from A and B (Theorem 25).

R

S

Hence, the points A and B lie on the circle whose centre is O and radius OA. Three different points A, B, C of a plane but not lying on a straight line are cyclic.

Any point on PQ, the perpendicular, bisector of AB is equidistant from A and B and A and C are equidistant from any point on R S the perpendicular bisector of AC (Theorem 25). As the lines containing the line segments AB and BC are different, the straight lines PQ and RS on the plane are not parallel. Hence, the straight lines PQ and R S have one and only one common point. The points A, B, C are equidistant from this unique point O and A, B, C are not equidistant from any other point on the plane. Henec, the circle whose centre is O and radius OA has the points A, B, C lying on it and there is no other circle passing through A, B and C. From the above discussion it is observed that, Hypothesis 10.1. Three points which are not collinear and lie on a plane determines one and only one circle on whose those three points lie.Interior and Exterior regions of a circle. Definition : If O be the centre of a circle and r its radius, then the set of all points on the plane whose distances from O are less than r is called the interior region of the circle and the set of all points on the plane whose distances from 0 are greater than r is called the exterior region of the circle. Remarks : The circle, its interior region and the exterior region are three subsets of the plane which are disjoint. Any point outside the circle is called an external point of the circle. The line segment joining two points inside a circle lies wholly inside the circle.


Two formulas relating to the continuity of circle. (A) Any line joining a point inside intersects a circle and another outside a circle intersects the circle at one and only one point. In fig, P is a point inside the circle and Q is a point outside the circle. The line segment PQ intersect the circle only at the point R. (B) If a point of a circle lies inside another circle and another point of the first circle lies outside the other circle, then the two circles have two and only two points of intersection.

10.2. Chord and Diameter of a circle. Definition : The line segment joining two different points of a circle is called a chord of the circle. If any chord, of a circle passes through the centre, then the chord is called a diameter of the circle. In the figure, AB and AC are two chords of the circle. AC is a diameter and O is the centre of the circle. Remarks : The centre of a circle is the middle point of every diameter. Hence, the length of every diameter is 2r, where r is the radius of the circle. Some Theorems : Theorem 33 The straight line joining the centre and the middle point of a chord other than the diameter is perpendicular to the chord. Let, ABC be a circle whose centre is O and AB be a chord which is not a diameter and let D be the middle point of this chord. O and D are Joined. It is required to prove that OD is perpendicular to AB.

A B

CO

P QR

PQ

O

DA B

C


Construction : Let us join O, A and O, B. Proof : In OAD and OBD, AD = BD [ D is the middle point of AB] OA = OB [ Both being radius of the same circle] and OD = OD [Common side]

OAD OBDODA = ODB.

As the angles are adjacent angles and they are of equal magnitude, hence ODA = ODB = 1 right angle. Hence, OD AB [ Proved]

Theorem 34 The perpendicular drawn from the centre of a circle on any chord other than the diameter bisects the chord. Let ABC be a circle with centre O and AB be a chord which is not a diameter. OD is the perpendicular from the centre O on this chord. It is required to prove that, OD bisects the chord AB at the point D. i.e. AD = BD.

Construction : Let O, A and O, B be joined. Proof : since OD AB,

ODA = ODB = 1 right angle. Hence, both ODA and ODB are right angled triangles. Now, in the right angled triangles ODA and ODB, Hypotenuse OA = Hypotenuse OB [both being the radius of the same circle.] and OD = OD [common side]

ODA ODBHence, AD = BD [Proved ] Remark : Theorem-33 and Theorem-34 are converse to each other. Corollary-1. The perpendicular bisector of any chord of a circle passes through the centre.

O

DA B

C


Corollary-2. Any straight line can not intersect a circle more than at two points.

Corollary-3. The line segment joining the centres of two intersecting circles bisects their common chord at right angles.

Hint : Two circles with centres A and B intersect at the points P and Q and C is the middle point of the chord PQ. A, C and B, C are joined. Since A is the centre and C is the middle point of the chord PQ, AC PQ.

Hence, ACP = 1 right angle. Similarly, PCB = 1 right angle. AC and BC lie in the same straight line. N.P. Hence, the straight line AB bisects the chord PQ at right angles. [Proved]

Exercise 10.1 1. Prove that, the centres of all circles passing through two fixed . points are

collinear.2. Prove that, if two chords of a circle bisect each other, their point of

intersection is the centre of the circle. 3. Prove that, the straight line Joining the middle points of two parallel

chords of a circle pass through the centre and is perpendicular to the chords.

4. Two chords AB and AC of a circle subtend equal angles with the radius passing through A. Prove that, AB = AC.

5. In the figure, O is the centre of the circle and chord AB = chord AC. Prove that BAO = CAO.

OA

B

C


6. A circle passes through the vertices of a right angled triangle. Show that, the centre of the circle is the middle point of the hypotenuse.

7. A chord AB of one of the two concentric circles intersects the other circle at points C and D. Prove that, AC = BD.

Theorem 35 Equal chords of a circle are equidistant from the centre, Let O be the centre of the circle and AB and CD be the two equal chords of the circle. It is to be proved that, AB and CD are equidistant from O.

Construction : Perpendiculars OE and OF are drawn from O on the chords AB and CD respectively. Let O, A and O, C be joined.

Proof : Since the perpendicular drawn from the centre of a circle on any chord other than the diameter bisects the chord and OE AB and OF CD. Hence, AE = BE and CF = DF.

AE = AB and CF = DF

AE = 12 CD.

But AB = CD [ by hypothesis]

AE = CF.

Now in the right angled OAE and OCF,hypotenuse OA = hypotenuse OC [ both being the radii of the same circle]

and AE = CF

OAE PCF

OE = OF. But OE and OF are the distances of the chords AB and CD respectively from the centre O. Hence, the chords AB and CD are equidistant from the centre. [Proved]


Theorem 36 All chords equidistant from the centre of a circle are equal. Let, O be the centre of the circle and AB and CD be two chords of it. OE and OF are perpendiculars drawn from O on AB and CD respectively and OE = OF. It is required to prove that, AB = CD.

Construction : Let us join O, A and O, C.

Proof : Since OE AB and OF CD. Hence, OEA = OFC = 1 right angle. Now, in the right angled triangles OAE and OCF, hypotenuse OA = hypotenuse OC [both are the radii of the same circle] and OE = OF [by hypothesis]

OAE = OCF. AE = CF

But the perpendicular drawn from the centre of a circle on any chord other than the diameter bisects the chord.

AE = 12 AB and CF =

12 CD

Hence,12 AB =

12 CD.

or, AB = CD [Proved ]Remark : Theorem -35 and theorem -36 are converse to each other.

Example 10.1 Prove that, of two chords of a circle, the chord which is nearer to the centre is greater than the other. Let AB and CD be two chords of a circle whose centre is O.


OE and OF are perpendiculars from O on AB and CD respectively and OE < OF. It is required to prove that, AB > CD. Construction : Let O, A and O, C be joined.

Proof : Since OE AB and OF CD.

Hence, AE = 12 AB and CF =

12 CD.

Now in the right angled triangles OAE and OCF, we get respectively OA2 = OE2 + AE2 and OC2 = OF2 + CF2

But, OA = OC OA2 = OC2

OE2 + AE2 = OF2 + CF2..................................... (1) Now as OE < OF, OE2 < OF2

Hence from (1) we get, AE2 > CF2

AE > CF

or,12 AB >

12 CD.

Hence, AB > CD [Proved]

Example 10.2 Prove that, the diameter is the greatest chord of a circle. Let O be the centre of the circle ABDC. AB is its diameter and CD is a chord other than the diameter. It is required to prove that, AB > CD. Construction : Let O, C and O, D be joined.Proof : OA = OB = OC = OD [each being radius of the same circledNow in OCD, OC + OD > CDor, OA + OB > CDor, AB > CD [Proved]

A


Exercise 10.2 1. If two equal chords of a circle intersect each other, show that two

segments of one are equal to two segments of the other. 2. Prove that, the middle points of equal chords of a circle are concyclic. 3. Show that, the two equal chords drawn from two ends of the diameter on

its opposite sides are parallel.4. Show that, the two parallel chords of a circle drawn from two ends of a

diameter on its opposite sides are equal. 5. Show that, of the two chords of a circle the greater chord is nearer to the

centre than the shorter. 6. In a circle, AB is a given chord and CD is another chord whose middle

point E lies on the chord AB. Show that the nearer is E to the middle point of AB, the greater is the length of the chord CD.

10.3 The arc of a circle Definition : If A and B are the two different points of a circle, then A, B and the set of points on one side of

AB is called an arc of the circle. A and B are the terminal points of this arc and all other points of the arc are its interior points. By taking a fixed point C on the arc, it is denoted as the arc ACB and is expressed by the symbol ACB.

Another arc

An arcC

BA

Remark : The arc of a circle is a sub-set of the circle. The two points A and B of the circle divide the circle into two arcs. Both the arcs have their terminal points at A and B and the arcs have no common point other than the terminal points. Definition : In a circle ACB is called (A) Semi-circle if AB passes through the centre, (B) Minor arc if the interior points of the arc lie on the side opposite to that of AB in which the centre lies.

another semicircle

BA

C Minor arc

a semi circle C

BA


(c) Major arc if the interior points of the arc and the centre lie on the same side of AB .

Definition : In a circle the arcs ACBan ADB having the, same terminal points are said to be conjugated to each other if the points C and D are on opposite of AB .

Remark : If ACB is a minor arc, then its conjugate arc ADB is a major arc and conversely,

A B

OMajore arc

C

C

A B

D

Arc intersected by an angle. Definition : An angle is said to intersect or cut an are of a circle if (1) each terminal point of the arc lies on the arm of the angle, (2) each arm of the angle contains at least on terminal point and (3) every interior point of the arc lies inside the angle.

C

O BA ABO

BABAPP P P

Figure-3Figure-1 Figure-2 Figure-4D

BAP

Every angle shown in each of the figures above intersects the arc APB of the circle with centre O. In fig 5, the angle also intersects the arc APB.

BD B

A BO

PP

P AFigure-5 Figure-7

AFigure-6


Angle in a circle Definition : If the vertex of an angle is a point on a circle and each arm of the angle contains a point of the circle other than the vertex, then the angle is side to be an angle in the circle or an angle inscribed in the circle. In the above Figs, no 2, 3, 4 are angles inscribed in the circle, Remark : Every angle inscribed in a circle intersects an arc. This arc may be major arc, semi circle or a minor arc. Definition : The arc subtended in a circle by an angle inscribed in a circle the angle stands on the arc and lies in the conjugate arc of the segmented arc or inscribed.In the Figs 3 or 4 above, the angle stands on the arc APB and is inscribed in the arc ACB. It is noted that arcs APB and ACB are conjugated. Remark : The angle inscribed in an arc of a circle is that angle whose vertex is a vertex in the arc and each of whose arms passes through a terminal point of an arc. An angle standing on an arc is the angle inscribed in the conjugate arc. Angle at the centre of a circle. Definition : The angle whose vertex is at the centre of a circle, is called an angle at the centre of that circle and is said to stand on the arc which it subtends.The angle in Fig. No. I above is an angle at the centre and it stands on the arc APB.Every angle at the centre subtends a minor arc in the circle. An angle at the centre and standing on a minor arc means such an angle whose vertex is at the centre of the circle and whose arms pass through the terminal points of that arc. (Fig-8).

For the discussion of angles at the centre standing on a semi-circle and a major arc, the above description is not meaningful. But by taking a point P in the arc APB, AOP and POB standing on the arcs AP and PB respectively may be

BA

A

O

P

A BO

P

B

O

PFigure-8 Figure-9 Figure-10


considered (Fig-9 and Fig-10). In the case of a semi-circle (Fig-9) AOP + POB = straight angle AOB and in the case of a major arc (Fig- 10) AOP + POB = reflex AOB. Such straight or reflex angles are considered to stand

on semi-circle and major arc respectively. Measure of arc in degree In many cases the measure of an arc is described with the help of the angle at the centre measured in degree.

Definition : The measure of an arc of a circle in degrees is as follows :

(A) If the arc is a minor arc its measure in degree is the angle in degree at the centre and standing on that arc.

(B) If the arc is a semi-circle its measure in degree is 180°.

(C) If the arc is a major arc its measure in degree is 360-d where d is the measure in degree of its conjugate arc. It may be noted that the sum of the four angles formed at the point of intersection of two straight lines measured in degrees is 360. By using the extended conception of an angle at the centre it may be said that the measure of an arc in degree = The angle at the centre measured in degree.

Remark : The measure in degree of an arc does not denote the length of the arc. In the adjacent figure the two arcs AB and A'B are two concentric circles with centre O have equal measures in degrees, but it is clear from the figure that their lengths are not the same.

Theorem 37 Standing on the same arc the angle at the circumference is half the angle at the centre.Let, ABC be a circle with centre 0 and let BAC be the angle at the circumference and BOC be the angle at the centre, standing on the same arc BC.

It is required to prove that, BAC = 12 BOC.

Proof : (1) At first, let the line AC pass through the centre (Fig-1)


In this case, in AOB OA = OB [ being radius of the same circle]

OAB = OBA.But in AOB, the exterior angle BOC= OAB + OBA = OAB + OAB= 2 OAB

OAB = 12 BOC

that is, BAC = 12 BOC

(2) Now suppose AC does not pass through the centre (Fig 2 and Fig-3). In this case the diameter AD through A is drawn. Now the sides AD of CAD standing on the arc CD passes through the centre.

Hence, according to (1), CAD = 12 COD at the centre.

Again the side AD of the BAD standing on the arc BD passes through the centre. Hence according to (1),

BAD = 12 BOD at the centre.

Now, in Fig.-2 (Where the points B and C lie on opposite sides of the line AD)

CAD + BAD = 12 ( COD + BOD)

or, BAC = 12 BOC,

Again in Fig.-3 (Where the points B and C lie on the same side of AD)

A

O

C

B

Figure-1


CAD BAD = 12 ( COD BOD)

or, BAC = 12 BOC.

In all cases, BAC = 12 BOC.

Theorem 38 Angles in the same segment of a circle are equal

Let, O be the centre of the circle and BAD, BED be two angles at the circumference standing on the arc BCD It is required to prove that, BAD = BED.Construction : Let O, B and O, D be joined.

Proof : Here the angle BOD is the angle at the centre O standing at the arc BCD. Since on the same arc the angle at the circumference is half the angle at the centre,

BAD = 12 BOD and BED =

12 BOD

BAD = BED

Theorem 39 If the line segment joining two points subtends equal angles at two other points on the same side of it the four points are concyclic. Let, A and B be two different points and the angles ACB and ADB subtended at the points C and D on the same side of the line AB are equal, that is, ACB = ADB.It is required to prove that, the four points A, B, D, C are concyclic.


Constructions : Let a circle be drawn through the points A, B and C (it is possible to draw the circle because they do not lie in the same straight line). Let it intersects the line AD at E. Let E, B be joined.

Figure-l Figure-2

Proof : ACB = AEB [in the same segment of the circle]

But ACB = ADB [given]

AEB = ADB.

But it is impossible because from BED in Fig I, exterior AEB > opposite interior ADB and from BED in fig 2.

Exterior ADB > opposite interior AEB.

Hence, the points E and D can not be different; F must coincide with the point D.

Hence, the four points A, B, C, D are concyclic.

Corollary : Triangles on the same base and on the same side of it having equal vertical angles have their verities concyclic.

Remark : Theorem-38 and Theorem-39 are converse hypothesis.

Exercise 10.3 1. ABCD is a quadrilateral inscribed in a circle with centre O. If the

diagonals AC and B intersect at the point E, prove that,

AOB + COD = 2 AEB.2. Two chords AB and CD of the circle ABCD intersect at the point E.

Show that, AED and BEC are equiangular.

A B

CDE

A B

C DE


3. In the figure, chord AB chord CD. The arc AC and BD subtend AOCand BOD respectively at the centre.

Prove that, AOC + BOD = 2 right angles.

[ Hint : Join A, D. AOC + BOD= 2 ( ADC + BAD)

Now ADC + BAD = right angle. Because in ADE, AED = 1 right

angle.AOC + BOD = 2 right angle

4. In the circle ABCD with centre 0. ADB + BDC = I right angle. Prove that. A, O and C lie in the same straight line.

5. In interior of a circle two chords AB and CD intersect at a point E. Prove that, the sum of the angles subtended by the arcs AC and BD at the centre is twice AEC.

