std. 9th perfect mathematics - 2 notes, english medium (mh ... · the study of these topics...

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© Target Publications Pvt. Ltd. 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.

Mathematics Part – II

STD. IX

Printed at: India Printing Works, Mumbai

Written as per the latest syllabus prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune.

P.O. No. 165901 Balbharati Registration No.: 2018MH0022 TEID: 13441

Written as per the new textbook.

Exhaustive coverage of entire syllabus.

Topic-wise distribution of textual questions and practice problems atthe beginning of every chapter.

Covers solutions to all Practice Sets and Problem Sets.

Includes additional activities for practice.

Includes additional problems and MCQs for practice.

Indicative marks for all problems.

Chapter-wise assessment for every chapter.

Includes ‘Challenging Questions (4 Marks)’.

Includes ‘One mark and Two marks questions (from Std. VIII)’.

Constructions drawn with accurate measurements.

Salient Features

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Preparing this ‘Mathematics Part - II’ book was a rollercoaster ride. We had a plethora of ideas, suggestions and decisions to ponder over. However, our basic premise was to keep this book in line with the new, improved syllabus and to provide students with an absolutely fresh material. Mathematics Part - II covers several topics including basic concepts in geometry, logical proofs, trigonometry, co-ordinate geometry and surface area and volume. The study of these topics requires a deep and intrinsic understanding of concepts, terms and formulae. Hence, to ease this task, we present ‘Std. IX: Mathematics Part - II’ a complete and thorough guide, extensively drafted to boost the confidence of students. For better understanding of different types of questions, Topic-Wise Distribution of Textual Questions and Practice Problems have been provided at the beginning of every chapter. Before each Practice Set, short and easy explanation of different concepts with illustrations for better understanding is given. Solutions and Answers to Textual Questions and Examples are provided in a lucid manner. ‘Smart Check’ given to enable the students to cross-check their answers. ‘Apply your knowledge’ covers all the textual activities and projects along with their answers. ‘Multiple Choice Questions’ and ‘Additional Problems for Practice’ include multiple unsolved problems for revision and in the process help the students to sharpen their problem solving skills. ‘Solved examples’ from textbook are also included in the book. Every chapter ends with a ‘Chapter Assessment’. This test stands as a testimony to the fact that the child has understood the chapter thoroughly. ‘Activities for practice’ includes additional activities along with their answers for the students to practice. ‘Challenging Questions’ include questions that fall out of the textbook, yet are core to the concerned subject. These questions would prepare the students for the Four Mark Questions that are supposed to be asked in the examination from outside the textbook. Marks are provided for each and every problem. However, marks mentioned are indicative and are subject to change as per Maharashtra State Board’s discretion. All the diagrams are neat and have proper labelling. The book has a unique feature that all the constructions are as per the scale. Since according to the paper pattern, weightage is accorded to the Std. VIII syllabus, we have aptly included them in a separate chapter i.e. ‘One mark and Two marks questions (from Std. VIII)’. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! From, Publisher Edition: Second

Disclaimer This reference book is transformative work based on 'Mathematics Part - II; First Reprint: 2018' published by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

No. Topic Name Page

No. No. Topic Name Page

No. 1 Basic Concepts in Geometry 1 7 Co-ordinate Geometry 167 2 Parallel Lines 19 8 Trigonometry 189 3 Triangles 42 9 Surface Area and Volume 209 4 Constructions of Triangles 82 Challenging Questions (4 Marks) 229 5 Quadrilaterals 101 1 Mark and 2 Marks

Questions (from Std. VIII) 248 6 Circle 137

PREFACE

Content

Note: Solved examples from textbook are indicated by “+”. Steps of construction are provided in Chapters for the students’ understanding.

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Chapter 2: Parallel Lines

Type of Problems Practice Set Q. Nos.

Parallel lines, interior angle theorem, corresponding angle theorem, alternate angle theorem

2.1 Q.1, 2, 3, 4, 5 Practice Problems

(Based on Practice Set 2.1) Q.1, 2, 3, 4, 5, 6

Problem Set- 2 Q.3, 5, 6, 7

Test for parallel lines (Interior angles test, alternate angles test, corresponding angles test)

2.2 Q.1, 2, 3, 4, 5, 6 Practice Problems

(Based on Practice Set 2.2) Q.1, 2, 3, 4, 5

Problem Set- 2 Q.4, 8 Complementary angles, Supplementary angles Problem Set- 2 Q.2

1. Parallel lines:

Lines in the same plane which do not intersect each other are called parallel lines.

In the given figure, line m || line n. 2. Transversal: If a line intersects two lines in two distinct points, then that line is

called a transversal of those two lines. In the given figure, line l intersects lines m and n at points P and

Q respectively. Hence, line l is a transversal of line m and line n. Angles formed by two lines and their transversal: 1. Corresponding angles: If the arms on the transversal of a pair of angles are in the same

direction and the other arms are on the same side of the transversal, then it is called a pair of corresponding angles.

In the adjacent figure, a and b is a pair of corresponding angles. 2. Alternate angles: A pair of angles which are on the opposite sides of the transversal

such that their arms on the transversal are in opposite directions is called a pair of alternate angles.

