Prime Mathematics 7

Prime

Mathematics
Series

7

Raj Kumar Mathema

Dirgha Raj Mishra Bhakta Bahadur Bholan

Uma Raj Acharya Yam Bahadur Poudel

Naryan Prasad Shrestha Bindu Kumar Shrestha

Prime

Mathematics
Series

7

Approved by
Government of Nepal, Ministry of Education, Science and
Technology Curriculum Development Centre Sanothimi,

Bhaktapur as and additional learning materials.

Raj Kumar Mathema Authors
Dirgha Raj Mishra Bhakta Bahadur Bholan
Uma Raj Acharya Yam Bahadur Poudel
Naryan Prasad Shrestha Bindu Kumar Shrestha

Editors
Anil Kumar Jha
Dhurba Narayan Chaudhary
Hari Krishna Shrestha

Language Editor
Mrs. Tara Pradhan

Pragya Books & Distributors Pvt. Ltd.

Printing history
First Edition 2074 B.S.
Second Edition 2077 B.S.

Authors
Raj Kumar Mathema
Dirgha Raj Mishra
Bhakta Bahadur Bholan
Uma Raj Acharya
Yam Bahadur Poudel
Naryan Prasad Shrestha
Bindu Kumar Shrestha

Editors
Anil Kumar Jha
Dhurba Narayan Chaudhary
Hari Krishna Shrestha

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 : 978-9937-9170-1-8

Printed in Nepal

Published by
Pragya Books & Distributors Pvt. Ltd.

Lalitpur, Nepal
Tel : 5200575
email : [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. Dinesh
phdera,Mr.Nara Bahadur Gurung,Mr.Ram Narayan Shah,Mr.Tulsi Kharel,Mr.Mani
Ram Khabas, Mr. Umesh Acharya, Mr. J. Phuldel, Mr. Kamal RajTripathee, Mr. Rudra
Prasad Pokharel, Mr. Uttam Prasad Panta, Mr. L.N. Upadhyaya, Mr. Shakti Prasad
Acharya,Mr.UpendraSubedi,Mr.KulNarayanChaudhary,Mr.BishonathLamichhane,
Mr. Harilal Lamichhane, Mr. Govinda Paudel, Mr. Krishna Aryal, Mr. Nim Bhujel, Mr.
Santosh Simkhada,Mr.Pashupati Upadhyaya,Mr.DipakAdhikari,Mr.MuktiAdhikari,
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

Unit Topic Pages

1 Geometry 1-54

1.1 Lines and angles 2

1.2 Angles in Pairs 6

1.3 Triangles, Quadrilaterals and Polygonals 14

1.4 Construction of Triangles 20

1.5 Construction of Special Quadrilaterals 22

1.6 Polygon 28

1.7 similarity and Congruency of Figures 33

1.8 Congruent Figures 38

1.9 Circle and its parts 42

1.10 Geometrical Solids 45

2 Co-ordinate Geometry 55 - 60

2.1 Introduction of Quadrants 56

2.2 Plotting the given points in the Graph 58

3 Mensuration 61 - 84

3.1 Perimeter of Plane Figures 62

3.2 Area of Plane Figures 67

3.3 Surface area of Cuboids and Cubes 77

3.4 Volume of Cuboids and Cubes 80

4 Transformation 87 - 110

4.1 Transformation 88

4.2 Symmetry, Tessellation of Design of Polygon 97

4.3 Bearing and Scale Drawing 104

5 Sets 111 - 126
5.1 Describing Sets 112
5.2 Set Operation 115
6 Whole Numbers 127 - 158
6.1 Squares and Square roots 128
6.2 Cubes and Cube roots 135
6.3 Highest Common factors (H.C.F.) 139
6.4 Lowest Common Multiple (L.C.M.) 143
6.5 Number System 148
7 Integers, Rational and Irrational Numbers 159 - 176
7.1 Integers 160
7.2 Rational Numbers 168
7.3 Order of Operation 172
8 Fraction and Decimals 177 - 196
8.1 Operations on Fractions 178
8.2 Word problem on Fraction 181
8.3 Decimals 183
8.4 Fundamental Operations with Decimals 169
9 Percentage, Ratio and Proportion 197 - 212
9.1 Percentage 198
9.2 Ratios 203
9.3 Proportions 206

10 Profit and Loss 213 - 222
10.1 Profit and Loss 214
10.2 Problems involving percentage in Profit and Loss 216
11 Unitary Method 223 - 228
11.1 Direct Variation 224
11.2 Indirect Variation 226
12 Simple Interest 229 - 236
12.1 Introduction 230
12.2 Finding, Time, rate of interest and principal 232
13 Statistics 237 - 252
13.1 Introduction 238
13.2 Graphical Representation of Statistical Data 242
13.3 Analysis of Data 247
14 Algebra 253 - 301
14.1 Algebric Expressions 254
14.2 Indices 275
14.3 Equation, Inequality and Line Graph 281
PRIME Questions for Practice 302
Model Question 308

UNIT Objectives:

1 GEOMETRY
Estimated periods 39

At the end of this unit students will be able to
● bisect angles and construct standard angles using compass.
● transform given angle using compass.
● identify angle pairs (adjacent angles, vertically opposite angles,

complementary and supplementary angles, alternate angles, corresponding
angles, co-interior angles and to verify their relations.
● know the properties of triangles and quadrilaterals.
● construct triangles and special quadrilaterals.
● calculate the interior and exterior angles of regular polygon.
● identify similar and congruent figures.
● identify different parts of circles.
● identify different polyhedra and draw nets and make models of polyhedra,
cylinders and cones.

D

Teaching Materials:
paper models of triangles, quadrilaterals and polygons, models of solids,
geometry box, geoboard.

Activities:

It is better to
● demonstrate construction of different angles, bisecting and transferring

angles.
● discuss about relation of angle pairs and demonstrate with experiment.
● discuss about properties of triangles and quadrilaterals and demonstrate

with experiments.
● demonstrate construction of triangles and quadrilateral under different conditions.
● discuss about interior and exterior angle of polygons.
● demonstrate the nets and models of special solid objects.

Lines and Angles Estimated Period - 15

1.1 Construction of angles using compass.

By using Euclidian tools we can draw standard angles and their halves

( )1 1 1
2 , 4 , 8 only. Construction of angles using compass is based on angle of 60° and
process of angle bisecting.

Construction of angle 60°. o

To construct ABC = 60° C
i. draw a line segment BA.
ii. Taking B as centre draw an arc to cut AB at M. r r
iii. Taking M as centre and keeping the extension of compass same, A
B 60o
take arc to cut the arc drawn in Step II at N.
iv. Join BN and produce to C. r

Thus ABC = 60° is constructed .

Definition of angle of 60o: Angle subtend by a chord of a circle, equal
to the radius of the circle, at the centre is 60°.

Historical fact: 60o
Babylonians subdivided the circle using the angle of an equilateral
triangle as basic unit and further subdivided the later into 60
parts following their sexagesimal numeric system.
Thus, angle of an equilateral triangle = 60° and six angle of
equilateral triangle = 1 complete angle = 360°

Note: Degree is not SI unit. In SI unit, radian is the unit of measurement of angle.