6. To the exterior of a circle, two chords AB and CD intersect at a point E. Prove that, the difference of the angles subtended by the two arcs AC and BD, at the centre is twice AEC.

7. Show that, the oblique sides of a cyclic trapezium are equal.

8. AB and CD are the two chords of a circle : P and Q are the middle points of the two minor arcs intersected by them. The chord PQ intersects the chords AB and AC at points D and E respectively. Show that, AD = AE.

Theorem 40 The angle in a semi circle is a right angle.

Let, AB the diameter of a circle with the centre at O and ACB is an angle in a semi circle.

It required to prove that, ACB = 1 right angle.

BC

O

A D


Construction : A point D is taken on the circle at the side opposite to that of AB, in which C lies.

Proof : Standing on the arc ADB, ACB at the circumference = 12 (the straight

angle AOB at the centre). But straight angle AOB = 2 right angles.

ACB = 12 (2 right angles) = 1 right angle.

Corollary 1. (A) The angle in a major arc of a circle is an acute angle and (B) the angle in the minor arc is an obtuse angle.

Figure-1 Figure-2 Let O be the centre of a circle and the arc BAC be a major arc (in Fig. I) and minor arc (in Fig. 2). It is required to prove that, (A) in Fig. 1, BAC is an acute angle and (B) in Fig. 2, BAC is an obtuse angle.Construction : Let diameter BD be drawn. Proof: In both the figures.

BAD is an angle in a semi-circle. Hence, BAD =1 right angle. (A) In Fig. 1, the points A and C are at the opposite sides of the straight line BD. Hence, BAC < BAD

BAC < 1 right angle. That is, BAC is an acute angle. (B) In fig-2. the points A and C lie in the same side of the straight line BD. Hence, BAC > BAD

BC

D


BAC > 1 right angle That is, BAC is an obtuse angle. Corollary 2. The circle drawn on the hypotenuse of a right- angled triangle as diameter passes through the vertex which is a right angle.

Example 1. Two circles intersect at the points A and B. If AC and AD are the diameters of the circles, show that, the three points C, B, D are collinear.

[ Hint : Since AC and AD are diameters., hence ABC and ABDin semicircles are each 1 right angle.

CBD = one straight angle that is, the three points C, B, D are collinear]

Exercise 10.4 1. If two circles are drawn on two equal sides of an isosceles triangle as

diameters, prove that, they will intersect each other at the middle point of the base.

2. Prove that, the line segment joining the middle point of the hypotenuse of a right-angled triangle and the opposite vertex is half the hypotenuse.

3. Prove that, the sum of the angles in the arcs AB, BC and CA of the circle drawn through the vertices of ABC is, equal to four right angles.

4. Given : AD BC and AE is a diameter of a circle. Prove that,

BAD = EAC.[Hint. ACE = I right angle, as AE is the diameter. EAC + AEC = 1 right angle.

Again, ABD + BAD = I right angle.


Since ADB =1 right angle But ABD = AEC, since they are angles at the circumference in the

same arc AC. BAD = EAC]

5. ABC is a triangle. If a circle described on AB as diameter meets the side BC at D, then prove that, the circle described on the side AC as diameter also passes through the point D.

Quadrilateral inscribed in a circle Definition : A Polygon is said to be inscribed in a circle if the vertices lie in circle. In that case, the circle is said to be circum-circle of the polygon.

In Fig-1, ABC and in Fig-2, the quadrilateral ABCD are inscribed in the circle. It may be mentioned that a circle can always be drawn through three non-collinear points. Hence every triangle can be inscribed in a circle, But each of other polygons except triangles cannot be inscribed in a circle.

Theorems related to quadrilaterals inscribed in a circle

Theorem 41 The sum of the two opposite angles of a quadrilateral inscribed in a circle is two right angles. Let ABCD be a quadrilateral inscribed in a circle with centre 0. It is required to prove that,

ABC + ADC = 2 right angles, and BAD + BCD = 2 right angles.

A

Figure-1

B C

A D

B C

Figure-2


Construction : Let O, A and O, C be joinedProof : Standing on the same arc ADC, the angle at centre AOC = 2 ( ABC at the circumference) that is

AOC = 2 ABC.Again standing on the same arc ABC, reflex AOC at the centre = 2 ( ADCat the circumference) that is, reflex

AOC = 2 ADCAOC + reflex AOC = 2 [ ABC + ADC]

But AOC + reflex AOC = 4 right angles 2 ( ABC + ADC) = 4 right angles

ABC + ADC = 2 right angles. In the same way, it can be proved that. BAD + BCD = 2 right angles.

Corollary 1 : If one side of cyclic quadrilateral is produced, the exterior angle formed is equal to the opposite interior angle.

Corollary 2 : A parallelogram inscribed in a circle is a rectangle.

Theorem 42 If two opposite angles of a quadrilateral are supplementary, the vertices of the quadrilateral are concyclic.

D DEA A E

B C

C

D

A

O

B

BC


In the quadrilateral ABCD, let ABC + ADC = 2 right angles.

It is required to prove that, the points A, B, C. are concyclic.

Construction : Since the three points A, B, C are not collinear, therefore one and only one circle can be drawn through these three points. Let the circle intersect the straight line AD at E. C. E are joined.

Proof : Since the quadrilateral ABCD is inscribed in a circle, therefore, ABC+ AEC = 2 right angles

But ABC + ADC = 2 right angles [given]

AEC = ADC.But that is impossible, because in the CED in Fig-1, the exterior AEC > opposite interior EDC or ADC and in CED, in Fig-2, exterior

ADC > opposite interior DEC.Hence the points E and D cannot be different. The point E must coincide with the point D. Hence, the four points A, B, C, D are concyclic.

Example 1 : Any straight line drawn parallel to the base BC of an isosceles triangle ABC intersects AB and AC at the points D and E respectively. Show that, the four points B, C, E, D are concyclic. Since ABC is an isosceles triangle.

Therefore, ABC = ACB.

Again since DE || BC and AC is their intercept. DEC + ECB = 2 right angles DEC + DBC = 2 right angles Hence, the four points B, C, F, D are

concyclic.

Exercise 10.5 1. In the ABC if the bisectors of B and C meet at P and their exterior

bisectors at Q, show that, the four points B, P, C. Q are concyclic.

A

B C

D E


2. Prove that, the bisector of any angle of a cyclic quadrilateral and the exterior bisector of its opposite angle meet on the circumference of the circle.

3. ABCD is a circle. If the bisectors of CAB and CBA meet at the point P and the bisectors of DBA and DAB meet at Q, prove that, the four points A, Q, P, B are concyclic.

4. The chords AB and CD of a circle with centre D meet at right angles at some point within the circle, prove that, AOD + BOC = 2 right angles.

5. If the vertical angles of two triangles standing on equal bases are supplementary, prove that their circum-circles are equal.

6. The opposite angles of the quadrilateral ABCD are supplementary to each other. If the line AC is the bisector of BAD, then prove that, BC = CD.

The length of circle and arc At this stage we admit that, Formula (A) Every circle has finite length and this length is proportional to its radians.Formula (B) The length of that circle of which the length of the diameter is unity = units. where = 3.141592653897932 .................. an irrational number. In different calculations the value of is used correct to give decimal places. Definition : The length of a circle is called its circumference. Formula (C) : If the radius of a circle is r its circumference c = 2 r.Proof : The length of the diameter d = 2r. From Formula (A) it can be said that c = kd, where k is a constant. But from Formula (B) it is seen that

= k c = d = (2r) = 2 r

Definition : Circles having equal radii are called equal circles. In the adjoining figures, the radius of each of the circles with centre P and Q.

r

Pr

Qr


They are called equal circles. The length of both the circles is c = 2 r.Remark : Equal circles have equal lengths. Definitions : Arcs of same or equal circles having equal measure in degrees are called equal arcs.

In the adjoining figures, arcs AB and CD of the circle with centre at P and chord EF with centre at Q are equal, because both the circles have radius r and the measure of each chord in degrees is x. Remark : The measure in degrees of each semi-circle of a circle is 180. Hence all semi-circles of equal circles are equal.

Theorem 43 Standing on equal arcs the angles either at the centre or at the circumference are equal.

Let the arc AB of the circle ABC with the centre M, the arc DE of the circle DEF with the centre N are equal. Let on the arcs AB and DE two angles at the centre are AMB and DNE respectively and two angles at the circumference are ACB and DFE respectively. It is required to prove that,

1) AMB = DNE and

2) ACB = DFE

A B

CD

P r

x

x E

F

Qr

x

A B

C

M

D E

F

N


Proof : Since arc AB = arc DE, Hence the radii of the two circles are equal and the measure in degrees of the two arcs are also equal. But by definition, measure of the arc AB in degrees = measure of AMB at the centre in degrees. And measure of arc DE in degree = measure of DNE at the centre in degree.

AMB = DNE ...........................................(1)Because on any arc the angle at the circumference is half the angle at the centre.

Hence. ACB = 12 AMB

DFE = 12 DNE

But from (1),

AMB = 12 DNE

ACB = DFE ......................................(2.)

Theorem 44 In equal circles, arcs which subtend equal angle, either at the centre or at the circumferences, are equal.

Let the circles ABC and DBF with centres at M and N respectively be equal. Standing on the arcs AB and DE, let AMB and DNE be two angles at the centres and ACB and DFE be two angles at the circumference respectively, where, AMB = DNE ...................................... (1) or, ACB = DFE ...........................(2) It is required to prove that, arc AB = arc DE.

Proof : Since on any arc, angle at the circumference is half the angle at the centre,

A B

C

M

D E

F

N


Hence, ACB = 12 AMB

and DFE = 12 DNE

Hence, if (2) is true, that is, if ACB = DFE.

then12 AMB =

12 DNE

AMB = DNEIf (1) is true.Hence, in both cases, AMB = DNE.or, measure of AMB in degree = measure of DNE in degree.

measure of arc AB in degree = measure of arc DE in egree. Since the two circles are equal, hence according to definition, arc AB = arc DE. Formula (D). The length S of an arc of measure x in degrees, of a circle with

radius r is equal to rx

180

Proof : Considering two mutually perpendicular diameters AB and CD of the circle it is found that 4 right angles have been formed at O, the centre of the circle, the measure of each in degree is 90. Dividing each of the 4 angles in 90 equal parts we get 4 × 90 = 360 angles of measure 1° each at the centre. As a result the circumference is divided into 360 equal arcs. If the length of each such arc is a, then we get 360 × a = circumference of the circle = 2 r

a = 2 r360 =

r180

Hence, the length of the arc of measure x in degrees is given by S = x a

= x r

180 =rx

180

C

D

O


Theorem 45 Equal chords in equal circles cut off equal arcs. Let the circles ABCD and EFGH with centres M and N are equal that is, their radii are equal.

Let chord AC = chord EG. It is required to prove that, the minor arc ABC = the minor arc EFG and the major arc ADC = the major arc EHG Construction : Let M, A; M, C; N, E and N, G be joined. Proof : since the radii are equal two circles are equal. Now in MAC and NEGMA = NE [ radii being equal] MC = NG [radii being equal] and AC = EG [ given ]

MAC NEG AMC = ENG

But, in equal circles, arcs which subtend equal angles at the centre are equal. The minor arc ABC = the minor arc EFG.

Again, since the two circles are equal, hence the circumference ABCD = the circumference EFGH

The remaining major arc excluding the minor arc ABCD = the remaining major arc excluding the minor arc EFG

The major arc ADC = the major arc EHG.

A

B

D

E

G

H

C

M

F

N


Theorem 46 In equal circles chords which cut off equal arcs are equal. Let ABCD and EFGH be equal circles with centres M and N, that is, their radii are equal.

Let the arc ABC = the arc EFG It is required to prove that, the chord AC = the chord EG.Construction : Let M, A; M, C; N, E and N, G be joined. Proof: In equal circles, equal arcs subtend equal angles at the centre.

AMC = ENG. Now in MAC and NEG MA = NE [ being radii of equal circles] MC = NG [ being radii of equal circles] and AMC = ENG

MAC NEG AC = EG

chord AC = chord EG. Example. If two chords AB and CD of a circle intersect each other at right angles, then prove that, arc AC + arc BD = half the circumference. Let the diameter AF be drawn and A, D, and D, F be joined. In ADE, the exterior AEC = DAE +

ADFDAE + ADE = 1 right angle.

[ AEC=1 right angle] Again, ADE + EDF = 1 right angle

A

B

D

C

M

E

F

H

G

N

A B

C

D

O

E

F


[ ADF is an angle in the semicircle ] Hence, DAE + ADE = ADE + EDFor, DAE = EDF. that is, DAB = CDFBut DAB and CDF are angles at the centre, standing on arc BD and CF respectively.

Arc BD = arc CF Arc AC + arc BD = arc AC + arc CF

= semi circle ACBF = half of the circumference. [ AF is the diameter]

Exercise 10.6 1. AB and CD are two parallel chords in the same circle. Prove that, the arc

AC = the arc BD. 2. A, B, C be three points in a circle with centre O. If AOC = K AOB,

then prove that, arc AC is K times the arc AB. 3. Given that in ABC the

perpendicular BD drawn from B on AC and the perpendicular CE drawn from C on AB meet the circumcircle at M and N respectively. Prove that, arc MA = arc N A.

4. If two chords AB and CD of a circle with centre O, intersect in the interior point E of the circle,

then prove that, AEC = 12 ( BOD + AOC).

5. If two chords AB and CD of a circle with centre O, intersect at the

exterior point E of the circle, then prove that, AEC = 12 ( BOD.

AOC).6. AB is the common chord of two circles with equal radii. If any straight

line drawn through the point B meets the circle at the points P and O, then prove that, AOP is an isosceles triangle.


7. If two circles of equal radii intersect in a way such that the centre of circle is a point of another circle, then prove that, one third of a circle lies interior to the other cricle.

Secant and tangent A circle and a straight line in a plane may at best have two points of intersection.Definition : If a circle and a straight line in a plane have two points of intersection, then the straight line is called a secant to the circle and if the point of intersection is one and only one, then the straight line is called a tangent to the circle. In the later case, the common points is called the point

of contact the tangent. In the figure, PA ,

PB , PC are the secants to the circle and PT is a tangent to the circle and P is the point of contact. Remark : All the points between two points of intersection of every secants of the circle lie interior to the circle.

Common tangent Definition : If a straight line be a tangent to two circles, then it is called a


common tangent to the two circles. In the adjoining figures, AB is a common tangent to both the circles. In Figs-1 and 3, the points of contact are different. In Figs, 2 and 4, the points of contact are the same. Definition : If the two points of contact of the common tangent to two circles be different, then the tangent is said to be (A) direct common tangent and if the two centres of the circles lie on the same side of the tangent and (B) transverse common tangent, if the two centres lie on opposite sides of the tangent. In Fig 1, the tangent is a direct common one and in Fig. 2, it is a transverse common tangent.Definition : If a common tangent to a circle touches both the circles at the same point, then the two circles are said to touch each other at that point. In such a case, the two circles are said to have touched internally or externally according to their centres lie on the same side or opposite side of the tangent. In Fig. 2, the two circles have touched each other internally and in Fig. 4 externally.

Chord of Contact and tangent line segment. P is a point outside a circle whose centre is O. At this stage we take it to be true

that, the circle has two and only two tangents passing through the point P and the points of contact of two tangents lie on opposite sides of the straight line PO. In the figure PA and PB are two such tangents. Definition : The straight line joining the points of contact of two tangents to a circle drawn from a point outside the circle is called the chord of contact. In the figure the line segment AB is the chord of contact of the point P.

A

B

O P


Definition : If A be the point of contact of the tangent passing through P, a point outside the circle, then the line segment PA is called a tangent line to the circle.In the figure, the line segments PA and PB are two tangent lines drawn from the point P to the circle.

Alternate segments of circle : Definition : If an angle be such that its vertex lies on a circle, one of its sides is a tangent to the circle and the other a secant, then they are conjugate to that intercepted by the angle or alternate arc is said to be the alternate segment of the angle. In the figure, P is a point on the circle, the line PT is a tangent to the circle at the point P and the line PQ intersects the circle at the point Q. The arc PBQ is the segment of circle alternate to TPQ.