In the adjacent figure, a and b is a pair of alternate interior angles, and d and e is pair of alternate exterior angles. 3. Interior angles:

A pair of angles which are on the same side of the transversal and lie between the two given lines is called a pair of interior angles.

In the adjacent figure, a and b is a pair of interior angles on the same sides of the transversal.

e

d

Let’s Recall

Parallel Lines2

l

P m

Qn

m

n

a

b

b

a

b

a

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Std. IX: Maths (Part - II)

1. Angles formed by two lines and their transversal. (Textbook pg. no. 13) When a transversal (line n) intersects two lines (line l and m) in two distinct points, 8 angles are formed as

shown in the figure. Pairs of angles formed out of these angles are as follows: Pairs of corresponding angles

Pairs of alternate interior angles

Pairs of interior angles on the same side of the transversal i. d, h i. c, e

ii. a, e ii. b, h i. c, h iii. c, g Pairs of alternate exterior

angles ii. b, e

iv. b, f i. d, f ii. a, g Some important properties: 1. When two lines intersect, the pairs of vertically opposite angles formed are congruent. Example: In the given diagram, line l and m intersect at point P. The pairs of vertically opposite angles that are congruent are: i. a b ii. c d 2. The angles in a linear pair are supplementary. Example: For the given diagram, a and c are in linear pair a + c = 180 Also, d and b are in linear pair d + b = 180 3. When one pair of corresponding angles is congruent, then all the

remaining pairs of corresponding angles are congruent. Example: In the given diagram, If a b then e f, c d and g h 4. When one pair of alternate angles is congruent, then all the

remaining pairs of alternate angles are congruent. Example: For the given diagram, If e d, then g b Also, a h and c f 5. When one pair of interior angles on one side of the transversal is

supplementary, then the other pair of interior angles is also supplementary.

Example: For the given diagram, If e + b = 180, then g + d = 180

f

n

l

mg e h

b c a d

Try This

ac

b d

l m

P

f h d b

e g c a l

m

n

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Theorem: If two parallel lines are intersected by a transversal, the interior angles on either side of the

transversal are supplementary. Given: line l || line m, and line n is their transversal.

a, b are interior angles formed on one side and c, d are interior angles formed on the other side of the transversal.

To prove: a + b = 180 c + d = 180 Proof: Indirect proof There are three possibilities with respect to sum of measures of a and b i. a + b < 180 ii. a + b > 180 iii. a + b = 180 Case I: Let us assume that a + b < 180 is true. Since a + b < 180,

then according to Euclid’s postulate

If the lines l and m are produced, they will intersect each other on the side of transversal where a and b are formed.

But, line l || line m, [Given] line l and line m do not intersect a + b < 180 is not possible. (i) Case II: Let us assume a + b > 180 is true. a + b > 180 Here, a + d = 180 (ii)

[Angles in a linear pair] and c + b = 180 (iii) a + d + c + b = 180 + 180 = 360 [Adding (ii) and (iii)] c + d = 360 – (a + b) (iv) But, a + b > 180 [360 – (a + b)] < 180 (v) c + d < 180 [From (iv) and (v)] If the lines l and m are produced, they will intersect

each other on the side of transversal where c and d are formed.

[Euclid’s postulate]

But, line l || line m, [Given] line l and line m do not intersect c + d < 180 is not posssible. i.e. a + b > 180 is not possible (vi) a + b = 180 is the only possibility. [From (i) and (vi)]

a + b = 180 and c + d = 180 Note: In this proof, because of the contradictions we have denied the possibilities a + b >180° and

a + b <180°. Therefore, this proof is an example of indirect proof.

n

l

m

b c a d

Interior angle theorem

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Theorem: The corresponding angles formed by a transversal of two parallel lines are of equal measure. Given: line l || line m, and line n is the transversal. To prove: a = b Proof: a + c = 180 (i) [Angles in a linear pair] line l || line m, and line n is their

transversal [Given]

b + c = 180 (ii) [Interior angle theorem] a + c = b + c [From (i) and (ii)] a = b Theorem: The alternate angles formed by a transversal of two parallel lines are of equal measures. Given: line l || line m, and line n is the transversal. To prove: d = b Proof: d + c = 180 (i) [Angles in a linear pair] line l || line m, and line n is their

transversal [Given]

c + b = 180 (ii) [Interior angle theorem] d + c = c + b [From (i) and (ii)] d = b Example: In the given figure, line m || line n. Find the measures of a,

b, c, d and e from the given measure of angle. Solution: i. a + 75 = 180 [Angles in a linear pair] a = 180 – 75 a = 105

ii. b = 75

(i) [Vertically opposite angles]

iii. line m || line n and line l is their transversal.

d = b [Alternate angles] d = 75 [From (i)] iv. line m || line n and line l is their

transversal.

b + c = 180 [Interior angles] 75 + c = 180 [From (i)] c = 180 – 75 c = 105

v. line m || line n and line l is their transversal.

e = b [Corresponding angles] e = 75 [From (i)]

75a b

m n

cd e

l

Alternate angles theorem

Corresponding angles theorem

l

m b c a

n

l

m

d

n

b c

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1. In the given figure, line RP || line MS and line DK is their transversal.