Bisecting an angle:

Given ABC

Taking vertex B as centre, take an arc to cut the arms BA

and BC at M and N respectively.

Taking the extension of compass more than half of the

chord MN, take arcs from M and N to intersect at P

interior to ABC.

Join B and P 1
2
Thus, ABP = CBP = of ABC

2 Geometry

Construction of angle 30o
1
Since 30° = 2 of 60°,

Draw a line segment BA.
Construct ABC = 60° (No need to draw the arm BC,
mark the point N, only)

Draw bisector BD of ABC 30o
Thus, ABD = 30° 120o

Construction of angle 120o 90o
45o
Since 120° = 2(60°)
Draw a line segment BA 135o
Taking B as centre, draw an arc, to cut AB at L.
Without changing the extension of compass, take arcs

LM = MN = as in the figure.
Join BN and extend to C.
Thus ABC = 120° is constructed.

Construction of angle 90o

Since 90° = 60° + 1 of 60°
2

draw a line segment OA

taking O as centre, draw an arc to cut OA at L.

mark M and N corresponding to 60° and 120° angles.
1
draw bisector OB of MON such that MOB = 2 of 60°
Thus, AOB = 90° is drawn.

Construction of angle 45°
Since 45o = 1
2 (90°)

construct angle 90°

draw bisector OB of angle 90°.
Thus AOB = 45° is constructed.

Construction of angle 135°.
1
Since 135° = 90+ 2 (90°)

draw a line segment OA

taking O as centre, draw an arc greater than half circle.

draw AOM = 90°, no need to draw arm OM, just mark
the point M on the arc.

produce AO to C. Now draw bisector OB of MOC = 90°
such that AOB = AOM + MOB = 90° + 45° = 135°.
Thus AOB = 135° is constructed.

Prime Mathematics Book - 7 3

Construction of angle 150o.
Sincec1o5n0st°ru==c1t12200A°°O++3M021°=(16200°°) and produce AO to N.
TdhrMuaOsw,Bth=AeO21bB(i6s=e0c°t)o=rA3OO0MB°o+f MON, such that 150o
MOB = 120° + 30° = 150° is
constructed.

Construction of angle 75°. 75o

SinacccsooeAnrO7rsLBe5tOsr°=puN=co=tnA6dO30i0°LnoAg+,+Odt1Nor5LaO°6w=0Bob=,i9s160e20co,0otoo+rma1On5aBdorko=9f07to5hLooe.OnNtphsoueicnahtrstch.aLt, M and N

SCaisonicnnestt1hr0eu5cocot=nios9t0nrou+oct1fi5oaonn=ogf9l0eaon+1gl012e5(73o5.0oo)explained above,
TNinhOsutBse,a=d21AoO(fB3b0=ios)ec=AtO1inN5go+. LON, draw bisector OB of MON such that
NOB = 90o + 15o = 105o is constructed. 105o

Note: Use the idea explained above to draw angles 15o, 2221 o 165o, 6712 o etc.

,

Request to the teachers: In the current curriculum, construction of angles is not in
class 8, 9 and 10. So the teachers are requested to explain the processes to construct
possible different angles.

Transferring angles (To construct an angle equal to a given angle).

Given ABC, let we need to construct PQR = ABC
Draw a line segment QP.
In ABC, taking B as centre, taken an arc to cut AB and BC at Y and X respectively.
Taking Q as centre, draw same arc cutting PQ at M.
Adjust the compass if necessary, put needle at X and extend to touch Y with
pencil, measure XY, (Chord).
Put the needle at M and take same arc to take arc MN = arc XY.
Join QN and produce to R.
Thus PQR = ABC is constructed.

4 Geometry

Exercise 1.1 d)

1. Bisect the given angles and measure each part.
(a) (b) c)

2 a. Constructs an angle of 30o and bisect it by compass.
b. Construct an angle of 75o and bisect it by compass.
c. Construct an angle of 45o and bisect it by compass.

3. Draw a line segment AB = 7cm. and constuct the angles as given below.
(a) 30o at point A and 45o at point B.
(b) 120o at point A and 135o at point B.

4. Construct the following angles by using compass. g) 135o h) 75o i) 105o
a) 60o b) 120o c) 30o d) 90o e) 45o f) 150o

5. Construct the following angles.

a) 15o b) 22.5o c) 67.5o d) 82.5o

6. Construct the angle equal to the following angles.

a) b)

c) d)

Prime Mathematics Book - 7 5

1.2 Angles in Pairs.

A. Angle pairs under ordinary conditions.

(i) Complementary angles: If the sum of the two angles is equal to a right angle (90°),
the two angles are said to be complementary angles to each other. If A and B are
two angles such that A+ B = 90°, A and B are complementary angles to each other.

Here B is complementary of A
But B = 90°- A
So, 90° - A is complementary of A.

Example 1: Find the complementary of angle 40°.
Solution.

Complementary of 40° is 90° - 40° = 50°

Alternate method.
Let the complementary of 40° is x, then
40° + x = 90°
or x = 90°- 40°
\ x = 50°

So, complementary of angle 40° is 50°

(ii) Supplementary angles : If the sum of two angles is two right angles (180°), the two
angles are said to be supplementary angles to each other. If A and B are two angles
such that A+ B = 180°, A and B are supplementary angles to each other.

Here B is supplementary of A
But B = 180°- A [ A + B = 180°]
So,180o - A is the supplementary angle of A

Example 2: Find the supplementary of angle 110°.
Solution.
Supplementary of angle 110° is 180°-110° = 70°

Alternate method.
Let x be the supplementary of angle 110°, then

110°+ x = 180°
or x = 180-110°
\ x = 70°

So, supplementary of 110° is 70°

6 Geometry

B. Angle pairs formed when two straight line intersect each others:

When two lines intersect each others, the following pairs of angles are formed.
i) Vertically Opposite Angles:
When two straight lines intersect each other, the opposite
angles with common vertex so formed are called vertically
opposite angles. Vertically opposite angles are equal. Here
straight lines AB and CD intersect at O, where

AOC = BOD and
AOD = BOC are vertically opposite angles.
ii) Supplementary Adjacent Angles:
When two straight lines intersect each other, the angle
pairs with a common vertex and a common arm between
them so formed are called supplementary adjacent angles
or linear pairs.
Here.
AOC+ AOD = 180o
AOC+ COB = 180o
AOD+ BOD = 180o and
BOC+ BOD = 180o are supplementary adjacent angles or linear pairs.

Note:Adjacent angles have a common vertex and a common arm, other arm lie opposite
of the common arm.

AOB and AOC are adjacent angles.
If the sum of two adjacent angles is 90o, they are said to be complementary adjacent
angles.
If the sum of two adjacent angles is 180o, they are said to be supplementary
adjacent angles.

AOB + BOC = 90 o AOB + BOC = 180 o supplementary

\ AOB and BOC are comple- \ AOB and BOC are

mentary adjacent angles. adjacent angles.

Prime Mathematics Book - 7 7

Example 3: Find the size of the unknown angle from the given figures.
Solution.

a) Being complementary adjacent angles
x + 40o = 90o

\ x = 90o- 40o = 50o

b) Being linear pair 40o
y + 50o = 180o 50o

or y = 180o - 50o
\ y = 130o

Some experimental verifications on pair of angles.