Theorem 47 The tangent drawn at any point of a circle is perpendicular to the radius through the point of contact. Let PT be a tangent at the point P to the, circle with centre O and OP is the radius through the point of contact. It is required to prove that, PT OP. Proof : Since PT is a tangent to the circle at the point P, hence every point on it except P will be outside the circle. Hence the point Q is outside the circle. OQ is greater than OP, the radius of the circle, that is, OQ > OP and it is true for every point Q on the tangent PT except P.

OP is the shortest distance from the centre O to PT. PT OP.

P

B

A

T

Q

P

O

Q T


Corollary 1. Only one tangent can be drawn to a circle at a given point on it.

Corollary 2. The perpendicular to a tangent at its point of contact passes through the centre

Corollary 3. The perpendicular drawn from the centre of a circle to any tangent to it passes through the point of contact.

Corollary 4. The perpendicular at a point to the radius of the circle is a tangent to the circle through that point.

Theorem 48

If two tangents are drawn to a. circle from an external point, the distances of the points of contact from that point are equal.

Let P be a point outside a circle ABC with centre O, and let two lines PA and PB be two tangents to the circle. It is required to prove that. PA = PB.

Construction : Let us join O, A; O, B and O, P. Proof : PAO = PBO = 1 right angle. Now in the right angled PAO and PBO, hypotenuse PO = hypotenuse PO OA = OB [ radius of the same circle]

APAO PBO [ Theorem 12] PA = PB.

P

B

A

OC


Exercise 10.7 1. From some exterior point P of a circle with centre O, two tangents are

drawn to the circle. Prove that, OP is the perpendicular bisector of the chord of contact.

2. Given : O is the centre of the circle and the two tangents PA and PB touch the circle at the points A and B respectively.

Prove that, PO bisects APB.

3. Prove that, the centres of those circles which touch each of two intersecting straight lines arc collinear.

4. Prove that, the centres of the circles through the same point and touching each other at that point are collinear.

5. Prove that, if two circles are concentric and if a chord of the greater circle touches the smaller, the chord is bisected at the point of contact.

6. AB is a diameter of a circle and BC is chord equal to its radius. If the tangents drawn at A and C meet each other at the point D, then prove that ACD is an equilateral triangle.

7. Given : O is the centre of a circle. Two parallel tangents PQ, RS and another tangent MN touch, the circle at points D. E and C respectively. The tangents PQ and RS meet MN at A and B respectively. Prove that, AOB = I right angle.

[Hint : In ADO and ACO AD = AC OD = OC

P O

A

B

A BC

OD E

M N

Q

P R

S


and AO =AO ADO ACP DAO = CAOCAO = DAC.

Similarly, CBO = 12 EBC.

Now, DAC + EBC = 2 right angles.

CAO + CBO = 12 ( DAC + EBC) = I right angle.

Again, in AOB,AOB = OAB + OBA = 1 right angle

AOB = 1 right angle. ]

8. Prove that, the two angles subtended by any two opposite sides of a cyclic quadrilateral are supplementary.

Theorem 49 If two circles touch each other, the centres and the point of contact are collinear. Let two circles whose centres are A and B touch each other at the point O. It is required to prove that. A. O and B are collinear.

Construction : Since the given circles touch at O, they have a common tangent at the point O. Now the common tangent POQ is drawn at the point O. O, A and O, B are joined.

P

A

Q

O B

P

B

Q

O A


Proof : Since OA and OB are radii through the point of contact, POA = 1 right angle

and. POB =1 right angle. But they are adjacent angles and each is equal to one right angle. Hence, A, O and B lie in one straight line that is. A, O and B are collinear.Corollary 1. If two circles touch each other externally, the distance between their centres is equal to the sum of their radii.Corollary 2. If two circles touch internally, the distance between their centres is equal to the difference of their radii.Corollary 3. if two circles touch each other externally, all the points of one excepting the point of contact will be outside the other circle.Corollary 4. If two circles touch each other internally, all the points of the smaller circle excepting the point of contact will lie inside the greater one.

Theorem 50 The angle made by a tangent to a circle with any chord drawn from the point of contact is equal to any angle in the alternate segment of the circle. Let PAQ be a tangent at the point A to a circle whose centre is O and let AC be a chord through that point. Let the chord AC divide the circle into arcs ABC and AEC such that, the point B is at side of Q and E at the side of P. Then

AEC is an angle in the alternate segment of QAC and ABC is an angle in the alternate segment of PAC. it is required to prove that, (1) QAC = the angle AEC in the alternate segment and (2) PAC = the angle ABC in the alternate segment.

Construction : Let the diameter AD be drawn through A and D, C be joined. Proof : Since PAQ is a tangent at A and AD is a diameter through the Point A, Hence DAQ = 1 right angle and ACD = 1 right angle [angle in a semi-circle ]

DAC + ADC = 1 right angle.

E

A Q

B

O

D

C


and DAC + QAC = DAQ = 1 right angle. Hence, DAC + ADC = DAC + QAC.

ADC = QAC.But ADC = AEC [ angles in the same segment of a circle]

QAC = AEC ............................(1)Since the quadrilateral ABCE is inscribed in the circle,

ABC + AEC = 2 right angles. Again, PAC + QAC = 2 right angles. Hence. PAC + QAC = ABC + AEC

PAC = ABC [ QAC = AEC] ................................. (2)

Exercise 10.8 1. P is the middle point of the arc APB of a circle. Prove that. the tangent to

the circle drawn at the point A is parallel to the chord AB. 2. Given : Two circles touch

externally at the point A. The tangent TT' touches the two circles at P and Q. Prove that.

PAQ = 1 right angle. 3. If an equilateral triangle be

inscribed in a circle, show that the tangents at the vertices of the triangle form are equilateral triangle.

4. Given : A and B are the centres of two circles and C is their point of contact. A straight line drawn through the point C intersects the circles at the points P and Q, Prove that, AP || BQ.

[Hint : In ACP, AC = AP

A

Q

C

P

B

Q

P

T

T

O OA


APC = ACP Similarly, BQC = BCQ But, ACP = BCQ

APC = BQC AP || BQ.]

5. AB and AC are two equal chords of a circle. Prove that, the straight line BC is parallel to the tangent to the circle at the point A.

6. Two circles touch internally at the point O, the radius of one is equal to the diameter of the other. The straight line OPQ intersects the smaller circle at P and the greater circle at Q. Prove that, OP = PQ.

7. Two circles intersect at the points A and B and the two straight lines PAC and PBD drawn from P, a point on one of the circles meet the other circle at the points C and D respectively. Prove that, the straight line CD is parallel to the tangent at P to the first circle. [Hint : TP is a tangent and the line segment PA is a chord of the circle.

TPA = the angle PBA in the alternate segment. Again, PBA = internal opposite

ACD.TPA = PCD. But they are

alternate angles. PT || CD]

8. The angle C of the ABC is one right angle and the straight line CD is perpendicular to AB. Prove that, BC bisects the angle formed by CD and the tangent at C on the circum-circle of ABC.

9. Two circles touch internally at the point P. The chord AB of the greater circle touches the smaller circle at C. Show that, the line PC bisects

APB.

Chapter Eleven Problem Related to Circle

Problem 11.1 Given circle or an arc of a circle; to determine its centre.

Given a circle. Fig. I or an arc of a circle, Fig-2. It is required to determine the centre of the circle or the arc. Construction : In the given circle or the arc of the circle three points A, B and C are taken. A, B and B, C are joined. Perpendicular bisectors EF and GH of the chords AB and CD respectively are drawn. Let them intersect each other at O. Then O is the required centre of the circle or the arc of the circle. Proof : EF and GH are the perpendicular of the chords AB and BC respectively. Hence both EF and GH pass through the centre and their common point is O. Hence, the point O is the centre of the circle or the arc of the circle.

Problem 11.2To bisect a given arc of a circle. Let AEB a given arc of a circle. It is required to bisect the arc.

Construction : A, B are joined and CD the perpendicular bisector of the chord AB is drawn. Let it intersect the arc at the point D. Then the arc AEB is bisected at the point D.


Proof : A, D and B, D are joined and let the point C lie on AB. Now in ACDand BCD. AC = BC [ by construction] DC = DC and the included ACD = included BCD [ each being a light angle]

ACD BCD. Hence, the chord AD = the chord BD.

The arc AD = the arc BD.

Problem 11.3 To draw, a tangent at a point on a circle. Let A be a point on a circle whose centre is O. It is required to draw a tangent to the circle at the point A. Construction : O, A are joined. Perpendicular AT is drawn to OA, at the point A. Then AT is the required tangent.Proof : The line segment OA is the radius passing through A and AT is perpendicular to it. Hence, AT is the required tangent.Note : Only one tangent can be drawn at a point on a circle.

Problem 11.4 To draw a tangent to a circle from a point outside it. Let T be a point outside a circle whose centre is O. A tangent is to be drawn to the circle from the point T.

Construction : T, O is joined. The middle point X of the line segment T is determined. Now with X as centre and XO as radius a circle is drawn. Let the new circle intersect the given circle at the points A and B. A, T and B, T are joined. Then AT or BT is the required tangent. Proof : A, O and B, O are joined. To is the diameter of the circle ATB.

TAO = 1 right angle.


Hence, the line segment OA is perpendicular to the line AT. AT is a tangent at the point A on the circle whose centre is O.

Similarly, the line BT is also a tangent. NB. Two and only two tangents can be drawn from a point outside a circle.

Problem 11.5 To draw a direct common tangent to two circles of unequal radii.

Let LMN and FGH be two circles whose centres are A and B and their radii are a and b respectively where a > b. It is required to draw a direct common tangent to the circles.

Construction : With centre A and radius equal to (a-b), a circle PQR is drawn. From B the tangent BC is drawn to the circle PQR and let it touch the circle at the point C. A, C are joined and produced such that it intersects the circle LMN at the point O. Through the point B, BE is drawn parallel to AO and let it intersects the circle FGH at the point E. O, E are joined. Then OE is the required tangent.

Proof : Since BC is a tangent to the circle PQR. ACB = 1 right angle, that is, BC AO OCB = 1 right angle.

Again, CO = AO AC = a (a b) = b and AO || BE [by construction] CO || BE [ CO lies on AO]

OCBE is a parallelogram.


Again, since OCB = 1 right angle, hence, OCBE is a rectangle. COE = BEO = 1 right angle.

Hence both CO and BE are perpendiculars to OE. The straight line OE is a direct common tangent to both the circles.

N. B. 1. If the radii of the two circles are equal, it is not possible to draw a direct

common tangent in the above method. In that case, by joining A and B it perpendiculars AO and BE drawn in the same direction on AB meet the circles at O and E respectively, then OE will be the required direct common tangent. Two such tangents can be drawn which are parallel to each other.

2. It is not possible to draw a common tangent to them if one circle lies wholly inside the other.

3. If the two circles touch each other internally, then the tangent drawn through the point of contact will be the common tangent.

4. If the two circles touch-externally, then one tangent at their point of contact and by the method described earlier, two common tangents more can be drawn.

Problem 11.6 To draw transverse common tangent to two given circles.

Let A and B be the centres of the two circles LMN and FGH and their radii be a and b respectively. We are to draw a transverse common tangent to the circles.


Construction : Let us draw a circle with A as centre and radius equal to (a + b). From B the tangent BC is drawn to the circle PQR and let it touch the circle at the point C. Let C, A be joined and let the line segment CA intersects the circle LMN at E. Through B. BK drawn parallel to CA meets the circle FGH at K. E, K are joined. Then, EK is the required transverse common tangent to the two circles.Proof : BC is a tangent to the circle PQR.

ECB = 1 right angle. Again, CE = b = BK and CE || BK [by construction]

CEKB is a parallelogram. Again, since the angle CEK of the parallelogram CEKB = 1 right angle. Therefore, CEKB is a rectangle.

BKE = CEK = 1 right angle. Again, AEK = I right angle [ AEC = 1 straight angle] Hence, the line EK is a common tangent to the two circles. Again, since the two circles lie on opposite sides of EK

EK is a transverse common tangent to the circles. NB. By adopting the method stated above, transverse common tangent be drawn even though the two circles have equal radii.

Problem 11.7 To draw a segment of a circle at one extremity of a line segment such that the line segment subtend an angle equal to a given angle in that segment. Let AB be a given line segment and C be a given angle. We are to describe a segment of a circle through A and B such that the angle in the segment is equal to the given angle C.Construction : At the end A of the line segment AB, the angle BAD is drawn equal to C. Perpendicular AH is drawn to AD at A and perpendicular bisector KF of the line segment AB is drawn. The line KF also intersects AH at the point G Taking G as centre and radius equal to GA, the arc ALB is drawn where L and D lie on opposite sides of AB. Then, the segment ALB is the required segment of the circle.

A

D

D

B

HG

c

F

KL


Proof : Since KF is the perpendicular bisector of the line segment AB, G is equidistant from the points A and B i.e. GA = GB.

The arc ALB passes through A and B. Since AD is a tangent and AB is a chord of the circle [by construction]

BAD or C is equal to any angle in the alternate segment ALB.

N.B. If the given angle is equal to one right angle, the semicircle drawn on the given straight line is the required segment of the circle.

Problem 11.8 To draw a circle circumscribing a given triangle. Let ABC be a given triangle. It is required to draw a circle circumscribing it that is, to draw a circle passing through the three vertices A, B and C of the triangle ABC.Construction : EM and FN the perpendicular-bisectors of AB and AC respectively are drawn. Let them intersect each other at O. A, O are

joined. Taking O as centre and radius equal to OA, a circle is drawn. Then the circle will pass through the points A, B and C and this circle is the required circum-circle of the ABC.

Proof : B, O and C, O are Joined. The point O stands on EM, the perpendicular bisector of AB

OA= OB. Similarly, OA = OC

OA = OB = OC.

Hence, the circle drawn with O as the centre and OA as the radius passes through the three points A, B and C.

Hence, this circle is the circum-circle of the triangle ABC.

B

E

A

F

C

MNO


Remark : In the above Fig, the circum-circle of an obtuse angles triangle is drawn. For acute angle and right angle it will be as in the adjoining figure. It is noticable that for obtuse angles triangle the incentre lies outside the triangle, for acute angletriangle the incentre lies within the triangle and for right angle triangle the incentre lies on the hypotenuse of the triangle.

Problem 11.9 To draw a circle inscribed in a triangle. Let ABC be a triangle. To inscribe acircle in it or to draw a circle in it such that it touches each of the three sides BC, CA and AB of the triangle. Construction : BL and CM, the bisectors of the angles ABC and

ACB respectively are drawn. Let them intersect at the point O, OD is drawn perpendicular to BC from O and let it intersect BC at D. With O as centre and OD as radius a circle is drawn. Then, this circle is the required in circle. Proof : From O, OE and OF are drawn perpendicular to AC and AB respectively. Let these two perpendiculars intersect the respective sides at E and F. The, point O lies in the bisector of ABC.OF = OD. Similarly, as 0 lies on bisector of ACB,

OD = OE OD = OE = OF

Hence, the circle drawn with centre as O and OD as radius passes through D, E and F.


Again, BC, AC and AB are perpendiculars to OD, OE and OF respectively at their extremities. Hence, the circle lying inside the triangle ABC touches its sides at the points D, E and F. Hence, the circle DEF is the required incircle of the ABC.

Problem 11.10 To draw an described circle of a given triangle. Let ABC be the given triangle. It is required to draw an escribed circle of the

ABC. That is, to draw a circle which touches one side of the ABC and the other two sides produced. Construction : Let AB and AC be produced to D and F respectively. BM and CN, the bisectors of DBCand FCB respectively are drawn. Let E be their point of intersection. From E, perpendicular EH is drawn on BC and let EH intersect BC at H. With E as centre and radius equal to EH, a circle is drawn. Proof : From E, perpendiculars EG and EL are drawn to line segments BD and CF respectively. Let the perpendicular intersect line segments. at G and L respectively, E lies on the bisector of DBC.

EH = EG Similarly, the point E lies on the bisector of FCB,

EH = EL EH = EG = EL

Hence, the circle drawn with E as centre and radius equal to EH passes through H, G and L. Again, the line segments BC, BD and CF are perpendiculars at the extremities of EH, EG and EL respectively. Hence, the circle touches the three line segments at the three points H, G and L respectively. Hence, the circle HGL is the described circle of the triangle ABC. Remark : Three described circles can be drawn of any triangle.


Exercise 11 1. Draw a circle which passes through two given points A and B and whose

centre lies at a distance 5 cm. from AB.

2. Draw a circle which passes through the points of intersection of two circles and whose centre lies on a giver straight line.

