Math 7

Approved by the Government of Nepal, Ministry of Education, Curriculum Development
Centre, Sanothimi, Bhaktapur as an additional material.

7

Author
Shyam Datta Adhikari

M.Sc. (Maths), T.U.

7

Name : ............................................................
Class : ....................... Roll No. : ................
Section : ..........................................................
School : ............................................

Publisher

Oasis Publication Pvt. Ltd.

Copyright
The Publisher
Edition
B.S. 2073 (2016 AD)
B.S. 2074 (2017 AD)
B.S. 2075 (2018 AD)
B.S. 2078 (2021 AD) (Completely revised)

Contributors
Shuva Kumar Shrestha
Yadav Shiwakoti
Laxmi Gautam

Layout
Oasis Desktop

Ramesh Bhattarai

Printed in Nepal

Preface

Oasis School Mathematics has been designed in compliance with
the latest curriculum of the Curriculum Development Center
(CDC), the Government of Nepal with a focus on child psychology
of acquiring mathematical knowledge and skill. The major thrust
is on creating an enjoyable experience in learning mathematics
through the inclusion of a variety of problems which are closely
related to our daily life. This book is expected to foster a positive
attitude among children and encourage them to enjoy mathematics.
A conscious attempt has been made to present mathematical
concepts with ample illustrations, assignments, activities, exercises
and project work to the students in a friendly manner to encourage
them to participate actively in the process of learning.

I have endeavored to present this book in a very simple and
interesting form. Exercises have been carefully planned. Enough
exercises have been presented to provide adequate practice.

I have tried to include the methods and ideas as suggested by the
teachers and subject experts who participated in the seminars,
and workshops conducted at different venues. I express my
sincere gratitude to my friends and well wishers for their valuable
suggestions.

I am extremely grateful to Man Bahadur Tamang, Shuva Kumar
Shrestha, Prakash Ghimire, Laxmi Gautam, Purushottam Bhatta,
Sunil Kumar Chaudhary, Ram Prasad Sapkota, Saroj Neupane and
Yadav Shiwakoti for their invaluable suggestions and contributions.

Shyam Datta Adhikari
Author

March 2021

7

Contents

Geometry ............................................................................................. 1

1. Lines and Angles ......................................................................... 2
1.1 Construction of Angles..................................................... 2
1.2 Construction of Equal Angle with Given Angle............. 5
1.3 Some special pairs of Angles .......................................... 6
1.4 Transversal and Parallel lines ......................................... 12
2. Triangles Quadrilaterals and Circles ...................................... 19
2.1 Triangles ............................................................................ 19
2.2 Construction of Triangles................................................. 28
2.3 Quadrilateral...................................................................... 30
2.4 Construction of Parallelogram ....................................... 37
2.5 Construction of Rectangle................................................ 39
2.6 Polygons ............................................................................ 41
2.7 Circles and Its Different Parts.......................................... 47
3. Congruency and Similarity........................................................ 50
4. Vertices, Edges and Faces of Solid Objects............................ 5 3
4.1 Vertices, Edges and Faces of Solid objects..................... 53
4.2 Construction of Some Models of Solids......................... 54

Co-ordinates...................................................................................... 59

5. Coordinates.................................................................................. 60

Mensuration...................................................................................... 65

6. Perimeter and Area..................................................................... 66
6.1 Perimeter ........................................................................... 66
6.2 Area of Plane Figure.......................................................... 71
6.3 Total Surface Area of Cube and Cuboid......................... 76

Transformation................................................................................ 83

7. Transformation............................................................................ 84

7.1 Reflection.......................................................................... 84
7.2 Rotation ............................................................................ 88
7.3 Translation ....................................................................... 94

8. Symmetry, Tessellation and Designs..................................... 97

8.1 Symmetry.......................................................................... 97
8.2 Tessellation....................................................................... 99
8.3 Designs Using Circles and Polygons........................... 102

9. Bearing and Scale Drawing..................................................... 103

9.1 Bearing ............................................................................. 103
9.2 Scale Drawing.................................................................. 107

Sets ..................................................................................................... 112

10. Sets............................................................................................... 113

10.1 Review............................................................................... 113
10.2 Set relation ....................................................................... 117
10.3 Subset ............................................................................... 119
10.4 Set Operation................................................................... 121
10.5 Venn diagram.................................................................. 123