In experimental verification, conditions for a geometric proposition are observed in
different figures and reach a conclusion from the common result so obtained. Geometry
in its beginning was experimental instead of logical.

Formal steps for an experimental verification.

I. Statement : Write the statement of the proposition or theorem.

II. Figure : Draw two or three figures representing the statement.

III. Experiment : Explain the steps of construction.

IV. To verify : Write mathematically, the condition to verify.

V. Verification : Tabulate the necessary measurements.

VI. Conclusion : Write the common result so obtained.

1. Verify experimentally that when a straight line meets another straight line, the

sum of the adjacent angles so formed is two right angles.
Solution:
Statement: When a straight line meets another straight line, the sum of the
adjacent angles so formed is two right angles.

i) ii)
8 Geometry

Experiment: Draw a straight line CD, meeting a straight line AB at C so that the
adjacent angles ACD and BCD are formed.

To verify: ACD + BCD = 180o

Verification: Measure ACD and BCD in each figure and tabulate the measures as:

Figure ACD BCD ACD+ BCD Result
(i) 130o 50o 130o+ 50o = 180o ACD+ BCD = 180o
(ii) 85o 95o 85o+ 95o = 180o

Conclusion: In the above experiment, we observed that ACD+ BCD = 180o in each
figure. Hence the sum of adjacent angles formed by two straight lines is 180o. Verified

2. Verify experimentally that the sum of all the angles around a point is equal to 360°.

Solution:
Statement: The sum of all the angles around a point is equal to four right angles.

i) ii)
Experiment: Straight lines AO, BO, CO, DO and EO from different directions meet at a
point O and AOB, BOC, COD, DOE and AOE are formed around the point O.

To Verify: AOB + BOC + COD + DOE + EOA = 360o
Verification: AOB, BOC, COD, DOE and EOA are measured and their measures
are tabulated as

Figure AOB BOC COD DOE EOA AOB+ BOC+ COD Result
+ DOE+ EOA

(i) 45o 86o 60o 64o 105o 360o AOB + BOC +
(ii) 60o 36o 84o 60o 120o 360o COD + DOE +

EOA = 360o

Conclusion: In this experiment, we observed that AOB + BOC + COD + DOE +
EOA = 360° in each figure. Hence the sum of all the angles around a point is equal to
Verified
360°.

Prime Mathematics Book - 7 9

3. Verify experimentally that when two straight lines intersect at a point, the vertically
opposite angles so formed are equal.

Experimental verifications.
Statement : When two straight lines intersect at a point, the vertically opposite angles
are equal.

i) ii)
Experiment: Draw two figures showing straight lines AB and CD meet at a point O

To verify : AOC = BOD and AOD = BOC

Verification: In each figure, AOC, BOD, AOD and BOC are measured and their
measures are tabulated as

Fig AOC BOD AOD BOC Result
(i) 55o 55o 125o 125o AOC = BOD and
(ii) 91o 91o 89o 89o
AOD = BOC

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

C. Angle pairs formed when two parallel lines are intersected by a
transversal.

Let parallel lines AB and CD are intersected by transversal EF at P and Q respectively.
Following special angle pairs one formed

i) Corresponding angles.
When a pair of parallel lines are intersected by a transversal the angle pairs formed to
same side of the transversal one interior and other exterior
non adjacent angles are called corresponding angles.
Corresponding angles are equal.

Here, in the figure alongside,
PQC = EPA,
APQ = CQF,
EPB = PQD and
BPQ = DQF are corresponding angles.

10 Geometry

ii) Alternate angles:

When a pair of parallel lines are intersected by a transversal a pair of angles formed
to opposite sides of transversal and non adjacent interior angles are called interior
alternate angles or simply alternate angles. Alternate angles are equal.

Here in the above figure,
APQ = PQD and BPQ = PQC are alternate angles.

Pair of angles formed to opposite sides of the transversal and exterior to the parallel
lines are called exterior alternate angles. Exterior alternate angles are equal, When
the pair of lines are parallel.

Here in the above figure, APE = DQF and EPB = CQF are exterior alternate angles.

iii) Co-interior angles:

Angle pairs lying to the same side of the transversal and interior of two parallel lines
are called co-interior angles. Co-interior angles are supplementary.

Here in the above figure, APQ and PQC; BPQ and PQD are co-interior angles.
where APQ+ PQC = 180o and BPQ+ PQD = 180o

The angle pairs lying to same side of the transversal and exterior of the two parallel
lines are called co-exterior angles. Co-exterior angles are supplementary.

Here, APE and CQP ; EPB and DQF are co-exteriors angles.
where APE+ CQF = 180o and EPB+ DQF = 180o

Note: When non-parallel lines are intersected by a
transversal, all the angles pairs explained above are
formed but they do not satisfy the given conditions.

Example 4: Find the value of x from the given figure. x

Solution:
APE = CQP = 110o (Being corresponding angles)

Again APE + EPB = 180o (Being a linear pair angles)

110o+ x = 180o
x = 180o-110o
\ x = 70o

Prime Mathematics Book - 7 11

Example 5. Find the values of angles x and y in the given figure.

Solution:

Here, in the figure,

y + 120o = 180o (being linear pair)

or y = 180o-120o \y = 60o 120o

Now, x = y (being alternate angle) \x = 60o

Exercise 1.2

1. Write down the complementary angle of each of the following angles.
(a) 38° (b) 74° (c) 88° (d) 90° (e) 22.5°

2. Write down the supplementary angle of each of the following angles.
(a) 105o (b)50o (c)120o (d) 65o (e) 160.5°

3. (a) If x and (x+15°) are complementary to each other, find measure of them.
(b) If 2x and x form a linear pair, find measure of them,
(c) A pair of supplementary angles are in the ratio 3:7. Find measure of them.
(d) Find the size of an angle which is four times its complementary.

4. Find the size of x, y and z. (c) (d)
(a) (b)

(e) (f) (g) (h)

55°

315° y

5 a. A straight line AB meets another straight line CD at the point C. Verify
experimentally that ACD+ BCD = 180°.
b. Verify experimentally that when two straight line intersect each other at a
point, the vertically opposite angle so formed are equal.
c. Straight lines OP, OQ, OR, OS and OT from different directions meet at O.
Verify experimentally that the sum of the POQ, QOR, ROS, SOT and
TOP is equal to four right angles.
d. Using relations of supplementary angles,
prove that
(i) the angles, a = c
(ii) the angle, b = d

12 Geometry

6. Name the pair of angles a and b and write the relation between them in each of
the following.

a) b) c)

d) e) f)

7. a. Which is the alternate angle of ∠PQR. P Y
b. Which is the co-interior angle of ∠SRQ. QU

c. Which is the corresponding angle of ∠QRT. S R T
W
8. From the given figure X
find the value of x + y. Y

9. a. If 3x + 10o and 4x - 5o are the angle of a linear pair, find their measurement.
b. If 3x - 3o and 5x + 7o are the angle of a linear pair, find their measurement.

10. a. Two supplementary angle are in the ratio of 4:5 find the angles.
b. Two complementary angles are in the ratio 2:3, find the angles.

c. Find the angle which is 10o more than its complement.
d. Find the angle which is 45o less than twice its supplement.