[Hint. If A and B be the points of intersection of two circles and LM be the given straight line, then draw the perpendicular-bisector to AB. It intersects LM at a point P. The circle drawn .with P as centre and radius equal to PA is the required circle.]

3. Draw a circle which touches two parallel straight lines and a given line intersecting one of them.

4. Draw a circle which touches a given circle and a given straight line at a given point.

5. Draw a tangent to a circle which is parallel to a given straight line.

6. Draw a tangent to a circle which is perpendicular to a given straight line.

7. Draw two tangents to a circle such that the angle between them is 60°.

8. Draw the circum-circle of the triangle whose sides are 3 cm, 4 cm and 5cm and determine the radius of this circle.

9. Draw an described circle to an equilateral triangle ABC touching the side CA of the triangle, the length of each side being 5 cm and determine the radius of this circle.

10. Draw inscribed and described circles of a square.

[Hint : Bisect A and B of the square ABCD. Let the bisectors meet at O. Draw OP, the perpendicular bisector of AB. Join O, A. Then the circle drawn with O as centre and radius equal to OP will be the inscribed circle of the square and the circle with O as centre and radius equal to OA will be the required described circle of the square.]


About Circle

Multiple Choice Questions (MCQ) : 1. Two equal circles touch each other externally. What is the distance

betwee the centres if one of their radius is 4 unit? A. 0 B. 4 C. 8 D. 12 2.

In the figure, ABC is an equilateral triangle inscribed in a circle of centre O. Which of the following is the value of AOB?

A. 300° B. 240° C. 180° D. 120°3. How many tangents can be drawn at any point on a circle? A. one B. two C. three D. four 4. An equilateral triangle is inscribed in a circle, the triangle that is drawn

by the tangents of three vertex of the triangle will be a A. Right angled triangle B. Equilateral triangle C. Scalene triangle D. Obtuse angled triangle 5. Look at the following statements: i. A perpendicular on a tangent line drawn at the tangent point bisects

the tangent. ii. Two and only two tangents can be drawn from a point outside of a

circle. iii. A parallelogram inscribed in a circle is a rectangle.


Which of the above statements are correct? A. i and ii B. ii and iii C. i and iii D. i, ii and in The following information refers to questions (6-8)

CO

BA D

AB is a chord other than a diameter of a circle ABC whose centre is 0. OD AB

6. If B = 45° then AOD = ? A. 90° B. 60° C. 45° D. 30°

7A. 315° B. 270°

. If A = 45° then reflex AOB = ?

C. 225° D. 200°

8. Which of the following relations is correct? A. OAD = OBD

B. OAB OAD C. OA = OB = AB D. ODA =

12 AOB

9. What type of triangle is ACP in the figure?

A. Equilateral triangle B. Isosceles triangle C. Acute angled triangle D. Right angled triangle


10. Sora moves from point B to point C in a circular way where arc BC creates 50° angle at the centre. Few times later she arrives at point A.

BAC = ? A. 25° B. 40° C. 60° D. 90°

11. Look at the figure at right side: i. AOC = 2 ADC ii. DCE = 2 BAD iii. ABC + ADC = 180° Which of the above relations are

correct? A. i and ii B. ii and iii C. i and iii D. i. ii and iii

E

The following information refers to questions (12-14)

C

BO

A

E

Which of the following is correct about triangle OCB? A. Isosceles triangle B. Scalene triangle C. Right angled isosceles triangle D. Equilateral triangle


13. BCE = ? A. OCB B. OBC C. BAC D. BOC

Creative Questions : 1. O is the centre of a circle shown in the

figure at the right side where AB > CD, AB CD

A. Find the length of AP B. Show that OH > OP C. Prove that the sum of these angles

subtended by the arcs AC and BD at the centre is 180 .

2. C is a centre of a circle whose radius is 3 cm. There is a pole 10 cm from the centre perpendicular on the ground at point T.

A. Draw the geometric figure of the above information. B. From the foot of the pole draw two tangents to the circle and show

that, both tangents points are in same distance from the foot of the pole.

C. Prove that consider as a side of an inscribed equilateral triangle by the line segment joining two tangent points and show that the tangent line at the vertex point of the triangle create another equilateral triangle.

5O

D

H

BC

A P

3

Chapter Twelve Trigonometry

12.1 Introduction Trigonometry originated as a consequence of the culture of geometrical concepts acquired from the lust for knowledge and indomitable curiosity of man to discover the unknown Geometrical concept of man is quite ancient in nature. In that ancient age, men learnt the method of determining the width of a river by standing at its bank. Even without climbing a tree they learnt how to measure its height accurately by composing its shadow with that of a stick. But that too could not bring an end to main curiosity to learn more. They were eager to know the distances of the moon. the sun and the stars from the earth. But they have not been able to solve these problems with the ordinary knowledge of geometry. This did not stop their endless efforts. As their efforts were not stopped, solution of those problems gave rise to the creation of Trigonometry, another branch of Mathematics. Its extensive application is seen in the survey of land and engineering in ancient Egypt. In the present days trigonometry is used in all branches of Mathematics. Trigonometry is extensively used in area of triangle, solution of triangle and in navigation.

12.2 Trigonometrical ratio of acute angle

Let, XOA be an acute angle. Let us take a point P on its arm OA. From P perpendicular PM is drawn on OX. Thus the right-angled triangle POM is formed. The six ratios of the sides PM, OM and OP of the POM are called trigonometrical ratios of the angle XOA and each of them is given a specific name.

OM

PA

XFigure-1


Referred to the angle XOA of the right -angled triangle POM, the side PM is taken as the perpendicular, the side OM as the base and the side OP as the hypotenuse. Taking the angle XOA = , the six trigonometrical ratios associated to the angle that can be obtained are described as follows.

Definition : In fig-1,

PMOP =

perpendicularhypotenuse = Sine of the angle or sin (in brief)

PMOP =

basehypotenuse = Cosine of the angle or cos (in brief)

PMOP =

perpendicularbase = Tangent of the angle or tan (in brief)

OMPM =

baseperpendicular = Cotangent of the angle or cot (in brief)

OPOM =

hypotenusebase = Secant of the angle or sec (in brief)

OPPM =

hypotenuseperpendicular = Cosecant of the angle or cosec (in brief)

Constancy of Trigonometrical ratios The trigonometrical ratios described above do not depend on the position of P lying on the side OA.

Figure - 2

Two similar right angled triangles POM and P1OM1 are formed by drawing perpendiculars PM and P1M1 to OX from any two points P and P1 on the side OA of the XOA.

A

XO M M1

P1

P


Now, POM and P1OM1 being similar.

PMP1M1

=OPOP1

= or, PMOP =

P1M1OP1

........................................ (i)

OM1OM =

OPOP1

= or, OMOP =

OM1OP1

........................................ (ii)

PMP1M1

=OMOM1

= or, PMOM =

P1M1OM1

........................................ (iii)

Hence, if XOA = , it is obseved from (i) (ii) and (iii) that,

sin =PMOP =

P1M1OP1

, cosec = OPPM =

OP1P1M1

cos = OMOP =

OM1OP1

, sec = OP

OM1=

OP1OM1

tan = PMOM =

P1M1OM1

, cot = OMPM =

OM1P1M1

That is, each of the trigonometrical ratios of the same angle is constant.

Limits of trigonometrical ratios Let, = POM be an acute angle and PM OM. (i) Each trigonometrical ratios of the angle is the ratio of two sides of the

POM. Hence each ratio is a positive real number, (ii) The hypotenuse OP of the POM is the greatest side. Hence, PM < OP and OM < OP

PMOP < 1 and

OMOP < 1

and OPPM > 1 and

OPOM < 1

sin < 1 and cos < 1 and cosec > 1 and sec > 1 O M

P

Figure-3


(iii) In the triangle POM, the sum of any two sides is greater than the third. Hence, PM + OM > OP

PMOP +

OMOP > 1 [by dividing both sides by OP ]

that is, sin + cos > 1.

12.3 Some relations between the trigonometrical ratios Let = XOA be an acute angle,

Referred to the above diagram, by definition,

sin = PMOP , cosec =

OPPM

cos = OMOP , cosec =

OPOM

tan = PMOP , cot =

OMPM

Hence, it is observed that.

(i) sin .cosec = PMOP .

OPPM = 1

sin = 1

cosec and cosec = 1

sin

(ii) cos .sec = OMOP .

OPOM = 1

cos = 1

sec and sec = 1

cos

A

O XM

P

Figure - 4


(iii) tan . cot = PMOM .

OMPM = 1

tan = 1

cot and cot = 1

tan

(iv) tan = PMOP =

PMOPOMOP

= [ dividing the numerator and denominators by OP]

tan = sincos and in a similar way,

cot = cossin

(v) (sin )2 + (cos )

2

=PMOP

2

+OMOP

2

=PM2

OP2 +OM2

OP2

=PM2 + OM2

OP2

But in the right angled triangle POM, the hypotenuse is OP.Hence, OP2 = PM2 + OM2.

(sin )2 + (cos )2 = OP2

OP2 = 1

or, sin2 + cos2 = 1 Remark : For integral index n, (sin )n is written as sinn , (cos )n as cosn etc.

(vi) sec2 = (sec )2 = OPOM

2

= OP2

OM2

=PM2 + OM2

OM2 [ OP is the hypotenuse of the right angled triangle POM

=PM2

OM2 + OM2

OM2 = 1 + PMOM

2

= 1 + (tan )2


sec2 = 1 + tan2

or, sec2 tan2 = 1

(vii) cosec2 = (cosec )2 = OPPM

2=

OP2

PM2

=PM2 + OM2

PM2 [Since OP is the hypotenuse of the right-angled triangle POM]

=PM2

PM2 + OM2

PM2

= 1 + OMPM

2

= 1 + (cot )2

cosec2 = 1 + cot2

or, cosec2 cot2 = 1

12.4. Some examples. Example 1. Prove that,

(i)1

1 + sin2A + 1

1 + cosec2A = 1;

(ii)1

1 + cos2A + 1

1 + sec2A = 1;

(iii)1

1 + tan2A + 1

1 + cot2A = 1;

Solution : (i) L.H.S. = 1

1 + sin2A + 1

1 + cosec2A

= 1

1 + sin2A + 1

1 + 1

sin2A

= 1

1 + sin2A + sin2A

1 + sin2A

= 1 + sin2A1 + sin2A = 1 = R.H.S.

L.H.S. = R.H.S.


(ii) L.H.S = 1

1 + cos2A + 1

1 + sec2A

= 1

1 + cos2A + 1

1 + 1

cos2A

= 1

1 + cos2A + cos2A

1 + cos2A

= 1 + cos2A1 + cos2A

L.H.S = R.H.S

(iii) L.H.S = 1

1 + tan2A + 1

1 + cot2A

= 1

1 + tan2A + 1

1 + 1

tan2A

= 1

1 + tan2A + tan2A

1 + tan2A

= 1 + tan2A1 + tan2A = 1 = R.H.S.

L.H.S. = R.H.S.

Example 2. Prove that, cosA

1 tanA + sinA

1 cotA = sinA + cosA

Solution : L.H.S = CosA

1 tanA + sinA

1 cotA = cosA

1sinAcosA

+sinA

1cosAsinA

= cos2A

cosA sinA + sin2A

sinA cosA

= cos2A

cosA sinAsin2A

cosA sinA

= cos2A sin2AcosA sinA = cosA + sinA = R.H.S.

L.H.S. = R.H.S.


Example 3. Prove that, tan 1 sin2 = sin

Solution : L.H.S. = tan 1 sin2

= tan cos2

= sincos cos

= sin = R.H.S. L.H.S. = R.H.S.

Example 4. Show that, cot + cosec 1cot cosec + 1 =

1 + cossin

Solution : L.H.S. = cot + cosec 1cot cosec + 1

= cot + coec (coec2 cot2 )

cot cosec + 1

= cot + cosec (cosec + cot )(cosec cot )

cot cosec +1

= (cot + cosec )(1 cosec + cot )

(cot cosec + 1) = cot + cosec

= cossin +

1sin

= 1 + cos

sin = R.H.S

L.H.S = R.H.S.Example 5. If tan A + sinA = m and tan A sinA = n, prove that,

m2 n2 = 4 mnSolution :

R.H.S = 4 mn

= 4 (tanA + sinA) (tanA sinA)

= 4 tan2A sin2A

= 4 tan2A 1sin2Atan2A


= 4 tan2A 1sin2 Acos2A

sin2A = 4 (tan2A) (1 cos2A) = 4 tan2A sin2A

= 4tanA sinA = (tanA + sinA)2 (tanA sinA)2 [ 4ab = (a + b)2 (a b)2] = m2 n2 = L.H.S.

m2 n2 = 4 mnExample 6. If sinA + cosA = a and secA + cosecA = b, then prove that, b(a2 1) = 2a Solution : L.H.S. = b(a2 1)

= (secA + coecA) {(sinA + cosA)2 1}

= 1

coA + 1

sinA (sin2A + cos2A + 2sinA cosA 1)

= sinA + cosA

sinAcoA (1 + 2 sinA cosA 1)

= sinA + cosAsinA cosA . 2 sinA cosA

= 2(sinA + cosA)= 2a = R.H.S.

b(a2 1) = 2a

Example 7. Prove that, sec tan = 1 sin1 + sin

Solution: L. H. S. = sec tan

= 1

cosinco

= 1 sin

co

= (1 sin )2

co2 =(1 sin )2

1 sin2

= (1 sin )2

(1 + sin ) (1 sin ) =(1 sin )1 + sin = R.H.S

L.H.S = R.H.S


Example 8. If cosA + sinA = 2 cosAthen prove that, cosA sinA = 2 sinASolution : cosA + sinA = 2 cosAor, sinA = ( 2 1) cosA

or, cosA =sinA2 1

=( )2 + 1 sinA

2 1or, cosA = ( 2 + 1) sinA

cosA sinA = 2 sinA (by transpoition)

Exercise 12.1 Prove that, (Question 1 12)

1. (i) 1

sec2A + 1

cosec2A = 1

(ii) 1

cos2A1

cot2A = 1

2. (i) sinA

cosecA + cosAsecA = 1

(ii) secAcosA

tanAcotA = 1

3.tanA

1 cotA + cotA

1 tanA = secA cosecA + 1

4.tan2A

1 + tan2A + tan2A

1 + cot2A = sin2A sec2A

5.1

2 sin2 + 1

2 + tan2 = 1

6.secA + tanA

cosecA + cotA = cosecA cotAsecA tanA

7.cosecA

cosecA 1 + cosecA

cosecA + 1 = 2sec2A

8.1

1 + sinA + 1

1 sinA = 2sec2A

9.1

cosecA 11

cosecA + 1 = 2tan2A

10.sinA

1 cosA + 1 cosA

sinA = 2 cosecA


11.tanA

secA + 1secA 1

tanA = 0

12. (tan + sec )2 = 1 + sin1 sin

13.cotA + tanBcotB + tanA = cotA. tanB

14.1 sinA1 + sinA = secA tanA

15.secA + 1secA 1 = cotA + cosecA

16. If sin2A + sin4A = 1, then prove that, tan4A tan2A = 1

17. If tanA = 13

, then find the value of cosec2A sec2Acosec2A + sec2A

18. If sec + tan = 25 , then find the value of sec tan

19. If sin + cossin cos =7, then fmd the value of tan .

20. cotA =ba , then find the value of

asinA bcosAasinA + bcosA

12.5 Trigonometrical ratios of the angles 30°, 45° and 60° The actual value of the trigonometrical ratios of some angles can be determined by geometrical method. Angles measuring 30°, 45° and 60° are such angles. Trigonometrical ratios of angles of 30° and 60° Let XOZ = 30° and P be a point on OZ, PM is drawn perpendicular to OX and produced to Q such that MQ = PM. Let us join O, Q. Now, in POM and QOM,PM = QM, OM common side and PMO = QMO = 600

POM QOMHence, QOM = POM = 30° and OQM = OPM = 60

OPQ is an equilateral triangle.

