Geometry: Std. 9th, Maharashtra Board - Target Publications

1191 downloads 12750 Views 969KB Size Report
It is widely used in the fields of science, engineering, computers, architecture etc. It ... we bring to you “Std. IX: Geometry” a complete and thorough guide critically ...
  Written as per the revised syllabus prescribed by the Maharashtra State Board   of Secondary and Higher Secondary Education, Pune.

STD. IX

Geometry    

Fifth Edition: March 2016                

Salient Features • Written as per the new textbook. • Exhaustive coverage of entire syllabus. • Topic-wise distribution of all textual questions and practice problems at the beginning of every chapter • Covers answers to all textual exercises and problem set. • Includes additional problems for practice. • Multiple choice questions for effective preparation. • Comprehensive solution to Question Bank. • Constructions drawn with accurate measurements.

         

 

   

Printed at: Dainik Saamana., Navi Mumbai No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

 

P.O. No. 15197

10046_10511_JUP

Preface Geometry is the mathematics of properties, measurement and relationships of points, lines, angles, surfaces and solids. It is widely used in the fields of science, engineering, computers, architecture etc. It is a vast subject dealing with the study of properties, definitions, theorems, areas, perimeter, angles, triangles, mensuration, co-ordinates, constructions etc. The study of Geometry requires a deep and intrinsic understanding of concepts. Hence, to ease the task we bring to you “Std. IX: Geometry” a complete and thorough guide critically analysed and extensively drafted to boost the students confidence. The question answer format of this book helps the student to understand and grasp each and every concept thoroughly. The book is based on the new text book and covers the entire syllabus. At the beginning of every chapter, topic-wise distribution of all textual questions and practice problems has been provided for simpler understanding of different types of questions. It contains answers to textual exercises, problems sets and Question bank. It also includes additional questions and multiple choice questions for practice. All the diagrams are neat and have proper labelling. The book has a unique feature that all the constructions are as per the scale. Another feature of the book is its layout which is attractive and inspires the students to read. There is always room for improvement and hence, we welcome all suggestions and regret any error that may have occurred in the making of this book. A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! Yours faithfully, Publisher

PER  

No. 1 2 3 4 5 6 7 8 9 10

Topic Name Lines and Angles Triangles Congruence of Triangles Circle Quadrilaterals Co‐ordinate Geometry Geometric Constructions Trigonometry Mensuration Question Bank

Page No. 1 39 72 109 143 185 198 215 232 256

01 

Lines and Angles

Type of Problems Axioms,

Collinear

Exercise Points,

Non-collinear Points and Parallel Planes

1.1 Practice Problems (Based on Exercise 1.1) 1.2

Co-ordinates of Points and Distance

Practice Problems (Based on Exercise 1.2) Problem set-1 1.2

Betweeness, Segment and Ray

Practice Problems (Based on Exercise 1.2) Problem set-1 1.3 Practice Problems

Types of Angles, Pairs of Angles

(Based on Exercise 1.3) Practice Problems (Based on Exercise 1.4) Problem set-1

For the given measures of angles find

1.3

their Complementary/Supplementary

Practice Problems

Angles

(Based on Exercise 1.3) 1.4

Parallel Lines, Corresponding Angles, Alternate Angles, Vertically Opposite Angles and Interior Angles

Chapter 01: Lines and Angles

Practice Problems (Based on Exercise 1.4) Problem set-1

Q. Nos. Q.1, 2, 3, 4 Q.1, 2, 3 Q.1, 2 Q.1 Q.6 Q. 3, 4, 5, 6, 7, 8, 9, 10, 11 Q. 2, 3, 4 Q.1, 2, 3, 4, 5 Q.1, 4, 5 Q.3, 4, 5, 6, 7

Q.5 Q.7, 8, 9, 11, 12, 13, 14, 16 Q.2, 3 Q.1, 2 Q.1, 2, 3, 4 Q.1, 2, 3, 4, 6, 7 Q.10, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25

1

Std. IX : Geometry  Introduction Basic Geometry: A point, a line and a plane are undefined terms which are basic concepts in geometry. i. Lines and planes are set of points. ii. Each line and each plane contain infinite number of points. Axioms / Postulates: The simple properties which we accept as true are called Axioms or Postulates. OR The terms or statements whose proofs are not to be asked are called Axioms. Theorem: Important and useful results derived from the axioms are called Theorems. OR The statements that we prove from the axioms are called Theorems. Euclid’s five postulates: 1. A straight line can be drawn from any point to any other point. 2. A terminated line can be produced indefinitely. 3. A circle can be drawn with any centre and any radius. 4. All right angles are equal to one another. 5. If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the angles taken together are less than two right angles. OR Two distinct intersecting lines cannot be parallel to the same line. 1.1

Some Axioms

1.

Infinite number of lines can be drawn through a given point.



m

P n

2.

There is one and only one line passing through two distinct points.

3.

When two distinct lines intersect, their intersection is exactly one point.



A

B

 P m

4.

E

A

There is exactly one plane passing through three non-collinear points.

C B A

5.

There is exactly one plane passing through a line and a point, not on the line. 

2

E

Chapter 01: Lines and Angles 6.

E

There is exactly one plane passing through two distinct intersecting lines.

 P m E

7.



When two planes intersect, their intersection is exactly one line.

F  E

8.

When a line intersects a plane but does not lie in it, then their intersection is a point.

9.

A line containing two given points of a plane lies wholly in that plane.

P

E  Q

P

10.

If the lines lie in the same plane, then they are called ‘Coplanar lines’, otherwise they are called ‘non-coplanar lines’.

1.2

Collinear Points, Non-collinear points, Parallel planes

1.

Collinear Points: If there exists a line containing all the given points, then those points are called collinear points. Points A, B, C are collinear points.

2.

Non-collinear Points: If there does not exist a line containing all the given points, then those points are called non-collinear points. Points A, B, C are non collinear points.

3.

A

B

C



B 

C

A



Coplanar Lines: Lines which lie in the same plane are called coplanar lines. line  and line m are coplanar lines.

m



4.

Non-coplanar Lines: Lines which do not lie in the same plane are called non-coplanar lines. line  and line m are non-coplanar lines.

m m

n



5.

Concurrent Lines: If three or more lines pass through one point, then the lines are called concurrent lines.

o P

Line , m, n and o are concurrent lines and P is their point of concurrence. Note: The common point of intersection is called the point of concurrence. 3

Std. IX : Geometry  6.

Parallel Lines: The lines in a plane which are not intersecting are called parallel lines. line  || line m

 m

D

7.

Parallel Planes: Two non-intersecting planes are said to be parallel planes. Plane ABCD and plane PQRS are parallel planes.

C

A

B S

R Q

P

Exercise 1.1 1.

Take any three non-collinear points A, B, C on a paper. How many lines in all can you draw through different pairs of the points? Name the lines. C Solution: We can draw three lines through three non-collinear points A, B and C. The lines are line AB, line BC and line AC. B

A

2.

Take four points P, Q, R, S in a plane. Draw lines by joining different pairs of points. How many lines can you draw in the following cases? i. No three of them are collinear. ii. Three of these points are collinear. S R Solution: i. We can draw six lines (line PQ, line QR, line SR, line PS, line QS and line PR). Q

P

S

ii.

We can draw four lines (line PR, line PS, line QS, line SR).

Q

P

3. Observe the given figure and write the sets of all the points which are collinear. Solution: Set of collinear points are P S i. P, S, T, Q ii. P, F, R, B F iii. A, F, S, D R A iv. A, R, E, Q v. B, E, T, D 4

R

D T Q E B

Chapter 01: Lines and Angles 4.

Observe the given figure and answer the following: i. Name the lines parallel to the line AB.

D

ii.

Can you say that line AD and the point R lie in the same plane? Why?

iii.

Are the points A, S, B, R coplanar? Why?

iv.

Name the three planes passing through point A.

v.

Name the points such that the plane containing them does not contain points P, Q, C and D.

C V

A

B S

R

P

Q

Solution: i. line DC, line PQ and line SR are parallel to line AB. ii. Yes, line AD and point R lie in the same plane. [There is exactly one plane passing through a line and a point not on it (axiom)] iii. Yes, points A, S, B, R are coplanar (since these points are contained in the plane ASRB) iv. Planes passing through point A are plane APSD, plane ABQP and plane ABCD. v. Points A, S, R, B, V (reason: Plane containing points P, Q, C and D is plane PQCD and it does not contain points A, S, R, B, V). 1.3

Co-ordinates of Points and Distance

Co-ordinates of a Point: The number associated with a point is called the co-ordinate of that point. In adjoining figure, co-ordinate of point P is 1 and that of point B is 2.