11. Find the values of unknown angles a, b, c, d from the following diagram.

a) b) c)

d) e) f)

g) h) i)

Prime Mathematics Book - 7 13

j) k) l)

Triangles, Quadrilaterals and Polygons Estimated Period - 12
1.3 Properties of triangles and quadrilaterals.

Properties of triangles.

1. The sum of angles
(interior) of a triangle is
always 180o

Draw three triangles namely ABC of different shape and sizes.
Measure the angles A, B and C and find their sum.

Figure A B C A+ B+ C Result
i 80o 70o 30o 180o A+ B+ C = 180o
ii 30o 100o 50o 180o
iii 110o 40o 30o 180o

Conclusion : We observed in each figure A+ B+ C = 180o.
Hence, the sum of angles of a triangle is 180o.

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

In DABC, side BC is produced to D forming exterior angle

ACD.
Then ACD = BAC+ ABC

14 Geometry

Draw three triangles namely ABC of different measures.
Produce side BC to D in each figure.

Measure ACD, BAC and ABC then complete the following table with their
measures and give conclusion.

Figure ACD BAC ABC BAC+ ABC Result

i

ii

iii
Conclusion:

Other properties of triangles
1. Base angles of an isosceles triangle are equal.
Here, in the figure, ABC is an isosceles triangle, where AB = AC
So, ABC = ACB.
Conversely, if base angles are equal,
the triangle is an isosceles triangle.
i.e. if ABC = ACB, then AB = AC.

2. Equilateral triangle is an equiangular.
In the figure AB = BC = AC.
So A = B = C = 60o

Properties of Quadrilaterals. O

1. Parallelogram : A parallelogram is a quadrilateral in
which opposite sides are parallel. In the given figure,
ABCD is a parallelogram where AB//DC and AD//BC
[//stands for parallel]

i) Opposite sides of a
parallelogram are equal.

Prime Mathematics Book - 7 15

Here in the given figure, Suggestion for teachers:
AB = DC and AD = BC. Drawing an attractive figure of a parallelogram
on the black/white board or showing a model
ii) Opposite angles of a parallelogram, let the students investigate
parallelogram are equal. i) About opposite sides
A = C and B = D ii) About opposite angles
iii) About diagonals, and list the properties.

Verify experimentally that opposite angles of a parallelogram are equal.

i) ii)
Draw parallelogram ABCD of different measures using set squares. Measures the angles.

Fig.No BAD ABC BCD CDA Result
i 66O 114O 66O 114O BAD = BCD
ii 140O 40O 140O 40O ABC = CDA

Conclusion: In this experiment, we observed that in parallelogram ABCD,
BAD = BCD and ABC = CDA. Hence opposite angles of a parallelogram are equal.

iii) Diagonals of a parallelogram bisect each other. Here, in the above figure, diagonals
AC and BD intersect at O where AO = OC and BO = OD.

The tree diagram of quadrilaterals.

Quadrilaterals

Convex quad Concave Quad

Trapezium Simple quad
Parallelogram Kite

Rectangle Rhombus

Square

16 Geometry

Venn - diagram of Quadrilateral

U

Convex Concave
TraPpaenragzllileuemlogram
Recta Rho
Square

mbus

Note : In any one angle of a quadrilateral is more B A
than 180o, the quadrilateral is known as concave
quadrilaeral. P C
D
If every angle of a quadrilateral is less than 180o, the
quadrilateral is known as convex quadrilateral. Q

S R
2. Rectangle : A rectangle is a parallelogram having each angle 90o

In the given figure ABCD is a rectangle where O
i. A = B = C = D = 90o
ii. AB//DC and AD//BC
iii. AB = CD and AD = BC
iv. Diagonals AC = BD
v. Diagonals bisect each other

AO = OC and BO = OD

3. Square: A square is a parallelogram in which all the sides are equal and each angle 90o.
In the given figure, ABCD is a square where
i) AB//DC, AD//BC and AB = CD = AD = BC.
ii) A = B = C = D = 90o O
iii) Diagonals AC = BD.
iv) Diagonals bisect each other i.e. AO = OC and BO = OD.
v) Diagonals AC and BD are perpendicularly bisect at O.

Prime Mathematics Book - 7 17

4. Rhombus: A rhombus is a parallelogram with all sides equal.
In rhombus ABCD
i) AB//DC, AD // BC
ii) A = C and B = D
iii) Diagonals AC ≠ BD.
iv) Diagonals bisect each other at O i.e. AO = OC and BO = OD.

Example 1: Find the values of x and y in the figure given alongside. A
Solution:
x
Here x = 65° [base angles of isosceles triangle are equal]

And y = x+65° [exterior angle of a triangle is equal to y

the sum of opposite interior angles.] 65o C

or y = 65°+65° DB x
y
\ y = 130°.

Example 2: Find the values of x and y in the figure given alongside.

Solution:
Here x = 70o [Opposite angles of parallelogram are equal]
\y = 70o
y = x [alternate angles] 70o

Exercise 1.3

1. Write the properties of (b) Isosceles triangle. (c) Triangle.
(a) Equilateral triangle.

2. Write the properties of (b) Rectangle (c) Square (d) Rhombus
(a) Parallelogram

3. Find the size of x and y c)
a) b)

2X-30o

2X-30o X+10o

d) e) f)

X-15o 4X
2X+25o

18 Geometry

g) h) i) j)

4. Find the size of x and y of following figures. c)
a) b) f)

y

d) e)

5. From the given figure, find the value of x, y and z.
a) A b)
B
120o 100o B A 3x 2x

DX 60o x 4x C
C D

c) A3x 2x-38o B d) A D
x y

D 3x + 20o xC 60o Z
B C

Prime Mathematics Book - 7 19

e) 4y 2x-60o f) x
y

2y x + 30o Z z 60o

6. a. If three angle of a quadrilateral are 60o, 115o, 105o, find the fourth angle.
b. If three angle of a quadrilateral are in the ratio 2:3:5 and fourth angle is 130o,
find the angles.

c. If 2x, 3x, 2x + 10 and x + 30 are the angle of a quadrilateral, find all angles.

7. Verify experimentally that
a) the sum of angles of a triangle is 180o.
b) exterior angle of a triangle is equal to sum of the non adjacent interior angles.
c) opposite angles of a parallelogram are equal.
d) diagonals of a rectangle are equal.

1.4 Construction of Triangles

There are different conditions for constructing triangles. Here, we will construct the
triangles under the following simple conditions.

1. When length of all the three sides are given (SSS condition).

For example, to construct a triangle ABC in which AB =
5cm, BC = 6cm and AC = 4cm.

Steps of construction.
i) Draw a line segment AB = 5 cm
ii) with centre B and radius 6cm, draw an arc.
iii) with centre A and radius 4cm, draw an arc to cut the

arc in step (ii) at C.
iv) Join B and C ; A and C.

Thus, the required DABC is constructed.

2. When length of two sides and angle made by them are given.
(SAS condition)

For example, to construct a triangle ABC in which
AB = 5.5cm, AC = 4.5 cm and BAC = 60o
Steps of construction.
i) Draw a line segment AB = 5.5cm
ii) At A, construct BAX = 60o
iii) On AX, take AC = 4.5cm, taking an arc of radius

4.5 cm.
iv) Join B and C.

