Prime Mathematics 8

Prime

Mathematics
Series

Approved by 8
the Curriculum Development Centre

(CDC/Government of Nepal)
Sanothimi, Bhaktapur

Authors

Raj Kumar Mathema

Bhakta Bahadur Bholan Uma Raj Acharya

Yam Bahadur Poudel Narayan Prasad Shrestha

Bindu Kumar Shrestha

Editors
Anil Kumar Jha
Dhurba Narayan Chaudhary
Dirgha Raj Mishra
Hari Krishna Shrestha

Language Editor
Mrs. Tara Pradhan

Pragya Books Pragya Books and Distributors Pvt. Ltd.
and
Distributors Pvt. Ltd.

Printing history
First Edition 2074 B.S.

Layout and design
Rabi Man Shrestha

© Publisher

All rights reserved. No part of this book, or designs and illustrations here within, may be
reproduced or transmitted in any form by any means without prior written permission.

ISBN : 978-9937-9170-0-1
Printed in Nepal

Pragya Books Published by
and
Pragya Books and Distributors Pvt. Ltd.
Distributors Pvt. Ltd.
Kathmandu, Nepal
E-mail : [email protected]

Preface

Prime Mathematics Series is a distinctly outstanding mathematics series
designed in compliance with Curriculum Development Centre (CDC) to meet
international standards. The innovative, lucid and logical arrangement of the
content makes each book in the series coherent. The presentation of ideas
in each volume makes the series not only unique, but also a pioneer in the
evolution of mathematics teaching.

The subject matter is set in an easy and child-friendly structure so that
students will discover learning mathematics a fun thing to do.A lot of research,
experimentation and careful gradation have gone into the making of the series
to ensure that the selection and presentation is systematic, innovative and both
horizontally and vertically integrated.

Prime Mathematics Series is based on child-centered teaching
and learning methodologies, so the teachers will find teaching this series
equally enjoyable. We are optimistic that this series shall bridge the existing
inconsistencies between the cognitive capacity of children and the course matter.

We owe an immense debt of gratitude to the publishers for their creative,
thoughtful and inspirational support in bringing about the series. Similarly, we
would like to acknowledge the tremendous support of teachers, educationists
and well-wishers for their contribution, assistance and encouragement in making
this series a success.

We hope the series will be another milestone in the advancement of teaching
and learning mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that we can refine and improvise the
series in the future editions.

Our team would like to express our special thanks to Mr. Nara Bahadur
Gurung, Mr. Ram Narayan Shah, Mr.Tulsi Kharel, Mr. Mani Ram Khabas, Mr. Umesh
Acharya, Mr. J. Phuldel, Mr. Kamal Raj Tripathee, Mr. Rudra Prasad Pokharel, Mr.
Uttam Prasad Panta, Mr. L.N. Upadhyaya, Mr. Shakti Prasad Acharya, Mr. Upendra
Subedi, Mr. Kul Narayan Chaudhary, Mr. Bishonath Lamichhane, Mr. Harilal
Lamichhane, Mr. Govinda Paudel, Mr. Krishna Aryal, Mr. Nim Bhujel, Mr. Santosh
Simkhada, Mr. Pashupati Upadhyaya, Mr. Dipak Adhikari, Mr. Mukti Adhikari,
Mr. Dipendra Upreti, Mr. Dipak Khatiwada, Mr. Narayan Nepal, Mr. Raj Kumar
Dahal, Mr. Bhim Raj Kandel, Mr. Prem Giri,Mr. Iswor Khanal, Mr. Balram Ghimire,
Mr. Om Kumar Chhetri, Mr. Ram Hari Bhandari, Mr. Krishna Kandel, Mr. Madhav
Atreya, Mr. Harihar Adhikari, Mr. Chura Gurung, Mr. Shiva Devkota, Mr. Chandra
Dev Tiwari, Jivan K.C., Raghu Kandel and Baikuntha Marhattha and Subash Bidari
for their Painstaking effort in peer reviewing of this book.

CONTENTS

Area Pages
1 - 68
1. GEOMETRY
3
Unit 1. Line and angles 15
36
Unit 2. Triangle, quadrilateral & polygons 47
54
Unit 3. Congruency of triangles 58
69 - 80
Unit 4. Similar triangles 70
81- 97
Unit 5. Circle 82
98 - 123
Unit 6. Solid shapes 99
111
2. COORDINATE GEOMETRY 124 - 148
125
Unit 1. Coordinates 149 - 191
150
3. MENSURATION 164
170
Unit 1. Perimeter, area and volume 176
192 - 208
4. TRANSFORMATION 193
209 - 220
Unit 1. Transformation 210
221 - 226
Unit 2. Bearing and Scale drawing 222
227 - 234
5. SETS 228
235 - 256
Unit 1. Sets 236
257 - 298
6. REAL NUMBER SYSTEM 258
299 - 305
Unit 1. Number system 300
306 - 324
Unit 2. Integers 307
325 - 327
Unit 3. Scientific notation 328

Unit 4. Rational & irrational number

7. RATIO, PROPORTION AND PERCENTAGE

Unit 1. Ratio proportion and percentage

8. PROFIT AND LOSS

Unit 1. Profit and Loss

9. UNITARY METHOD

Unit 1. Unitary Method

10. SIMPLE INTEREST

Unit 1. Simple interest

11. STATISTICS

Unit 1. Statistics

12. ALGEBRAIC EXPRESSION

Unit : 1. Algebraic Expression

13. INDICES

Unit: 1. Indices

14. EQUATION, INEQUALITY AND GRAPH

Unit: 1. Equation, Inequality and Graph

Model question

Grid

At the end of this unit the students will be able to Estimated periods - 42
Objectives:

● know the relations between the angle pairs formed when two straight lines intersect each
other.

● know the relations between the angle pairs formed when parallel lines are intersected by a
transversal.

● investigate and verify that the sum of the angles of a triangle is two right angles.
● investigate and verify different properties of isosceles triangle.
● investigate and verify different properties of squares, rectangles and parallelograms.

Area● construct regular polygons.

● know the conditions for similarity and congruency of triangles.
● investigate the relation of diameter and circumference of circle and find the area of circle.
● identify solids like prisms, pyramids, cubes and cuboids, tetrahedrons, cylinders and cones.

Teaching Materials:

Paper models of triangles, quadrilaterals and polygons, models of solids, geometry box,
geoboard.

Activities:

It is better to
● discuss and demonstrate relation between the different angle pairs.
● discuss and demonstrate different properties of triangles.
● discuss and demonstrate different properties of special quadrilaterals.
● demonstrate the construction of regular polygons.
● discuss about the condition for similarity and congruency of triangles.
● discuss about the relation between diameter (radius) and circumference of circle and find the

area of circle.
● demonstrate the nets and models of solids.

M P A P
B hl
A o rA

C?

CN D E 6m

Q Alternate angles 8m D 12 m B

Prime Mathematics Book - 8

Geometry

Introduction:

Euclid (300 B.C., Greek) is regarded as the father of geometry. He made the study
of geometry so systematic that even after more than two millennia, his work has dominated
all the teaching of geometry. He organized all the results of geometry and put them in a
logical order in his book 'The Elements'. This pioneering work consists definitions, axioms,
postulates and propositions or theorems derived from them by deductive reasoning. This
book contains:

(i) 23 definitions (ii) 5 postulates iii) 5 - axioms (iv) 48 propositions

The axioms or common notions:

These are the general truths (assertions) accepted without proof, Which are not
limited to the content of geometry but applicable in other fields as well. In the Elements,
Euclid has listed the following axioms.

(i) Things which are equal to the same thing are equal to one another.
(ii) If equals be added to the equals, the wholes so formed are equal.
(iii) If equals be subtracted from equals, the remainders are equal.
(iv) The things which coincide with one another are equal to one another.
(v) The whole is greater than the part.

The postulates:

These are the general truths accepted without proof specially concerned with
geometric relations. In the Elements, Euclid has listed the following 5 postulates.

i. A straight line can be drawn from any point to any point.

ii. A finite straight line can be produced continuously in a straight line.

iii. A circle may be described with any centre and distance.

iv. All right angles are equal to one another.

v. If a straight line falling on two straight lines make the interior angles on the
same side together less than two right angles, the two straight lines, if produced
indefinitely, meet on that side on which the angles are together less than two right
angles.

2 1 Geometry

Prime Mathematics Book - 8

Unit - 1 Lines and angles Estimated period : 8

Some special pairs of angles:

A. Complementary angles:

A pair of angles are said to be complementary, if their sum is a right angle (90o).
For example, 50o and 40o, 60o and 30o, 75o and 15o etc. If ∠ A and ∠ B are complementary,

then ∠ A + ∠ B = 90o
∴ ∠ B = 90o - ∠ A
Here, ∠ B is complementary of ∠ A
∴ Complementary of ∠ A = 90o - ∠ A

Example: Find the complementary of 36o.
Solution:
Here, given angle ∠ A = 36o
We know, Complementary of ∠ A = (90o - ∠ A) = (90o - 36o) = 54o
∴ Complementary of 36o is 54o.