DHP = 85°. Find the measures of following angles. [3 Marks] i. RHD ii. PHG iii. HGS iv. MGK Solution: i. DHP = 85 (i) [Given] DHP + RHD = 180 [Angles in a linear pair] 85 + RHD = 180 RHD = 180 – 85 RHD = 95 (ii) ii. PHG = RHD [Vertically opposite angles] PHG = 95 [From (ii)] iii.

line RP || line MS and line DK is their transversal.

HGS = DHP [Corresponding angles] HGS = 85 (iii) [From (i)] iv. MGK = HGS [Vertically opposite angles] MGK = 85 [From (iii)] 2. In the given figure, line p || line q and line l and line m are

transversals. Measures of some angles are shown. Hence find the

measures of a, b, c, d. [3 Marks] Solution: i. 110 + a = 180 [Angles in a linear pair] a = 180 – 110 a = 70 ii. consider e as shown in the figure line p || line q, and line l is their transversal. e + 110 = 180 [Interior angles] e = 180 – 110 e = 70 But, b = e [Vertically opposite angles] b = 70 iii. line p || line q, and line m is their

transversal.

c = 115 [Corresponding angles]

Practice Set 2.1

H R P 85

M S

D

G

K

p

c

a 110

115

b

l

m

q

d

e

p

c

a 110

115

b

l

m

q

d

There is more than one method of finding measure of these angles. Example: If you have DHP = 85 PHG + 85 = 180 …[Angles in linear pair] PHG = 180 85 PHG = 95 Now, HGS = DHP …[Corresponding angles] HGS = 85 Now, MGK = DHP …[Alternate exterior angles] MGK = 85

Smart Check

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iv. 115 + d = 180 [Angles in a linear pair] d = 180 – 115 d = 65 3. In the given figure, line l || line m and line n || line p.

Find a, b, c from the given measure of an angle. [3 Marks] Solution: i. consider d as shown in the figure line l || line m, and line p is their

transversal.

d = 45 [Corresponding angles] Now, d + b = 180 [Angles in a linear pair] 45 + b = 180 b = 180 – 45 b = 135 (i) ii. a = b [Vertically opposite angles] a = 135 [From (i)] iii. line n || line p, and line m is their

transversal.

c = b [Corresponding angles] c = 135 [From (i)] 4. In the given figure, sides of PQR and XYZ are parallel to each

other. Prove that, PQR XYZ. [3 Marks] Given: Ray YZ || ray QR and ray YX || ray QP To prove: PQR XYZ Construction: Extend ray YZ in the opposite direction. It intersects ray QP at point S. Proof: Ray YX || ray QP [Given] Ray YX || ray SP and seg SY is their

transversal [P–S–Q]

XYZ PSY (i) [Corresponding angles] ray YZ || ray QR [Given] ray SZ || ray QR and seg PQ is their

transversal. [S–Y–Z]

PSY SQR [Corresponding angles] PSY PQR (ii) [P–S–Q] PQR XYZ [From (i) and (ii)] 5. In the given figure, line AB || line CD and line PQ is transversal. Measure

of one of the angles is given. Hence find the measures of the following angles. [3 Marks]

i. ART ii. CTQ iii. DTQ iv. PRB

A B 105

C D

P

T

Q

R

Q

P

Z

R

X

Y

Q

P

Z

R

X

S Y

n

m

l

b c

45

a

p

d

n

m

l

b c

45

a

p

Finding angles using other methods. b = a = 70 …[Alternate exterior angles]

Smart Check

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Solution: i. BRT = 105 (i) [Given] ART + BRT = 180 [Angles in a linear pair] ART + 105 = 180 ART = 180 – 105 ART = 75 (ii) ii. line AB || line CD and line PQ is their

transversal.

CTQ = ART [Corresponding angles] CTQ = 75 [From (ii)] iii.

line AB || line CD and line PQ is their transversal.

DTQ = BRT [Corresponding angles] DTQ = 105 [From (i)] iv PRB = ART [Vertically opposite angles] PRB = 75 [From (ii)] Theorem: The sum of the measures of all the angles of a triangle is 180. Given: ABC is any triangle. To prove: ABC + ACB + BAC = 180. Construction: Draw a line parallel to seg BC and passing through A. On the line take points P and Q such that, P A Q. Proof: line PQ || side BC and seg AB is their

transversal. [Construction]

ABC = PAB (i) [Alternate angles] line PQ || side BC and seg AC is their

transversal. [Construction]

ACB = QAC (ii) [Alternate angles] Adding (i) and (ii), we get ABC + ACB = PAB + QAC ABC + ACB + BAC

[Adding BAC to both sides] = PAB + QAC + BAC = PAB + BAC + QAC = PAC + QAC [ PAB + BAC = PAC]

= 180 [Angles in a linear pair] ABC + ACB + BAC = 180 The sum of measures of all angles of a

triangle is 180.