R

Q

P

O

A B C

3 2 1

0

1

2

3



Distance between Two Points: If x and y are the co-ordinates of point P and Q respectively, then the distance between P and Q is defined as the absolute value of the difference between the number associated with those points. P y

Q x (x > y)



1.4



P Q x y (x  y)



d(P, Q) = x  y if x > y d(P, Q) = y  x if x < y d(P, Q) = | x  y | Thus, distance between any two distinct points is a unique non-negative real number. Betweenness

If points P, Q and R are three distinct collinear points and if d(P, Q) + d(Q, R) = d(P, R), then the point Q is said to be between points P and R, written as P-Q-R or R-Q-P. 1.5

P

Q

R



Segment and Ray

Line Segment: The set consisting of points A and B and all the points between A and B is called segment AB, written as seg AB.

A

B

5

Std. IX : Geometry  Note: i. seg AB and seg BA denote the same line segment. ii. The points A and B are called the end points of seg AB. iii. A line segment is a subset of a line. Length of a Line Segment: The distance between the end points of a line segment is called as the length of the segment. It is denoted by (AB). Note: AB = (AB) = d(A, B) Congruent Segments: Two line segments are said to be congruent, if they are of the same length.

A

B

If (AB) = (CD), then we write seg AB  seg CD.

C

D

Note: i. If we have to consider the length of segment AB, we write only AB or l(AB). ii. If we have to consider the set of points between A and B, we write seg AB or side AB. Properties of Congruent Segments: i.

Reflexivity: seg AB  seg AB (Every segment is congruent to itself).

ii.

Symmetry: If seg AB  seg CD, then seg CD  seg AB.

iii.

Transitivity: If seg AB  seg CD and seg CD  seg PQ, then seg AB  seg PQ.

Midpoint of a Segment: The point M is said to be the midpoint of seg AB, if A-M-B and d(A, M) = d(M, B).

M

A

B

1 AM = BM = AB 2



Note: Every line segment has one and only one midpoint. Comparison of Segments: Suppose seg AB and seg CD are given. If AB < CD, then we say that seg AB is smaller than seg CD. This is denoted by seg AB < seg CD. Ray: Suppose A and B are two points, then set of all the points of seg AB together with all the points P on the line AB for which B is between A and P is called ray AB.

A

B

C

A

D

B

P

Note: i. Point A is called as the origin of ray AB. ii. The ray is a subset of a line. Opposite Rays: Two rays which lie on a line having same origin and opposite direction are called opposite rays. ray OA and ray OB are opposite rays. 6

A

O

B

Chapter 01: Lines and Angles Exercise 1.2 1.

Observe the number line in the figure and answer the following questions: E D C B A O P Q R S T

5 4 3 2 1 0 1 2 3 4 5 i. Write the co-ordinates of the points C, S, Q, D. ii. Name the points whose co-ordinates are 4, 5, 0, 2. iii. Find d(Q, T), d(E, B), d(O, C), d(O, R). iv. Name the points which are at a distance of 4 from the point O. Solution: i. Co-ordinates of the points C, S, Q, D are 3, 4, 2 and 4 respectively. ii. The points whose co-ordinates are 4, 5, 0, 2 are S, T, O and B respectively. iii. a. d(Q, T) Co-ordinate of point Q is 2 and co-ordinate of point T is 5. 2 5  d(E, B) = 2 (5) = 2 + 5 = 3 d(E, B) = 3 c. d(O, C) Co-ordinate of point O is 0 and co-ordinate of point C is 3. 0 > 3   d(O, C) = 0  (3) = 0 + 3   d(O, C) = 3 d. d(O, R) Co-ordinate of point O is 0 and co-ordinate of point R is 3. 3>0   d(O, R) = 3  0   d(O, R) = 3 iv. There are two possibilities: a. The point can be towards the positive side i.e. point S -----[ d(O, S) = 4  0 = 4] b.

The point can be towards the negative side i.e. point D

-----[ d(O, D) = 0  (4) = 0 + 4 = 4]

2.

The co-ordinates of two points P and Q are x and y respectively. Find d(P, Q) in the following cases: i. x = 7, y = 10 ii. x = 2, y = 11 iii. x =  8, y = 3 iv. x = 5, y = 9 Solution: i. Co-ordinate of point P is x = 7. Co-ordinate of point Q is y = 10. 10 > 7  d(P, Q) = 10  7 = 3  d(P, Q) = 3 ii.  

Co-ordinate of point P is x = 2. Co-ordinate of point Q is y = 11. 11 >  2 d(P, Q) = 11  (2) = 11 + 2 = 13 d(P, Q) = 13 7

Std. IX : Geometry  iii.

  iv.

 

Co-ordinate of point P is x = 8. Co-ordinate of point Q is y = 3. 3 > 8 d(P, Q) = 3  (8) = 3 + 8 = 5 d(P, Q) = 5 Co-ordinate of point P is x = 5. Co-ordinate of point Q is y = 9. 5 > 9 d(P, Q) = 5  (9) = 5 + 9 = 14 d(P, Q) = 14

3.

In each of the following, decide whether the relation of betweenness exists among the points A, B and D. Name the point which lies between the other two. i. d(A, B) = 5, d(B, D) = 8, d(A, D) = 11 ii. d(A, B) = 11, d(B, D) = 6, d(A, D) = 5 iii. d(A, B) = 2, d(B, D) = 15, d(A, D) = 17 Solution: i. d(A, B) + d(B, D) = 5 + 8 = 13 d(A, D) = 11  d(A, B) + d(B, D)  d(A, D)  The relation of betweenness does not exist among the points A, B and D. ii.  

iii.  

4.

d(B, D) + d(A, D) = 6 + 5 = 11 d(A, B) = 11 d(B, D) + d(A, D) = d(A, B) The relation of betweenness exists among the points A, B and D, such that point D lies between A and B. d(A, B) + d(B, D) = 2 + 15 = 17 d(A, D) = 17 d(A, B) + d(B, D) = d(A, D) The relation of betweenness exists among the points A, B and D, such that point B lies between A and D. Draw the figures according to the given information and answer the questions: i. When A-B-C, (AC) = 12, (BC) = 7.5, then (AB) = ? ii.

When R-S-T, (ST) = 3.75, (RS) = 2.15, then (RT) = ?

iii.

When X-Y-Z, (XZ) = 5 2 , (XY) = 2 2 , then (YZ) = ?

Solution: i. A

B

7.5 12 (AB) + (BC) = (AC)



(AB) + 7.5 = 12



(AB) = 12  7.5



(AB) = 4.5 8

C ----[A-B-C]

Chapter 01: Lines and Angles ii.

R

2.15

T

S 3.75

(RS) + (ST) = (RT) 

2.15 + 3.75 = (RT)



(RT) = 5.90

iii.

X

2 2

Y

----[R-S-T]

Z 5 2

(XY) + (YZ) = (XZ)

----[X-Y-Z]



2 2 + (YZ) = 5 2



(YZ) = 5 2  2 2



(YZ) = 3 2

5.

In the adjoining figure, (LN) = 5, (MN) = 7, (ML) = 6, (NP) = 11, (MR) = 13, (MQ) = 2, then find (PL), (NR), (LQ).

Solution: (PL) +(LN) = (PN) 

P

----[P-L-N]

(PL) + 5 = 11



(PL) = 11  5



(PL) = 6 (MN) + (NR) = (MR)



7 + (NR) = 13



(NR) = 13  7



(NR) = 6 (LM) +(MQ) = (LQ)

11 L 6

----[M-N-R] Q

2 M

N 13

7

R

----[L-M-Q]



6 + 2 = (LQ)



(LQ) = 8

6.

In the adjoining figure, (AC) = 8, (BC) = 5.

Seg BD  seg CE  seg AC, then determine whether the segments in each of the following pairs are congruent or not. i. seg BC and seg DE ii. seg AB and seg CD. Solution: i. seg BD  seg CE  seg AC ----[Given]  (BD) = (CE) = (AC) = 8 ----(i) [  (AC) = 8] 

5

(BC) + (CD) = (BD)

----[B-C-D]

5 + (CD) = 8

----[From (i) and Given]

R

P A

B

C

E

D

Q S

9

Std. IX : Geometry  

(CD) = 8  5



(CD) = 3

----(ii)

(CD) + (DE) = (CE)

----[C-D-E]



3 + (DE) = 8

----[From (i) and (ii)]



(DE) = 8  3



(DE) = 5

----(iii)

But (BC) = 5

----(iv) [Given]



(BC) = (DE)

----[From (iii) and (iv)]



seg BC  seg DE

ii.

(AB) + (BC) = (AC)

----[A-B-C]



(AB) + 5 = 8

----[From (iv) and Given]



(AB) = 8  5



(AB) = 3

----(v)

(AB) = (CD)

----[From (ii) and (v)]



seg AB  seg CD

7.