B. Supplementary angles:

A pair of angles are said to be supplementary, if their sum is two right angles (180o).
For example, 120o and 60o, 90o and 90o, 100o and 80o etc.

If ∠ A and ∠ B are supplementary angles, then ∠ A + ∠ B = 180o. So, ∠ B = 180o - ∠ A
Here, ∠ B is supplementary angle of ∠ A.
\ Supplementary of ∠ A = 180o - ∠ A.

Example: Find supplementary angle of 64o.

Solution:
Here, given angle ∠ A = 64o
We know, supplementary of ∠ A = (180o - ∠ A) = (180o - 64o) = 116o
\ Supplementary of 64o is 116o.

C. Adjacent angles: C B

Two angles are said to be adjacent angles if they have O A
a common vertex and a common arm, two angles lying
either side of it. In the given figure, ∠ AOB and ∠ BOC
have a common arm OB and two angles lie on either
side of the common arm OB. So ∠AOB and ∠ BOC are
adjacent angles.

Lines and angles 3

Prime Mathematics Book - 8 C B
O A
D. Complementary adjacent angles :

If the sum of two adjacent angles is a right angle (90o), the
two angles are said to be complementary adjacent angles.

In the given figure, ∠ AOB and ∠ BOC are adjacent
angles and ∠AOB + ∠BOC = 900, so ∠ AOB and ∠ BOC are
complementary adjacent angles.

E. Supplementary adjacent angles (Linear pair): A B
OC
If the sum of two adjacent angles is two right angles (180o),
the two angles are called supplementary adjacent angles or
linear pair. In the given figure, ∠ AOC = 180o and a straight
angle ∠ AOC = ∠ AOB + ∠ BOC.

\ ∠ AOB + ∠ BOC = 180o.

So, ∠ AOB and ∠ BOC are supplementary adjacent angles or
linear pair.

F. Vertically opposite angles : O D
B
When two straight lines intersect each other, the equal
opposite angles formed with a common vertex and having D
no common arm are called vertically opposite angles. In A B
the given figure, two straight line segments intersect each
other at point O. ∠AOC = ∠BOD and ∠AOD = ∠COB are C D
vertically opposite angle pairs.

Experimental Verification: A O
C (i)
Experiment : Two straight line segments AB and CD are
drawn which intersect at O. Vertically A
opposite angles ∠ AOC and ∠ BOD; ∠ AOD and
∠ BOC are formed. O

To verify : ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC. C B
(ii)
Verification : ∠ AOC, ∠ BOD, ∠ AOD and ∠ BOC are measured
and their measures are tabulated as,



Figure ∠ AOC ∠ BOD ∠ AOD ∠ BOC Result

(i) 35o 35o 145o 145o ∠ AOC = ∠ BOD

(ii) 90o 90o 90o 90o ∠ AOD = ∠ BOC

Conclusion: In the above experiment we observed ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC in
each figure. Hence, vertically opposite angles are equal.

4 1 Geometry

Prime Mathematics Book - 8

Theoretical Proof:

Statement : When two straight line segments intersect each other, the vertically opposite

angles so formed are equal. D

Given : Line segments AB and CD intersect at O. ∠ AOC A

and ∠ BOD, ∠ AOD and ∠ BOC are vertically O B
opposite angles.
C

To prove : ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC

Proof : Statements S.N. Reasons
S.N. 1.
1. ∠ AOC + ∠ AOD = 180o 2. Being linear pair.
2. ∠ AOD + ∠ BOD = 180o 3. Being linear pair.
3. ∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD From statements 1 and 2.
\∠ AOC = ∠ BOD 4. (by equal axiom)
4. ∠ AOD = ∠ BOC By similar reasonings.

Q.E.D.

Note: Q.E.D. stands for quod erat demonstration in Latin which means 'that is to be proved'.

Demonstration with paper model: D
To show vertically opposite angles are equal.
O
⇒ Draw line segments AB and CD to intersect at O A B
on a chart paper.

⇒ Cut and take out the part of ∠ BOD.

⇒ Put ∠ BOD over ∠ AOC. They match (coincide) C
exactly. Hence vertically opposite angles are
equal.

Different pairs of angles formed when a transversal cuts parallel lines :

A straight line drawn across a set of given lines P B
(parallel or not) is called a transversal. M D

A straight line PQ is drawn across a set of lines AB A N
and CD to cut them at M and N. The straight line Q
PQ is Transversal. The angles formed outside the C
pair of lines ∠ AMP, ∠ BMP, ∠ CNQ and ∠ DNQ
are called exterior angles and the angles formed
inside the pair of lines ∠ AMN, ∠ BMN, ∠ CNM and
∠ DNM are called interior angles.

Lines and angles 5

Prime Mathematics Book - 8

A pair of interior angles on opposite sides of P

transversal (non adjacent) ∠ AMN and ∠ MND; A M
∠ BMN and ∠ MNC are called alternate angles. B

A pair of angles on same sides of transversal,

one exterior and other internal non adjacent, C N D

∠ AMP and ∠ CNM; ∠ AMN and ∠ CNQ; ∠ BMP Alternate angles
and ∠ MND; ∠ BMN and ∠ DNQ are called
QP

corresponding angles.

Pair of interior angles on the same side of A M B

transversal ∠ AMN and ∠ MNC; ∠ BMN and ∠ MND CN D

are called co-interior angles. Q Corresponding angles

P

M When a transversal cuts two parallel lines,
N B pair of alternate angles are equal. pair of
A
C corresponding angles are equal and pair of co-
D interior angles are supplementary.

Q Co-interior angles

Experimental Verification

1. To verify experimentally that alternate angles formed by a transversal with parallel

lines are equal.

Experiment: (i) A pair of parallel lines AB and CD are drawn by the help of scale and set square.

(ii) A transversal PQ is drawn to cut AB and CD at M and N respectively.
P

P B P MB AM B
AM D A D
C D C
CN N N
Q (i) (ii) Q
Q
(iii)

To verify: ∠ AMN = ∠ MND; ∠ BMN = ∠ MNC.
Verification: Interior angles ∠ AMN, ∠ BMN, ∠ MNC and ∠MND are measured and their

measures are tabulated.

Fig. ∠ AMN ∠ MND ∠ BMN ∠ MNC Result
(i)
∠ AMN =∠ MND

(ii) ∠ BMN =∠ MNC

(iii)

Conclusion: In this experiment, we observed ∠ AMN = ∠ MND and ∠ BMN = ∠ MNC in each
figure. Hence, the pairs of alternate angles formed by transversal with the
parallel lines are equal.

6 1 Geometry

Prime Mathematics Book - 8

2. To verify experimentally that the pairs of corresponding angles formed by a
transversal with parallel lines are equal.

Experiment : (i) A pair of parallel lines AB and CD are drawn with the help of scale and
set square.

(ii) A transversal PQ is drawn to cut AB and CD at M and N respectively.

To verify : ∠ AMP = ∠ CNM, ∠ BMP = ∠ DNM. P
∠ AMN = ∠ CNQ, ∠ BMN = ∠ DNQ. AM

P B P MB C B
AM D A N D
C D
CN N Q (iii)
Q (i) (ii) Q

Verification : The exterior and interior angles are measured and their measures are
tabulated.

Fig. ∠ AMP ∠ CNM ∠ BMP ∠ DNM ∠ AMN ∠ CNQ ∠ BMN ∠ DNQ Result

(i) ∠AMP =∠ CNM
(ii) ∠BMP =∠ DNM

(iii) ∠AMN =∠ CNQ
(iv) ∠BMN =∠ DNQ

Conclusion: In this experiment, we observerd that ∠ AMP = ∠ CNM, ∠ BMP = ∠ DNM,
∠ AMN = ∠ CNQ, ∠ BMN = ∠ DNQ in each figure.

Hence, corresponding angles formed by a transversal with the parallel lines are equal.
Verified.

3. To verify experimentally that the sum of the co-interior angles formed by a transversal
with the parallel lines is 180o.

Experiment: (i) A pair of parallel lines AB and CD are drawn with the help of scale

and set square.

(ii) A transversal PQ is drawn to cut AB and CD at M and N respectively.

P B P B P B
M D M AM D
A A C
C D
CN N
Q (i) N
Q (iii)
(ii) Q

Lines and angles 7

Prime Mathematics Book - 8

To verify : ∠ AMN + ∠ MNC = 1800, ∠ BMN + ∠ MND = 1800
Verification : ∠ AMN, ∠ MNC, ∠ BMN and ∠ MND are measured and the values are tabulated.