Thus, the required triangle ABC is constructed.

20 Geometry

3. When two angles and their common side are given. (ASA conditon)

For example, to construct a triangle ABC in which
AB = 5.8 cm, CAB = 60o and ABC = 45o

Steps of construction. ABY = 45o
(i) Draw a line segment, AB = 5.8 cm,
(ii) At A, draw BAX=60o and at B draw

towards same side to intersect at C.

Thus DABC is the required triangle.

4. When hypotenuse and one side of a right angled triangle are given
(R.H.S condition).

For example, to construct a right angled triangle in which AB
= 5cm, B = 90o and AC = 6.8cm
Steps of construction:
i) Draw a line segment, AB = 5cm.
ii) At B, draw ABX = 90o
iii) With centre A and radius 6.8 cm, take an arc to cut BX at C.

Thus, ABC is the required triangle.

Note: RHS condition for construction of triangle is not same
as SSA condition. In SSA condition two triangles
( ABC or ABC’) may form so, SSA is not a simple condition
for construction of a triangle.
e.g. To construct ABC with AB = 6cm,
AC = 5cm and ABC = 45o

Exercise 1.4

1. Construct DABC with the following condition.
(a) AB = 5 cm, BC = 5.8 cm and AC = 6.4 cm
(b) AB = 5 cm, BC = 5 cm and AC = 5 cm
(c) AB = 5 cm, BC = 7 cm and AC = 3.5 cm
(d) AB = 6 cm, BC = 5.2 cm and AC = 5.2 cm

2. Construct triangle XYZ in which,
(a) XY = 4.8 cm, X = 45o and XZ = 6 cm.
(b) XY = 5 cm , Y = 60o and YZ = 7 cm
(c) YZ = 6 cm, XYZ = 75o and XY = 7.2 cm
(d) XY = 6.5 cm, YXZ = 120o and ZX = 4 cm.

Prime Mathematics Book - 7 21

3. Construct DABC under the following conditions.
A = 45o and B = 60o
(a) AB = 5 cm, A = 30o and B = 75o

(b) AB = 6 cm, B = 90o and C = 45o

(c) BC = 6 cm, A = 45o and C = 45o

(d) AC = 5.4 cm, B = 120o and C = 30o

(e) BC = 5 cm,

4. Construct the right angled triangle PQR in which
(a) PQ = 5.2 cm and hypotenuse PR = 7.2 cm
(b) QR = 4.5 cm and hypotenuse PQ = 6 cm
(c) PR = 6 cm and hypotenuse QR = 10 cm

1.5 Construction of special quadrilaterals
Construction of parallelograms.

A. When two adjacent sides and angle between them are given.
For example, to construct a parallelogram ABCD in which AB = 5 cm, BC = 4cm and

ABC = 60o
Steps of construction:

i) Draw a line segment, AB = 5cm

ii) At B, draw ABX = 60o and take BC = 4cm along
BX.

iii) Since opposite sides of parallelogram are
equal with centre A and radius BC = AD = 4
cm take an arc and with centre C and radius
AB = CD = 5 cm take another arc to cut at D.

iv) Join A, D and C, D.

Thus, the required parallelogram ABCD is constructed.

B. When two adjacent sides and a diagonal are given.

For example, to Construct a parallelogram
ABCD in which AB = 6cm, BC = 4cm and
diagonal AC = 8cm.
Steps of construction:
i. Draw a line segment, AB = 6cm
ii. With centre B and radius BC = 4cm take

an arc and with centre A and radius
AC = 8cm to cut the two arcs at C.

22 Geometry

iii. With centre A and radius BC = AD = 4 cm take an arc and with centre C and radius
AB = CD = 6cm to intersect the two arcs at D.

iv. Join B and C, A and C, C and D, A and D.
Thus, the required parallelogram ABCD is constructed.

C. When a base, a diagonal and angle made by the diagonal and the base are given.
For example, to construct parallelogram ABCD is which AB = 5.2cm, diagonal AC =
4.5cm and BAC = 30o.

Steps of construction :
i. Draw a line segment, AB = 5.2 cm.
ii. At A, draw BAX = 30o and take AC = 4.5 cm on

AX.
iii. With centre A and radius AD = BC, draw an arc

and with centre C and radius CD = AB draw an
arc to intersect previous arcs at D.
iv. Join B and C, C and D, A and D.
Thus, the required parallelogram ABCD is
constructed.
Note: There are some more conditions to construct parallelogram like
i) a base and length of two diagonals are given.
ii) length of two diagonals and angle between them given which you will be learning in higher
classes.

D. When two diagonals and angle between them are given.
For example, to construct a parallelogram ABCD, in which diagonal AC = 6.1cm, BD = 8.3
cm and angle between them 30o.

30o

Step of construction: 23
i. Draw a line segment AC = 6.1 cm & locate its mid point O.
ii. At O, draw CON = 30o and produce NO back to M.
iii. Draw a line segment XY = 8.3cm and locate its mid point P.
iv. With centre O, take OB = OD = PX = PY on MN.

Prime Mathematics Book - 7

v. Join A and B; B and C; C and D; A and D.
Thus, ABCD is the required parallelogram.

E. When one side and two diagonals are given.
For example, to construct a parallelogram in which
AB = 4.5 cm, AC = 6.4 cm and BD = 7.8 cm.

Steps of construction.
i. Draw a line segment AB = 4.5 cm.

ii. With centre A and radius = AC = 6.4cm = 3.2cm, draw an o
2 2

arc and with centre B and radius = BD = 7.8 cm = 3.9 cm
2 2

draw an arc to cut these arcs at O
iii. Produce AO and take AO = OC. Similarly produce BO and take BO = OD.
iv. Join B and C; C and D; A and D.

Thus, ABCD is the required parallelogram.

Construction of Rectangles

A. When adjacent sides of a rectangle are given.
For example: to construct a rectangle ABCD in which AB = 6cm and BC = 4.5 cm.

Steps of construction

i. Draw a line segment AB = 6 cm.

ii. At B, draw ABM = 90o and take BC = 4.5 cm on

BM.

iii. With centre A and radius AD = BC take an arc

and with centre C and radius AB = CD take 4.5 cm

another arc to cut at D.

iv. Join A and D, C and D. 6 cm
Thus, ABCD is the required rectangle.

B. When the lengths of a base and a diagonal are
given.
For example, to construct a rectangle ABCD in which AB = 5.5 cm and diagonal AC = 7cm.

Steps of construction:
i. Draw a line segment, AB = 5.5cm.

24 Geometry

ii. At B, draw ABM = 90o
iii. With centre A and radius AC = 7cm , take a

arc to cut BM at C.
iv. With centre A and radius BC = AD take an

arc and with centre C and radius AB = CD
take another arc to cut the two arcs at D.
v. Join A and C, A and D, C and D.
Thus, the required rectangle ABCD is
constructed.

C. When length of diagonals and angle between them are given.
For example, to construct a rectangle ABCD in which length of diagonal is 6.5cm and
angle between the diagonals is 30o.