If OP = 2a, then PM = 12 PQ =

12 OP = a [ PQ = OP]

and OM = OP2 PM2 = 4a2 a2 = 3a


sin 30 = PMOP =

a2a =

12

cos 30 = OMOP =

3a2a =

32

tan 30 = PMOM =

a3a

=13

cot 30 = OMPM =

3aa = 3

sec 30 = OPOM =

2a3a

=23

cosec 30 = OPPM =

2aa = 2

again,

sin 60 = OMOP =

3a2a =

32

cos 60 = PMOP =

a2a =

12

tan 60 = OMPM =

3aa = 3

cot 60 = PMOM =

a3a

=13

sec 60 = OMPM =

2aa = 2

cosec 60 = OPOM =

2a3a

=23

Trigonometrical ratios of the angle 45° Let XOZ = 45° and P be a point on OZ.PM is drawn perpendicular to OX.In the right angled triangle OPM,

POM = 45Hence, OPM = 45

PM = OMNow, OM2 + PM2 = OP2

or, 2. OM2 = OP2 [ PM = OM]

or, OM2 = 12 OP2

P Z

Q MX

45

2a

2a

Figure-5Q

Z

Z

MX

3a

P

2a

30

30

60

60

O

a


or, OM = 12

OP.

If = 2a, then PM = OM = 12

2a = 2 .a.

sin 45 = PMOP =

2a2a =

12

cos 45 = OMOP =

2a2a =

12

tan 45 = PMOM =

2a2a

= 1

cot 45 = OMPM =

2a2a

= 1

sec 45 = OPPM =

2a2a

= 1

sec 45 = OPPM =

2a2a

= 2

12.6 Trigonmetrical ratios of complementary angles. Definition : If the sum of two acute angles is 90°, then one of these two angles is said to be the complementary angle to the other. For example, 30° and 60°, 15° and 75° are complementary to each other. Generally, ° and (90- )° are complementary to each other. Trigonometrical ratios of complementary angles.Let XOY = and P be a point on the arm OY of this angle. Let PM be drawn perpendicular to OX. Since the sum of three angles of a triangle is two right angles, hence, in the right- angled triangle POM,

OPM + POM = 1 right angle = 90°OPM = 90° POM = 90° POM = XOY =

sin (90° ) = OMOP = cos POM = cos

cos (90 ) = PMOP = sin POM = sin

tan (90 ) = OMPM = cot POM = cot

P

O XM

90 -

Y

Figure-7


cot (90 ) = PMOM = tan POM = tan

sec (90 ) = OPPM = cosec POM = cosec

cosec (90 ) = OPOM = sec POM = sec

The above formula can be expressed in the following way :

The sine of the complementary angle = cosine of the angle.

The cosine of the complementary angle = sine of the angle.

The tangent of the complementary angle = cotangent of the angle, etc.

12.7. Trigonometrical ratios of 90° and 0° In the plane of coordinates, let us consider XOZ of which one side lies along OX, the positive X-axis and the other side OZ lies in the first (i.e, positive) quadrant (Fig-8). If XOZ be ° XOZ is called the standard position of the angle . In such a position the ray OX is called the initial side of the angle and the ray OZ is called its terminal side. Now, the circle drawn by taking the origin 0 as centre and 1 unit as radius intersects the ray OX at A, the ray OY at B and the ray OZ at P. PM is drawn perpendicular to OX. Let the co-ordinates of P be (x, y). Then, in the adjoining figure, OP = l, OM = x, PM = y.

Hence, cos = OMOP = x

and sin = PMOP = y

Here it is seen that,Formula : In the standard position of the angle ° the coordinates of the points where the unit circle (circle with origin as the centre and radius 1 unit)

Figure-8

Y

Y

X XO M

B(0, 1) P

Z

(cos , sin )

A(1, 0)

1


intersects its terminal side are (cos °, sin °). By extending the formula stated above the sines and cosines of the angles 0° and 90° are defined. (A) We notice that in the standard position, the terminal side of the angle 90° remains in the position OY and the unit circle intersects this side at the point B whose coordinates are (0,1). Hence keeping consistency with the formula stated earlier, it is said that, Definition : cos 90° = 0 sin 90° =1, (B) In geometry, two different rays with the same vertex produce an angle and the measure of every angle is a positive number. For the convenience of trigonometrical discussion, the angle ° is introduced and in the standard position, the initial and terminal sides of the angle ° are taken as the same ray OX. Since the unit circle intersects this ray at the point A whose coordinates are (1,0), hence inconsistent with the formula stated earlier it is said that, Definition : cos0° = 1 sin0° = 0. We have seen in case of being one acute angle (Art -12. 3) that,

tan = sincos , cot =

cossin

sec = 1

cos , cosec = 1

sin

Keeping in view that these relations may be true in possible cases of the angles 0° and 90° it is said that,

Definition : tan 0° = 0

sec 0° = I

cot 90° = 0

cosec 90° = l.

Remark : As the division by zero is not possible, hence cosec 0°, cot 0° and tan 90°, sec 90° cannot be defined.

Note : For the convenience of application, the values of the trigonometrical ratios of the angles 0°, 30°, 45°, 60° and 90° (which are defined) are shown in the table below :


angle ratio 0° 30° 45° 60° 90"sine 0 1

212

32

1

cosine 1 32

12

12

0

tangent 0 13

1 3 undefined

cotangent undefined 3 1 13

0

secant 1 23

2 2 undefined

cosecant undefined 2 2 23

1

We notice : Easy method for remembering the values of the trigonometrical ratios of some selected angles: i) By dividing 0,1, 2, 3 and 4 by 4 and then taking the square roots of the

quotients, we get the values of sin 0°, sin 30°, sin 45°, sin 60° and sin 90° respectively.

ii) By dividing 4, 3, 2, 1 and 0 by 4 and taking the square roots of these quotients, the values of cos 0°, cos 30°. cos 45°, cos 60° and cos 90 respectively are obtained.

iii) By dividing each of the numbers 0, 1, 3 and 9 by 3 and taking the square roots of these quotients, the values of tan 0°, tan 30°, tan 45° and tan 60° respectively are obtained. (It may be noted that tan 90° is undefined).

12.8 Some Examples

Example 1. Find the value of : 1 cot2 601 + cot2 60

Solution : The given expression = 1 cot2 601 + cot2 60

1 ( )cot 60 2

1 + ( )cot 60 2 = 1

13

2

1 + 13

=1

13

1 + 13

=

2343

=12


Example 2. Find the value of: tan2 45°, sin 60° tan 30° tan2 60°Solution : The given expression = tan2 45° sin 60° tan 30°. tan2 60°

= (1)2.3

2 .13

. ( )32

= 1. 3

2 .13

. 3 = 32

Example 3. Solve : cosA sinAcosA + sinA =

3 13 + 1

Solution : Here, cosA sinAcosA + sinA =

3 + 13 1

or,cosA sinA + cosA + sinAcosA sinA cosA sinA =

3 1 + 3 + 13 1 3 3

.

[By componendo and dividendo]

or,2cosA 2sinA =

2 32

or, cotA = 3 = cot 30 A = 30

Example 4. Solve : tan2 (1 + 3 ) tan + 3 = 0.

Solution : Here, tan2 (1 + 3 ) tan + 3 = 0

or, tan2 tan 3 . tan + 3 = 0

or, tan (tan 1) 3 (tan 1) = 0

or, (tan 1) (tan 3 ) = 0 tan = 1 = tan 45° that is, = 45° or, tan = 3 tan 60° that is, = 60°

= 45° and 60°

Example 5. Show that, cos 3A = 4 cos3 A- 3 cos A, if A = 30° Solution : L.H.S. = cos 3A = cos 3. 30° = cos 90° = 0 R.H.S. = 4 cos3A 3 cosA

= 4 cos3 30° 3 cos 30°


= 4 3

23

3.3

2

= 4.3 3

83 3

2

= 3 3

23 3

2 = 0

L.H.S = R.H.S cos3A = 4cos3A 3cosA.

Exercise 12.2Show that, 1. cos2 30° sin2 30° = cos 60° 2. sin 60°. cos 30° + cos 60°. sin 30° = sin 90° 3. cos 60° cos 30° + sin 60° sin 30° = cos 30° 4. sin 3A= cos 3A, if A= 15°

5. sin 2A = 2tanA

1 + tan2A , if A = 30

6. tan 2A = 2tanA

1 tan2A , if A = 30

7. cos 2A = 1 tan2A1 + tan2A , if A = 45

8. If 2 cos (A + B) = 1 = 2 sin (A B) and A, B are acute angles, then A = 45° and B = 15°.

9. 2 cos(A B)= l, 2 sin(A + B) = 3 and A, B is an acute angle find A, B. 10. Find the values of A and B when A and B are acute angles and cot

(A + B) =1 and cot (A B) = 311. Solve : sin + cos = 1. when 0° < < 90°. 12. Solve : cos2 sin2 = 2 5 cos , when is an acute angle.13. Solve : 2 cos2 + 3 sin 3 = 0, is an acute angle.


14. Show that, 3tan230 + 14 sec 60° + 5cot 452 0 2

3 sin 60° = 6; 2

15. Find the values of, 3 cot260 + 14 cosec 30° + 5 sin 45 4cos 602 0 2 0

12.9 Problems relating to distance and height.From very ancient times trigonometrical ratios are applied to find the distance and height of distant objects. Even in modern age its importance is boundless. The heights of those hills and mountains and the width of those rivers which cannot be measured in ordinary methods. They are measured with the help of trigonometry. For this reason, it is very useful to know the measurement of angles. Angles can be measured by using an instrument called "sextant," Horizontal line and Vertical line and Vertical planeThe horizontal line is any straight line lying on the horizontal plane. A straight line parallel to horizon is called a horizontal line.Again, vertical line is any line perpendicular to the horizontal plane. It is called normal line. A horizontal line and a vertical line intersection at right angles on the horizontal plane, determines a plane. It is known as vertical plane. Angle of elevation and angle of depression. Let, XOX be a line parallel to the horizontal line. In the figure, the points 0, P, X lie in the same vertical plane and lies above XOX . The angle of elevation of the point P at the point O is POX. Similarly, the angle of elevation of P' at the point O is

P'OX'. In the figure, the points 0, P, X lie in the same vertical plane and the point P is below the line XOX parallel to the horizontal line. Then the angle of depression of the point P at the point O is POX. Similarly, the angle of depression of the point P" at the point O is P OX .

P P

OX XFigure-9

X X

O

P PFigure-10


Some examples relating to height and distanceExample 1. The angle of elevation of the top of a Minar at a point on the ground 30 metres from the foot of the Minar is 60°; find the height of the Minar.A Solution : Let P be the foot of the Minar, the given point on the ground be O and the top of the Minar be A. Hence, POA = 60° and PO = 30 metres. Let AP, the height of the Minar = h metres.

Now, tan 60° = APOP

or, 3 =h30

or, h = 30 3 = 51.962

The height of the Minar = 30 3 metres 51.962 meters Example 2. The angle of elevation of the top of a tree at a point from its foot on the ground is 30°. How far away is the point from the tree if the tree is 26 metres?Solution : Let the point B be the foot of the tree, O, the given point on the ground and A be the top of the tree. Let, the distance of the given point from the tree be BO = x metre.

AOB= 30 and BA = 26 metres. AOB = 30 , and BA = 26 metres.

Now, tan 30 = ABOB

or,13

=26x

or, x = 26 3 metres = 45.033. The distance of the given point

from the tree = 26 3 metres = 45.033 metres. Example 3. A ladder 18 metre long touches the roof of a wall and makes an angle 45° with horizon. What is the height of the wall? Solution : Let the point of contact of the ladder and the roof be B and the height of the wall AB = h metre.

A

h

P30 m

O60

Figure-11

A

BO

30x

26m

Figure-12


The length of the ladder OB = 18 metres and AOB = 45°

sin 45 = ABOB

or,12

=h18

or, h = 18

2=

18 22 2

= 9 2 = 12.728

The height of the wall = 9 2 metres

= 12.728 metres.

B

h18m

45

AOFigure : 13

Example 4. A man standing at a place or the bank of a river observes that the angle of elevation of a tower exactly opposite to him on the other bank is 60°. On moving 25 metres in the backward direction he observed that the angle of elevation of the tower is 30°. Find the height of the tower and the width of the river.Solution : Let the height of the tower AB = h metre and the width of the river BP = x metre. Here, BPA = 60°, BOA = 30° and OP = 25 metre.

BO = (BP + PO) = (x + 25) metre.

Now, tan AOB = ABOB

or, tan 30 = h

x + 25

or,13

=h

x + 25

or, x + 25 = h 3 ...................(i)

Again, tan BPA = ABBP

tan60 = hx

or, 3 =hx

or, h = x 3 ..............(ii)

A

h

30 60O BP x25m


Hence, from (i) and (ii) we get, x + 25 = x 3 . 3or, x + 25 = 3xor, 2x = 25

or, x = 252 = 12

12 = 12.5

h = x 3 =252 3 = 21.651

the height of the tower = 25 3

2 metres = 21.651 metres and width of the

river = 12.5 metres. Example 5. From a helicopter above a point between two kilometre posts, the angle of depression of the two .posts are 60° and 30° respectively. Find the height of the helicopter. Solution : Let O be the position of the helicopter and A and B be the tops of the two posts one kilometre away. The angle of depression of A and B from O are 60° and 30° respectively. Hence, A'OA = 60° and B'OB = OBA = 30° A' B' and AB being parallel,

A'OA = OAB = 60° and B'OB = OBA = 30°. Here, AB =1000 metres. Let OP be drown perpendicular to AB from O. Let AP = x metre, OP = h metre Hence, BP = (1000 x) metre.

Now, tan OAP = OPAP

or, tan 60 = OPAP

or, 3 =hx

or, 3 .x = h

Again, tan OBP = OPBP

or, tan 30 = OPBP

or,13

=h

1000 x

Figure-15

O

A BPx

30

30

60

h

1000 m

60

A B


or, 1000 x = 3.h or, 1000 x = 3. 3.xor, 1000 x = 3xor, 4x = 1000 or, x = 250 or, h = 3 . x = 250 3 metres = 433.013 metres.

Exercise 12.31. If the angle of elevation of the top of the minar to a point on the ground

20 metres from the foot of the minar is 60°, then find the height of the minar.

2. The angle of elevation of the top of a to a point on the ground 90 metres from the foot of a tall tree is 300, find the height of the tree.

3. The angle of elevation of the top of a tree of height 150 metres standing on the bank of a river is 60° to a point on the bank of other side, what is the width of the river?

4. The length of the shadow of a minar is 240 metres when the angle of elevation of the sun is 60°. What is the height of the minar?

5. The angle of depression of a point on the ground, 15 metres from the top of a minar is 45°, find the height of the minar.

6. The angle of elevation of a point of the roof of a building is 30° to a point on the ground on moving 60 metres towards the building, the angle of elevation becomes 45°. Find the height of the building.

7. The shadow of a tower on the ground is increased by 24 metres, when the angle of elevation of the sun is changed from 60° to 45° What is the height of the tower?

8. A pole of 48 metres long breaks such that the two parts are not completely seperated and the upper part makes an angle 30° with the ground. At what height did the pole break?

9. A tree is broken by storm such that the broken part makes an angle of 30° with the other and touches the ground at a distance of 10 metres from it. Find the length of the tree.

10. If the angle of elevation at the top of the minar is from 45° to 60°, from a point on moving 60 metres towards the minar, find the height the minar.


TrigonometryMultiple Choice Questions (MCQ) :1. The figure at the right side represents

a proportional picture of a garden: Where, A = 90° and B = Which of the following is the ratio of tan ?

A. ABAC B.

ABBC

C. ACAB D.

BCAB

2. A flag stand pole has been broken and the broken part made 30° angle with the ground. If the length of the broken part of the pole is 16 cm, how mar meters will be the part remains standing?

A. 6 B. 8 C. 8 3 D. 16 33. A right angled triangular shaped iron sheet has sides 6cm and 8cm

adjacent to right angle. If the opposite angle of the side of 6 cm is , The value of Cot is

A. 34 B.

43

C. 45 D.

53

4. Which of the following relations is correct? A. Sin + Cos = 1 B. Cos + Sin > 1 C. tan2 Sec2 = 1 D. Sin2 + Cos2 < 1 5. Look at the following relation:

i. Sin2 + Cos2 = 1

ii. Sec = sincos

iii. Cos + Sin > 1

C

A B


Which of the above relations are correct? A. i B. ii C. ii and iii D. i and iii 6. In the figure, A = 90° and B = 60°

i. tan60° = CAAB

ii. Cosec2600 Cot2600 = 1 iii. Sin60° = Cos600

Which of the above equations are correct? A. i B. ii C. ii and iii D. i and iii The following information refers to questions (7-9)As of the figure, a person reaches at point B by passing 3km straight west side from point A and from point 6 after passing 1km directly to the north he arrives at point C. where, BAC = .