Fig. ∠ AMN ∠ MNC ∠ AMN + ∠ MNC ∠ BMN ∠ MND ∠ BMN + ∠ MND Result

(i) ∠ AMN + ∠ MNC= 1800

(ii) ∠ BMN + ∠ MND= 1800

(iii)

Conclusion : In this experiment we observed that ∠ AMN + ∠ MNC = 1800 and ∠ BMN + ∠
MND = 1800 in each figure. Hence the sum of the co-interior angles formed
by a transversal with parallel lines is 1800.

Verified.

Note: Proof of theorems on parallel lines and transversal are given in proposition
27, 28 and 29 of Euclid’s 'The Elements'.

We use the following facts.

- If a pair of parallel lines are intersected by a transversal,
(i) pair of alternate angles are equal.
(ii) pair of corresponding angles are equal.
(iii) pair of co-interior angles are supplementary.

- A pair or lines intersected by a transversal are parallel if any of the following
conditions hold.

(i) If a pair of alternate angles are equal.
(ii) If a pair of corresponding angles are equal.
(iii) If a pair of co-interior angles are supplementary.

Example 1: From the given figure, find the size of unknown angles.

Solution: 1200 + x = 1800 (sum of co-interior angles)

\ x = 1800 - 1200 = 600

v = x (Being alternate angles equal as AB//CD.)

\ v = 600 P
Mt u
w = 1200 (being alternate angles) A B
t = w = 1200 (being corresponding angles) v 120o D

u = x = 600 (being corresponding angles) C wx
z = 1200 (being corresponding angles) y zN

y = v = 600 (being corresponding angles) Q

8 1 Geometry

Prime Mathematics Book - 8

Example 2: Find the values of the variables x, y and z from the figure given below.

Solution: Here, 3x + 2x = 1800 (sum of co-interior angles as AB//CD) M

5x = 1800

∴ x = 360 A y 3x B
z 2x D
Now, y = 2x (Being alternate angles as AB//CD)

= 2 × 36o = 720 C
And Z = 3x (Being an alternate angle.)

= 3 × 360 = 1080 N

Example 3: Find the size of unknown angles from the given figure.

Solution: Here,

4x = 1800 - x (being corresponding angles as AB//DC)
A
or, 4x + x = 1800 1800 - x rB
q
or, 5x = 1800
C
∴ x = 1800 = 360 p
5 D

∠ CDE = ∠ BAD 4x

= 1800 - x = 1800 - 360 = 1440 E

Again, p + (1800 - x) = 1800 (sum of co-interior angles as AB//DC)

or, p + 1440 = 1800

∴ p = 1800 - 1440 = 360
Again, p + q = 1800 (sum of co-interior angles as AD//BC)

or, 360 + q = 1800

∴ q = 1800 - 360 = 1440
And, q+r = 1800 (Sum of co-interior angles as AB//DC)

or, 1440 +r = 1800

∴ r = 1800-1440 =360.

Example 4 : Find the size of angle x in the figure alongside. M
440
Solution: Lets draw XY//AB//CD from P. A B
Px
Here, D
B
a = 440 (alternate angles as AB//XY) C 560 Y
N D
And b = 560 (alternate angles as XY//CD)

Now, A M
X 440
x = a + b (whole part axiom)
a
= 440 + 560 Pb x

∴ x = 1000 C 560
N

Lines and angles 9

Prime Mathematics Book - 8

Example 5: Find the size of the unknown angles from the given figure.

Solution: ∠ AOD = 1800 (being a straight angle) C

∠ AOB + ∠ BOC + ∠ COD = ∠ AOD B
A
(by whole part axiom) 2x

or, x + 2x + 3x = 1800 D 3x x

or, 6x = 1800 O

∴ x = 1800 = 300
6

Now, ∠ AOB = x = 300,

∠ BOC = 2x = 2×300 = 600 and

∠ COD = 3x = 3×300 = 900

Example 6: Find the value of the unknown angles in the given figure.

Solution: Here, 2y 600
x + 600 = 900 (being complementary adjacent angles) 3y x
∴ x = 900 - 600 = 300

Again 2y + 3y = 900 (being complementary adjacent angles)

∴ 5y = 900 y = 180, 2y = 2 × 180 = 360 and 3y = 3 ×180 = 3 ×180 = 540

Example 7: Find the size of the angles represented by x,y and z.

Solution: From the given figure,

x + 2x = 1800 (being linear pair) 2x
y
or, 3x = 1800 x z

∴ x = 600

Now, x = z= 600 (being vertically opposite angles)

And x + y = 1800 (being a linear pair)

or, 600 + y = 1800

∴ y = 1800 - 600 = 1200

10 1 Geometry

Prime Mathematics Book - 8

Exercise 1

1. Find the complementary angles of the following angles.

(a) 320 (b) 680 (c) 450 (d) 750

(e) 00 (f) 900 (g) 500 (h) 100

2. Find the supplementary angles of the following angles:

(a) 400 (b) 700 (c) 900 (d) 1000

(e) 1200 (f) 1350 (g) 1600 (h) 1720

3. (a) If 2x and 4x are complementary angles, find them.
(b) If x and 5x are supplementary angles, find them.
(c) If x and (x + 500 ) are complementary angles, find the angles.
(d) If (2x + 100) and (3x + 100) form a linear pair, find the angles.

4. (a) If (x + 300) and (1800 - 2x) are vertically opposite angles, find the value of x.
(b) Find the size of an angle which is two third of its complement.
(c) Find the size of an angle which is one fifth of its supplement.
(d) Two supplementary angles are in the ratio 7:3, find them.

5. Find the size of unknown angles in the following figures.

(c)

(a) (b) x + 100

4x 3x 2x 2x + 20 0
3x 3x + 300

2x

(d) (e) (f)

2x 1100 a + 300
3x y xz a

z y

(g) (h) (i) c (j)

3x 2x 1400 1200 a b 1100 1200 800
4x x 2x xy
z
5x

Lines and angles 11

Prime Mathematics Book - 8

6. In the given figure, a pair of parallel lines PQ and RS are intersected by a transversal

MN. Write down, with their relations, a pair of: M

(a) alternate angles P ab Q
(b) corresponding angles dc
(c) co-interior angles.
R ef S
hg

7. Give reason for the following conditions. N l

(a) a = s M
(b) c + p = 1800
(c) c = s ab
dc

(d) d = r sp m
(e) d = p rq
(f) d + s = 1800

8. State whether the given pair of lines are parellal or not and justify thNe answer.

(a) E B (b) E B
G 50° D
A A G
130°
CH D C
50° 60°
F 50°

F

(c) X (d) C

MAN A P B
128° 110° Y
X
OB P 110°
52° Q
Y
D

G (f) M
(e) 78° P
I
CI D J
82° F L

EJ K 76° Q
102° 102°
H N

12 1 Geometry

9. Find the unknown angles in the following figures. Prime Mathematics Book - 8

(a) (b) (c)

zy 1300 x
3x x 1100 y
y
2x z

(d) z (e) (f) x
x
y 30 0 1500
3x x 35 0
(i) 50 0
y
65 0 x
65 0
(g) y (h) 80 0 x 65 0
(l) 1400
1150 x zy
x
(j) (k) 620 1500
x
y 1450 z
x y
85 0
1200
z
(m) 1400 (n) y 40 0

x x

800
500

(o) 1100 (p) y

x

zy x
30 0

Lines and angles 13

Prime Mathematics Book - 8 A 35 0 B

10. (a) In the adjoining figure,
AB//DC. Show that AD//BC.

75 0 70 0 C
D B

(b) In the given figure, E C
show that : AB//DC. 550 750

500
DA

11. (a) Verify experimentally that when two parallel lines are intersected by a
transversal, pair of alternate angles are equal.

(b) Verify experimentally that when a pair of parallel lines are intersected by a
transversal, co-interior angles are supplementary.

(c) Verify experimentally that when a pair of parallel lines are intersected by a
transversal, pairs of corresponding angles are equal.

(d) Verify experimentally that when two straight line segments intersect each
other, the vertically opposite angles so formed are equal.

14 1 Geometry

Prime Mathematics Book - 8

Unit - 2 Triangles, Quadrilaterals & PolygonsEstimated period : 18

2.1 Triangles A
Introduction:

A triangle is a plane figure bounded by three straight lines.

A triangle consists- B C

- Three corners or vertices and are denoted by capital
letter A,B,C or any other letters. A triangle is represented
by the letters representing its vertices as ∆ ABC.