The co-ordinates of the points on the number line are as follows: Points Co-ordinates

P 3

Q 5

R 2

S 7

T 9

Find the lengths of: seg PQ, seg PR, seg PS, seg PT, seg QR, seg QS, seg QT, seg RS, seg RT, seg ST. Solution: i. Co-ordinate of point P is 3 and co-ordinate of point Q is 5. 5 > 3  d(P, Q) = 5  (3) = 5 + 3 = 8  (PQ) = 8 ii.   iii.   iv.   10

Co-ordinate of point P is 3 and co-ordinate of point R is 2. 2 > 3 d(P, R) = 2  (3) = 2 + 3 = 5 (PR) = 5 Co-ordinate of point P is 3 and co-ordinate of point S is 7. 3 > 7 d(P, S) = 3  (7) = 3 + 7 = 4 (PS) = 4 Co-ordinate of point P is 3 and co-ordinate of point T is 9. 9 > 3 d(P, T) = 9  (3) = 9 + 3 = 12 (PT) = 12

Chapter 01: Lines and Angles v.   vi.   vii.  

Co-ordinate of point Q is 5 and co-ordinate of point R is 2. 5>2 d(Q, R) = 5  2 = 3 (QR) = 3 Co-ordinate of point Q is 5 and co-ordinate of point S is 7. 5>7 d(Q, S) = 5  (7) = 5 + 7 = 12 (QS) = 12 Co-ordinate of point Q is 5 and co-ordinate of point T is 9. 9>5 d(Q, T) = 9  5 = 4 (QT) = 4

viii. Co-ordinate of point R is 2 and co-ordinate of point S is 7. 2 > 7  d(R, S) = 2  (7) = 2 + 7 = 9  (RS) = 9 ix.   x.  

Co-ordinate of point R is 2 and co-ordinate of point T is 9. 9>2 d(R, T) = 9  2 = 7 (RT) = 7 Co-ordinate of point S is 7 and co-ordinate of point T is 9. 9>7 d(S, T) = 9  (7) = 9 + 7 = 16 (ST) = 16

8. If P is the midpoint of seg AB and AB = 7 cm, find AP. Solution: P is the midpoint of seg AB.  d(A, P) = d(P, B) ----(i) d(A, P) + d(P, B) = d(A, B) ----[A-P-B]  d(A, P) + d(P, B) = 7 ----[ AB = 7 cm]  

d(A, P) + d(A, P) = 7 2[d(A, P)] = 7



d(A, P) =



(AP) = 3.5 cm

A

B

P

----[From (i)]

7 = 3.5 2

9. If Q is the midpoint of seg CD and d(C, Q) = 4.5, find the length of CD. Solution: Q is the midpoint of seg CD. C  d(C, Q) = d(Q, D)  d(C, Q) = d(Q, D) = 4.5 ----(i) [ d(C, Q) = 4.5]

Q

D

11

Std. IX : Geometry    

d(C, Q) + d(Q, D) = d(C, D) 4.5 + 4.5 = d(C, D) d(C, D) = 9 (CD) = 9

----[ C-Q-D] ----[From (i)]

10. If AB = 7 cm, BP = 4 cm, AP = 5.4 cm, then compare the segments. Solution: AB = 7 cm, BP = 4 cm, AP = 5.4 cm  4 < 5.4 < 7 

BP < AP < AB

11. In the adjoining figure, lengths of the segments are shown. Write the congruent segments. Solution: A (AB) = (AC) = 5 cm 

seg AB  seg AC (BC) = (DE) = 5.5 cm



seg BC  seg DE

B

4 cm 5.5 cm

D

C

(CD) = (CE) = 4 cm 

seg CD  seg CE

1.6

Plane Separation Axiom

E

Given a line in a plane, the points in the plane that do not lie on the line form two disjoint sets H1 and H2. Each set H1 and H2 is called a half plane and the line is called the edge of the half plane. If P is any point in any of the half plane, then that half plane is called as the P side of the half plane. 1.7

1.8

H2

A

O

B

Angles in terms of Rotation

An angle is obtained by rotating a ray about its end point. The rotation of a ray may be in clockwise direction or in anticlockwise direction. When the rotation of the ray is in clockwise direction, the angle is regarded as negative. [See adjoining figure.]

When the rotation of the ray is in anticlockwise direction, the angle is regarded as positive. [See adjoining figure.] 12

P

Angle

An angle is the union of two rays having the same origin. The origin is called the vertex of the angle and each of the rays is known as arm or side of the angle. AOB and BOA are considered as same angles.

l

H1

Terminal side

Initial side

Chapter 01: Lines and Angles Measure of an angle: The amount of rotation of the ray from its initial position to the terminal position is called the measure of the angle. mAOB =  1.9

B



O

A

Directed Angle

B

The ordered pair of rays (ray OA, ray OB) together with rotation of ray OA to occupy the position of ray OB is called directed angle AOB, denoted by AOB. O

In the ordered pair (ray OA, ray OB), the first element ray OA is called initial arm and second element ray OB is called terminal arm.

A

Initial side

Note:The directed angle BOA is not the same as directed angle AOB. i.e. BOA  AOB

B

Positive angle: Anticlockwise direction of ray OA about ‘O’ is regarded as positive angle. AOB is a positive angle. O

Negative angle: Clockwise direction of ray OA about ‘O’ is regarded as negative angle.

Initial arm

O

A

Initial arm

A

AOB is a negative angle. B

1.10 One Complete Rotation Suppose an initial arm OA is rotated about its end point O in the anticlockwise direction and takes final position OA again for the first time, then ray OA is said to have formed one complete rotation.

O

A

Note: The angle traced during one complete rotation in anticlockwise direction is 360°. The angle traced during two complete rotations in the anticlockwise direction is 2  360 = 720° and so on. 1.11 Zero Angle, Straight Angle, Reflex Angle, Co-terminal Angle Zero Angle: If there is no rotation of initial ray OA, then the directed angle so formed is called a zero angle, i.e. the amount of rotation of ray OA about point O is zero. Straight Angle: If the ray OA and OB of the directed angle AOB are opposite rays, i.e. by rotating the ray OA about O, it occupies the position of the ray OB as shown in the adjoining figure, then the directed angle so formed is called the straight angle. mAOB = 180

B O

A

B

O

A

13

Std. IX : Geometry  Reflex Angle: If the initial ray OA rotates about O in the anticlockwise direction and takes final position OB before coinciding the ray OA again, such that its degree measure lie between 180 and 360 then we get the directed angle which is called a reflex angle. Co-terminal Angle: The directed angles of different measures having same position of initial ray and terminal ray are called co-terminal angles. There are infinitely many directed angles co-terminal with a given directed angles. The measures of co-terminal directed angles differ by an integral multiple of 360.

A

O B

B

O 40

400

A

1.12 Perpendicularity C

The two lines are said to be perpendicular to each other when a right angle is formed at the point of intersection of the two lines. line CD and line AB are perpendicular to each other at the point O and this is writen as line CD  line AB.

A

O

B D

Perpendicularity of Segments and Rays: Two rays or two segments or a ray and a segment are said to be perpendicular to each other, if the lines containing them are perpendicular to each other. C

A

B

O

C

C

D

ray OC  ray OB

A

O

B

A

D

seg CD  line AB

O

B D

line CD  ray OB

Note: Point O is called as the foot of the perpendicular drawn from a Point C to line AB. 1.13 Congruent Angles P

A

If the measures of two angles are equal, then the angles are called congruent angles. Since, mABC = mPQR = 30 

B

30

ABC  PQR

Properties of Congruent Angles: i.

Reflexivity: ABC  ABC (Every angle is congruent to itself).

ii.

Symmetry: If ABC  PQR, then PQR  ABC.

iii.

Transitivity: If ABC  PQR and PQR  XYZ, then ABC  XYZ.

14

C

Q

30 R

Chapter 01: Lines and Angles Inequality of Angles: If the measure of one of the angles is greater than that of the other, then the angle with the greater measure is said to be greater than the angle with smaller measure. Since, mABC = 110 and mPQR = 60  ABC > PQR

P

A

110°

60°

B

Q

C

1.14 Types of Angles

R

A

Acute Angle: If the measure of an angle is less than 90°, then it is called an acute angle.

50°

O

B A

Right Angle: If the measure of an angle is 90°, then it is called a right angle.

O

Obtuse Angle: If the measure of an angle is greater than 90°, then it is called an obtuse angle.

B

A 110° O

Straight Angle: If the measure of an angle is 180°, then it is called a straight angle.