Steps of construction 30o
i. Draw a line segment AC = 6.5cm and locate

its mid point O.
ii. At O, draw COM = 30o and produce MO back

to N.
iii. On MN, take the points B and D such that

OB = OD = OA = OC.
iv. Join A and B; C and D and and D.

Thus ABCD is the required rectangle.

D. When a side and length of diagonal are given.
For example, to construct a rectangle ABCD in which AB = 5cm and length of
diagonal = 7cm.

Steps of construction.

i. Draw a line segment AB = 5cm. 7cm
2
ii With centre A and radius = AC = =
2
3.5 cm draw an arc and

with centre B and radius = 3.5 draw an arc to

intersect these arcs at O.

iii. Produce AO and BO and take AO = OC and BO
= OD.

iv. Join B and C; C and D; A and D
Thus ABCD is the required rectangle.

Prime Mathematics Book - 7 25

Construction of Squares.

A. When a side of a square is given.
For example, to construct a square of side 4.5 cm.

Steps of construction.
i. Draw line segment AB = 4.5 cm
ii. At B, draw ABM = 90o and take BC = AB = 4.5 cm on

BM ( by using simply scale or taking arc with centre
B and radius BC = AB)
iii. With centre A and C and radius AB = 4.5 cm take arcs
to cut at D.
iv. Join C and D, A and D,
Thus, the required square ABCD is constructed.

B. When length of diagonal is given,
For example, to construct a square ABCD having diagonal
5.5 cm.
Steps of construction:
i. Draw a line segment AC = 5.5 cm
ii. Draw perpendicular bisector XY of AC which cuts AC at O.
iii. Take OA = OB = OC = OD with centre O and radius OA = OC
iv. Join A and B; B and C; C and D, A and D.

Thus ABCD is the required square.

Construction of Rhombus.

A. When side and an angle are given.
For example , to construct a rhombus ABCD in which AB = 4.5 cm and DAB = 45o

Steps of construction:
i. Draw line segment, AB = 4.5cm
ii. At A, draw BAM = 45o.
iii. Take AD = 4.5 cm on AM.
iv. With B as centre and radius = 4.5 cm

take an arc and with D as centre and
radius = 4.5 cm take an arc to cut the
previous arc at C.
v. Join B and C; D and C.
Thus, ABCD is the required rhombus.

26 Geometry Applied Mathematics Book - 7

B. When lengths of two diagonals are given
For example, to construct a rhombus ABCD in which diagonal AC = 7.3cm and BD = 4.7cm.

Steps of construction:
i. Draw a line segment, AC = 7.3cm.
ii. Draw perpendicular bisector MN of AC which

cuts AC at O.
iii. Draw line segment XY = 4.7cm = BD and

locate its mid point at P. (separately)
iv. With centre O and radius = PX = PY draw arcs

to cut MN at B and D opposite to O [It is not
easy to measure half of 4.7cm]
v. Join A and B; B and C; C and D; A and D.

Thus, ABCD is the required rhombus.

Exercise 1.5

1. Construct a parallelogram ABCD in which
(a) AB = 4.5cm, BC = 3.5cm and B = 45o
(b) AB = 5 cm, AD = 3.5cm and BAD = 60o
(c) Base AB = 4.5cm, diagonal AC = 6.5cm and BAC = 30o
(d) BC = 5cm, BD = 4cm and CBD = 30o
(e) AC = 5cm, BD = 6.5cm and angle between AC and BD = 60o
(f) AC = 6cm, BD = 8cm and angle between AC and BD= 75o
(g) AB = 4.8cm, AC = 4..2cm and BD = 6.6cm
(h) BC = 4.8cm, AC = 7.2cm and BD = 5cm

2. Construct the rectangle PQRS in which
(a) PQ = 5cm and QR = 4cm.
(b) QR = 6.2 and RS = 4.5 cm
(c) length of diagonals = 6 cm and angle between the diagonals = 30o
(d) length of diagonals = 5.8 and angle between the diagonals = 60o
(e) PQ = 4.5 and length of diagonals = 6.2 cm
(f) QR = 5cm and length of diagonals = 8 cm

3. Construct a square ABCD in which. (b) BC = 5.5 cm
(a) AB = 4.6 cm (d) diagonal AC = 5.8 cm
(c) diagonal = 6 cm

4. Construct the rhombus ABCD in which.
(a) AB = 4.5 cm and ABC = 30o
(b) AB = 5.4 cm and BAD = 60o
(c) Diagonal AC = 6cm and diagonal BD = 4.8cm

(d) AC = 5cm and BD = 4cm

Prime Mathematics Book - 7 27

1.6 Polygons

Polygons are the closed plane figures bounded by three or more than three line segments.

A triangle A quadrilateral A pentagon A hexagon

A triangle is a polygon with the least number of sides.
The figures shown alongside seem to be polygons. But these are open figures.

Polygons are named according to the number of Sides.

Number sides Name
3 Triangle (Trigon)
4 Quadrilateral (Tetragon)
5 Pentagon
6 Hexagon
7 Heptagon
8 Octagon
9 Nonagon
10 Decagon etc.

If non of the angles of a polygon are reflex angle, they are called convex polygons and
if at least an angle of a polygon is reflex angle, it is called concave polygon.

Convex Polygons:

Concave Polygons:

28 Geometry

Note: Polygons generally known to

us as Stars are called polygrams.

A pentagram A hexagram

Regular Polygons.

Polygons having all the sides and angles equal are called regular
polygons.

A regular trigon (equilateral triangles) has all three sides equal and
all the three angles equal (60o).

A regular tetragon (square) has all four sides
equal and all four angles equal (each 90o).

A regular pentagon has all the five sides equal and all the five angles 29
equal (each 108o)
The angle formed by two adjacent sides of a polygon towards
interior of the polygon is called its interior angle.
If we produce a side of a polygon, the angle formed by the
produced part of the side and the adjacent side is called
exterior angle.
We know that in a triangle ABC, A + B + C = 180°.
Drawing diagonal AC, the quadrilateral ABCD can be divided into
two triangles.
From the figure , b + c + d = 180o and a + e + f = 180o.
Now the sum of the angles of the quadrilateral

A+ B+ C+ D
=a+b+c+d+e+f
= (b + c + d) + (a + e + f)
= 180o + 180o = 360o = 2 × 180o
Again consider a pentagon ABCDE and draw diagonals AC and
AD to divide the pentagon into three triangles. From the figure
a+b+c=180o, d+e+f=180 and g+h+i=180o
Now,

A+ B+ C+ D+ E
=a+b+c+d+e+f+g+h+i
= (a+b+c)+(d+e+f)+(g+h+i)
= 180o+180o+180o = 540o = 3 × 180o
Thus, if we devide a polygon into n non intersecting triangles,
then sum of the interior angles of the polygon = (n-2) × 180o

Prime Mathematics Book - 7

Observe the following

Polygons No. of No of non intersecting Sum of interior angles
side triangles
Triangle 1=3-2 1 × 180o = (3-2) 180o
Quadrilateral 3 2=4-2 2 × 180o = (4-2) 180o
Pentagon 4 3=5-2 3 × 180o = (5-2) 180o
Hexagon 5 4=6-2 4 × 180o = (6-2) 180o
Heptagon 6 5=7-2 5 × 180o = (7-2) 180o
Octagon 7 6=8-2 6 × 180o = (8-2) 180o
n-gon 8 n-2
n (n-2) 180o

We observed that No. of non-intersecting triangle in a polygon having n sides = (n-2).
And sum of interior angles of the polygon = (n-2) x 180o

Consider a regular polygon of number of sides n .