- Three sides: Here in the figure, AB, BC and AC are three sides which are also
represented by c,a and b respectively.

- Three angles: Here in the figure, three angles are called interior angles of the
triangle which are ∠ BAC, ∠ ABC and ∠ ACB or simply ∠ A, ∠ B and ∠ C.

Other parts of a triangle: B A exterior angle
CD
When a side of a triangle is extended, the angle formed by A Altitude
the extended part and the other side is called an exterior C
angle. D
A Median
The perpendicular drawn from a vertex to the opposite B
side is called altitude (or height). Here AD BC so AD
is the altitude of the triangle over the base BC. Altitude
can be drawn from any vertex to the opposite side.
Altitude may fall inside or outside the base.

The line joining vertex and the mid point of the opposite B D C
side is called a median.

A

Ortho centre

Three altitudes of a triangle are concurrent (intersect at C
a point) and the point of intersection is called an ortho- B D A
centre.
Centroid
Three medians of a triangle are concurrent and the F E
point of intersection of medians is called centroid.
B DC

Triangles, Quadrilaterals & Polygons 15

Prime Mathematics Book - 8 A

Classification of triangle

A. Classification of triangle with regard to their sides:
A triangle is said to be
(i) equilateral, when all its sides are equal.
(ii) isosceles, when two of its sides are equal.
(iii) scalene, when non of its sides are equal.

A
A

B An equilateral C B An isosceles CC B
triangle triangle
A scalene
AB=BC=CA AB=AC triangle
AB≠BC≠CA
B. Classification of triangle with regard to their angles:
A triangle is said to be: A
(i) right angled, when one of its angles is right angle.
(ii) obtuse angled, when one of its angles is obtuse angle.
(iii) acute angled, when all of its angles are acute.

AA

BC CB BC
A right angled triangle An obtuse angled triangle An acute angled triangle

∠ AMN=900 ∠ ABC = an obtuse angle

Special properties of triangles.

A. Sum of three interior angles of a triangle is two right angles (1800).

Experimental verification:

Experiment: A

C B

C BB (ii) A A (iii) C
(i)

Three triangles of different sizes namely ∆ ABC are drawn:

To Verify: ∠ A + ∠ B + ∠ C = 1800

16 1 Geometry

Prime Mathematics Book - 8

Verification: The interior angles of each triangle are measured and the measures are
tabulated:

Fig. ∠ A ∠ B ∠ C ∠ A + ∠ B + ∠ C Result

(i) ∠ A + ∠ B + ∠ C = 1800

(ii)

(iii)

Conclusion: In this experiment we observed that ∠ A + ∠ B + ∠ C = 1800 in each figure.

Hence the sum of the angles of a triangle is two right angles.

Verified.

AE

B. Exterior angle of a triangle is equal to the sum of two
interior non-adjacent angles.

Experimental Verification:
Bo o D

Experiment: C
A

A
A

B (i) C DB (ii) C D B (iii) C D

(i) Three triangles of different sizes, namely ∆ ABC are drawn.

(ii) Side BC is produced to D in each figure.

To verify: ∠ ACD = ∠ ABC + ∠ BAC.
Verification: ∠ ABC, ∠ BAC and ∠ ACD are measured in each figure and their measures are

tabulated as:

Fig. ∠ ABC ∠ BAC ∠ ACD ∠ ABC + ∠ BAC Result
(i)
(ii) ∠ ABC + ∠ BAC = ∠ ACD
(iii)

C. Angles opposite to the equal sides of an isosceles triangle are equal. C
or
Base angles of an isosceles triangle are equal.
Experimental Verification:
Experiment:

A (i) B

Triangles, Quadrilaterals & Polygons 17

Prime Mathematics Book - 8 C

C

A (ii) B A (iii) B

(i) Three line segments namely AB of different lengths are drawn.
(ii) From A and B arcs with radius more than half of AB are drawn towards same

side to cut at C.
(iii) A,C and B,C are joined.

Thus isosceles triangle ABC with AC = BC are constructed in which base angles
are ∠ CAB and ∠ CBA.

To verify: ∠ CAB = ∠ CBA

Verification: ∠ CAB and ∠ CBA are measured in each of triangle and their measures are
tabulated as:

Fig. ∠ CAB ∠ CBA Result
(i)
(ii) ∠ CAB = ∠ CBA
(iii)

Conclusion: In this experiment, we observed ∠CAB = ∠CBA in each isosceles triangle. Hence
base angles of an isosceles triangle are equal.
Verified.

D. Base angles of a right angled isosceles triangle are each 450 .

Experimental Verification:

Experiment: y

y C
yC

C

A (i) B x A (ii) B x A (iii) B x

(i) ∠ XAY = 900, with three different figures are drawn.
(ii) On AX and AY, AB = AC are taken with different lengths in different figures.
(iii) B and C are joined. So, ABC is an isosceles right triangle with AB = AC.

18 1 Geometry

Prime Mathematics Book - 8

To verify: ∠ ABC = ∠ ACB = 450
Verification: ∠ ABC and ∠ ACB are measured in each triangle and their measures are

tabulated as:

Fig. ∠ ABC ∠ ACB Result

(i) ∠ ABC = ∠ ACB = 450

(ii)

(iii)

Conclusion: In each figure, we observed that ∠ ABC = ∠ ACB = 450. Hence the base angles
of a right angled isosceles triangle are each 450. Verified.

E. Median drawn from the vertex of an isosceles triangle to its base is perpendicular
to the base.
Experimental Verification:

Experiment: (ii) C (iii) B
(i) C
C

M

A M B AMB A

(i) Three line segments namely AB of different lengths are drawn.
(ii) From A and B, arcs are drawn to same side with different radii to cut at C.
(iii) A, C and B, C are joined. Thus, isosceles triangles ABC with AC = BC are formed.
(iv) Mid point of AB is located at M.
(v) C and M are joined.

To verify: CM ⊥ AB
Verification: ∠ CMA and ∠ CMB are measured in each figure and their measures are tabulated.

Fig. ∠ CMA ∠ CMB Result

(i) ∠ CMA = ∠ CMB = 900

(ii)

(iii)

Conclusion: In this experiment, we observed that ∠ CMA = ∠ CMB = 900 in each triangle
which shows that CM ⊥ AB. Hence the median drawn from vertex of an
isosceles triangle to is base it's perpendicular to the base.

Verified.

Triangles, Quadrilaterals & Polygons 19

Prime Mathematics Book - 8

Note: Experimental verification must be carried out in the following steps
- (i) Experiment (which is construction of figure).
(ii) To verify (where we write the statement to be verified)
(iii) Verification (this step is for measurements and tabulation)
(iv) Conclusion (Give the conclusion from the result of the experiment)
- Student must write the experiment in present passive instead of imperative

sentence.
- Theoretical proof of the properties of isosceles triangle verified experimentally

above require the knowledge of congruency. So, they are not given here.

Example 1: Find the size of angles represented by x, y and z in the given figure.

Solution: Here,

x = 500 (base angles of isosceles triangle as AC = BC)

In ∆ ABC, ∠ A + ∠ B + ∠ C = 1800 (sum of angles of triangle) D
z
or, x + 500 + y = 1800 A
or, 500 + 500 + y = 1800 x E
or, y = 1800 - 1000
yC

∴ y = 800 500
Again in ∆ CDE, B

∠ DCE = y = 800 (vertically opposite angles)

∠ CDE = ∠ CED = z (base angles of isosceles triangle as CD = CE)

Now, ∠ C + ∠ D + ∠ E = 1800 (sum of angles of triangle)

or, 800 + z + z = 1800

or, 2z = 1800 - 800
or, z = 10200­
∴ z = 500

Therefore x = 500, y = 800 and z = 500.

Example 2: Find the value of x and y from the figure given alongside.

Solution: Here, x
y = 720 (being base angles of isosceles triangle)
Now, x + y + 720 = 1800 (Sum of angles of a triangle) y 720
or, x + 720 + 720 = 1800
or, x = 1800 -1440
∴ x = 360

20 1 Geometry

Prime Mathematics Book - 8

Example 3: Find the value of unknown angle x from the figure given below.

Solution: Here, A

∠ ABC = ∠ BAC = 550 550

(being base angles of a triangle as AC = BC)

Now, ∠ ACD = ∠ BAC + ∠ ABC x
(Relation of exterior and interior opposite angles) B CD

or, x = 550 + 550 ∴ x = 1100

Example 4: From the given figure find the size of the unknown angles x and y.