A

B

O

1.15 Pairs of Angles

A

Adjacent Angles: Two angles are called adjacent angles, if they have a common side and their interiors are disjoint. AOB and COB are adjacent angles. Angles in a Linear Pair: Two adjacent angles are said to form a linear pair of angles, if their noncommon sides form a pair of opposite rays. AOB and AOC form angles in a linear pair.

B

B O

C

A

B

O

C

Note: Sum of the measures of the angles in a linear pair is 180°. mAOB + mAOC = 180° Vertically Opposite Angles: Two angles are called vertically opposite angles, if their sides form two pairs of opposite rays. AOD and BOC, AOC and BOD form pairs of vertically opposite angles.

A

C O

D

B

15

Std. IX : Geometry  Complementary Angles: If the sum of the measures of the two angles is 90°, then these angles are called complementary angles. mAOB + mPQR = 30 + 60 = 90°.

A

30 O

O

B

1. Answer the following questions and justify. i. Can two acute angles be complement to each other? Solution: Yes, consider two acute angles 30° and 60°. Their sum is 90°. ii. Can two obtuse angles be complement to each other? Solution: No, consider two obtuse angles 100 and 120 (greater than 90°). Their sum is also greater than 90°. iii. Can two right angles be complement to each other? Solution: No, consider two right angles (equal to 90°). Their sum is 180°  90°. iv. Can two acute angles be supplementary? Solution: No, consider two acute angles 40 and 60 (less than 90°). Their sum is less than 180. v. Can two obtuse angles be supplementary? Solution: No, consider two obtuse angles 120 and 100 (greater than 90°). Their sum is greater than 180°. vi. Can two right angles be supplementary? Solution: Yes, consider two right angles (equal to 90°). Their sum is 180°. vii. Can two adjacent angles be supplementary? Solution: Yes, if the sum of the measures of the two adjacent angles is 180°. viii. Can two adjacent angles be complementary? Solution: Yes, if the sum of the measures of the two adjacent angles is 90°.

16

R P

120°

Exercise 1.3

A 100° B

60°

Q A

Supplementary Angles: If the sum of the measures of the two angles is 180°, then these angles are called supplementary angles. mAOB + mPQR = 120 + 60 = 180°

ix. Can two obtuse angles be adjacent angles? Solution: Yes, AOB and BOC are obtuse and adjacent angles.

P

B

O

120°

C

60° Q

R

Chapter 01: Lines and Angles x. Can an acute angle be adjacent to an obtuse angle? Solution: Yes, AOB is obtuse and AOC is acute. Together they form adjacent angles.

A C 20° 110° O

B

xi. Can two acute angles form a linear pair? Solution: No, the sum of the measures of the two acute angles is not equal to 180° (will be less than 180°). xii. Can two obtuse angles form a linear pair? Solution: No, the sum of the measures of the two obtuse angles is not equal to 180° (will be greater than 180°). xiii. Can two right angles form a linear pair? Solution: Yes, the sum of the measures of the two right angles is equal to 180°. 2.

Measures of some angles are given below. Find the measures of their supplementary angles. i.

60

ii.

138

iii.

 3 5  

th

of a right angle iv.

v. (90 + r) vi. 87 vii. 124 Solution: i. Measure of the supplementary angle = 180  60 = 120.  Measure of the supplementary angle is 120. ii. 

Measure of the supplementary angle = 180  138 = 42 Measure of the supplementary angle is 42°.

iii.

3 3   of a right angle =  90 = 3  18 = 54 5 5  

 

Measure of the supplementary angle = 180  54 = 126 Measure of the supplementary angle is 126.

iv.

Measure of the supplementary angle = 180  (180  r) = 180  180 + r = r Measure of the supplementary angle is r.

(180  r)

viii. 108

th





Measure of the supplementary angle = 180  (90 + r) = 180  90  r = (90  r) Measure of the supplementary angle is (90  r).

vi. 

Measure of the supplementary angle = 180  87 = 93 Measure of the supplementary angle is 93.

vii. 

Measure of the supplementary angle = 180  124 = 56 Measure of the supplementary angle is 56.

v.

viii. Measure of the supplementary angle = 180  108 = 72  Measure of the supplementary angle is 72 17

Std. IX : Geometry  3.

Measures of some angles are given below. Find the measures of their complementary angles. rd

i.

58

ii.

16

iii.

 2  3  of a right angle iv.  

v. (90  r) vi. 78 vii. 68 Solution: i. Measure of the complementary angle = 90  58 = 32  Measure of the complementary angle is 32. ii. 

Measure of the complementary angle = 90  16 = 74 Measure of the complementary angle is 74.

iii.

2 2  90 = 2  30 = 60   of a right angle = 3 3

 

Measure of the complementary angle = 90  60 = 30 Measure of the complementary angle is 30.

iv. 

Measure of the complementary angle = 90  (a + b) = (90  a  b) Measure of the complementary angle is (90  a  b).

v. 

Measure of the complementary angle = 90  (90  r) = 90  90 + r = r Measure of the complementary angle is r

vi. 

Measure of the complementary angle = 90  78 = 12 Measure of the complementary angle is 12.

vii. 

Measure of the complementary angle = 90  68 = 22 Measure of the complementary angle is 22°.

(a + b)

viii. 56

rd

viii. Measure of the complementary angle = 90  56 = 34  Measure of the complementary angle is 34. 4. In the adjoining figure, find the measures of AOC and BOC. Solution: AOC + BOC = 180° ----(Angles in a linear pair)  (3x + 5) + (2x  25) = 180°  3x + 2x + 5  25 = 180  5x  20 = 180°  5x = 180 + 20  5x = 200 200 = 40° 5



x=



Now, AOC = (3x + 5) = (3  40 + 5) = (120 + 5) = 125° BOC = (2x  25) = (2  40  25) = (80  25) = 55° AOC = 125° and BOC = 55°

18

C

(2x  25)°

(3x + 5)° A

O

B

Chapter 01: Lines and Angles In the adjoining figure, COD = 90°, BOE = 72° and AOB is a straight angle. Find the measures of the following angles. AOC, BOC, AOE Solution: AOC = BOE ----[Vertically opposite angles] But, BOE = 72 ----[Given]  C = 72 AOC + BOC = 180 ----[Angles in a linear pair]  72 + BOC = 180  BOC = 180  72  BOC = 108 ----(i) AOE = BOC ----[Vertically opposite angles]  AOE = 108 ----[From (i)]  AOC = 72, BOC = 108 and AOE = 108.

5.

C

A

D O

72 B E

1.16 Geometric Sentences and their Proofs If-then form of Statement: Any statement stated in the ‘if-then’ form is said to be a conditional statement. The part of ‘if-then’ form of statement, which follows, ‘if’ is called antecedent and that which follows ‘then’ is called consequent. Example: General Statement: Two intersecting lines are contained in one plane. Conditional Statement: If two lines intersect each other then they are contained in one plane. Broadly speaking: ‘If p then q.’ If -------p---------antecendent Data (Given) sufficient for consequent

then -----------q ---------Consequent to be proved necessary for antecedent

Converse of a Statement: A statement obtained by interchanging the antecedent and consequent is called the converse of the original statement. Example: General statement: If two lines intersect, then they are in a plane. Converse: If two lines are in a plane, then they intersect each other. Note: If a property is true, then its converse may or may not be true. (For the above example its converse is not true.) Proof: The process of establishing the conclusion by logical reasoning on the basis of axioms and previously proved theorems is called proof. There are two types of proofs. i. Direct proof: If from an antecedent, we reach upto the consequent using axioms or previously proved theorems, then it is called a direct proof. ii. Indirect proof: In this method, we suppose that the consequent is false and proceed logically and arrive at a step which contradicts what is given (antecedent) or some well known fact, then we agree that the consequent is true. Corollary: A property that can be readily derived from a theorem is called a corollary. 19

Std. IX : Geometry  1.17 Parallel Lines



Parallel Lines: Non intersecting coplanar lines are called parallel lines.

m n

line  || line m || line n P

Transversal: A line intersecting two or more coplanar lines in distinct points is called as a transversal.

m

A

n

B

Line P is a transversal intersecting line m, line n and line  at points A, B, C respectively.



C

m



Intercept: A segment cut off on a transversal by two distinct lines is called as an intercept. n AB is an intercept on transversal n. AC is an intercept on transversal .

p

A

B D

C

Angles formed by the transversal Alternate Angles:

A

B

ABC and BCD C

D

A

Interior Angles:

P

B

PBC and QCB C D

Q

Corresponding Angles:

ABC and BPQ

A

A

A B

C

P

Q

B

C

C

Q

Q

P

Properties for Parallel Lines: 1.

2.