As n sides make 360o at the centre.
360o
1 side makes n at the centre.

\ In the figure, AOB = 360o . Here OAB = OBA = OBC = a(say)
n

then in DOAB,

a+a+ 360o = 180o
n
360o
or 2a = 180o- n

But 2a = ABO + OBC is interior angle

\Interior angle = 180o - 360o = 180o x (n-2)
n nn

Again from the figure,
Exterior angle = CBX = 180o - 2a

30 Geometry

( )= 180o - 360o = 180o - 180o + 360o 360o
180o - n n = n

\ Exterior angle = 360o
n

Example 1: Find the number of non-intersecting triangles inside the given polygon and
find the sum of interior angles.

Solution
Here number of sides (n) = 10
\ No of non-intersecting triangles = n-2 = 10-2 = 8
Now, the sum of interior angles = (n-2) 180o

= (10-2) 180o
= 8×180o
= 1440o

Example 2: Find the size of an interior angle of regular pentagon.

Solution:

No of sides of a regular pentagon (n) = 5

\ Interior angle of a pentagon = 180o x (n-2)
= n
180o x (5-2)
5 = 36o x 3 = 108o

Example 3: Find the number of sides of a polygon whose sum of interior angle is 1080°
Solution:
Sum of interior angle of polygon = 1080o
let n be the number of sides, then (n-2)180o = 1080o

or n-2 = 6
or n = 6+2
\ n =8

\ The polygon has 8 sides.

Example 4: The interior angles of a pentagon are 100, x+50°, 2x+30°, x+40° and

6x+30°, find the value of x.
Solution: Here, angles of a pentagon are 100o, x+50o, 2x+30o x+40o and 6x+30o
We know,

the sum of angles of a pentagon = ((x5--22))118800oo
=

= 534×01o80o
=

Prime Mathematics Book - 7 31

\100o+(x+50o)+(2x+30o)+(x+40o)+(6x+30o) = 540o
or 10x+250o = 540o

or 10x = 540o-250o

or x= 290o \ x = 29o
10

Exercise 1.6
1. Find the number of non-intersecting triangles inside the given figure.

(a) (b)

(c) (d)

2. Find the sum of interior angles of the following polygons.
(a) Pentagon (b) Decagon (c) Quadrilateral (d) Hexagon
(e) Nonagon (f) Heptagon (g) Octagon (h) Triangle

3. Find the size of an interior angle of the following regular polygons.

(a) Pentagon (b) Decagon (c) Octagon (d) Hexagon

4. Find the size of an exterior angle of the following polygons.
(a) Equilateral triangle (b) Square (c) regular pentagon
(d) regular hexagon (e) regular octagon (f) regular decagon.

5. a) The interior angles of a pentagon are 70o, 90o, 105o, x+60o and x + 25o, find
the value of x.

b) The interior angles of a hexagon are 70o, 95o, 105o, x+20o, x+30o and x+40o. Find
the value of x.

c) The interior angles of a pentagon are in the proportion 1:2:3:4:5. Find the
angles

d) The interior angles of a heptagon are a+20o, a+30o, a+40o, a+50o, a+60o,
2a+100o and 3a. Find the value of a.

6. Find the number of sides of the polygon where sum of whose interior angles are
(a) 720o (b) 900o (c) 1440o (d) 2160 o (e) 1620o

32 Geometry

1.7 Similarity and Congruency of figures. Estimated Period - 4

Similarity of figures:

Look at the following figures. (ii)
(i)

Two similar A’s Two similar hexagrams
(iii) (iv)

Two similar circles Two similar crosses
Let’s consider two right angled triangles

If we put DABC over DDEF, A coincides with D, AB overs DE and AC overs DF.

\ AB = 4 cm = 1 , BC = 3 cm = 1 , AC = 5 cm = 1
DE 8 cm 2 EF 6 cm 2 DF 10 cm 2

Prime Mathematics Book - 7 33

Two figures with same shape and corresponding sides in proportion are called similar figures.
Here, DABC ~ DDEF. [ ~ is the symbol of similarity]

A = D, B = E and C = F are corresponding angles and sides AB and DE, BC and
EF, AC and DF are corresponding sides.

Corresponding Vertices Corresponding angles Corresponding sides

Notes
Any two line segments are similar.
Any two circles are similar.
Any two regular polygons with same number of sides (n-gons) are similar.

Similar figures may or may not be equal.

Similar Triangles.

Two triangles with corresponding angles equal and corresponding sides proportional
are called similar triangles.

Here, in DABC and DDEF, 2cm 4cm
6cm
A = D, B = E, C= F and 1cm 2cm
AB BC AC 3cm
DE = EF = DF

\DABC~DDEF

Necessary conditions for two triangles to be similar

i. AAA condition: If three angles of a triangle are respectively equal to three angles
of other, then two triangles are similar.

ii. AA condition: If two angles of a triangle are respectively equal to two angles of
other, then the two triangles are similar. [ If two angles are equal, the remaining
angles will be equal. So, AA condition is same as AAA condition].

iii SSS condition: If three sides of a triangle are proportional to the three side of
other, the triangles are similar.

iv. SAS condition : If two sides of one triangle are proportional to two sides of other
and angle included by the proportional sides are equal, then the triangles are
similar.

34 Geometry

Note:
Any one of the above condition is sufficient for two triangles to be similar.
Two similar triangles satisfy all the above conditions.

Example 1: Under which conditions are the following pairs of triangles similar.
(i) (ii) P

T

8 cm 3 cm 4 cm

Q 6 cm R U S

(iii) (iv)

6 cm

Solution:

(i) In DABC and DDEF

A = D, B = E, C = F

\DABC~DDEF by AAA condition.

(ii) In DPQR and DSTU.

PQ = 8cm = 2 (S)
TS 4cm 1

PQR = STU (A)

QR = 6cm = 2 (S) \DPQR~DSTU by SAS condition.
TU 3cm 1

(iii) In DUVW and DXYZ.
UV 2cm 1
XY = 6cm = 3 (S)

VW = 4cm = 1 (S)
YZ 12cm 3

UW = 3cm = 1 (S)
XZ 9cm 3

\ DUVW~DSTU by SSS condition.
(iv) In DABC and DMBN.

ABC = MBN (A) (Common angle)

ACB = MNB (A) (Corresponding angles)

\ DABC~DMBN by AA condition.

Prime Mathematics Book - 7 35

Example 2. From the similar triangles DABC and DDEF, find the length of side DF.