1. In the given figure, how will you decide whether line l and line m are

parallel or not? (Textbook pg. no. 19) Ans: In the figure, we observe that line l and line m are coplanar and do not

intersect each other. Line l and line m are parallel lines.

l

m

Try This

Use of properties of parallel lines

Let’s Study

A P Q

B C

A

B C

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Whether given two lines are parallel or not can be decided by examining the angles formed by a transversal of the lines. i. If the interior angles on the same side of a transversal are supplementary, then the lines are parallel. ii. If one of the pairs of alternate angles is congruent, then the lines are parallel. iii. If one of the pairs of corresponding angles is congruent, then the lines are parallel. Theorem: If the interior angles formed by a transversal of two distinct lines are

supplementary, then the two lines are parallel. Given: line XY is the transversal of line AB and line CD. BPQ + PQD = 180 To prove: line AB || line CD Proof: Indirect proof Suppose line AB is not parallel to line

CD.

They intersect at point T (say). Since, points P, Q, T are not collinear,

they form a triangle.

In PQT, TPQ + PQT + PTQ = 180° [Sum of the measures of the angles of a triangle is 180] (BPQ + PQD) + PTQ = 180 [P – B – T and Q – D – T] 180 + PTQ = 180 [Given] PTQ = 0 Lines PT and QT are not distinct lines

i.e., lines AB and CD are not distinct lines

This contradicts the given, that line AB and CD are two distinct lines.

So, our assumption is wrong. line AB || line CD Theorem: If a pair of alternate angles formed by a transversal of two lines is congruent, then the two lines

are parallel. Given: Line n is the transversal of line l and line m. a and b is a congruent pair of alternate angles. i.e., a = b To prove: line l || line m Proof: a + c = 180 (i) [Angles in a linear pair] a = b (ii) [Given] b + c = 180 [From (i) and (ii)] But b and c are interior angles on

lines l and m when line n is the transversal.

line l || line m [Interior angles test]

Alternate angles test

Tests for parallel lines

Interior angles test

A B

C DQ

Y

P

X

Y

A P

X

B T

D C Q

n

l

m b

c a

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Theorem: If a pair of corresponding angles formed by a transversal of two lines is congruent, then the two

lines are parallel. Given: Line n is the transversal of line l and line m. a and b is a congruent pair of corresponding angles. i.e., a = b To prove: line l || line m Proof: a + c = 180 (i) [Angles in a linear pair] a = b (ii) [Given] b + c = 180 [From (i) and (ii)] But b and c are interior angles on

lines l and m when line n is the transversal.

line l || line m [Interior angles test] Corollary I: If a line is perpendicular to two lines in a plane, then the two lines are parallel to each other. Given: Line n line l, and line n line m To prove: line l || line m Proof: line n line l [Given] a = 90° (i) line n line m [Given] c = 90° (ii) a = c [From (i) and (ii)] But a and c are corresponding

angles on lines l and m when line n is the transversal.

line l || line m [Corresponding angles test] Corollary II: If two lines in a plane are parallel to a third line in the plane, then those two lines are parallel

to each other. Given: Line l, line m and line n are coplanar lines. line l || line m, and line l || line n To prove: line m || line n Construction: Draw a transversal q intersecting lines l, m and n Proof: line l || line m and line q is their

transversal.

a = b (i) [Corresponding angles] line l || line n and line q is their

transversal.

a = c (ii) [Corresponding angles] b = c [From (i) and (ii)] But b and c are corresponding

angles on lines m and n when line q is their transversal.

line m || line n [Corresponding angles test]

c

q

l

m

n

a

b

n

l

m b

c a

n

a

c

l

m

Corresponding angles test

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1. In the given figure, y = 108 and x = 71. Are the lines m and n parallel? Justify? [1 Mark] Solution: y = 108, x = 71 …[Given] x + y = 71 + 108 = 179 x + y 180 The angles x and y are not supplementary. The angles do not satisfy the interior angles test for parallel lines line m and line n are not parallel lines. 2. In the given figure, if a b then prove that line l || line m. [2 Marks] Given: a b To prove: line l || line m Proof: consider c as shown in the figure a c (i) [Vertically opposite angles] But, a b (ii) [Given] b c [From (i) and (ii)] But, b and c are corresponding

angles on lines l and m when line n is the transversal.

line l || line m [Corresponding angles test] 3. In the given figure, if a b and x y, then prove that line l || line n. [3 Marks] Given: a b and x y To prove: line l || line n Proof: a b [Given] But, a and b are corresponding

angles on lines l and m when line k is the transversal.

line l || line m (i) [Corresponding angles test] x y [Given] But, x and y are alternate angles on

lines m and n when seg PQ is the transversal.

line m || line n (ii) [Alternate angles test] From (i) and (ii), line l || line m || line n i.e., line l || line n 4. In the given figure, if ray BA || ray DE, C = 50° and

D = 100°. Find the measure of ABC. [3 Marks] (Hint: Draw a line passing through point C and parallel to line AB.) Solution: Draw a line FG passing through point

C and parallel to line AB

line FG || ray BA (i) [Construction]

Practice Set 2.2

l

m

n

x y

m l

b a

x y

n

k m l

a k b

m

P a

x y

n

k

Q

c

b

a

n

l

m

100°

D E

C

50 A B

b

a

n

l

m

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E

AB

F

DC

x x

y y

A B

F

C

x y

Ray BA || ray DE (ii) [Given] line FG || ray BA || ray DE (iii) [From (i) and (ii)] line FG || ray DE [From (iii)] and seg DC is their transvensal DCF = EDC [Alternate angles] DCF = 100 [ D = 100]

Now, DCF = BCF + BCD [Angle addition property] 100 = BCF + 50 100 – 50 = BCF BCF = 50 (iv) Now, line FG || ray BA and seg BC is

their transversal.