Arithmetic ........................................................................................ 131

11 Operation on whole number .................................................. 132

11.1 Binary Number System.................................................. 132
11.2 Quinary Number System ............................................... 135
11.3 Highest Common Factor (H.C.F).................................. 138
11.4 Lowest Common Multiple (L.C.M).............................. 142
11.5 Relation between H.C.F and L.C.M.............................. 147
11.6 Square and Square Root ................................................ 150
11.7 Cube and Cube Root ...................................................... 157

12 Operation on Integers .............................................................. 160

12.1 Integers ............................................................................ 160
12.2 Operations on Integers .................................................. 164
12.3 Multiplication and Division of Integers ...................... 169
12.4 Order for Operations for Simplification....................... 173

13 Rational Numbers .................................................................... 177

14 Fractions and Decimals ........................................................... 180

14.1 Fractions........................................................................... 180
14.2 Fundamental Operations on Fractions........................ 184
14.3 Verbal Problems on Fractions........................................ 188

14.4 Multiplication of Fractions ........................................... 191
14.5 Decimal ............................................................................ 197
14.6 Multiplication of Decimals ......................................... 200
14.7 Division of Decimals ................................................... 202
15. Ratio and Proportion.............................................................. 208
15.1 Ratio................................................................................ 208
15.2 Proportion ..................................................................... 212
16. Percentage ................................................................................ 217

17. Unitary Method....................................................................... 223

18. Simple Interest......................................................................... 226

19. Profit and Loss.......................................................................... 231

Statistics........................................................................................... 242

20. Statistics.................................................................................... 243
20.1 Review ........................................................................... 243
20.2 Bar Graph ....................................................................... 248
20.3 Arithmetic Mean or Average....................................... 252

Algebra ............................................................................................ 257

21 Algebraic Expression.............................................................. 258
21.1 Algebraic Expression.................................................... 258
21.2 Addition and Subtraction of Algebraic Expressions..... 259
21.3 Indices ............................................................................ 263
21.4 Multiplication of Algebraic Expressions ................... 266
21.5 Factorization.................................................................. 269
21.6 Factors of Algebraic Expression in the Form of
Difference of Two Squares.......................................... 272
21.7 Some Special Product and Formulae......................... 274
21.8 Cubes of the Sum of Two Quantities.......................... 279
21.9 Division of Algebraic Expressions ............................. 282
21.10 Simplification of Rational Expressions...................... 284

22 Equations and Inequality....................................................... 286

22.1 Equation......................................................................... 286
22.2 Problems of Linear Equation of one Variable........... 290
22.3 Graph of Linear Equations ......................................... 294
22.4 Function Machine ........................................................ 296
22.5 Inequality....................................................................... 298
Model Test Paper............................................................................... 308

Geometry

39Estimated Teaching Hours

Contents
• Lines and Angles
- Types of angles
- Transversal and parallel lines
• Triangles, Quadrilaterals and Circles
- Types of triangle
- Properties of triangle
- Types of quadrilateral and their properties
- Construction of quadrilaterals
- Polygons - Vertices, edges and faces of solid object
• Similarity and Congruency

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• Use the properties of pair of angles to solve the problems related to

them
• Use the properties of alternate angles, corresponding angles and

co-interior angles to solve the problems related to them
• Apply the properties of triangle and quadrilateral to solve the

problems related to them
• Identify whether the given figures are congruent, similar or none
• Identify different parts of a circle
• Solve the problems related to polygons

Teaching Materials

• Model of polygons
• A4 size paper, scissors, etc.

Oasis School Mathematics – 7 1

Unit

1 Lines and Angles

1.1 Construction of Angles

Angle:

In the given figure, two rays OA and OB meet at a point O. Then an angle is formed

at O. 'O' is the vertex of this angle. A

Arms of this angle are OA and OB. O arm B
Name of this angle is ∠AOB or ∠BOA Vertex arm

Construction of angles using compass

I. Construction of an angle 120° Steps :
• Draw a line segment AB.
F • Taking A as the centre, draw an arc

which cuts AB at C.
• Taking C as the centre, cut the arc at D.
ED • Taking D as the centre, cut the arc at E.
• Join AF through E.
A CB Then ∠FAB = 120°

II. Construction of an angle 135° Steps :

• Draw a line segment AB.
G

• At A, draw ∠GAB = 90°.