Solution: Here, A

∠ ADB = 900 (Being ∆ABC is an isosceles triangle 350
and AB is a median. So, AD ⊥ BC)

In ∆ ABD

∠ A + ∠ B + ∠ D = 1800

(sum of angles of triangle) Bx D yC
or, 350 + x + 900 = 1800

∴ x = 1800 - 1250 = 550

Now, y = x = 550 (base angles of isosceles triangle as AB = AC)

∴ X = 550, y = 550

Exercise 2.1

1. Find the values of the unknown variables x, y and z from the following figures.

(a) A (b) P (c)

800 750

Bx C Qx R 650 x
(e) 1400 1500
(d) (f)
x-300
1000 2x0
x 3x0
3x x

Triangles, Quadrilaterals & Polygons 21

Prime Mathematics Book - 8 (h) x (i) 400

(g)

1800 - x 500 1420 x y
300 y
(j) (k) (l)

x

xy z
x

2. Find the values of x, y, a, b and c from the given figures.

(a) P (b) (c) A

x 850 R 120 0 B 720 600 30 0
40 0 a b D
T 75 0
y S c b 110 0 E y
Q
a
C

(d) c (e) B D (f)

820 b 720 Fx 300 G x
E
54 0

400 y
AC
480 a
(h) b
(g) x
b

650 a c 2a
55 0 y

3. Solve the following problems.
(a) Three angles of a triangle are in the ratio 2:3:4. Find the angles.
(b) If two acute angles of a right angled triangle are in the ratio 4:5.
Find the measures of the acute angles.
(c) One of the equal angles of an isosceles triangle is 400. Find the other angles.
(d) The vertical angle and the base angle of an isosceles triangle are in the ratio
4:3. Find the size of all the angles.

22 1 Geometry

Prime Mathematics Book - 8

4. Verify experimentally that
(a) The sum of interior angles of a triangle is two right angles.
(b) Exterior angles of a triangle is equal to the sum of interior opposite angles.
(c) Base angles of an isosceles triangle are equal.
(d) Bisector of vertical angles of an isosceles triangle is perpendicular bisector of

the base.
(e) Angles of an equilateral triangle are equal.
(f) The sum of any two sides of a traingle is greater than third side.

(g) Draw three triangles ABC having AC>AB. Draw a Table given below and measure

the given sides and angles then calculate them.

Figures AC AB ∠ ACB ∠ ABC Results
(i)
(ii)

(iii)
Conclusion :

5. (a) In the adjoining figure, AB || CD, ∠ AEC = 620 A E B
620
780

∠ BEF = 780, then, find ∠ ECF, ∠ EFC and ∠ EFD.

CF D

AEB

(b) In the figure given alongside, XE and XF are X
bisectors of ∠ AEF and ∠ EFC respectively.

Prove that ∠ EXF is a right angle. C FD

A B
(c) In the given figure, ABCD is a parallelogram.

Bisector of ∠ BAD and ∠ ABC meet at a point

P on CD. Prove that ∠ APB = 900

2.2 Quadrilaterals: DPC

Introduction: A B
D C
A quadrilateral is a rectilineal figure bounded by four sides.
Given figure is a quadrilateral ABCD with four vertices A,B,C
and D; four sides AB, BC, CD and DA and four angles (interior
angles) ∠ A, ∠ B, ∠ C and ∠ D. In a quadrilateral, the lines
joining the opposite vertices are called diagonals. Here, AC and
BD are two diagonals.

Triangles, Quadrilaterals & Polygons 23

Prime Mathematics Book - 8

You are already acquainted with the following.
– Sum of angles of a quadrilateral is 360o.
– Types of quadrilaterals.

1. Parallelogram, a quadrilateral with opposite sides parallel.
2. Rectangle, a quadrilateral with opposite sides equal and all angles 900.
3. Square, a quadrilateral with all sides equal and all angles 900.
4. Rhombus, a parallelogram with all sides equal.
5. Trapezium, a quadrilateral with any two opposite sides parallel.
6. Trapezoid, a quadrilateral with non of the sides parallel.
7. Kite, quadrilateral with two pairs of adjacent sides equal separately.

In this section we will learn in detail about (d) Rhombus. B
(a) parallelogram (b) Rectangle (c) Square C
A
Parallelogram: B

A parallelogram is a quadrilateral with opposite sides D
parallel. In the figure, ABCD is a parallelogram where
AB//DC and AD//BC. C

Properties of Parallelogram:

A. Opposite sides of a parallelogram are equal.
Experimental Verification:
Experiment:

DA
AB

D

D (i) C C BA

(ii) (iii)

Three parallelograms, namely ABCD of Fig. AB CD BC AD Result
different sizes are drawn by using set (i) AB = CD
squares. (ii) AD = BC
To verify: AB = DC and AD = BC (iii)
Verification: Sides AB, BC, CD and DA are
measured and their measures are tabulated.

24 1 Geometry

Prime Mathematics Book - 8

Conclusion: In this experiment we observed AB = CD and AD = BC in each figure.
Hence opposite sides of a parallelogram are equal.

Verified.

B. Opposite angles of a parallelogram are equal.

Experimental Verification:

Experiment: A A

A B BD

D B
DC
CC

Three parallelograms namely ABCD are drawn by using set squares where ∠A and ∠C;

∠B and ∠D are opposite angles.

To verify: ∠A = ∠C and ∠B = ∠D.

Verification: ∠A, ∠B, ∠C and ∠D are measured and their measures are tabulated.

Fig. ∠ A ∠ C ∠ B ∠ D Result

(i) ∠A = ∠C

(ii) ∠B = ∠D

(iii)

Conclusion: In this experiment we observed that in each parallelogram ∠A = ∠C and
∠B = ∠D. Hence opposite angles of a parallelogram are equal.
Verified.

C. Diagonals of a parallelogram bisect each other.

Experimental Verification:

Experiment: A DB C
A
B

O OO

D (i) C B C A (iii) D
(ii)

(i) Three parallelograms namely ABCD of different sizes are drawn by using set

squares.

(ii) A, C and B, D are joined where they intersecting at O.

To Verify: AO = CO and BO = DO.

Verification: In each figure, AO, CO, BO and DO are measured and their measurements
are tabulated.
Triangles, Quadrilaterals & Polygons 25

Prime Mathematics Book - 8

Fig. AO CO BO DO Result

(i) AO = CO, BO = DO

(ii) AO = CO, BO = DO

(iii) AO = CO, BO = DO

Conclusion: In this experiment we observed that in each parallelogram, AO = CO and

BO = DO. Hence diagonals of a parallelogram bisect each other.

Verified.

Rectangle A B
D C
A rectangle is a parallelogram in which each angle is right
angle (900). In the given figure, ABCD is a parallelogram
where ∠A = ∠B = ∠C = ∠D = 900. So it is a rectangle.

Being a parallelogram, a rectangle holds all the properties of parallelogram.

a. Each angle of a rectangle is 900.

Draw three rectangles having different size. Measure the angles of the rectangle

using Protractor and tabulate them.

A BA B

D D (ii) C
C
(i)

AB

Figures ∠DAB D (ii) C Results
(i) ∠ABC ∠BCD ∠CDA
(ii)
(iii)
Conclusion :

26 1 Geometry

Prime Mathematics Book - 8

b. Diagonals of a rectangle are equal. C D D
C C
Experimental Verification:
Experiment:

D

A (i) BB (ii) AB (iii) A

(i) Three rectangles, namely ABCD of different sizes are drawn by using set square (or
compass).

(ii) Diagonals AC and BD are drawn in each rectangle.

To Verify: AC = BD
Verification: Lengths of diagonals AC and BD are measured in each figure and their
measures are tabulated in the following table.

Fig. AC BD Result
(i) AC = BD
(ii)
(iii)

Conclusion : In this experiment, we observed that in each rectangle, diagonals AC B
= BD. Hence diagonals of a rectangle are equal.
Square
Verified.

A

A square is a rectangle with all sides equal. In the given
figure, ABCD is a rectangle where AB = BC = CD = DA. So,
ABCD is a square. A square holds all the properties of
parallelogram and rectangle.

D C

a. The diagonal of the square bisects the vertical angle.
Draw a square ABCD with the diagonals AC & BD.

AB

O

DC
Triangles, Quadrilaterals & Polygons 27

Prime Mathematics Book - 8

Measure the angles given in the table and tabulate them in the table.

Vertical angles Divided angles Results

∠ABC = ... ... ... ... ∠ABD = ... ... ... ... ∠CBD =

∠BCD = ... ... ... ... ∠BCA = ∠ACD =
∠CDA = ... ... ... ... ∠ADB = ∠BDC =
∠BAD = ... ... ... ... ∠DAC = ∠CAB =

Conclusion :

b. Diagonals of a square bisect each other at right angle.

Experimental Verification: C D C
Experiment:
D B
C

O O
O

B AB AA D

(i) Three squares, namely ABCD are drawn with different sizes by using set squares (compass)

(ii) Diagonals AC and BD are drawn which cut each other at O.
To Verify: AC = BD and AC ⊥ BD

Verification: Lengths of AO, CO, BO, DO and size of ∠AOB are measured in

each square and their measures are tabulated.