20

For two lines (intersected by a transversal) to be parallel, i.

at least a pair of alternate angles should be congruent (alternate angles test).

ii.

at least a pair of corresponding angles should be congruent (corresponding angles test).

iii.

at least a pair of interior angles should be supplementary (interior angles test).

If a transversal intersects two parallel lines, then i.

all pairs of alternate angles are congruent.

ii.

all pairs of corresponding angles are congruent.

iii.

all pairs of interior angles are supplementary.

B P

Chapter 01: Lines and Angles Theorems and Corollaries Theorem 1.1 If two lines intersect each other, then the vertically opposite angles are congruent. Given: line AB and line CD intersect each other at point O. To prove: AOC  BOD and AOD  BOC Proof: line AB and line CD intersect each other at point O.  mAOC + mBOC = 180° ----(i) [Angles in a linear pair] mBOC + mBOD = 180° ----(ii) [Angles in a linear pair]  mAOC + mBOC = mBOC + mBOD ----[From (i) and (ii)]  mAOC = mBOD  AOC  BOD Similarly, it can be proved that AOD  BOC.

A

D O B

C

Theorem 1.2 (Alternate angles test)

n

If a pair of alternate angles formed by a transversal of two coplanar lines is congruent, then the lines are parallel. Given: Transversal n intersects coplanar lines, line  and line m in points P and Q respectively. APQ  PQD To prove: line  || line m

E A

P

B

C

Q

D

Proof: F (Indirect proof) Suppose line  is not parallel to line m. It means that the coplanar lines AB and CD are not parallel. Then, they intersect in say point R. Now, points P, Q, R are non-collinear, so we get PQR. n APQ + RPQ = 180° ----(i) [Angles in a linear pair] E RPQ + PQR + PRQ = 180° ----(ii) [Sum of measures of angles A of a triangle] P  APQ + RPQ = RPQ + PQR + PRQ ----[From (i) and (ii)] B  APQ = PQR + PRQ  APQ > PQR D Q C i.e., APQ > PQD ----[Q-D-R] F This contradicts the given, APQ  PQD. So, our assumption is wrong.  line  || line m.



m

R

Theorem 1.3 (Corresponding angles test) If a pair of corresponding angles formed by a transversal of two coplanar lines is congruent, then the lines are parallel. Given: Transversal n intersects coplanar lines, line  and line m at points P and Q respectively. n

EPB  PQD To prove: line  || line m Proof: EPB  PQD EPB  APQ  PQD  APQ  line  || line m

E

----(i) [Given] ----(ii) [Vertically opposite angles] ----[From (i) and (ii)] ----[Alternate angles test]

A

C

P

Q

B

D



m

F

21

Std. IX : Geometry  Theorem 1.4 (Interior angles test) If a pair of interior angles formed by a transversal of two coplanar lines is supplementary, then the lines are parallel. Given: Transversal n intersects coplanar lines, line  and line m respectively. APQ + PQC = 180° Proof: APQ + PQC = 180° ----(i) [Given] PQC + PQD = 180° ----(ii) [Angles in a linear pair]  APQ + PQC = PQC + PQD ---- [From (i) and (ii)]  APQ = PQD  APQ  PQD 

line  || line m

E

P

A

To prove: line  || line m

n

B

Q

C

D



m

F

----[Alternate angles test for parallel lines]

Theorem 1.5 (Converse of alternate angles test) If the two lines are parallel, then the alternate angles formed by a transversal are congruent. Given: Transversal n intersects line  and line m at points P and Q respectively. line  || line m

n

To prove: APQ  PQD Proof: (Indirect proof) Suppose APQ is not congruent to PQD. Then draw a line r passing through point P such that SPQ  PQD.  line r || line m ----[Alternate angles test for parallel lines] But, line  || line m

 

r

E

S

A

P

B



T C

----[Given]

Q

D

m

F

This means there are two parallel lines, parallel to line m, passing through the same point P which is outside the line m. This contradicts the uniqueness of parallel lines. Our assumption is wrong. APQ  PQD.

Theorem 1.6 (Converse of corresponding angles test) If the two lines are parallel, then the corresponding angles formed by a transversal are congruent. Given:

Transversal n intersects line  and line m at points P and Q respectively. n

line  || line m

E

To prove: EPB  PQD Proof: line  || line m   22

PQD  APQ EPB  APQ EPB  PQD

A

P

B

----[Given] ----(i) [Alternate angles] ----(ii) [Vertically opposite angles] ----[From (i) and (ii)]

C

Q F

D



m

Chapter 01: Lines and Angles Theorem 1.7 (Converse of interior angles test): If two lines are parallel, then the interior angles formed by a transversal are supplementary. Given:

n

Transversal n intersects line  and line m in points P and Q respectively.

E

line  || line m To prove: BPQ + PQD = 180° Proof:  

line  || line m

----[Given]

PQD  APQ BPQ + APQ = 180° BPQ + PQD = 180°

---- (i) [Alternate angles] ----(ii) [Angles in a linear pair] ----[From (i) and (ii)]

A

P

B

C

Q

D

To prove:

q

line , line m, line n are coplanar lines. line  || line m, line  || line n

G A

line m || line n

Construction: Draw a transversal q intersecting lines , m and n in points P, Q and R respectively.

m

F

Corollary 1: If two lines are parallel to the same line, then they are parallel to each other. Given:



C E

B

P Q

D

R

F

 m n

H

Proof: line  || line m and line q is a transversal. GPB  PQD

----(i) [Corresponding angles]

line  || line n and line q is a transversal.  

GPB  QRF PQD  QRF line m || line n

----(ii) [Corresponding angles] ----[From (i) and (ii)] ----[Corresponding angles test for parallel lines]

Corollary 2: If a line coplanar with two parallel lines is perpendicular to one of them, it is also perpendicular to the other. n Given:

line  || line m and line n intersects them at P and Q respectively.

E

line n  line  To prove: Proof:

A

line n  line m

line  || line m and line n is a transversal.

  

EPB  PQD

----(i) [Corresponding angles]

line n  line 

----[Given]

EPB = 90° PQD = 90° line n  line m

----(ii) ----[From (i) and (ii)]

C

B

P

Q

D



m

F

23

Std. IX : Geometry  Corollary 3: If a line is perpendicular to two coplanar lines, then those two lines are parallel to each other. n Given: line n  line    line n  line m To prove: line  || line m Proof: 

line n  line  EPB = 90° line n  line m PQD = 90° EPB  PQD line  || line m

 

E

----[Given]

A

----(i) ----[Given] C ----(ii) ----[From (i) and (ii)] ----[Corresponding angles test for parallel lines]

B

P

D

Q



m

F

Properties of Corresponding, Alternate and Interior Angles When a line intersects two coplanar lines, i. If one pair of corresponding angles is congruent, then the remaining pairs of corresponding angles are congruent. ii. If one pair of alternate angle is congruent, then the other pair of alternate angles is also congruent. iii. If one pair of interior angle is supplementary, then the other pair of interior angles is also supplementary. iv. When two lines are parallel and a transversal intersects then a. A pair of corresponding angles is congruent. b. A pair of alternate angles is congruent. c. A pair of interior angles is supplementary. Exercise 1.4 1.

i. ii. iii. iv. v.

If a transversal intersects two parallel lines, then state the relation between alternate angles. If each of the two lines are parallel to the third line, then what is the relation between them? If a transversal intersects two parallel lines and the corresponding sides of two angles are parallel, then what is the relation between these angles? If AB is a line and P is a point outside it, then how many lines can be drawn through P and parallel to AB? If a transversal intersects two parallel lines such that the ratio between the interior angles on one of its side is 2 : 7, then find the measure of the greater angle.

Solution: i. Both the pairs of alternate angles are congruent. ii. The three lines are parallel to each other. iii. The two angles are congruent. iv. One and only one line can be drawn. v. Let the measures of the interior angles be 2x and 7x. Since, interior angles are supplementary,  2x + 7x = 180  9x = 180    24

x =

180 = 20 9

The measure of the greater angle is 7x = 7  20 = 140°. Measure of the greater angle is 140°.

Chapter 01: Lines and Angles 2.

In the adjoining figure, line  || line m and line p is the transversal.

If r = 20°, then find a : b. Solution: r=a  a = 20°  

Now, a + b = 180 20 + b = 180 b = 180  20 = 160

p r

b

a 20 2 1 = = = 160 16 8 b



a:b=1:8

3.

In the given figure, If line  || line m || line n, then find A.

P 

Solution: line  || line n and seg OC is a transversal. 

   4.

In the adjoining figure, line  || line m and line PQ is the transversal.