ABC + BCF = 180 [Interior angles] ABC + 50 = 180 [From (iv)] ABC = 180 50 ABC = 130 5. In the given figure, ray AE || ray BD, ray AF is the bisector

of EAB and ray BC is the bisector of ABD. Prove that line AF || line BC. [3 Marks]

Given: Ray AE || ray BD, and ray AF and ray BC are the bisectors of EAB and ABD respectively. To prove: line AF || line BC Proof:

Ray AE || ray BD and seg AB is their transversal.

EAB = ABD (i) [Alternate angles]

FAB = 12EAB

[Ray AF bisects EAB]

2FAB = EAB (ii)

CBA = 12ABD [Ray BC bisects ABD]

2CBA = ABD (iii) 2FAB = 2CBA [From (i), (ii) and (iii)] FAB = CBA

But, FAB and ABC are alternate angles on lines AF and BC when seg AB is the transversal.

line AF || line BC [Alternate angles test] 6. A transversal EF of line AB and line CD intersects the lines

at points P and Q respectively. Ray PR and ray QS are parallel and bisectors of BPQ and PQC respectively.

Prove that line AB || line CD. [3 Marks] Given: Ray PR || ray QS Ray PR and ray QS are the bisectors of BPQ and PQC respectively. To prove: line AB || line CD

E

F

P

B

D

A

C

S R

Q

100°

D E

C

50

F G

A B

C

A B

F

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Proof: Ray PR || ray QS and seg PQ is their

transversal.

RPQ = SQP (i) [Alternate angles] RPQ = 1

2BPQ (ii) [Ray PR bisects BPQ]

SQP = 12PQC (iii) [Ray QS bisects PQC]

12BPQ = 1

2PQC [From (i), (ii) and (iii)]

BPQ = PQC But, BPQ and PQC are alternate

angles on lines AB and CD when line EF is the transversal.

line AB || line CD [Alternate angles test] 1. Select the correct alternative and fill in the blanks in the following statements. [1 Mark each] i. If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal

is _______. (A) 0 (B) 90 (C) 180 (D) 360 ii. The number of angles formed by a transversal of two lines is _______. (A) 2 (B) 4 (C) 8 (D) 16 iii. A transversal intersects two parallel lines. If the measure of one of the angles is 40°, then the measure of its

corresponding angle is _______ (A) 40 (B) 140 (C) 50 (D) 180 iv. In ABC, A = 76°, B = 48°, then C = _______. (A) 66 (B) 56 (C) 124 (D) 28 v. Two parallel lines are intersected by a transversal. If measure of one of the alternate interior angles is 75°

then the measure of the other angle is _______. (A) 105 (B) 15 (C) 75 (D) 45 Answers: i. (C) ii. (C) iii. (A) iv. (B) v. (C) Hints: iv. In ABC, A + B + C = 180 C = 180 – 76 – 48 = 56 2. Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of

QPR respectively. Ray PB and ray PA are perpendicular to each other. Draw a figure showing all these rays and write - i. A pair of complementary angles [1 Mark] ii. A pair of supplementary angles [1 Mark] iii. A pair of congruent angles. [2 Marks] Solution:

Problem Set – 2

P Q

R A

B

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i. Complementary angles: RPQ = 90 [Ray PQ ray PR] RPB + BPQ = 90 [Angle addition property] RPB and BPQ are pair of

complementary angles

APB = 90 [Ray PA ray PB] APR + RPB = 90 APR and RPB are pair of

complementary angles

ii. Supplementary angles: APB + RPQ = 90 + 90 = 180° APB and RPQ are a pair of

supplementary angles.

iii. Congruent angles: a. APB RPQ [Each is of 90°] b. APB = RPQ APR + RPB = RPB + BPQ [Angle addition property] APR = BPQ APR BPQ 3. Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the

other line also. [3 Marks] Given: line AB || line CD and line EF intersects them at P and Q respectively. line EF line AB To prove: line EF line CD Proof: line EF line AB [Given] EPB = 90° (i) line AB || line CD and line EF is their

transversal.