7. Which of the following direct the value of Cos ?

A. 13

B. 3

2

C. 23

D. 12

8. How many km is the length of AC? A. 1 B. 3

C. 2 D. 4 9. Which of the following relations is correct?

A. Sin Cos = 12 (1 + 3 )

B. Sec + Sin = 12 (1 + 3 )

C. Sin Cos = 12 (1 + 3 )

D. Sec Sin = 12 (1 3 )

C

A B

C

B A


Creative Questions :

1. Above figure is the proportional picture of frontside garden of the office

of our school. BD is a path accross the garden, where, AB BC and

BD AC.

A. Find the value of Sin and Cos .

B. Show that, AC2 AB2 = 4BD2

C. Using the value of AD, find the value of BAC and BCA then

from that justify Sin BAC and Cos BCA.

2. A pole was broken by a storm and the broken part creates an angle of

with the ground as it touches the ground.

A. Show the above description in a picture and find the length of

broken part and the part remains standing.

B. Deduct the theorem Sin2 + Cos2 = 1 geometrically.

C. Justify the equation (Sec + tan )2 = 1 + Sin1 Sin by using the theorem

Sin2 + Cos2 = 1

2

A

BC

D

2

1

Chapter ThirteenMensuration

13.1. Unit and measurement. The length of a line, the area of a plane, the volume of a solid etc. are determined for practical purposes. In the measurement of any such quantity another quantity of the same kind having some definite magnitude is taken as a unit, The ratio of the quantity measured and the unit defined in the above process is the amount of the quantity.

i.e. =quantity measured

unit quantity = magnitude

It is observed that referred to a fixed unit, every magnitude is a number which expresses how many times of the unit is the magnitude of the quantity measured. For example, when it is said That a table is 3 metres long, it is understood that metre is a definite length which is taken as a unit and in comparison to that the table is 3 times in length. Measurement of length; In the measurement of length, generally metre and the other units that arise out of it are used. Foot, cubit etc. units are also used. Measurement of region : The unit of the measurement of area is fixed on the basis of the unit of length. The area of the square region is supposed Isq. unit (such 1 sq. cm.) if the length of its side is I unit (such

1 cm.) and it used as unit in determining area. as1 cm

13.2. Areas of some rectilinear regions (A) Area of rectangular region: LetABCD be a rectangular region whoselength AB = 7 metres and breadth AD = 4 metres.

1 sq. cm

D

A B

C


AB and AD are divided respectively into 7 and 4 equal parts such that the length of each part is 1 metre. Through the points of section, straight lines are drawn, parallel to AB and AD respectively. Thus the rectangular region is divided totally into (7 4) equal square regions. The length of the side of each square region is 1 metre. Hence the area of each square region is 1 square metre.

ad

b

Area of the rectangular region = (7 4) sq. m. = 28 sq. metres. In general it can be said that if a rectangular region has length 'a' units and breadth 'b' units, then the area of the rectangular region A = a b square units. It is to be noted that the perimeter of the rectangular region S = 2 (a + b) units, and the diagonal of the rectangular region d = a2 + b2 units.

(B) Area of a square region; Let the length of each side of the square region ABCD = a units Then area of the square region = a2 sq. units. It is to be noted that the perimeter of the square region S = 4a units and the diagonal of the square region

d = a2 + a2 units = 2a2 units = 2a units

A

D C

d

a B

(C) Area of a parallelogram region : (1) The base and height of a parallelogram region are given : Let the base of the parallelogram ABCD be AB = a units and its height DE = h units.Let us consider the rectangular-region ABGH formed by drawing perpendiculars to CD from A and B. The parallelogram region ABCD and the rectangular region ABGH are equal in area as they stand on the same base and lie between the same parallels.


The area of the parallelogram region ABCD = area of rectangular ABGH= AB BG sq. units= AB DE sq. units ( BG = DE)= ah sq. units. This formula can be stated as follows, Area of a parallelogram region = base of the region x height of the region.(2) Two adjacent sides of a parallelogram-region and the angle included between them are given : Let AB = a units, AD = b units, and

DAB = ° of the parallelogram ABCD DE is drawn perpendicular to AB from D. Therefore, the area of the parallelogram region ABCD = AB DE sq. units . = AB AD sin DAB sq. units [ DE = AD sin DAB]= ab sin ° sq. units [ DAB = °] (D) Area of a triangular region : (1) The base and altitude of a triangle are given : Let the base of the triangle ABC be BC = a units and its altitude AD = h units. The rectangular region BCEF is drawn with its adjacent sides equal to the base and altitude of the triangle. We know that, the area of the A region ABC

=12 area of rectangular region BCEF

=12 BC EC sq. units

=12 BC AD sq. units, ( EC = AD)

=12 ah square units.

This fromula can be stated as follows,

Area of a triangular region = 12 base altitude.

D GC

BaEA

H

A BE

D C

a

b

AF E

D h

B D Ca


Remark : In a right angled triangle if one of the sides adjacent to the right angle is taken as the base then the other is the altitude.

Area of a right angled triangular region

= 12 base altitude

= 12 ab sq. units.

(2) Two sides of a triangular region and the angle included between them are given : Let ABC be a triangle. AD is drawn perpendicular from A to BC. The altitude of the triangle AD (h) = AC sin C = b sin C.

Area of the region ABC = 12 BC AD

= 12 ab sin C

Similarly, area of the region ABC

= 12 bc sin A =

12 ac sin B

(3) Given three sides of a triangle It should be noted that the sum of the lengths of the three sides of a triangle is called its perimeter. The perimeter of a triangle is denoted by 2S.Let the sides of the ABC be BC = a, CA = b, AB = c.AD is drawn perpendicular to BC from the vertex A. Let BD = x, then CD = a xFrom ABD and ACD we get, AD2 = AB2 BD2 = AC2 CD2

or, c2 x2 = b2 (a x)2

2

or, 2ax = c2 + a2 b2

x = c2 + a2 b2

2a

Agin, AD2 = c2 x2 = c2 c2 + a2 b2

2a

= c + c2 + a2 b2

2a cc2 + a2 b2

2a

b

a

A

B CDa

bc

A

B CD a

bc


={(a + c)2 b2}{b2 (a c)2}

4a2

=(a + c + b) (a + c b) (b + a c) (b a + c)

4a2

=2s(2s 2b) (2s 2c) (2s 2a)

4a2 [ 2s = a + b + c]

=4s(s a) (s b) (s c)

a2

AD = 2a s(s a) (s b) (s c)

Area of -region ABC = 12 BC AD square units

= 12 a

2a s(s a) (s b) (s c) square units

= s(s a) (s b) (s c) square units(E) Area of trapezium region : The area of a trapezium region can be determined if its two parallel sides and the perpendicular distance between them are given. Let ABCD be a trapezium, whose parallel sides are, AB = a units, and a DC = b units. A, C are joined and from C, perpendicular CE is drawn on AB Then the perpendicular distance between the sides AB and DC is CE. Let CE = h units Therefore, area of the trapezium region ABCD = (area of -region ABC) + (area of -region ADC.)

= (12 AB CE) sq. units + (

12 DC CE) sq. units.

[Since the altitude of -ADC is also CE]

=12 CE (AB + DC) square units.

=12 h(a + b) square units.

A BE

CD b

a


(F) Some particular regions (1) Area of equilateral triangular region: Let the length of each side of the equilateral triangle ABC = a unit. From A, perpendicular AD is drawn to BC.

Then BD = 12 BC =

12 a

AD2 = AB2 AD2 = a2 a2

4 =3a2

4

AD = 32

Area of equilateral -region ABC

=12 a

3a2 square units =

34 a square units2

(2) Area of isosceles triangular region : Let ABC be an isosceles triangles of which AB = AC = a and BC = b From A, perpendicular A D is drawn to BC.

Then AD2 = AB2 BD2 = a2 b2

4 = 4a2 b2

4

AD = 4a2 b2

2 Area of the isosceles triangle

=12 b

4a2 b2

2 sq. units

=b4 4a2 b2 square units.

(3) The length of a diagonal and the perpendicular distance of the diagonal from the opposite vertex of a parallelogram are given : Let ABCD be a parallelogram. Its diagonal AC = d units and the perpendicular from the vertex D to the diagonal AC is DE = h units Area of the parallelogram region ABCD= 2 area of -region ADC

= 2 12 AC DE square units.

= dh square units.

A

D

a

a2

A

D

a

b2

a

b

CD

h

E

d

X a


(4) Two diagonals of a rhombus region are given : We know that the diagonals of a rhombus bisect each other at right angles. Let the diagonals of the rhombus ABCD be AC = d1 units and BD = d2 units. Let the diagonals intersect each other at O. Therefore area of rhombus region ABCD = (area of -region ADC) + (area of -region ABC)

=12 AC OD sq. units +

12 AC OB sq. units

=12 AC (OD + OB) sq. units.

=12 AC BD square units

=12 d d , square units.1 2

13.3. Measurement related to circle(A) Length of circle and arc of a circle. The length of circle is called its circumference. In the discussions of circle in the ninth chapter, we have seen that, (1) if the radius of a circle is r, then its circumference c = 2 r. where = 3.141592653897932 ............... an irrational number. Hence if the radius r is known, the approximate value of the circumference of the circle can be determined by using the approximate value of . The approximate value of is generally taken to be 3-1416;(2) The length of that arc of a circle of radius r whose measure in degree of x is

given by s = rx

180It is noted that the measure of an arc in degrees is the measure in degrees of the angle subtended by that arc at the centre. Hence if the radius r of a circle and the measure in degrees of the angle subtended by any arc at the centre are known, the approximate value of the length of the arc obtained by using the approximate value of .

A B

CD

d2

d1

O

r

c

r

sx


(B) Area of circular region and circular segment. Definition : The subset of the plane formed by the union of a circle and its interior is called a circular region and, the circle is called the boundary of the such circular region. If A and B are two points on a circle with centre O, then the subset of the plane formed by the union of the intersection of

AOB and the interior of the circle with the line segments OA, OB and the arc AB is called a circular segment. At this stage we assume it to be true that,

O

BA

Formula 1. The area of the circular region bounded by a circle of radius r units = r2 sq. units. Formula 2. The areas of two circular segments of the same circle are proportional to the measure in degrees of the arcs on which they stand. In the adjoining figure, area of the circular segment AOBarea of the circular segment COD

=measure of the arc APB in degreemeasure of the arc CQD in degree

It is observed that the approximate value of the area of the circle can be determined by using the approximate value of it in formula 1, if the radius of the circle is known.Circular area. Let O be the centre and r units be the radius of a circle. Let the circular segment AOB stand on the arc APB whose measure in degrees is O. Let OC be drawn perpendicular to OA. area of circular segment AOBarea of circular segment AOC =

measure in degree of AOBmeasure in degree of AOC

or, area of circular segment AOB = 90 area of circular segmet AOC

[Since AOC 90°]

= 9014 area of the circular region


= 9014 r square units. 2

= 360 r square units 2

Putting the approximate value of in this formula, the approximate area of the circular segment can be determined.

13.4. Measurement related to rectangular parallelepiped and cube, (1) Rectangular parallelepiped The solid bounded by three pairs of parallel planes or faces is called a rectangular parallelepiped. Let ABCDEFGH be a rectangular parallelepiped whose length AB = a. breath AD = b and height AH = c units. (i) The diagonal of the rectangul parallelpiped is AF

= AC2 + CF2

= AC2 + BC2 + CF2

E F

H G

D C

BA

= a2 + b2 + c2 [since BC =AD = b, CF = AH = c)It can be easily shown that all diagonals are of the same length, (ii) Area of the whole surface of the rectangular parallelepiped = 2 (area of the plane ABCD + area of the plane ABGH + area of the plane BCFG) = 2 (AB AD + AB AH + BC BG) square units. = 2 (ab + ac + be) square units [ Since BC = AD = b, BG = AH = c] = 2 (ab + be + ca) square units.(iii) Volume of the rectangular parallelepiped= (length × breath × height) of the rectangular parallelepiped.AB AD AH cubic units = abc cubic units. (2) Cube or cuboid. The rectangular parallelepiped whose length, breath and height are equal is called a cube or cuboid. Let OABCDEFG be a euboid whose length = breadth = height = a units


(i) The diagonal OE of the cube = OB2 + BE2

= OA2 + AB2 + BE2

= a2 + a2 + a2 units [Since AB = OC = a and BE = OG = a]

= 3a2 units = 3 . a units. (ii) Area of the whole surface of the cube = 2(a2 + a2 + a2) square units = 6a2 square units. (iii) Volume of the cube = a a a cubic unit = a3 cubic unit.

13.5. Measurement related to cone, cylinder and sphere.(1) The cone : The solid formed by a complete revolution of a right angled triangle about one of its fixed side adjacent to the right angle, is known as a right circular cone. The side about which the triangle is revolved is called the axis of the cone. The base of a right circular cone is a circle and its radius is equal the length of other side, other than the axis adjacent to the right angle. The end of the axis containing the right angle is the centre and the other end is the vertex of the cone. The length of the axis is the cone. The length of the line joining the vertex and any circumference of the base is the oblique height of the cone. [Note : By cone generally we mean right circular cone].

Area of a cone : Let ABCD be a cone. The radius of its base is BC = r, height AB = h and oblique height AC = From the right angled triangle ABC, we have AC2 = AB2 + BC2

or, 2 = h2 + r2

= h2 + r2

A B

C

DG

F

O

E

Ventex

Slaut

Axis

HeightRadius of the base

Center

A

D

h


(i) Area of the curved surface of a cone

=12 (circumference of base) oblique height

=12 2 r squre units

= r squre units= r h2 + r2 square units(ii) Area of the whole surface of a cone = area of the curved surface + area of base= r + r = ( + 1) r square. units. (iii) Volume of a cone

=13 (area of base) height

=13 r h = cubic units. 2

2. Cylinder : The solid formed by a corn plete revolution of a rectangle about one of its sides as axis is called a right circular cylinder. The two ends of a right circular cylinder are circles. The length of the axis of a cylinder is called its height. The side of the rectangle which is parallel to the axis and revolves about the axis is called of the generator of the cylinder. [Note : By cylinder generally we mean right circular cylinder] Area of cylinder : Let ABOC be a cylinder. The radius of its base OB = r units and height OC = h units (i) Area of the curved surface of a cylinder . = (circumference of the base) height = 2 r2 square units.(ii) Area of the whole surface of a cylinder= Area of the curved surface + area of the end faces = (2 rh + 2 r2) square units = 2 r (h + r) square units.(iii) Volume of the cylinder = Area of the base height = r2h cubic units.

Height

Axis

Generator axis

Radius of the base

AC

h

O r B


(3) Sphere : A sphere is a solid generated by the complete revolution of a semi circle about its diameter as axis. The centre of the semicircle is the centre of the sphere. The surface generated by the revolution of the semi-circle about its diameter is the surface of the sphere. The diameter of the semi-circle is the diameter of the sphere. Area of a sphere : Let ABPQ be a sphere. Its centre is O and radius r units (i) The area of surface of the sphere, = × (diameter) square units = × (2r)2 square units = 4 r2 square units. (ii) Volume of a sphere

= 43 r cubic units. 3

Centre

Axis

P

13.6. Miscellaneous problems on mensuration. (A) Rectangular region. Example 1. The length of a rectangular room is double its breadth. What is the perimeter if its area is 512 square metres? Solution : Let the breadth of the room = x metres The length the of the room = 2x metres The area of the room = 2x × x sq. metres = 2x2 sq. metres By the question, 2x2 = 512 or, x2 256 x = 16

Breadth of the room = 16 metres. and length of the room = 2 × 16 metres = 32 metres.

Perimeter of the room = 2 (length + breadth) = 2 (32 + 16) metres = 96 metres.

Example 2. The area of a rectangular region is 160 sq. metres. The region becomes a square if its length is reduced by 6 metres. Find the length and breadth of the rectangular region.