I • Bisect ∠GAH, from J and H

J E then ∠GAI = 90° = 45°
F 2

• Join AI.

Then ∠IAB = 45° + 900 = 1350

H A CB I already know, how to
draw the bisector of given

Analysis! ∠IAB = ∠IAG + ∠GAB angle.

= 45° + 90° = 135°

2 Oasis School Mathematics – 7

III. Construction of an angle 75°.

EF Steps :

• Draw a line segment AB.

DG C • At A, draw ∠EAB = 90°.
• Cut arcs from G and C to get point

F.

• Join AF.

Now, ∠FAB = 75°.

A CB

Analysis! ∠EAB = 90°, ∠EAF = 30° = 15° ∴∠ FAB = 90° – 15° = 75°
2

IV. Construction of an angle 105º

GF Steps :

• Draw a line segment AB.

H • At A, draw ∠FAB = 90°.
ED • Cut arcs from E and H to get point

G.

• Join AG.

Now, ∠GAB = 105°.


A CB

Analysis! ∠GAB = ∠GAF + ∠FAB = 15° + 90° = 105°

V. Construction of an angle 15°

E Steps :
D • Draw a line segment AB.

F • At A draw ∠EAB = 30°.

G

• Bisect ∠EAB from F and C.

A CB Then ∠GAB = 30° = 15°
2


Oasis School Mathematics – 7 3

VI. Construction of an angle 45º F Steps :
• Draw a line segment AB.
E • At A, draw an angle 90°, such

H that ∠EAB = 90°.
DC • Bisect the angle EAB cutting at

A GB F from the points G and H.
• Join AF.
∴ ∠FAB = 45°.

Exercise 1.1

1. Construct the following angles with the help of compass.

(a) 60° (b) 120° (c) 90°

2. Construct the following angles with the help of compass.

(a) 30° (b) 45° (c) 75° (d) 105° (e) 135°

3. Draw the following angles at the given point. P

(a) 15° at A (b) 105° at P

A B Q

(c) 75° at X Y (d) 135° at M.

NM
X

4. (a) Draw an angle of 60°. Bisect it. Measure each angle. What is the measurement of
each angle?

(b) Draw an angle of 75°. Bisect it. Measure each angle. Write the measurement of
each angle.

(c) Draw a line segment AB. Draw an angle of 30° at A. Bisect the angle. Measure
each angle and write their measurement.

5. (a) Draw a line segment AB. Draw an angle of 45° at A and 75° at B. If the point of
intersection of two lines is C, measure angle ACB.

(b) Draw a line segment MN. Draw an angle of 60° at M and 30°at N. If two lines
meet at P, measure the angle MPN.

Answer
Consult your teacher.

4 Oasis School Mathematics – 7

1.2 Construction of Equal Angle with Given Angle

Given, an angle ABC.
Let's draw an angle equal in magnitude to the ∠ABC.

A

Steps

M • Draw a line PQ.
B • Taking P as the centre draw an arc which

N cuts PQ at R.
C • With the same arc, take B as the centre,

draw an arc which cuts AB and BC at M
T and N respectively.

• Measure arc MN and from R cut at point S.

S • Join P through S.

∠TPQ is the required angle where

P ∠ABC = ∠TPQ.

RQ

Exercise 1.2 A

1. (a) Draw ∠ABC in your copy. Draw angle

PQR equal in magnitude to ∠ABC. B C
Y
X

(b) Draw an angle XYZ in your copy. Draw

an angle PQR equal in magnitude to

∠XYZ. Z

2. (a) Draw an angle ABC. Draw another angle PQR equal in magnitude to angle ABC.
Measure both angles using a protractor. Are both angles equal in magnitude?

(b) Draw an angle PQR of any measurement. Draw an ∠XYZ equal in magnitude
to ∠PQR. Measure both angles ∠PQR and ∠XYZ. Are both angles equal in
magnitude?

(c) Draw ∠ CDE = 80° at point D with the help of a protractor. Take any point Q and
draw ∠PQR equal in magnitude to ∠CDE. Measure ∠PQR. Is ∠PQR = 80°?

Answer
Consult your teacher.