EPB PQD (ii) [Corresponding angles] PQD = 90° [From (i) and (ii)] line EF line CD 4. In the given figure, measures of some angles are shown. Using

the measures find the measures of x and y and hence show that line l || line m. [2 Marks]

Proof: x = 130

[Vertically opposite angles] y = 50 Here, mPQT + mQTS = 130 + 50 = 180 But, PQT and QTS are a pair of

interior angles on lines l and m when line n is the transversal.

line l || line m [Interior angles test]

A B

C D

E

P

Q

F

y

x

m

130

50

l

n

Q P

S T

y

x 130

50

l

n

m

SAMPLE C

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32


5. In the given figure, Line AB || line CD || line EF and line QP is their

transversal. If y : z = 3 : 7 then find the measure of x. [3 Marks] Solution: y : z = 3 : 7 [Given] Let the common multiple be m y = 3m and z = 7m (i)

line AB || line EF and line PQ is their transversal [Given]

x = z [Alternate angles] x = 7m (ii) [From (i)]

line AB || line CD and line PQ is their transversal [Given]

x + y = 180 [Interior angles] 7m + 3m = 180 [From (i) and (ii)] 10m = 180 m = 18 x = 7m = 7(18) [From (ii)] x = 126 6. In the given figure, if line q || line r, line p is their transversal and if

a = 80°, find the values of f and g. [2 Marks] Solution: i. a = 80 [Given] g = a [Alternate exterior angles] g = 80 (i) ii.

Now, line q || line r and line p is their transversal.

f + g = 180 [Interior angles] f + 80 = 180 [From (i)] f = 180 80 f = 100 7. In the given figure, if line AB || line CF and line BC || line ED then

prove that ABC = FDE. [3 Marks] Given: line AB || line CF and line BC || line ED To prove: ABC = FDE Proof: line AB || line PF and line BC is their

transversal.

ABC = BCD (i) [Alternate angles] line BC || line ED and line CD is their

transversal.

BCD = FDE (ii) [Corresponding angles] ABC = FDE [From (i) and (ii)]

A B

C D

E F

Q

x

z

P

y

P

E

F

D

C B

A

p

r

q

e f d c

h g

a b

SAMPLE C

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33


8. In the given figure, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that QXRY is a rectangle.

[4 Marks] Given: line AB || line CD Rays QX, RX, QY, RY are the bisectors of AQR, QRC, BQR and QRD respectively. To prove: QXRY is a rectangle. Proof: XQA = XQR = x (i) [Ray QX bisects AQR] YQR = YQB = y (ii) [Ray QY bisects BQR] XRQ = XRC = u (iii) [Ray RX bisects CRQ] YRQ = YRD = v (iv) [Ray RY bisects DRQ]

line AB || line CD and line PS is their transversal.

AQR + CRQ = 180 [Interior angles]

(XQA + XQR) + (XRQ + XRC)

= 180 [Angle addition property]

(x + x) + (u + u) = 180 [From (i) and (iii)] 2x + 2u = 180 2(x + u) = 180 x + u = 90 (v) In XQR XQR + XRQ + QXR = 180 [Sum of the measures of the angles of triangle is 180] x + u + QXR = 180 [From (i) and (iii)] 90 + QXR = 180 [From (v)] QXR = 180 90 QXR = 90 (vi) Similarily we can prove that, y + v = 90 Hence QYR = 90 (vii) Now, AQR + BQR = 180 [Angles is linear pair]

(XQA + XQR) + (YQR + YQB)

= 180 [Angle addition property]

(x + x) + (y + y) = 180 [From (i) and (ii)] 2x + 2y = 180 2(x + y) = 180 x + y = 90 i.e. XQR + YQR = 90 [From (i) and (ii)] XQY = 90 (viii) [Angle addition property] Similarly we can prove that, XRY = 90 (ix) In QXRY QXR = QYR = XQY = XRY = 90 [From (vi), (vii), (viii) and (ix)] QXRY is a rectangle.

A

P

B

C

Q

D S

R

X Y

A

P

B

C

Q

D

S

R

X

x x

u u

A

P

B

C

Q

D S

R

Y y

y

v v

A B Q

R X Y

x x y

y

SAMPLE C

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34


1. To verify the properties of angles formed by a transversal of two parallel lines. (Textbook pg. no. 14) Take a piece of thick coloured paper. Draw a pair of parallel lines and a transversal on it. Paste straight

sticks on the lines. Eight angles will be formed. Cut pieces of coloured paper, as shown in the figure, which will just fit at the corners of 1 and 2. Place the pieces near different pairs of corresponding angles, alternate angles and interior angles and verify their properties.

[Students should attempt the above activity on their own.] 1. In the given figure, line l || line m and line n || line p. To find d, b, a from the given measure of an angle complete

the following activity. Solution: line l || line m, and line p is their

transversal.

d = [Corresponding angles]

line n || line p, and line l is their transversal.

b = [Corresponding angles]

Now, d + a = 180 [Angles in a linear pair]

+ a = 180

a =

2. In the adjoining figure, sides of PQR and XYZ are parallel to each other. Prove that, PQR XYZ.

Given: Ray YZ || ray QR and ray YX || ray QP To prove: PQR XYZ Construction: Extend ray YZ in the opposite direction. It intersects ray QP at point S. Proof: Ray YX || ray QP [Given]

Ray YX || ray SP and seg SY is their transversal [P–S–Q]

(i) [Corresponding angles] ray YZ || ray QR [Given]

ray SZ || ray QR and is their


PSY SQR

PSY (ii) [P–S–Q]