If PEB = 70°, find the measures of each of the remaining angles. Solution: AEF  PEB ----[vertically opposite angles]  AEF = 70° ----[PEB = 70]        

PEB + PEA = 180° 70° + PEA = 180° PEA = 180°  70° PEA = 110° BEF  PEA BEF = 110° DFQ  BEF DFQ = 110° CFE  DFQ CFE = 110° CFQ  AEF CFQ = 70° EFD  CFQ EFD = 70°

O 110°

A

----[Angles in a linear pair]

B

m

30

----[Alternate angles] ----[POC = 110]

OCA + ACD = OCD ----[Angle addition property] 30° + ACD = 110° ACD = 110°° ACD = 80° line m || line n and seg AC is a transversal. ACD + CAB = 180° ----[Interior angles] 80° + CAB = 180° CAB = 180  80  CAB = 100° A = 100°

  

m

----[Interior angles]



POC  OCD OCD = 110°



a

----[Vertically opposite angles] ----[ r = 20°]

n

C

D

P 

m

B

A E C

D F Q

----[Vertically opposite angles] ----[Corresponding angles] ----[Vertically opposite angles] ----[Corresponding angles] ----[Vertically opposite angles] 25

Std. IX : Geometry  Problem Set - 1 1.

Points P and Q are on line PS, but not on seg RS. Point Q is a point on ray RS but not on ray SR. Point P is a point of ray SR, but not of ray RS. Draw the figure. Solution: Point Q is on ray RS but not on ray SR.  R-S-Q Point P is on ray SR but not on ray RS. P R S Q  P-R-S Points P and Q are not on seg RS.  P-R-S-Q 2.

In the adjoining figure, A, B, C, D and E are the points of line .

i. Name the opposite rays with point A as origin. D ii. Name the ray opposite to ray BE. iii. Find the intersection of ray CE and ray EC. Solution: i. Ray AB and ray AC are the opposite rays with point A as their origin. ii. Ray BA is the opposite ray of ray BE. iii. Intersection of ray CE and ray EC is seg CE.

C

A

B

E



3.

Given that a point X is between the points Y and Z and the point Y is in between the points R and X. Draw the figure and write the betweenness of the points R, X, Y and Z on the line. Solution: Point X is between the points Y and Z.  YXZ Point Y is between the points R and X. Y X Z R  RYX The betweeness on the line is RYXZ. 4.

In a queue for a bus, 25 persons are standing. How many of them do not stand in between any two of them? How many of those 25 are standing between any two of them? If the number of persons in the queue is M, write down the answers for the above questions. Solution: 1. If there are 25 persons in a queue. i. 2 persons are not standing in between any two of them [Reason: since the first and the last person are not between any two persons] ii. The remaining 23 persons are standing in between any two of them.

2.

5.

If there are M persons in a queue. i. 2 persons are not standing in between any two of them. ii. The remaining (M  2) persons are standing in between any two of them.

The co-ordinate of a point R on a number line is 8. The point S is on the same line which is to the left of R and at a distance of 7 units from R. Find the co-ordinate of S. If P is the midpoint of seg RS, find the co-ordinate of point P. Solution: 3.5 Let the co-ordinate of point S be x. S P R The co-ordinate of point R is 8, 8 x y Since, S is to the left of point R, 7  8>x  d(S, R) = 8  x 26

Chapter 01: Lines and Angles   

7=8x x=87=1 Co-ordinate of point S is 1. P is the midpoint of seg SR.



PR =

1 1 SR = × 7 2 2



PR = 3.5 units Let the co-ordinate of point P be y. The co-ordinate of point R is 8. Since, P lies to the left of point R,  8>y  d(P, R) = 8  y  3.5 = 8  y y = 8  3.5 = 4.5  Co-ordinate of point P is 4.5.



Given A-B-C, the co-ordinate of point A is the least. If d(A, B) = 7.5, d(B, C) = 13.8 and the co-ordinate of point B is 8.5, then find the co-ordinates of points A and C. Solution:    C B A  Let the co-ordinate of point A be x. y x 8.5 The co-ordinate of point B is 8.5. 7.5 13.8 Since, A is to the left of point B  8.5 > x  d(A, B) = 8.5 – x  7.5 = 8.5  x  x = 8.5  7.5 = 1  Co-ordinate of point A is 1. Let the co-ordinate of point C be y. The co-ordinate of point B is 8.5. Since, B lies to the left of point C  y > 8.5  d(B, C) = y  8.5  13.8 = y  8.5  13.8 + 8.5 = y  y = 22.3  Co-ordinate of point C is 22.3. In the given figure, BAC = 75°, ABC = 35°. Find the measures of BAZ, CBX and ACY. Z Solution: A BAZ + BAC = 180° ----[Angles in a linear pair]  BAZ + 75 = 180° 75  BAZ = 180  75  BAZ = 105° 35 B C Y CBX + ABC = 180° ----[Angles in a linear pair]  CBX + 35 = 180° X  CBX = 180  35  CBX = 145° In ABC, BAC + ABC + ACB = 180° ----[Sum of the measures of the angles of a triangle is 180]

7.

27

Std. IX : Geometry        

75 + 35 + ACB = 180° 110° + ACB = 180° ACB = 180°  110° ACB = 70° ACY + ACB = 180° ACY + 70 = 180° ACY = 180  70 ACY = 110°

----[Angles in a linear pair]

In the given figure, XLN = 100°, YMN = 165°, ZNM = 95°. Find the measures of MLN, LMN and LNM. Solution: MLN + XLN = 180° ----[Angles in a linear pair]  MLN + 100° = 180°  MLN = 180°  100°  MLN = 80° LMN + YMN = 180° ----[Angles in a linear pair]  LMN + 165° = 180°  LMN = 180°  165°  LMN = 15° LNM + ZNM = 180° ----[Angles in a linear pair]  LNM + 95° = 180°  LNM = 180°  95°  LNM = 85°

8.

X

L

100 N

M Y

95

165

Z

In the given figure, if ABC = 135°, ABX = 90°, XCD = 55°, BCD = 100°, then determine whether XBC and XCB are congruent to each other. A Solution:

9.

D

       

XBC + ABX = ABC XBC + 90° = 135° XBC = 135°  90° XBC = 45° XCB + XCD = BCD XCB + 55° = 100° XCB = 100  55 XCB = 45° XBC = XCB XBC  XCB

----[Angle addition property] X

----(i) ----[Angle addition property]

B

C

----(ii) ----[From (i) and (ii)]

In the adjoining figure, lines RS, VP and TQ intersect in point O, VOS = 100°, ROT = 122°. Find ROQ, QOP and POS. T V Solution: VOR + VOS = 180° ----[Angles in a linear pair] S R  VOR + 100° = 180°  VOR = 180°  100° O  VOR = 80° ----(i) Q P POS  VOR ----[Vertically opposite angles]  POS = 80° ROT + SOT = 180° ----[Angles in a linear pair]

10.

28

Chapter 01: Lines and Angles        

122° + SOT = 180° SOT = 180°  122° SOT = 58° ROQ  SOT ROQ = 58° VOR + VOT = ROT 80° + VOT = 122° VOT = 122° – 80° VOT = 42° QOP  VOT QOP = 42°

----[Vertically opposite angles] ----[Angle addition property] ----[From (i) and Given]

----[Vertically opposite angles]

In the given figure, if AOC  BOD, AOB = 45°, BOD = 60° and AOB  COD, then find BOC. Solution: AOC  BOD ----[Given]  C = 60° ----[ BOD = 60°]

11.

  

BOC + AOB = AOC BOC + 45° = 60° BOC = 60°  45° BOC = 15°

B

A

C 45 60 O

D

----[Angle addition property]

12. If XPY = 135°, XPZ = 175° and the point Y is in the interior of ZPX, then find YPZ. Solution: Y YPZ + XPY = XPZ ----[Angle addition property]  YPZ + 135° = 175° 175°  YPZ = 175°  135° 135° Z  YPZ = 40° P

X

If POR = 120° and P-O-L, the points S and T be on the R-side of line PL, such that ROS  SOL and ROT  TOP. Draw the figure and find TOS. Solution: T P POR + LOR = 180° ----[Angles in a linear pair]  120° + LOR = 180° 120°  LOR = 180°  120° °  LOR ° ----(i) O °  R ROS  SOL ----[Given]   ray OS bisects LOR.

13.

  

1 LOR 2 1 ROS =  60° 2

S

ROS =



ROS = 30° ROT  TOP ray OT bisects POR.