Solution : Let the length of the rectangular region = x metres and its breadth = y metres

Area of the rectangular region = xy sq. metres.By the question, xy = 160 ................................ (i) Again by the given condition, x 6 = y or, x = y + 6 ................................................. (ii) Putting the value of x from (ii) in (i). we get (y + 6) y = 160 or, y2 + 6y 160 = 0 or, y2 + 16y + 10y 160 = 0 or, (y + 16)(y 10) = 0

y + 16 = 0 or, y 10 = 0. y = 10, 16 y = 10, since y = 16 , a negative value is not admissible. Then from (ii), we get x = 10 + 6. or, x = 16

the length of the rectangular region = 16 metres and its breadth = 10 metres. Example 3. A garden is 21 metres in length and 15 metres in breadth has a road 2 metres in width around and outside the garden. What will be the total expenditure of planting grass on the road at the rate of Tk.25 per square metre? Solution : Length of the garden = 21 metres Breadth of the garden =15 metres

Area of the garden = 21 × 15 sq.m. = 315 sq.m. Length of the garden including the road = (21 + 4) metres = 25 metres. Breadth of the garden including the road = (15 + 4) metres = 19 metres.

19 15

21

25

Area of the garden including the road = 25 × 19 sq. metres = 475 sq. metres.

Area of the road = (475 315) sq. metres = 160 sq. metres. Since of planting grass per sq. metre = Tk. 25

The cost of planting grass in 160 sq. metres = Tk (160 × 25) = Tk. 4000 The total cost of planting grass on the road is Tk. 4000.

Example 4. A square garden has a road 5 metres wide around and outside the garden. If the area of the road is 500 sq. metres, find the area of the garden.


Solution : Let the length of the garden = x metre. Area of the garden = x2 sq. metres. Area of the road = 500 sq. metres.

Area of the garden including the road = (x2 + 500) sq. metres..........(i) Again length of the garden including the road = (x + 10) metres.

Area of the garden including the road = (x + 10)2 sq. metres = (x2 + 20x + 100) sq. metres ........................(ii) From (i) and (ii) x2 + 20x + 100 = x2 + 500 or, 20x = 400 or, x = 20.

Area of the garden = x2 sq. metres = 202 sq. metres = 400 sq. metres.

Example 5 : The perimeter of a square region is equal to that of rectangular region. The length of the rectangular region is three times its beadth and its area is 768 sq.metres. How many stones each of 40 cm square is required to cover the square?

Solution : Let the breadth of the rectangular region be x metres. Then its length = 3x metres.

Area of the rectangular region = 3x2 square metres. By the question, 3x2 = 768 or x2 = 256 or, x = 16

The breadth of the rectangular region = 16 metres Length of the rectangular region = (3 × 16) metres = 48 metres. Perimeter of the square region

= 2 (length + breadth) = 2 (48 + 16) metres = 128 metres. Therefore, perimeter of the square region =128 metres

Length of a side of the square region = (128 4) metres = 32 metres. Area of the square region = (32)2 sq. metres = 1024 sq. metres. Area of a piece of stone = (0.4) 2 sq. metre = 0.16 sq. metre. Total number of stones required = (1024 0.16) pieces = 6400 pieces.


Exercise 13.1

1. The length of a plot of land is 80 metres and its breadth is 60 metres. A tank is dug in the plot. If the breadth of the bank of each side of the tank is 4 metres, find the area of the bank of the tank.

2. To cover the floor of a room by carpet, the total cost is Tk. 800.00 If the length of the room is reduced by 1 metre, the cost becomes Tk. 700.00. Find the length of the room.

3. The length of a garden is 40 metres and its breadth is 30 metres, There is a pond in the garden having bank of equal width on all sides. If the area of the pond is

12 that of the garden, find the length and breadth of the pond.

4. Inside a square field, there is a road around it of width 4 metres. If the area of the road is 1 hectare (10,000 square metres), then what is the inner area excluding the road?

5. The area of a rectangular regions is 2000 square metres. If the length is reduced by 10 metres, then it becomes a square region. Find the length and breadth of the rectangular region.

(B) Parallelogram and Trapezium. Example 1. The diagonals of a rhombus are respectively 40 cm, and 60 cm. Find its area, perimeter and height. Solution : Let ABCD be a rhombus and let its diagonals AC and BD intersects each other at O,

Area of the rhombus = 12 AC × BD

= 12 × 60 × 40 sq. cm. = 1200 sq. cm.

D C

810

A F 12 B EFrom the right angled triangle ABO. We get,

AB2 = AO2 + BO2 = (30)2 + (20)2 [since AO = 12 AC = 30, BO =

12 = BD = 20]

= 900 + 400 = 1300.

Length of the side AB of the rhombus = 1300 cm. = 36.05 cm. Perimeter of the rhombus = 4 × AB = 4 × 36.05 cm. = 1442 cm. (nearly).

and the height of the rhombus = (1200 36.05) cm. = 33.28 cm. (nearly).

O


Example 2. The length of the sides of a parallelogram are 12 metres and 8 metres. If one of its diagonals is of lenght 10 metres, find the length of the other diagonal. Solution : Let the sides of the parallelogram ABCD be AB =12 metres, and AD = 8 metres, and the diagonal BD = 10 metres. From C and D, perpendiculars DE are drawn on AB produced. Let A, C and B, D bejoined.

D C

810

A F 12 B E

In the triangle ABD, AB = 12 metres, AD = 8 metres and BD = 10 metres, Perimeter of triangle ABD = 2S = (12 + 10 + 8) metres = 30 metres S = 15 metres. Area of ABD = s(s a)(s b)(s c) sq. units

= 15(15 12)(15 10)(15 8) sq; units

= l5 × 3 × 5 × 7 sq. m = 1575 sq.m = 39.68 sq.m (nearly).

Again area of triangle ABD = 12 × AB × DF

39.68 =12 × 12 × DF

DF = 39.68 6 = 6.61 CE = 6.61 metres.

Again BC = AD = 8 metres.

Now from right angled triangle ABCE we get CE2 + BE2 = BC2 or, BE2 = BC2 CE2 = 82 (6.61 )2 = 64 43.69 = 20.31

BE = 4.5 Therefore, AE = AB + BE = 12 + 4 = 16-5 Hence from the right angled triangle ACE, we get AC2 = AE2 + CE2 = (16.5)2 + (6.61)2 = 272.25 + 43.69 = 315.94

AC = 315.94 metres = 17.77 metres (approx)

Example 3. One of the two parallel sides of a trapezium is greater than the other by 1 metre and the perpendicular distance between them is 2 metres. If the area of the trapezium is 27 square metres, find the lengths of the parallel sides.


Solution : Let the lengths of the parallel sides of the trapezium be a and b, and the perpendicular distance between them is h. Let a = x metres, then b = (x + 1) metre.

Area of the trapezium, A = 12 (a + b)h.

or, 27 = 12 (x + x + l) × 2

or, 2x = 26 x = 13 Length of one side of the trapezium = 13 metres and the length of the

other parallel side = (x + 1) metres = (13 + 1) metres = 14 metres.

Example 4. The difference of the lengths of the two parallel sides of a trapezium is 8 cm. and their perpendicular distance is 24 cm. If the area of the trapezium be 13 times the perpendicular distance, then find the lengths of the parallel sides. Solution : Let a and b be the lengths of the parallel sides of the trapezium and let h be the perpendicular distance between them.

Area = 13 × 24 square cm. = 312 sq. cm.

Therefore, 312 = 12 (a + b) × h

or, 312 = 12 (a + b) × 24

a + b = 26 .............................(i) By the question (a b) = 8 ............. (ii) Now from (i) and (ii) by addition 2a = 34

a = 17 cm. Again from (i) and (ii) by subtraction 2b = 18

b = 9 cm. Hence the lengths of the parallel sides are 17 cm. and 9 cm.

Exercise 13.2 1. The perimeter of a rhombus is 180 cm. and one of its diagonals is 54 cm.

Find its other diagonal and area.

2. The area of a parallelogram is 120 sq. cm. and one of the diagonals is 24 cm. Find the length of the perpendicular on the diagonal from the opposite vertex.


3. The lengths of the parallel sides of-a trapezium are 91 cm. and 51 cm and the length of the other/two sides are 37 cm, and 13 cm. Find its area.

4. The area of a parallelogram region is equal to that of a square region. If the base of the parallelogram region be 125 metres and its height be 5 metres, find the length of the diagonal of the square region.

5. One of the two parallel sides of a trapezium is greater than the other by 4 metres and the perpendicular distance between them is 8 metres. If the area of the trapezium is 112 sq. metre, find the lengths of the parallel sides.

6. The lengths of the sides of a parallelogram are 30 cm. and 26 cm. If one of its diagonals is 28 cm., find the length of the other diagonal.

(C) The Triangle Example 1. A ladder of length 20 metres stands vertically against a wall. How much further should the lower end of the ladder end be removed so that its upper end descends 4 metres? Solution : Let if the foot of the ladder AC is removed from C to D, then the upper ends descends from A to B. Then length of the ladder AC = BD = 20 metres,

D

20

B A

and AB = 4 metres, BC = 20 4 =16 metres,

Now BC2 + CD2 = BD2 16 or, CD2 = BD2 BC2

= (20)2 (16)2 = 400 256 = 144 CD = 12 metres. C

Example 2. The perimeter of an isosceles triangle is 16 metres, if the length of its equal sides is th of the base, then find the area of the triangular region. Solution : Let ABC be an isosceles triangle and its BC = x metres.

AB = AC = 5x6

By the question, x + 5x6 +

5x6 =16

or, 16x = 96 or, x = 6

x 56

x B

A

C

BC = 6 metres, AB = AC = 56 × 6 = 5 metres


Perimeter of the -region ABC = 2s = (6 + 5 + 5) metres = 16 metres S = 8 metres. Area of the - region ABC = s(s a)(s b)(s c) square metres

= 8(8 6) (8 5) (8 5) square metres

= 8 × 2 × 3 × 3 square metres

= 144 sq metres = 12 sq. metres.

Example 3. If the length of each side of an equilateral triangle be increased by 2 metres its area is increaed by 3 3 sq metres. Find the length of the side of the equilateral triangle. Solution : Let, the length of a side of the equilateral triangle = a metres.

Therefore, the area of the equilateral triangle A = 3a2

4 sq.m

If the length of each side be increased by 2 metres, the area of the triangle

B = 3(a + 2) 2

4 = 3(a2 + 4a + 4)

4 sq. metres.

By the questions, 3

4 (a2 + 4a + 4) = 3a2

4 + 3 3

or, 3 (a + 4a + 4) = 2 3 a + 122 3 or, a2 + 4a + 4 = a2 + 12 or, 4a = 8 or, a = 2

Length of the side of the equilateral triangle = 2 metres.

Example 4. Two roads at a given place make an angle 1200 between their directions. Two men from the given place move in the opposite directions with velocities 10 km. per hour and 8 km. per hour respectively. What will be the direct distance between them after 5 hours? Solution : Let, two men start from A with velocities 10 km/hour and 8 kml hour respectively and reach B and C after 5 hours. Then after 5 hours, the direct distance between them is BC, From C, perpendicular CD is drawn on BA produced.

AB = 10 × 5 km. = 50 km. AC = 8 × 5 km. = 40 km.


BAC = 120° CAD = 60°

C

40

120 60

From the right-angled triangle CAD, we get CDAC = sin 60 and

ADAC = cos 60

CD = AC sin 60 = 40 × 3

2 = 20 3

and AD = AC cos 60° = 40 × 12 = 20. B A D50

Therefore, from the right angled triangle CBD, We get, BC2 = BD2 + CD2 = (BA + AD)2 + CD2 = (50 + 20)2 + (20 3 ) = 4900 + 1200 = 6100 2

BC = 6100 = 78.1 km. (approximately)

Exercise 13.3 1. The hypotenuse of a right angled triangle is 25 metres. If one of its sides

is 43 th of the other, find the lengths of the sides.

2. The length of the base of an isosceles triangle is 60 cm. If its area is 1200 sq. cm. find the length of the equal sides.

3. If the length of the perpendiculars from a point interior of an equilateral triangle to three sides 6, 7, 8 cm. respectively, then find the length of sides of the triangle and the are a of the triangular region.

4. The lengths of the sides of a triangle are 25, 20, 15 units. Find the areas of the triangles in which it is divided by the perpendicular drawn from the vertex opposite to the greatest side.

5. Two roads from a given place make an angle of 1350 between their directions, Two men from the given place move in opposite direction place velocities 7 km./hour and 5 krn./hour respectively. What will be the direct distance between them after 4 hours?

6. The perpendicular of a right angled triangle is less than 1112 times of 4 the

base by 6 cm, and the hypotenuse is less than 43 times of die base by 3

cm. Find the length of the base of the triangle.


7. When the length of each side of an equilateral triangle is increased by 1 metre, its area increases by 3 sq. metres. Find the length of the side and the area of the triangle.

8. The length of the sides of a largest right angled triangle are respectively 28, 45, 53 cm. Find the length of a side of the largest square region in it such that one angular point of the square region lies on the hypotenuse.

(D) The circle and the arc. Unless other wise mentioned, the approximate values of all 22 measurements

related to circle will be determined by assuming 227 as the approximate value of

. For more accurate result, the approximate value of it may be taken as 3.1416. Example 1. If the circumference of a circle of diameter 28 cm. is equal to the perimeter of a square region^ then find the diagonal of that square. Solution : Given that the diameter of the circle = 28 cm.

Radius of the circle = 14 cm. Circumference of the circle = 2 r cm. = 2 × 3.1416 × 14 cm

= 87.9648 cm. By the question, perimeter of the square region 87.9648 cm.

One side a' of the square = (87.9648 4) cm = 21.9912 cm. Diagonal of the square a 2 = 21.9912 2 cm. = 31.1003 cm.

Example 2. The circumference of a circle is 220 metres. Find the length of a side of the inscribed square region. Solution : Let the radius of the circle be r metres. Let ABCD be the square inscribed in the circle. We know circumference of the circle = 2 r By the question = 2 r = 220 or, 2 × 3.1416 × r = 220 or, r = 35.0140

Radius of the circle = 35.0140 metres. Diameter of the circle AC = 2 × 35.0140 metres = 70.028 metres. Now from the isosceles right angled triangle ABC, we gets

D C

BA


AB2 + BC2 = AC2

or, 2AB2 = AC2 [ BC = AB] or, 2 AB = AC

or, AB = 1

2 = 70.028 = 35.0140 = 49.5173 (nearly)

The Required length of the side 49.5173 metres. Example 3. The radius of a circle is 10 cm and the length of an arc is 11 cm. How much angle in degrees does the arc subtend at the centre? Solution : Let 0 be the required angle in degrees,

We know that the length of an arc = 360 × 2 r

11 = 360 × 2 r × 2 × 3.1416 × l0

or, = 11× 360

2 × 3.1416 × l0

r

s

= 63.0252 The angle subtended at the centre = 63.02520

Example 4. The diameter of the front wheel of a car is 28 cm. and that of the back wheel is 35 cm. To describe a distance of 88 metres, how many times the front will revolve more than the back wheel?

Solution : Radius of the front wheel of the car = 282 cm.

Radius of the back wheel of the car = 352 cm.

Circumference of the front wheel = 2 × 3.1416 × 282 cm = 87.9648 cm.

and circumference of the back wheel = 2 × 3.1416 × 352 cm. 109.956 cm.

Hence to describe a distance of 88 metres, the front wheel revolves 88 × 10087.9648

or, 100.04 or, 100 times nearly.

and the back wheel revolves 88 × 100109.956 or, 80.032 or 80 times nearly.

Therefore, the front wheel moves (100 80) or, 20 times more than the back wheel.


Exercise 13.4 1. If the difference between the circumference and diameter of a circle is 60

cm. in the radius of the circle.

2. An arc of a circle subtands an angle 30° at the centre. If the diameter of the circle is 126 cm, find the length of the arc.

3. The diameter of a wheel is 4.2 metres. How many times will the wheel move to describe a distance of 330 metres.

4. A horse moves for 112 minute at the rate of 66 metre per minute in a

circular field. Find the diameter of that circular field. 5. To describe 211 metre 20 cm. the two wheels revolve 32 and 48 times

respectively. What is the difference of their radius of two wheels? (E) Area of circular region and its parts. Example 1. The diameter of a circular field is 100 metres. There is a road of breadth 5 metres along its boundary and outside it. Find the area of the road. solution : Radius of the field (OA) = (100 2) =50 metres. Width of the road (AB) = 5 metres. In this case, the road is a circular ring whose inner radius(r) = 50 metres., and outer radius (R) = (50 + 5) metres = 55 metres.

Area of the road = area of the outer circular- area of the inner circular

= ( R2 r2) sq. units. = (R2 r2) sq. units. = 3.1416 (552 502) sq metres. = 3.1416 (55 + 50) (55 50) sq. metres. = 3.1416 × 105 × 5 sq. metres. = 1649.34 sq. metres.