Fig. AO CO BO DO ∠AOD Result

(i) AO = CO

(ii) BO = DO

(iii) ∠AOD = 900

Conclusion : We observed in each square, AO = CO, BO = DO and ∠AOD = 900. Hence

diagonals of a square bisect each other at right angle.

Verified.

Rhombus A B

A rhombus is a parallelogram in which all the sides are equal.

In the given figure, ABCD is a rhombus where sides

AB = BC = CD = AD.

A rhombus holds all the properties of parallelogram.

Extra properties: D C

a. Diagonals of a rhombus bisect each other at right angle.
Experimental Verification:
Experiment:

28 1 Geometry

B CC Prime Mathematics Book - 8
CD D

OO O

B (i) A A (ii) DB (iii) A

(i) Three rhombus namely ABCD of different sizes are drawn.

(ii) In each figure, diagonals AC and BD are drawn which intersect each other at O.

To Verify: AO = CO, BO = DO and AC ⊥ BD
Verification: AO, CO, BO, DO and ∠AOB are measured and their measures are tabulated as

Fig. AO CO BO DO ∠AOB Result

(i) AO = CO, BO = DO, ∠AOB = 900

(ii) AO = CO, BO = DO, ∠AOB = 900

(iii) AO = CO, BO = DO, ∠AOB = 900

Conclusion : In this experiment, we observed that AO = CO, BO = DO and ∠AOB =
900. Hence diagonals of a rhombus bisect each other at right angle.

Verified.

b. Diagonals of a rhombus bisect opposite angles. Complete experimental verification
left for you.

Example 1 : Calculate the unknown angles of the given figure. Ay zB

Solution : In the figure

y = 1150 (Being opposite angles of a parallelogram)

x + 1150 = 1800 (Being a co-interior angles) Dx 1150 C
or, x = 1800 - 1150

∴ x = 650

And z = x (Being opposite angles of a parallelogram)

∴ z = 650

Example 2 : Find the values of x and y from the given figure. (3x - 5) cm (x + 8) cm
Solution : Here, (x + 3) cm3y cm

3x - 5 = x + 3 (Being opposite sides of a
parallelogram)
or, 3x - x = 3 + 5
or, 2x = 8

Triangles, Quadrilaterals & Polygons 29

Prime Mathematics Book - 8

8
∴ x=2 =4

Similarly, 3y = x + 8

or, 3y = 4 + 8 = 12
12
∴y= 3 =4

Example 3 : From the given alongside figure, find the size of C D E
the angles represented by a, b and c. a 800

∠ADC + 800 = 1800 (Being linear pair)

Solution : ∴ ∠ADC = 1000

bc

Now, a = 1 ∠ADC BA
2
(Diagonal of a rhombus bisects opposite angles)

or, a = 1 × 1000 = 500
2

∴ b = a = 500 (Base angles of isosceles triangle)

∠DAB = ∠ADE = 800 (Being alternate angles)

Now, c = 1 ∠DAB (Diagonal of a rhombus bisects the opposite angles)
2
1 Exercise 2.2
= 2 × 800

∴ c = 400

1. Find the values of x, y, z, a, b and c from the following figures:

(a) a b (b) b c (c) d 2a a

650 c 75 0 a d c b
x w 2x + 30 0
(d) z (e) z (f) y x + 640

y x + 40 0 y 3x 5x - 8 0 4x + 260

(g) 50 0 c (h) x y (i) a b
40 0
d

70 0 ab 60 0
75 0 d c
30 1 Geometry

Prime Mathematics Book - 8

2. From the following parallelograms, find the values of x and y.

(a) (4y - 8) cm (b) (c)

3x cm 3x y 5 cm y
(4x - 5) cm5o x 5 cm
4.2 cm(16-x)cm

2y cm

3. Verify experimentally that:
(a) the opposite angles of a parallelogram are equal.
(b) the opposite sides of a parallelo gram are equal.
(c) the diagonals of a parallelogram bisect each other.
(d) the diagonals of a rectangle are equal.
(e) the diagonals of a rhombus bisect each other at right angle.
(f) the diagonals of a square are at right angle.

2.3 Construction of polygons:

2.3.1 Construction of rectangles :
A. Construction of rectangle when length and breath are given:

Example :
Construct a rectangle ABCD in which length AB = 6.4 cm and breadth BC = 4.2 cm.

Steps of Construction: D X
(i) Draw a line segment AB = 6.4 cm. C

(ii) At B, draw ∠ABX = 90o and take B
X
BC = 4.2 cm along BX. CY

(iii) From A, take an arc of radius AD = BC B

and from C take an arc of radius CD= AB to A 6.4 cm
intersect at D.

(iv) Join A and D, C and D. Thus, ABCD is the required rectangle.

B. Construction of rectangle when a side and angle D
made by a diagonal with the side are given: 30o

Example : A 6 cm
Construct a rectangle ABCD in which side
AB = 6 cm and ∠BAC = 30o

Steps of Construction:
(i) Draw a line segment AB = 6 cm.
(ii) At B, draw ∠ABX = 90o and at A draw ∠BAY = 30o.

Triangles, Quadrilaterals & Polygons 31

Prime Mathematics Book - 8

(iii) AY and BX cut at C. Now, from A take an arc with radius AD = BC and from C
take an arc with radius CD = AB to cut at D.

(iv) Join A,D and C,D. Thus, ABCD is the required rectangle.

C. Construction of rectangle when lengths of diagonals and an angle between them are given:

Example :

Construct a rectangle ABCD when AC = BD = 7 cm angle between AC and BD is 60o.

Steps of Construction: Y
D
(i) Draw a line segment AC = 7 cm and locate its
mid point say O. A 60o C
(ii) Since angle between the diagonals is 60o , at O
draw ∠COY = 60o. B O
(iii) Produce YO to OX and take OB = OD, OC = OA. X
(iv) Join A and B; B and C; C and D; A and D.
Thus, ABCD is the required rectangle.

Exercise 2.3.1

1. With the given measurements construct a rectangle.
(a) ABCD where AB = 5.8 cm and BC = 4.2 cm.
(b) PQRS where PQ = 7.2 cm and PS = 4 cm.
(c) WXYZ where WX = 6 cm and XY = 4 cm.
(d) ABCD where AB = 6.5 cm and AD = 4.5 cm.
2. With the given measures construct a rectangle
(a) ABCD where AB = 6.4 cm and ∠BAC = 30o
(b) PQRS where PQ = 7 cm and ∠PQS = 45o
(c) WXYZ where XY = 5.8 cm and ∠YXZ = 60o
(d) KLMN where KL = 7.2 cm and ∠MKL = 75o
3. Construct the following rectangles with the given measures.
(a) ABCD where length of diagonal is 7.5 cm and angle between the the diagonals is 30o.

(b) PQRS where length of diagonals is 6.3 cm and angle between them is 45o.

(c) WXYZ where WY = XZ = 5.6 cm and angle between the diagonls is 60o.

(d) EFGH where EG = FH = 6 cm and angle between the diagonals is 75o.

4. Construct ABCD in which

(a) diagnol AC = BD = 7cm. and AB = 5cm.

(b) diagnol AC = BD = 8cm. and side BC = 6cm.

5. Construct a rectanguars having the given conditions :
a) A rectangle PQRS having diagonal PR = 6cm, PQ = 4cm, ∠QPR = 60°
b) A rectangle ABCD having diagonal AC = 8cm, AB = 5cm , ∠CAB = 45°.
c) A rectangle WXYZ having diagonal XZ = 6.7cm , XY = 4.2cm , ∠ZXY = 75°.

32 1 Geometry

Prime Mathematics Book - 8

2.3.2 Construction of regular polygons : C

A.. Construction of an equilateral triangle.
Example :
Construct an equilateral triangle of side 5 cm.

Solution: Steps of Construction:

(i) Draw a line segment AB = 5 cm.

(ii) From A and B, take arc of radius A 5 cm B
AC and BC = AB = 5 cm to cut at C.

(iii) Join A,C and B,C. Thus, ABC is the required equilateral triangle.

B. Construction of a square: D X
When length of side is given. C

Example : B
Construct a square ABCD of side AB = 5 cm

Solution: Steps of Construction:

(i) Construct a line segment AB = 5 cm A
and at B, draw ∠ABX = 90o.

(ii) Along BX, take BC = AB = 5 cm.

(iii) Take AD = CD = 5 cm taking arcs of radius 5 cm.

(iv) Join A, D and C, D.

Thus, ABCD is the required square.