PQR XYZ [From (i) and (ii)]

1 2 2 1

Apply your knowledge

Activities for Practice

Q

P

Z

R

X

Y

d

n

m

l b 50

a

p

SAMPLE C

ONTENT

35


3. In the given figure, if line m || line n and line p is a transversal, then find x. Solution:

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

BFG + = 180

= 180

x = 4. In the adjoining figure, if line q || line r, line p is their transversal. Find values of

c, e and h. Solution: i. c = [Vertically opposite angles]


e = [Alternate angles]

e = ii. Also, h + e = 180 [Angles in a linear pair]

h = 1. line l || line m, and line p is their

transversal.

d = 50 [Corresponding angles]

line n || line p, and line l is their transversal.

b = 50 [Corresponding angles] Now, d + a = 180 [Angles in a linear pair] 50 + a = 180 a = 180 – 50

a = 130

2. Ray YX || ray QP [Given] Ray YX || ray SP and seg SY is their




PSY SQR Corresponding angles

PSY PQR (ii) [P–S–Q] PQR XYZ [From (i) and (ii)]

Q

P

Z

R

X

S Y

Answers

d

n

m

l b 50

a

p

p

r

q

e f d c

h g

70 b

p

m

n

E

G

3x x

D C

A

H

F B

SAMPLE C

ONTENT

36


3. line m || line n and line p is a transversal.

BFG + FGD = 180 Interior angles

3 x x = 180 4x = 180 x = 180

4

= 45

4. i. c = 70 [Vertically opposite angles]

Now, line q || line r and line p is theirtransversal.

e = c [Alternate angles] e = 70

ii. Also, h + e = 180 [Angles in a linear pair] h + 70 = 180 h = 180 70 h = 110

[1 Mark each]

1. In the given figure, which of the following is apair of alternate exterior angles?

(A) a, f (B) e, j

(C) d, e (D) f, k

2. In the figure in Q.1 which of the following isnot a pair of corresponding angles?(A) c, e (B) q, h(C) c, k (D) d, h

3. In the figure in Q.1, if line l || line m || line n,then which of the following options is correct?(A) d + e = 180

(B) h + j = 180

(C) g + f = 180

(D) i + h = 180

4. If the interior angles formed by a transversalof two distinct line are _______, then the twolines are parallel.(A) congruent(B) complementary(C) supplementary(D) none of these.

5. If a and b are angles in a linear pair, anda and d are corresponding angles of twoparallel lines, then b + d =

(A) 2b (B) 90

(C) 180 (D) < 180

6. The sum of the measures of any two alternateangles is equal to _______ times the measureof any of the those angles.(A) one (B) two(C) three (D) half

7. In a right angled triangle ABC, if B = 90,then A + C =

(A) 45 (B) 90

(C) 180 (D) none of these

i

p

l

m

n

a

e

b c d

f g h

j k q

Multiple Choice Questions

p

r

q

e f d c

h g

70 b

p

m

n

E

G

3x x

D C

A

H

F B

SAMPLE C

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37


a b

c

l

m

(11x 30)

(5x + 42)

m

n

8. If in a pair of interior angles on the same sideof transversal, the measure of one angle is lessthan 90, then the measure of the other anglewill be(A) equal to 90(B) greater than 90 but less than 180(C) less than 90(D) greater than 180

9. If the ratio of the measures of the interiorangles on the same side of the transversal is5 : 4, then the measure of the smaller angle is(A) 40 (B) 50(C) 80 (D) 100

10. In the given figure, if ABE : BED =3 : 7, then, CBE : BEF =

(A) 3 : 7 (B) 3 : 10(C) 7 : 3 (D) 10 : 3

11. 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.

12. Two lines intersect to form four verticallyopposite angles. If all the angles are equal,then(A) the two lines intersect at 30(B) the two lines intersect at 60(C) the two lines are perpendicular to each

other(D) the two lines do not intersect

13. In the given figure, what is the value of x if mis parallel to n?

(A) 10 (B) 12(C) 11 (D) 13

14. In the given figure l || m. If a : b = 7 : 5, thenthe ratio a : c will be

(A) 5 : 7 (B) 7 : 5(C) 1 : 1 (D) 7 : 12

15. In the given figure, find the value of QRPwhen QP || TR.

(A) 113 (B) 129(C) 49 (D) 94

Based on Practice Set 2.1

1. In the figure below, if line AB || line CD, thenfind the values of x and y.

[2 Marks]

2. In the figure below, AB || CD, then find thevalues of x, y and z.

[3 Marks]

B

A

O

D

C

y

x

75°

z

35°

A

G

C

D

B

F H

E

Additional Problems for Practice

P

Q R S

64

67

T

A R

C

Q

D B

P

85° x

H

y F

115°

E G

SAMPLE C

ONTENT

38


3. In the figure below, if AB || CD, then find themeasures of PCD and CPD.

[3 Marks]

4. In the figure below, if line AB || line CD, thenfind the value of x.

[3 Marks]

5. In the figure below, AB CD, then find the

value of x.

[3 Marks] 6. In the figure below, line AB line CD, then

find the values of x, y and z.