Solution:

Here DABC~DDEF

Where A = D (given)

B = E (given)

C = F (remaining angles)

\ BC = AC
EF DF

or 8cm = 7cm
4cm
x

or x = 7cm = 3.5cm.
2

\DF = 3.5cm

Exercise 1.7
1. Under which condition are the following pairs of triangles similar ?

a) b)

c) d)

DPQR and DSQP

e) f)

36 Geometry

2. From the following similar triangles. Find the value of x.
(a) (b)

3 (a) In the given figure, MN//BC.
Prove that: DABC~DAMN

(b) In the given figure, PRQ = PST = 60°.
Prove that: DPQR~DPTS

(c) In the given figure, AB = 3cm,
BC = 4 cm, DE = 6cm, EF = 8cm
and ABC = DEF = 60°.
Prove that : DABC~DDEF

(d) In the given figure, AB = 4cm, BC
= 8cm, AC = 6 cm,
DE = 6cm, EF = 12cm and
DF = 9cm.
Prove that: DABC ~ DDEF

Prime Mathematics Book - 7 37

1.8 Congruent Figures.

The mathematical term for “same in size and shape” is
“congruent”. Thus two geometrical figures having same size
and shape are called congruent figures. If we place a figure on
the other, they exactly fit.

Two lines segments AB and CD are congruent if AB
and CD are of equal length. We denote it as AB

@ CD
or AB @ CD

Two angles ABC and DEF are congruent if they
have equal measure. We denote it as

ABC @ DEF.

Two triangles ABC and DEF are
congruent if their corresponding
sides are congruent and
corresponding angles are
congruent.

We denote it as DABC @ DDEF.

The symbol @ means “is congruent to” which is combination of similar (~) and equal
(=) in which similar (~) stands for same shape and equal (=) stands for same size.

The parts of the figure which fit to
each other are called corresponding
parts. If

DABC @ DDEF, correspondence of

vertices.

A ↔ D, B ↔ E, C ↔ F

The symobl ↔ Stands for Correspondence of sides

corresponding
Correspondence of angles

A↔ D, B↔ E, C↔ F AB ↔ DE, BC ↔ EF, AC ↔ DF.

So the order of letter is important.

Corresponding sides are the sides opposite to the congruent vertices

(angles) and corresponding angles are the angles opposite to congruent sides.

38 Geometry

Conditions for congruency of triangles.

1. SAS congruency axiom. If two sides and included angle of one triangle respectively
equal to two sides and included angle of the other triangle, the two triangles are
congruent.
In DABC and DDEF

(i) AB = DE (S)
(ii) ABC = DEF (A)
(iii) BC = EF (S)

\ DABC@DDEF (by SAS congruency axiom)

2. ASA congruency theorem: If two angles and included side of one triangle respectively
equal to two angles and included side of other triangle, the two triangles are congruent.
(i) ABC = PQR (A)
(ii) BC = QR (S)
(iii) ACB = PRQ (A)

\DABC @ DPQR (by ASA congruency theorem)

3. S.S.S. Congruency theorem: If three sides of one triangle are respectively equl to
the three sides of other triangle, the two triangles are congruent.
In DPQR and DXYZ.

(i) PQ = XY (S)
(ii) QR = YZ (S)
(iii) PR = XZ (S)

\ DPQR @ DXYZ (by SSS congruency theorem)

4. RHS congruency theorem: If right angle, hypotenuse and any other side of one
triangle are respectively equal to right angle, hypotenuse and any other side of other
triangle, the two triangle are congruent.
In DABC and DLMN.
(R)
(i) ABC = LMN = 90o (H)
(ii) AC = LN
(iii)AB = LM (S)

\ DABC @ DLMN (by RHS congruency theorem)

5. AAS congruency theorem: If two angles and one side of a triangle are
respectively equal to two angles and one side of other triangle, the two triangles are
congruent.
In DDEF and DGHI
i) DEF = GHI (A)
ii) DFE = GIH (A)
iii) DF = GI (S)

\DDEF @ DGHI (by AAS congruency theorem)

Prime Mathematics Book - 7 39

Note: Among the five conditions SAS, ASA, SSS, RHS and AAS, only SAS is the axiom others
are all theorems which can be proved using SAS which you will study in higher classes.

Example 1. Under which condition, the following triangles are congruent ?
(a) (b)

Solution.

(a) In DABC and DPQR (S)
(A)
(i) AB = QR (S)
(ii) ABC = QRP
(iii) BC = PR

\ DABC @ DQRP (by SAS congruence axiom)

(b) In DXYZ and DPQR.

(i) XYZ = QPR (R) (both 90o)
(ii) XZ = QR (H)
(iii) YZ = PR (S)

\ DXYZ @ DQPR (by RHS congruency theorem)

Example 2: In the given figure, AD^BC and AB = AC. Prove that DABD@DACD.

Solution:

Given: AD^BC

AB = AC

To prove: DABD @ DACD

Proof

Statements Reasons

1 In D ABD and D ACD 1

(i) ADB = ADC (R) (i) Both 90o as AD^BC.

(ii) AB = AC (H) (ii) Given
(iii) Common side.
(iii) AD = AD (S)

2 \DABD @ DACD 2. By RHS congruency theorem

Proved

40 Geometry

Exercise 1.8
1. Under which conditions (test) are the following pairs of triangles congruent ?

a) b)

c) d)

e) f)

2. From the following pairs of congruent triangles find the values of x, y and z.
a) b) (c)

3 (a) In the given figure, AB = AD and

BC = CD. Prove that DABC @ DADC.

(b) In the given figure, AD^BC and BAD = CAD.
Prove that DABD @ DACD.

(c) In the given figure, AB and CD bisect
each other at O. Prove that DAOC @ DDOB.

O

Prime Mathematics Book - 7 41

1.9 Circle and its parts Estimated Period - 3

A circle is The fixed point is called the centre and the boundary
a closed plane curve line is called the circumference.
figure bounded by
regularly curved line Circumference
every point of which
is equidistant from O Diameter O
a fixed point
Centre

Radius Chord

A straight line drawn from centre to the an arc
circumference is called radius. It follows that
all radii of a circle are equal. Segment
A straight line joining any two points on the
circumference is called a chord. Sector
A straight line drawn through the centre and
terminated both ways at the circumference
is called a diameter. A diameter is the
longest chord of the circle. Centre bisects a
diameter.

\Diameter = 2 radius.

A part of the circumference is called an arc.
The region bounded by a chord and its
corresponding arc is called a segment.
Region bounded by two radii and
corresponding arc is called a sector.

A closed region bounded by a diameter and
its corresponding arc is called semi circle
(half circle)

Semicircle

42 Geometry

Construction of a circle
For example, to construct a circle of radius 4cm.

Extend the compass to 4cm.
Put the needle of the compass at a point o and rotate the compass
keeping centre ‘o’ fixed tracing circumference with pencil of the
compass.
Thus, circle of radius OA = 4cm is constructed.

4cm

O

0 12 34 5

Precaution :
Never play with compass.
Never use pencil with compass for writing and drawing other lines.

To locate the centre of a circle.

Given a circle whose centre is unknown
Consider any two arcs AB and BC and draw
perpendicular bisectors of chord AB and BC
The two bisectors intersect at a point ‘O’ which is the
centre of the circle.

To locate the centre of an arc.

Given arc is AB.
Take a point X on the arc AB
Draw perpendicular bisectors of chord AX and BX which
intersect at point O. That point O is the centre of the
circle having arc AB.

Prime Mathematics Book - 7 43