C. When length of diagonals are given: Diagonals of a square
Example : bisect each other at
Construct a square ABCD where AC = BD = 6 cm. right angles.

Solution: Steps of Construction:

(i) Draw a line segment AC = 6 cm. X
B
(ii) Draw a perpendicular bisector XY of AC OC

which cuts AC at O. 1
2
(iii) Along OX and OY take OB = OD = BD.

i.e. OB = OD. A

(iv) Join A and B; B and C; A and D; C and D.

Thus, ABCD is the required square.

D
Y

Triangles, Quadrilaterals & Polygons 33

Prime Mathematics Book - 8

D. Construction of a regular pentagon:

Example :

Construct a regular pentagon of side 4 cm.

Solution: For a regular pentagon, number of sides (n) = 5

) )∴ Size of an interior angle = 180o n-2 D
n
) )= 180o 5 - 2 108 o
5
E C
= 36o × 3 = 108o 108o

Steps of Construction:

(i) Draw a line segment AB = 4 cm. A 108o B
(ii) At B, draw ∠ABC = 108o and take BC = 4 cm.
4 cm

(iii) At C, draw ∠BCD = 108o towards the same side and take CD = 4 cm.

(iv) Repeat the same process at D and join A and E.

Thus, the required regular pentagon ABCDE is constructed.

E. Construction of a regular hexagon: ED

Example :

Construct a regular hexagon of side 4 cm.

Solution: F 120o C
Steps of Construction:
4 cm
(i) Draw a line segment AB = 4 cm. 120o
(ii) Draw ∠ABC = 120o and take BC = AB = 4 cm. A
4 cm B

(iii) Draw ∠BCD = 120o towards same side and take CD = 4 cm

(iv) Proceed the same process at D and E.

(v) Join A and F. Thus, ABCDEF is the required regular hexagon.

Alternative method:

For a regular hexagon, E D
A
) ) ) )interior angle = n-2 180o = 6-2 180o = 120o OC
n 6
120o
Steps of Construction: F
B
(i) Draw a line segment AB = 4 cm.

(ii) At B, draw ∠ABC = 120o and take

BC = AB = 4 cm

(iii) Draw the perpendicular bisectors of AB

and BC which intersect at O.

34 1 Geometry

Prime Mathematics Book - 8

(iv) Taking O as centre and OA, OB & OC as radius, draw a circle.
(v) Take arcs AB = BC = CD = DE = EF = AE and join C, D; D, E; E, F; F, A.
Thus, ABCDEF is the required regular hexagon.

Note: For higher polygon, the alternative method is preferable.

F. Construction of a regular octagon:

Example: Construct a regular octagon of side 3 cm.

Solution: For a regular octagon,
) )interior angle =
n-2 180o where n = 8 FE
n o

) )= 8-2 180o G D
8 C

= 36 × 180o 45 135o
8 2 =

Steps of Construction: H
(i) Draw a line segment AB = 3 cm.
(ii) At B, draw ∠ABC = 135o and take 3 cm 1350
BC = AB = 3 cm.
AB

(iii) Draw the perpendicular bisectors of AB and

BC which cut at O.

(iv) Taking O as centre and OA = OB = OC as radius, draw a circle.

(v) Now take equal arcs AB = BC = CD = DE = EF = FG = GH = AH.

(vi) Join C, D; D, E; E, F; F, G; G, H; H, A.

Thus, ABCDEFGH is the required regular octagon.

Exercise 2.3.2

1. Construct an equilateral triangle of side

(a) 5 cm (b) 6 cm (c) 4.8 cm (d) 6.4 cm
(d) 6.5 cm
2. Construct a square of side. (d) 7 cm
(d) 4.5 cm
(a) 5 cm (b) 4.6 cm (c) 6 cm (d) 5 cm

3. Construct a square having length of diagonal

(a) 5 cm (b) 6 cm (c) 6.4 cm

4. Construct a regular pentagon of side

(a) 4 cm (b) 5 cm (c) 6 cm

5. Construct a regular hexagon of side

(a) 3.5 cm (b) 4 cm (c) 4.4 cm

6. Construct a regular octagon of side

(a) 3 cm (b) 3.5 cm

Triangles, Quadrilaterals & Polygons 35

Prime Mathematics Book - 8

Unit - 3 Congruency of Triangles Estimated period : 5

Introduction A

We compare two triangles (or figures) in the following ways

(i) In regard to area only: If two triangles ABC and DEF are

equal in area, we write ∆ ABC = ∆DEF. 6cm

Area of ∆ABC = 1 BC × AX X
2 8 cm
1 B C
= 2 × 8 cm × 6 cm D
8 cm
= 24 cm2
1 F
Area of ∆DEF = 2 CD × DY

= 1 × 6 cm × 8 cm = 24 cm2
2
∴ ∆ABC = ∆DEF Y
E 6 cm

Note: Triangles equal in area may or may not be of same shape.

(ii) If two triangles are equal in all respect A D

(shape and size) they are said to be

congruent or identically equal. In the

given figure, ∆ABC and ∆DEF are congruent.

In symbol, we write, ∆ABC ≅ ∆DEF B CE F

The symbol ≅ stands for “is congruent to”. The symbol ≅ carry

two symbols ~ (similar) and = (equal). A

(iii) If two triangles have same shape, they are said DB C
to be similar. In the given figure ∆ABC and ∆DEF F
are similar. We write ∆ABC ~ ∆DEF. The symbol ~ E
stands for “is similar to”.

Congruency of two triangles:

Congruent figures: Two figures are said to be congruent if both of them are of same shape

and size. When one figure is placed over other figure, congruent figures fit to each other.

Congruent triangles: A P

800 7.2 cm 800 7.2 cm

400 C
5.2 cm
5.2 cm

600 8 cm Q 600 8 cm 400 R

B

36 1 Geometry

Prime Mathematics Book - 8

Here, ∆ABC and ∆PQR have same shape and size. If we put ∆ABC over ∆PQR, they fit
exactly. Vertices A, B and C coincide with vertices P, Q and R; ∠A, ∠B and ∠C coincide
respectively with ∠P, ∠Q and ∠R; sides AB, BC and AC coincide respectively with
sides PQ, QR and PR. We say ∆ABC is congruent to ∆PQR, Symbolically, ∆ABC ≅ ∆PQR.

The parts which coincide to each other are called the corresponding parts.

Points correspond Angles correspond Sides correspond
AP ∠A = ∠P AB = PQ

BQ ∠B = ∠Q BC = QR

CR ∠C = ∠R AC = PR

In congruent triangles, the angles opposite to the equal sides are corresponding angles
which are equal and the sides opposite to the equal angles are corresponding sides
which are also equal,

It is better to write the corresponding vertices in order.

i.e. A P, B Q and C R

We write ∆ABC ≅ ∆PQR

Condition for congruency of triangles:

A) S.A.S. Congruency: A

If two sides a triangle respectively equal to the two sides of C
other triangle and the angles contained by the two sides are F
equal, then the triangles are congruent.

B

Given: AB = DE; AC = DF and ∠ BAC = ∠ EDF. D
To prove: ∆ABC ≅ ∆DEF

Proof: Applying ∆ABC to the ∆DEF so that the point A
falls on the point D and the side AB along DE.
Then, since AB falls on DE and ∠ BAC = ∠ EDF, E
AC must fall along DF and since AC = DF, thus
the point C must coincide with the point F.

∴ BC must coincide with EF.
Hence, ∆ABC coincides with the ∆DEF.

∴ ∆ABC ≅ ∆DEF

Note: Congruence of certain parts implies that correspondence ABC DEF is
congruence. However we can not prove this congruence from the facts that
we have in the system. So we assume S.A.S. as a congruence axiom.

Congruency of Triangles 37

Prime Mathematics Book - 8

B) S.S.S. Congruency theorem:

If two triangles have three sides of one equal to the three sides of the other, each to
D
each, they are equal in all respect.

AD

EF

B CE FG

Given : Let ABC and DEF are two triangles in which AB = DE, BC = EF and AC = DF.

To prove : ∆ABC ≅ ∆DEF

Proof : Applying the ∆ABC and ∆DEF so that B falls on E and BC along EF. And A is
on D which is opposite of the side EF. then since BC = EF, C must coincide
with F. Let GEF is the new position of the ∆ABC. Joining G and D.

Since, ED = EG ∴ ∠EDG = ∠EGD .............. (I)

Again, since DF = GF ∴ ∠FDG = ∠FGD .............. (II)

Hence, ∠EDG + ∠FDG = ∠EGD + ∠FGD adding (I) & (II)

∴ ∠EDF = ∠FGE i.e. ∠EDF = ∠BAC

Now, in ∆s ABC and DEF

AB = DE, AC = DF and included angle ∠BAC = EDF

∴ ∆ABC ≅ ∆DEF by SAS congruency axiom.