A O 50

B

Example 2. The circumference of a circle is equal to the perimeter of an equilateral triangle. Find the ratio of their areas.

5


Solution : Let the radius of the circle = r, then area of the circle = r2 and the circumference of the circle = 2 r

r By the question, perimeter of the equilateral triangle = 2 r

length of a side = 2

3If a be the length of a side of an equilateral triangle, then its area

= 3

4 a2 sq. units = 3

4 × 2

3 2 = 3

4 × 4 2r2

9 sq. units = 4 2r2

3 3 sq. units

Area of the circle t Area of the equilateral triangle

= r2 t 2r2

3 3 = 3 3 t

Example 3. The area of a circular segment of a circle is 77 square metres and the radius of the circle is 21 metres. Find the angle which the circular segment subtends at the centre.

Solution : We know, that the area of a circular segment = 360 r mr sq. 2

units Where the radius of the circle = r and the measure of the arc in degrees =

77 = 360 × 3.1416 × (2l)2

= 360 × 77

3.1416 × 21 × 21 = 20.008

The required angle 20.008°

r

s

Exercise 13.5

1. The radius of a circle is 14 cm. and a circular segment subtends an angle 75° at the centre. Find the area of the circular segment.

2. The radius of a circle is 14 cm. The area of a square is equal to that of the circle. Find the length of a side of the square.

3. There is a road surrounding a circular field. The outer circumference of the road is greater than the inner circumference by 44 metres. Find the width of the road.

4. The diameter of a circular park is 26 metres. There is a road of breadth 2 metres arround the park. Find the area of the road.

5. A cow which is tied with a rope such a way that it can eat grass from a grass field of area 3850 sq. metre. Find the length of the rope.


(F) Rectangular parallelepiped and cube Example 1. The area of the whole surface of a rectangular parallelepiped is 2368 sq. cm. Find its length, breadth and height if they are in the ratio 6 t t 5 4. Solution : Let the length of the solid (a) = 6x cm. and the breadth of the solid (b) = 5x cm. and the height of the solid (c) = 4x cm. We know that, the area of the whole surface of the solid = 2(ab + be + ca)

2368 = 2(6x.6x + 5x.4x + 4x.6x) or, 2368 = 2 × 74x2 or, x2 = 16 or, x = 4

Length of the rectangulafparallelopiped (a) = 6x = 6 × 4cm = 24 cm. Breadth of the rectangular parallelepiped (b) = 5x = 5 × 4 cm = 20 cm. Height of the rectangular parallelepiped (c) = 4x = 4 × 4cm = 16 cm.

Example 2. A rectangular parallelepiped stands on a base of area 48 sq. cm. Its height is 3 metres and its diagonal is 13 metres. Find the length and breadth of the rectangular parallelepiped. Solution : Let, the length of the rectangular parallelepiped a metres and the breadth of rectangular, parallelepiped = b meters,

Area of the base = ab sq.metres = 48 sq.metres. We know that the diagonal of a rectangular parallelepiped (d)

a2 + b2 + c2 Here, height (c) = 3 metres. 13 = a2 + b2 + 32

or, 169 = a2 + b2 + 9 or, a2 + b2 = 169 9 = 160 ............................(i)

(a + b)2 = a2 + b2 + 2ab = 160 + 2 × 48 = 256 [Since a2 + b2 = 160 and ab = 48]

a + b = 256 = 16 .................................(ii) Again (a b)2 = a2 + b2 2ab = 160 96 = 64 a b = 8 ..................................................(iii) Adding (ii) and (iii), we get 2a = 24 or, a = 12. Subtracting (iii) from (ii), we get 2b = 8 or, b = 4 Hence, length =12 metres and breadth = 4 metres.

Example 3. The edges of three cubes are respectively 3 cm, 4 cm and 5 cm. They are ftielted and formed into a single cube. Find the edge and diagonal of the new cube.


Solution : We know that, if a is the edge of the cube, then Volume of the cube = a3 cubic units Diagonal of the cube = a 3 units Volume of the new cube = (33 + 43 + 53) cubic cm. = (27 + 64 + 125) cubic cm = 216 cubic cm.

Edge of the new cube = 3

216 cm = 6 cm. Diagonal of the new cube = a 3 = 6 3 cm. = 10.3923 cm. Example 4. The outer measurements of a rectangular box are 8 cm, 6 cm and 4 cm, respectively and the area of the whole inner surface is 88 sq.cm. Find the thickness of the wood. Solution : Let the thickness of the wood = x. cm.

the inner length of the box (a) = (8 2x) cm. the inner breadth of the box (b) = (6 2x) cm.

the inner height of the box (c) = (4 2x) cm. Hence the area of the whole inner surface of the box = 2(ab + be + ca) sq. units = 2{(8 2x)(6 2x) + (6 2x) (4 2x) + (+ (4 2x)(8 2x)} sq. cm. = 2(48 28x + 4x2 + 24 20x + 4x2 + 32 24x + 4x sq) sq. cm. = 2(12x2 72x + 104) sq.cm. By the question, 2(12x2 72x + 104) = 88 or, l2x2 72x + 104 = 44 or, 12x2 72x + 60 = 0 or, x2 6x + 5 = 0 or, (x 5)(x l)=0

x = 5 or, 1 Since the outer height of the box is 4 cm, hence the thickness cannot be 5cm,. Hence, the thickness of the wood 1 cm.

Exercise 13.6 1. The volume of a rectangular parallelepiped is 220 cubic metres. If its

diagonal is 15 metres and length is 11 metres, find its breadth and height. 2. The length, breadth and height of a rectangular parallelepiped are in the

ratio 21 t 16 t 12 and its diagonal is 87 cm. Find the area of the whole surface of the solid.

3. The outer measurements of a box with its, top are 10 cm. 9 cm. and 7 cm, and the area of the whole inner surface is 262 sq. cm. Find thickness of its wall if it is uniform.

4. The area of the whole surface of a cube is 48 square metres. Find the length of its diagonal.


(G) Cone, cylinder and sphere Example 1. The radius of the base of a right circular cone of height 4 cm. is 3 cm. Find its volume and the area of the curved surface. Solution : We know, if the radius of the base of the cone is r, the height is h and the oblique height is h. then

Volume of the cone (V) = 13 r h cubic units. 2

Area of the curved surface of the cone = 2 r sq units Here, r = 3 cm. h = 4 cm.

= 32 + 42 = 9 + 16 = 5 = 5 cm.

h

r

Volume of the cone (v) = 13 × 3.1416 × 3 × 4 cubic cm 2

= 3.1416 × 3 × 4 cubic cm. = 37.6992 cubic cm.

and area of the curved surface of the cone = 3.1416 × 3 × 5 sq. cm. = 47.124sq.em.

Example 2 : A cylinder of height 10 cm. has a base of radius 4 cm. Find the area of the whole surface and its volume. Solution : We know, if the radius of the cylinder is r and the height is h, then the area of the whole surface of the cylinder = 2 r (h + r) sq. units. and volume of the cylinder = r2h cubic units. Here, r = 4cm. and h = 10cm. Hence, the area of the whole surface of the cylinder. = 2 × 3.1416 × 4 (10 + 4) sq. cm. = 2 × 3.1416 × 56 sq. cm. = 351.8592 sq. cm

r

r

h

and volume of the cylinder = 3.1416 × 42 × 10 cub cm. 502.656 cubic cm. Example 3. The area of the curved surface of cylinder is 100 sq. cm. and its volume is 150 cubic cm. Find the height and the radius of the base of the cylinder. Solution : Let the radius of the base of the cylinder be r cm. and its height h cm. Then area of the curved surface = 2 rh sq.cm and volume = r2h cubic cm. By the question, r2h = 150 ............(i) and 2 rh = 100 ............................... (ii)


Dividing (i) by (ii) r2h

2 r = 150100

or, r = 3 Radius of the base = 3 cm.

Substituting the value of r in (ii) we get 2 × 3.1416 × 3 × h = 100

or, = 100

2 × 3.1416 × 3 = 5.3052 cm

Example 4. Three solid spheres of radii 6 cm. 8 cm. and 10 cm. are melted and formed into a new solid sphere. Find the radius and area of the surface of the new sphere. Solution : Let r cm. be the radius of the new sphere.

We know that if the radius of the sphere is r then its volume (v) = 43 r cu.cm. 3

Volume of the first sphere (v1) = 43 8 cu.cm. = 3 4

3 × × 216 cu.cm.

Volume of the 2nd sphere (v2) = 43 8 .cu.cm. = 3 4

3 × × 512 cu.cm.

Volume of the 3rd sphere (v3) = 43 10 cu.cm. = 3 4

3 × × 1000 cu.cm.

By the question, 43 a = 3 4

3 × (216 + 512 +1000); where a is the radius

of the sphere. or, a3 = 1728 a =12

Radius of the new sphere = 12 cm. Again area of the surface of the sphere = 4 r2 sq. cm.

= 4 × 3.1416 × 122 cm. = 1809.5616 sq. cm.

Example 5. Radii of the base of right circular cone and a cylinder are equal. If heir heights are in the ratio 3 t 2 show that their volumes are in the ratio l t 2. t

Solution : Let radius of the base of the cone = r radius of the base of the cylinder = r height of the cone = 3h height of the cylinder = 2h

volume of the cone = 13 r (3h) = r h cubic unit. 2 2

r

r

h1 h

r

Volume of the cylinder = r2(2h) = 2 r2h cubic unit. [since the radius of the base of the cylinder = r ] Hence, volume of the cone t volume of the cylinder = r2h t 2 r2h = 1 t 2.

h


Exercise 13.7 [Taking =3.1416]

1. The height of a right circular cone is 8 cm. and the radius of the base is 6 cm. Find the area of the whole surface and the volume.

2. The radius of the base of a right circular cylinder is 5 cm. and its oblique height is 13 cm. Find its volume and the area of the curved surface.

3. The inner and outer diameters of an iron pipe are respectively 12 cm. and 14 cm, and its height is 5 metres. What is the weight of the iron contained in the pipe, if the weight of 1 c.c. of iron is 7.2 gm.

4. The depth of a well is 14 metres and diameters 28 metres. What will be cost of digging the well if the rate is TK. 5.00 per cubic metre.

5. The height of a right circular cylinder and a cone is h and they stand on the same base. If the areas of their curved surfaces are in the ratio 4 t 3, show

that the radius of the base is 5

2 h.

6. A metallic solid sphere of diameter 6 cm. is melted and formed into a solid cylinder rod of radius 6 cm. Find the length of the rod.

Mensuration

Multiple Choice Questions (MCQ) : 1. The lengths of the parallel sides of a trapezium shaped iron plate are 3cm

and 1cm. respectively. Their perpendicular distance is 2 cm. How many sqr. cm is the area of that iron plate?

A. 1 B. 2 C. 3 D. 4 2. The length and breadth of a garden are 50 m and 40 m respectively.

There is a road whose breadth is 5 m inside and around the garden, What is the length of the garden excluding the road?

A. 30 B. 40 C. 50 D. 60 3. The lengths of two adjacent sides of a parallelogram are 3cm and 5cm

respectively. What is the half of the perimeter of that in cm? A. 4 B. 8 C. 15 D. 16


4. If perimeter of an equilateral triangle is 6 cm, what is the area of that triangle in sqr. cm?

A. 9 3 B. 3

4

C. 3 3

2 D. 3

5. The diameter of a circular field is 26 metre. There is a road adjacent, outside and around the field with breadth of 2 metre. What is the area of the field including the road?

A. 225 B. 169 C. 121 D. 526. If the height of a cone is 4cm and radius of the base is 3cm then what is

the length of its oblique height? A. 1 B. 5 C. 6 D. 7 7. Look at the following statements : i. Perimeter of a 4 cm square shaped stone is 16 sqr. cm. ii. The area of a circular plate with radius 3cm is 3TT sqr. cm. iii. The volume of a cylinder shaped bar of height 5cm and radius 2cm

is 20 cc. Which of the above statements are correct? A. i and ii B. ii and iii C. i and iii D. i, ii and iii

8. Look at the following information : i. If radius of a sphere is P cm then the area of the surface of that

sphere is 4 P2 sqr. cm. ii. The angle caused by the diagonals of a rhombus is 90°. iii. The area of a triangle is 30 sqr. cm if the base of that triangle is 6cm

and height is 5cm. Which of the above statements are correct? A. i and iii B. ii and iii C. i and iii D. i, ii and iii

The following information refers to questions (9 11). The height and base of a right angled triangle shaped copper plate are

4cm and 3cm respectively.


9. What is the perimeter of the plate in cm? A. 5 B. 6 C. 7 D. 12

10. How much sqr. cm is the area of that plate? A. 6 B. 7 C. 12 D. 13

11. What is the volume of a parallelepiped that is supposed to be created by spinning the plate around the largest side?

A. 4 B. 12 C. 24 D. 36

Creative Questions : 1. The length of a rectangular field is one and a half times of its breadth and

the area is 2400 sqr. metre. (Breadth of the field is metre) A. With a short description draw the proportional picture of that field. B. Find the length and breadth of the field, C. There is a pond inside the field having bank of equal width. If the

area of the pond is 800 sqr. metre, what is the width of the bank.

2. The area of a rectangular field is 1200 sqr. metre. It will be a square if the length is reduced by 10 metre. (length of the field is metre)

A. After drawing the proportional figure of above information, present the length and breadth of the field in algebraic expression.

B. Find the length and breadth of the field. C. How many square shaped stones of 50 sqr. cm is required to cover a

square shaped Eidgah whose perimeter is as long as the field mentioned?

3. The area of a rectangular iron plate is 0.125 sqr. metre and the length is two times of its breadth.

A. If the breadth of the plate is x. express the length and breadth of the plate in algebraic expression by drawing proportional figure.

B. Find the perimeter of the plate. C. Calculate the total surface area of a parallelepiped that is supposed

to be created by spinning the plate around the largest side?

AnswersExercise 5

1. 17 m. 2. 25 06 m (approx) 3. 25km 4. 7 2cm; 15 12 sq. cm

5. 3

4 a2 sq. cm

Exercise 7 1. (i) 14 cm. (ii) 2 4 cm. 2. 5 6 cm.

Exercise 12.1 17.

12 18.

25 19.

43 20.

a2 b2

a2 + b2

Exercise 12.2 9. A = 52

12 , B = 7

12 10. A = 37

12 , B = 7

12

11. = 0 , = 90 , 12. = 60 , 13. = 30 , 15. 3

Exercise 12.3 1. 34 641 metres 2. 51 9620 metres 3. 86 603 metres 4. 415 692 metres 5. 10 607 metres 6. 81 962 metres 7. 56 785 metres 8. 16 metres 9. 37 321 metres 10. 141 962 metres 2. 51 9620 metres 3. 86 603 metres

Exercise l3.1 1. 1056 square metres 2. 8 metres 3. 30 metres, 20 metres 4. 38.56 hectare (approx.) 5. 50 metres, 40 metres.

Exercise l3.2 1. 72cm, 1944 sq, cm. 2. 5 cm. 3. 852 sq. cm. 4. 35-35 (approx) 5. 16 metres, 12 metres 6. 48 cm. (approx)


Exercise 13.3 1. 20 metres, 15 metres. 2. 50 cm. 3. 24 249 cm. (approx.), 254 611 sq. cm. (approx.) 4. 54 sq. units, 96 sq. units. 5. 44 44 km. (approx) 6. 36. cm. or 12 cm. 7. 1 5 metres, 0-974 sq. metres 8. 17 26 cm. (approx)

Exercise 13.4 1. 14 008 cm. 2. 32 cm. 3. 25 times 4. 31 513 metres 5. 0 35 metres.

Exercise 13.5 1. 128 282 sq.cm. (approx) 2. 24 814 cm. (approx) 3. 7 003 metres. 4. 175 93 sq.m. 5. 35 007 m.

Exercise 13. 6 1. 10 metres, 2 metres. 2. 14040 sq cm.3. 1 cm. 4. 4.899 metres

Exercise 13.7 1. 30 1594 sq.cm. 301 594 cubic cm. 2. 314 16 cubic cm, 204 204 sq. cm. 3. 147 027 kg 4. Tk. 43102 75 5. 1cm.

Secondary Geometry - e-Book

Documents

Transcript of Secondary Geometry - e-Book