ROT =

L

----[From (i)] ----[Given]

1  POR 2

29

Std. IX : Geometry  1  120° 2



ROT =



ROT = 60° TOS = ROT + ROS TOS = 60° + 30° TOS = 90°

 

----[Angle addition property]

Ray PR and ray PQ are perpendicular to each other. The point B is in the interior of QPR and the point A is in the exterior of RPQ such that ray PB and ray PA are perpendicular to each other. Draw the figure and write down the pairs of (i) Complementary angles (ii) Supplementary angles (iii) Congruent angles. R Solution: A i. Complementary angles: B APR + RPB = APB = 90° and RPB + BPQ = RPQ = 90°  a. APR and RPB are complementary angles. P Q b. RPB and BPQ are complementary angles.

14.

ii.  iii.

Supplementary angles: APB + RPQ = 90 + 90 = 180° APB and RPQ are supplementary angles. Congruent angles: a. APB  RPQ b. APB = RPQ  APR + RPB = RPB + BPQ  APR = BPQ  APR  BPQ

----[Each is 90°] ----[Given] ----[Angle addition property]

In the given figure, coplanar lines , m and n intersect each other in the same point. p, q, r, s, t and u denote the measures of the angles as shown.  If p = 97°, r = 29°, then find the values of q, s, t and u. Solution: m s=p ----[Vertically opposite angles]  s = 97° u=r ----[Vertically opposite angles]  u = 29° u + p + q = 180° ----[Angles in a linear pair]  29° + 97° + q = 180°  126° + q = 180°  q = 180°  126°  q = 54° t=q ----[Vertically opposite angles]  t = 54° 15.

16.

In the given figure, ray OC bisects AOB. Prove that: COD =

1 [BOD  AOD] 2

Proof: BOC  AOC COD + BOC = BOD 30

p

u t

C

B

D

O

.

n

A

----(i) [Ray OC bisects AOB] ----[Angle addition property]

q s

r

Chapter 01: Lines and Angles     

COD = BOD  BOC COD + AOD = AOC COD = AOC  AOD COD + COD = BOD  BOC + AOC  AOD 2COD = BOD  BOC + BOC  AOD 2COD = BOD  AOD



COD =

---- (ii) ----[Angle addition property] ---- (iii) ----(iv) [Adding (ii) and (iii)] ----[From (i) and (iv)]

1 [BOD  AOD] 2

17. In the given figure, AB || CD. Find the values of x, y and z. Solution: seg AB || seg CD and seg AD is a transversal.  CDA = BAD ----[Alternate angles] 

B

D

x

z

O

y

75°

 

z = 75° seg AB || seg CD and seg BC is a transversal. A ABC = DCB ----[Alternate angles] x = 35° In DCO, D + C + O = 180° ----[Sum of the measures of angles of a triangle is 180°] z + 35 + y = 180° 75° + 35° + y = 180° ----[ z = 75]

 

110° + y = 180° y = 70°

 



C

y = 180°  110°

18.

In the given figure, measure of some angles are given. Which two lines are parallel and which two are not. Justify. Solution: 105° + x = 180° ----[Angles in a linear pair]  x = 180°  105° = 75 Transversal k intersects line  and line m. 

35°

Corresponding angles so formed are congruent. line  || line m

k 75° 105°



x

m

65°

----[Each is 75°] ----[Corresponding angle test

n

for parallel lines] Transversal k intersects line m and line n. Corresponding angles so formed are not congruent. 

Line m is not parallel to line n. Transversal k intersects line  and line n. Corresponding angles so formed are not congruent.



----[ 

----[75  65]

Line  is not parallel to line n.

19. In the given figure , AB || CD. Find the values of x and y. Solution: line AB || line CD and ray QP is a transversal PGH = PEF ----[Corresponding angles]

B P

D

E

x

85°

H



 

x = 85° DHF = QHC DHF = 115°

----[Vertically opposite angles]

G

F R

y

Q 115° C

A

31

Std. IX : Geometry 

 

line AB || line CD and ray QR is a transversal. BFH + DHF = 180° ----[Interior angles] y + 115° = 180° y = 180°  115° y = 65°

20.

In the given figure, a  b, x  y. Prove that line  || line n.



Proof: a and b are the corresponding angles formed by the line  and line m. 



a  b line  || line m

a

----[Given] ---- (i) [Corresponding angles test for parallel lines]

b

m

n

y x

x and y are alternate angles formed by the line m and line n. x  y ----[Given) line n || line m ----(ii) [Alternate angles test for parallel lines] From (i) and (ii), line  || line n ----[If two lines are parallel to the same line, then they are parallel to each other]

In the figure, if x = 70° and y = 71°, state with reason whether the line m || line n or not. Justify. Solution: x = 70° and y = 71° 70  71  x is not congruent to y.  The corresponding angles are not congruent.  Line m is not parallel to line n.

21.

 m x n

y

22.

A In the given figure, measures of two angles are given. If line ED || seg AB 65° and E-C-D, then find the values of x, y and z. Solution: line ED || seg AB and seg AC is a transversal. y x z ACE  BAC ----[Alternate angles] E C  x = 65° line ED || seg AB and seg BC is a transversal. BCD  ABC ----[Alternate angles]  z = 35° In ABC, A + B + C = 180° ----[Sum of the measures of the angles of a triangle is 180°]  65° + 35° + y = 180°  100° + y = 180°  y = 180  100  y = 80°

23. If AB || CD, then find PCD and CPD from the adjoining figure. Solution: CPD + CPA = 180° ----[Angles in a linear pair] CPD + 100° = 180°  CPD = 180°  100°  CPD = 80° ----(i) Ray AB || ray CD and seg AD is a transversal. ADC = BAD ----[Alternate angles] 32

A

B 35°

D

B

56° 100° P

C

D

Chapter 01: Lines and Angles 

ADC = 56°

----[BAD = 56]



PDC = 56° In  PCD, CPD + PDC + PCD = 180° 80° + 56° + PCD = 180° 136° + PCD = 180° PCD = 180°  136° PCD = 44°

----(ii) [A-P-D]

   

----[Sum of the measures of the angles of a triangle is 180°] ----[From (i) and (ii)]

24.

If a transversal intersects two parallel lines, then show that the bisectors of any pair of alternate angles are also parallel. Given: Transversal n intersects line  and line m at points P and Q respectively. line  || line m. n

Ray PM and ray QN bisects APQ and PQD respectively. To prove: ray PM || ray QN Proof: 1 APQ 2 1 NQP = PQD 2

MPQ =

E

----(i) [ray PM bisects APQ]

° P °

A

----(ii) [ray QN bisects PQD]

  

1 1 APQ = PQD 2 2

N

 

----[Alternate angles) ----(iii) [Multiply both sides by



M

line  || line m and line n is a transversal. APQ = PQD

B

1 ] 2

C

Q

D

m

F

MPQ = NQP ----[From (i), (ii), (iii)] These are alternate angles on ray PM and ray QN, when line n is the transversal. ray PM || ray QN ----[Alternate angles test for parallel lines]

25. In the adjoining figure, AB || CD. Find the value of x. C Solution: 100° Construction: Extend ray AB to point S to intersect seg CE in point R. A B AB || CD ----[Given] S R  line AS || ray CD ----[A-B-S] 120° x line AS || ray CD and seg CR is a transversal. SRE = DCR ----[Corresponding angles] P Q E  SRE = 100° BRE + SRE = 180° ----[Angles in a linear pair]  BRE + 100° = 180°  BRE = 180  100  BRE = 80° ----(i) RBE + ABE = 180° ----[Angles in a linear pair]  RBE + 120° = 180°  RBE = 180  120  RBE = 60° ----(ii) In  BER, BRE + RBE + BER = 180° ----[Sum of the measures of the angles of a triangle is 180°]  80° + 60° + x = 180° ----[From (i) and (ii)]  140° + x = 180°  x = 180  140 = 40  x = 40

D

33

Std. IX : Geometry 

1.

Draw figure related to following axiom: Line AB and line AC have one and only one common point. B Solution: A

8.

C

2.

The co-ordinates of the points P and Q are 2 and 0 respectively. Find d(P, Q). Solution: 0 > 2  d(P, Q) = 0  (2) = 0 + 2  d(P, Q) = 2 3.

Measure of an angle is (60  r). Find the measure of its supplementary angle. Solution: Supplementary angle of (60  r) = 180  (60  r) = 180  60 + r = 120 + r 7.

One-Mark Questions

When PQR, (PQ) = 3 2 , (PR) = 5 2 ,

If d(A, B) = 6, d(B, D) = 5, d(A, D) = 11, then state the betweenness among the points A, D and B. Solution: d(A, B) + d(B, D) = 6 + 5 = 11 = d(A, D)  d(A, D) = d(A, B) + d(B, D)  ABD. 9.

Draw a figure showing two adjacent angles as well as complementary angles. Solution:

find (QR).