C) Angle side angle (A.S.A.) Congruency theorem: C

If correspondence of two triangles are such that the A B
two angles and the side common to them of one are G E
respectively equal (congruent) to the two corresponding
angles and the side common to them of other then the F
correspondence is a congruence of the triangles.

Let ∆ABC ↔ ∆DEF denote the correspondence of triangles in
which ∠A = ∠D, AB = DE and ∠B = ∠E.

In DF, there be a point G such that AC = DG. In the correspondence
∆CAB ↔ GDE, CA = GD, ∠A = ∠D and AB = DE.

∴ ∆CAB ≅ ∆GDE (From SAS congruous axiom) D

∴ ∠ABC = ∠DEG ......... (i) (Corresponding angles of congruent triangles)

∠ABC = ∠DEF ......... (ii) (Given)

∴ ∠DEF = ∠DEG [from (i) and (ii)]

⇒ EF = EG i.e. G and F coincide
∴ ∆CAB ≅ ∆FDE

38 1 Geometry

Prime Mathematics Book - 8

D. AAS Congruency theorem: B A
C
If two triangles have two angles of one equal to two
angles of the other and any side of the first equal to the D
corresponding side of the other, the triangles are equal in
all respect. F

Here, ∠A = ∠D, ∠B = ∠E E
∴ ∠C = ∠F (Remaining angles)
Now in ∆ABC ↔ ∆DEF,
∠B = ∠E, BC = EF and ∠C = ∠F
∴ ∆ABC ≅ ∆DEF (by ASA congruence theorem)

A

E. Right angle, hypotenuse and side (R.H.S.) Congruence theorem:

Two right angled triangles which have their BC
hypotenuses equal and one side of one equal D
to one side of other, are equal in all respect.

Let ABC and DEF are two right angled triangles
in which ∠ABC and ∠DEF are right angles,
hypotenuse AC = hypotenuse DF and AB = DE.

It is to prove that the triangles ABC and C' E F
DEF are equal in all respects.

Proof: Applying ∆ABC to the ∆DEF so that AB falls on DF and C on which is opposite
of the side DE F.

Let C′ be the point on which c falls.

Then DEC′ represents the ∆ABC in its new position.

Since ∠DEF = ∠DEC′ = 900,
∴ EF and EC′ are in one straight line.
Again in ∆C′DF, Since DF = DC′ = AC
∴ ∠DFC′ = ∠DC′F (base angles of isosceles triangle)
In ∆ DEF and ∆DEC′,
∠DEF = ∠DEC′, ∠DFE = ∠DC′E, DE = DE.
∴ ∆DEF ≅ ∆DEC′ (by AAS congruence theorem)
∴ ∆ABC ≅ ∆DEF.

Note: Under SSA condition, the triangles need not to be congruent.

Congruency of Triangles 39

Prime Mathematics Book - 8

AB = DE , ∠ABC = ∠DEF, B A D F'
CE F
AC = DF or DF′

angle opposite to AB are
equal or supplementary. Such
case is called ambiguous
case. If ∠ABC = ∠DEF = 900,
the ambiguity disappears.

Example 1: Under which condition, the following pairs of triangles are congruent?

(a) A 5 cm D
C
40 0 5 cm

6 cmB400 E
F 3 cm
(b) 6 cm
A 3 cm PQ

B C
(c) L R
P

4 cm 4 cm

M 5 cm NQ 5 cm R
(d) N
W 600 M

700

3 cm 3 cm

X 600 700

YL

40 1 Geometry

Prime Mathematics Book - 8

Solution : (a) Here, AB = EF = 3cm, ∠BAC = ∠DEF = 400 and AC = DE = 5cm

∴ ∆ABC ≅ ∆EFD by SAS congruence axiom.

(b) Here, ∠ABC = ∠QPR (R)

AC = QR (H)

AB = QP (S)

∴ ∆ABC ≅ ∆QPR by R.H.S. congruence theorem.

(c) Here, LM = PQ = 4cm (S)

MN = QR= 5cm (S)

LN = PR = 6cm (S)

∴ ∆LMN ≅ ∆PQR by SSS congruence theorem.

(d) Here, ∠XWY = ∠MLN = 700 (A)

XW = ML = 3cm (S)

∠XWY = ∠LMN = 600 (A)

∴ ∆WXY ≅ ∆LMN by ASA congruence theorem.

Example 2 : Find the values of x, y and z from the given congruent triangles.

5 cm A D x+3

y + 2 cm 800

Z

B 600 400 C E 9 cm 400 F
(2x + 1) cm

Solution :

Here, ∆ABC ≅ ∆DEF (given)

∠A = 1800 - (600 + 400) = 800

∠E = 1800 - (800 + 400) = 600
∴ ∠A ↔ ∠D, ∠B ↔ ∠E and ∠C ↔ ∠F

∴ BC = EF (corresponding sides of congruent triangles.)

or, 2x + 1 = 9

or, 2x = 8

∴ x = 4

AC = DF (corresponding sides of congruent triangles.)

(y + 2)cm = x + 3cm

or, y +2 = 4 + 3

or, y = 7-2

∴ y = 5

And AB = DE (corresponding sides of congruent triangles.)

∴ z = 5cm

Congruency of Triangles 41

Prime Mathematics Book - 8

Example 3 : Prove that the opposite sides of a parallelogram are equal.

Solution :

Given : ABCD is a parallelogram, A B
D C
where AB//DC and AD//BC.

To prove : AB = DC and AD = BC

Construction : A and C are joined.

Proof :

S.N. Statements S.N. Reasons
1.
1. In ∆s ABC and CDA (i) Being alternate angles as AB//DC.
(ii) Being common side.
(i) ∠BAC = ∠ACD (A) (iii) Being alternate angles as AD//BC.
2. From statement 1, by ASA congruence theorem.
(ii) AC = AC (S) 3. Corresponding sides of congruent triangles.

(iii) ∠ACB = ∠CAD (A)

2. ∆ABC ≅ ∆CDA

3. AB = DC and AD = BC

Q.E.D.

Example 4 : In the given figure, CP ⊥ AB, BQ ⊥ AC and AB = AC. Prove that CP = BQ.

Solution : A
Given : CP ⊥ AB and BQ ⊥ AC.

or, ∠APC = 900 and ∠AQB = 900 PQ
AB = AC

To prove : CP = BQ

Proof : BC

S.N. Statements S.N. Reasons
1. 1. Being both angles are
(i) In ∆s APC and AQB (i) 900 as CP ⊥ AB and BQ ⊥ AC.
(ii) (ii) Being common angle.
(iii) ∠APC = ∠AQB (A) (iii) Given.
2. 2. By AAS congruence theorem.
3. ∠PAC = ∠QAB (A) 3. Corresponding sides of congruent triangles.

AC = AB (S)

∆ADC ≅ ∆AQB

CP = BQ

Q.E.D.

42 1 Geometry

Prime Mathematics Book - 8

Exercise 3

1. Under which conditions are the following pairs of triangles congruent?
(a) A P

10 cm 10 cm

C 8 cm B Q 8 cm R

(b) R QW

600

5.4 cm4 cm
4 cm6 cm

5.4 cm
6 cm
600 X
S
PY

(C) A

400 700 400 700 R

B C T 7 cm
(d) G
J

7 cm
5 cm
5 cm

H 6.8 cm IK 6.8 cm L
(e)
C P

A 300 7.2 cm 500 B Q 500 7.2 cm 300 R

Congruency of Triangles 43

Prime Mathematics Book - 8

2. State the conditions under which the following pairs of triangles are congruent.
Also write down the corresponding sides and angles in each case.

(a) P RA

Q B C
(b) A J

B CK L
(c) D EP R

F Q
(d) L W

M NY X

3. From the following pairs of congruent triangles, find the values of x and y.

(a) A (y + 5) cm CD (x + 3) cm 400 F

600 (x - 1) cm
600 (y + 1) cm
6 cm
(x + 2) cm

800 E

B

44 1 Geometry

(b) P x cm RS 8 cm Prime Mathematics Book - 8
T
300 U

(y + 1) cm 2ycm x-2 (x +2)cm

600

Q

(c) A 860 4.2 cm C D
E
540

(x + 5) cm (x + 2) cm

400 5.6 cm 860 F

B

(d) K P x R

1250 300

4 cm (1.7x - 1.3)cm (2x + 2) cm

1250 250 M
L (x + 3) cm Q

(e) The given triangles ABC and XYZ are Congruent triangles. Find the values of x.

A 4.3cm. C X (3x - 3.9)cm.
35° Z
4.9cm.
123° 22° 123°
B
Y

Congruency of Triangles 45