[3 Marks]


1. Prove that the bisector of a pair of interior anglesformed on the same side of the transversal areperpendicular to each other. [4 Marks]

2. In the figure below, mABC = 65,mDCE = 35, mCEF = 145,mBCE = 30, then prove thatray AB ray EF.

[3 Marks]

3. In the adjacent figure, line AB line ED and

line EF line DC, then prove that ABC andDEF are supplementary.

[3 Marks]

4. In the given figure, prove that l || m.

[3 Marks]

5. In the given figure, PQR is isosceles inwhich Q = R. If PS bisects TPR, thenprove that PS || QR.

[4 Marks]

Total marks: 20

1. Choose the correct alternative for each of the following questions. [4] i. 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.

A B

O

D

135

145

x

C

BP

C Q R y

D

110

A

65z

x

EA

F

B D C

A B

C D

EP Q

x120°

100°

A B 65

3035

145E F

C D

Chapter Assessment

80 35

45

p

ml

P

T

S

R Q

A B

DC

56° P 100°

SAMPLE C

ONTENT

39


ii. In the figure given below if line l || line m, and line n is their transversal, then which of the following statement is incorrect?

(A) a + b = 180 (B) d + f = 180 (C) e + g = 180 (D) c + f = 180 iii. Find the measure of the alternate angle of the angle which is in linear pair with the angle of measure of 65. (A) 65 (B) 25 (C) 115 (D) 130 iv. Which of the following pairs is a pair of interior angles formed by two parallel lines and on the same side of

the transversal? (A) 30, 60 (B) 30, 150 (C) 60, 130 (D) 50, 120 2. Attempt the following. (any two) [4] i. In the figure below, if x = 84° and y = 83°, then state whether the statement ‘line m || line n’ is true or

false. Justify. ii. In the given figure, if line q || line r, line p is their transversal and if a = 80°, find the values of f and g. iii. In the given figure, write down the pairs of alternate exterior angles. 3. Complete the following activity. (any one) [2] i. In the adjoining figure, sides of PQR and XYZ are parallel to each

other. Prove that, PQR XYZ. Given: Ray YZ || ray QR and ray YX || ray QP To prove: PQR XYZ Construction: Extend ray YZ in the opposite direction. It intersects ray QP at point S. Proof: Ray YX || ray QP [Given]

Ray YX || ray SP and seg SY is their transversal [P–S–Q]

(i) [Corresponding angles]

ray YZ || ray QR [Given]

l

x

y

m

n

f e d c

h g

b a

n

l

m

g h f e

c db a

p

r

q

e f d c

h g

a b

Q

P

Z

R

X

Y

SAMPLE C

ONTENT

40


Q

L

M

B

A

15 35x

P

10

ray SZ || ray QR and is their


PSY SQR

PSY (ii) [P–S–Q]

PQR XYZ [From (i) and (ii)] ii. In the adjoining figure, if line q || line r, line p is their transversal. Find values of c, e and h. Solution: c = [Vertically opposite angles]


e = [Alternate angles]

e =

Also, h + e = 180 [Angles in a linear pair]

h = 4. Attempt the following. (any two) [6] i. In the given figure, if x y and x + z = 180, then prove that line m || line n. ii. In the given figure, line AB || line CD and line PQ is

transversal. Measure of one of the angles is given. Hence find the measures of the following angles.

a. ART b. CTQ c. DTQ d. PRB iii. Prove that the sum of the measures of all the angles of a triangle is 180. 5. Attempt the following: (any one) [4] i. In the figure, AB || PQ, BC || QR. Prove that ABC PQR.

ii. In the given figure, QP || ML. Find the value of x.

z

m yx

n

l

A B 105

C D

P

T

Q

R

p

r

q

e f d c

h g

70 b

B A

R

Q P

C

SAMPLE C

ONTENT

41


1. (D) 2. (A) 3. (B) 4. (C) 5. (C) 6. (B) 7. (B) 8. (B) 9. (C) 10. (C)11. (B) 12. (C) 13. (B) 14. (B) 15. (C)


1. x = 85, y = 65 2. x = 35, y = 70, z = 753. PCD = 44, CPD = 80 4. x = 405. x = 80 6. x = 65, y = 115, z = 45

1. i. B ii. D iii. C iv. B

2. i. False ii. g = 80, f = 100 iii. a and h; b and g

3. i.

Ray YX || ray QP [Given] Ray YX || ray SP and seg SY is their




PSY SQR Corresponding angles

PSY PQR (ii) [P–S–Q]

PQR XYZ [From (i) and (ii)] ii.

c = 70 [Vertically opposite angles] Now, line q || line r and line p is their transversal.

e = c [Alternate angles] e = 70

Also, h + e = 180 [Angles in a linear pair] h + 70 = 180 h = 180 70 h = 110

4. ii. a. 75 b. 75 c. 105 d. 75

5. ii. 30

Multiple Choice Questions

Additional Problems for Practice

Answers

Chapter Assessment

Q

P

Z

R

X

S Y

p

r

q

e f d c

h g

70 b

SAMPLE C

ONTENT

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