P

Solution: d(P, R) = d(P, Q) + d(Q, R)  5 2 = 3 2 + d(Q, R)  d(Q, R) = 5 2  3 2  d(Q, R) = 2 2  (QR) = 2 2

----[P-Q-R]

Measure of an angle is (60  r). Find the measure of its complement. Solution: Complement angle of (60  r) = 90  (60  r) = 90  60 + r = 30 + r 4.

5.

Observe the figure and find the opposite ray of ray QR. P

Q

R

S

Solution: Ray QP Point B has co-ordinate 3 on the number line. Find the co-ordinates of the points which are at a distance 6 from B. Solution: Let A and C be the points which are at a distance 6 from B as shown in the figure below. B A C 3  The co-ordinate of A is 3  6 =  9 and the co-ordinate of C is 3 + 6 = 3. 6.

34

S 60 Q

30 R

Ray AD is the bisector of BAC. Name the two angles which are congruent. Solution: BAD and DAC.

10.

11.

Two complementary angles are of equal measures. State their measures. Solution: Measure of each angle is 45. 12.

If initial arm of the directed angle is rotated 20 in anticlockwise direction, then state the measure of this angle. Solution: Measure of angle is 20. 13.

Draw the figure related to statement given below: Two angles are adjacent but they are neither complementary nor supplementary. Solution: A

B

25 30

C D

Chapter 01: Lines and Angles 14.

iii. iv.

Name the alternate angle of i.  PTU ii. RUT m n

P

3.

V Q

4. U

S

W

1.

Measures of some angles are given below. Find the measures of their supplementary angles. i. 45 ii. (90 – x) iii. r iv. 75

2.

Measures of some angles are given below. Find the measures of their complementary angles. i. 35 ii. (a – b)

Additional Problems for Practice

Based on Exercise 1.1  Draw the figure related to the following axiom : i. When two planes intersect, their intersection is exactly one line. ii. When a line intersects a plane but does not lie in it, then their intersection is a point. iii. Infinite number of lines can be drawn through a given point.

2.

Take any five points in a plane such that any 3 points are non-collinear. How many lines can be drawn through these points taking two at a time?

3.

Draw the figure related to the following statement : i. A plane F is determined by two parallel lines l and m. ii. Line AB and a plane E have one and only one common point R. iii. Lines p and q are in a plane E and they have one and only one common point R.

iv.

53

3.

Draw the figures satisfying the following conditions. i. Two angles are adjacent as well as supplementary. ii. Two angles are adjacent but not supplementary. iii. Two angles are supplementary of each other but not adjacent.

4.

The measure of an angle exceeds four times the measure of its supplement by 30. Find its measure.

5.

The measure of an angle is less by 5 than four times the measure of its complement. Find its measure.

6.

In the given figure, find the measures of BOC and BOA. B

The co-ordinates of some points on a line are given in the following table : Points Co-ordinates

P 8

Q 6

R 2

S 5

In each of the following, decide whether the relation of betweenness exists among the points. Name the point which lies between the other two. i. d(P, Q) = 10, d(Q, R) = 3, d(P, R) = 7 ii. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6

(5x + 6)

(13x + 30)

T 1

Find : d(P, Q), d(Q, R), d(Q, S), d(Q, T) 2.

2th of a right angle 5

iii.

Based on Exercise 1.2  1.

If B is the midpoint of seg PQ and (PQ) = 7 cm, find (PB).

Based on Exercise 1.3 

Solution: i. The alternate angle of PTU is TUS. ii. The alternate angle of RUT is QTU.

1.

When X–Y–Z, (XY) = 17, (YZ) = 8 then find (XZ).

T R

d(A, B) = 16, d(C, A) = 9, d(B, C) = 7 d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8

C

7.

O

A

Ray QT is bisector of PQR mPQR = 130. Find mPQT.

and

T P

Q

R

35

Std. IX : Geometry 

Based on Exercise 1.4  1.

2.

6.

If two parallel lines are intersected by a transversal, then prove that the bisectors of any two corresponding angles are parallel.

In the figure, line AB  line CD. Find the values of x, y and z. A 65

In the figure, EAB  DBA. Ray AF and ray BC are bisectors of EAB and ABD respectively. Prove that ray AF  ray BC

7.

z

x

y C

F

E

B

P

110 R

Q

D

In the figure, line AB  line ED and line EF  line DC. Prove that ABC and

B

A

DEF are supplementary angles. A

C

3.

In the figure, AB  CD. Find the value of x. A

F

E

D

B

B

C

D

135 x

O

Multiple Choice Questions

145 D

C

4.

In the figure, m ABC = 65, m DCE = 35, mCEF = 145, m BCE = 30. Prove that ray AB  ray EF. A

B

1.

How many lines can be drawn passing through two distinct points? (A) Two (B) Only two (C) Only one (D) Infinite

2.

There is exactly one plane passing through (A) two distinct points (B) a line and a point on that line (C) three collinear points (D) two distinct intersecting lines

3.

The intersection of two planes is exactly ____. (A) a point (B) a line (C) a plane (D) a line and a point not on the line

4.

In the given figure which of the following set of points are collinear?

65 E

F

145 30 35 D

C

5.

In the figure, side QP and side QR of PQR are produced to points S and T respectively. If SPR = 135, PQT = 110. Find PRQ.

A

S P

H B

36

Q

E

K

I

C

J D

110 T

L

G

F 135

R

(A) (C)

C, K, L, A G, L, K, C

(B) (D)

B, I, J, K D, E, G, F

Chapter 01: Lines and Angles 5.

If the points P, T and A are collinear and d(P, A) = 13, d(T, A) = 5 and d(T, P) = 8, then which of the following can be true? (A) A  P  T (B) T  A  P (C) A  T  P (D) All of the above

6.

The angle traced during one complete rotation in anticlockwise direction is _______. (A) 90 (B) 180 (C) 270 (D) 360

7.

The directed angle POR, as shown in the figure, is also called as P (A) (B) (C) (D)

zero angle straight angle reflex angle co-terminal angle

9.

The directed angle TOR, as shown in the figure, is regarded as T negative angle positive angle O reflex angle co-terminal angle

Terminal arm R

If the directed angles with measures 50, 410, 770 have same initial arm and same terminal arm, then such angles are called (A) reflex angles (B) straight angles (C) zero angles (D) co-terminal angles

11.

Which of the following is not a property of congruent angles? (A) Reflexivity (B) Symmetry (C) Transitivity (D) Inequality

13.

O

(A) (C)

M

50 70

(B) (D)

110 180

15.

If two lines intersect each other then (A) the pair of alternate angles are congruent (B) the pair of vertically opposite angles are congruent (C) the pair of corresponding angles are congruent (D) the pair of interior angles are supplementary

16.

If a transversal intersects two parallel lines, then which of the following is not true? (A) a pair of corresponding angles is congruent (B) a pair of interior angles is complementary (C) a pair of alternate angles is congruent (D) a pair of interior angles is supplementary

Answers to additional Problems for Practice

10.

12.

(2x  30)

L

R

If there is no rotation of initial ray OA, then the directed angle so formed is called as (A) one complete rotation (B) zero angle (C) straight angle (D) reflex angle

In the given figure, if L  O  M, then the measure of LON is N

(2x + 10)

O

8.

(A) (B) (C) (D)

14.

The pair of angles (90  x) and (90 + x) are (A) supplementary (B) complementary (C) congruent (D) vertically opposite angles The complementary angle of 34 is (A) 146 (B) 56 (C) 90 (D) 180

Based on Exercise 1.1  1.

i.

l

F

E

l

ii.

E •P

iii.

r

n

m l

O

37

Std. IX : Geometry  2. 3.

iii.

Ten F

i.

P •

B

Q

l m • C

•A

ii.

E

•R •B

q

R • p

Based on Exercise 1.2  d(P, Q) = 14, d(Q, R) = 4, d(Q, S) = 11, d(Q, T) = 5

2.

i. iii.

3.

25 units

4.

3.5 cm

Yes, P–R–Q Yes, A–C–B

ii. iv.

No Yes, X–Z–Y

Based on Exercise 1.3  1.

i. iii.

135 (180 – r)

ii. iv.

(90 + x) 105

2.

i. iii.

55 54

ii. iv.

(90 – a + b) 37

3.

i. •D

120

60 • C

B P

ii. • T 40 30 Q

150

5.

71

6.

BOC = 134 and BOA = 46

7.

65

3.

80

5.

65

6.

x = 65, y = 115, z = 45 Answers to Multiple Choice Questions

1.

• A

4.

• R

Based on Exercise 1.4 

E

iii.

38

•A

R

1. 5. 9. 13.

(C) (C) (A) (B)

2. 6. 10. 14.

(D) (D) (D) (B)

3. 7. 11. 15.

(B) (C) (D) (B)

4. 8. 12. 16.

(A) (B) (A) (B)