Oasis School Mathematics – 7 5

1.3 Some Special Pairs of Angles

Vertically opposite angles

In the given figure, two straight lines AB and CD intersect A D

each other at a point O. The angles AOC and BOD are

vertically opposite angles. Angles AOD and BOC are also C O B
vertically opposite angles.

When two straight lines intersect each other at a point, the pair of opposite angles
so formed are called vertically opposite angles.

Note : Vertically opposite angles are always equal i.e. ∠AOC = ∠BOD and
∠AOD = ∠BOC



Experimental Verifications

Experiment I:

If two straight lines meet at a point, vertically opposite angles so formed are always

equal.

A

D AD

A C D
O O O

B CB

C
B

Fig (i) Fig (ii) Fig (iii)

Draw three pairs of lines AB and CD intersecting at O.
∠BOC and ∠AOD are vertically opposite angles.
Similarly,
∠AOC and ∠ BOD are vertically opposite angles.
To verify: ∠BOC = ∠AOD
∠ AOC = ∠BOD


6 Oasis School Mathematics – 7

Observations
Measure ∠BOC, ∠AOD, ∠AOC and ∠BOD and plot in the given table.

Figure ∠BOC ∠AOD ∠AOC ∠BOD Remarks

i. ..................0 ..................0 ..................0 ..................0 ∠BOC = ∠AOD

∠AOC = ∠BOD

ii. ..................0 ..................0 ..................0 ..................0 ∠BOC = ∠AOD

∠AOC = ∠BOD

iii. ..................0 ..................0 ..................0 ..................0 ∠BOC = ∠AOD

∠AOC = ∠BOD

Conclusion: Hence, if two lines intersect at a point, vertically opposite angles so formed
are always equal.

Adjacent angles A B
O C
In the given figure, ∠AOB and ∠BOC have a common

vertex O and a common arm OB. They lie on either sides
of the common arm OB.

So, ∠AOB and ∠BOC are adjacent angles.

Hence angles are said to be adjacent angles if;

(i) they have a common vertex, (ii) they have a common arm, and

(iii) the other arms of the two angles lie on either side of the common arm.

Note : • If the sum of two adjacent angles is 1800, then they are called linear pair.
• If the adjacent angles forming the linear pair are equal, each angle is 900.

Experiment II:
If two straight lines meet at a point, the sum of two adjacent angles forming a linear pair
is 1800.

AD D
A
D

CC

B AC B B

Fig. (i) Fig. (ii) Fig. (iii)

Draw three pairs of straight lines AB and CD intersecting at C.

Now, ∠ACD and ∠BCD are a pair of adjacent angles forming a linear pair.

To verify: ∠ACD + ∠BCD = 1800

Oasis School Mathematics – 7 7

Observations:
Measure each of the angles ACD and BCD. Then tabulate their values below:

Figure ∠ACD ∠BCD Remarks
i. ..................0 ..................0 ∠ACD + ∠BCD = 1800
ii.
iii.

Conclusion: Hence, the sum of adjacent angles forming a linear pair is 180°.

Note: B
C
• Sum of the angles around a point is 3600. In the given figure A
∠AOB + ∠BOC + ∠COD + ∠AOD = 3600 O
D
• In the given figure, ACB is a straight line, E D

Hence, ∠ACE + ∠ECD + ∠BCD = 1800

AC B

Complementary angles D
A
In the given figure,

(i) ∠ABC = 60° and ∠ABD = 30°

∴ ∠ABC + ∠ABD = 60° + 30° = 90° 300

∠ABC and ∠ABD are complementary angles. ∠ABC is C 600
B

the complement of ∠ABD and ∠ABD are also called the

complement of ∠ABC.

Hence, two angles are complementary angles if their sum is 90°.

Supplementary angles A

In the given figure,

∠AOC = 40° and ∠AOB = 140° 1400
∴ ∠AOB + ∠AOC = 40° + 140° = 180°
B 400 C

O

∠AOC and ∠AOB are supplementary angles. ∠AOC is supplement of ∠AOB and

∠AOB is also the supplement of ∠AOC.

Hence, two angles are supplementary angles if their sum is 180°.

Remember !

• Vertically opposite angles are always equal.
• Sum of adjacent angles forming a linear pair is always 1800.
• Sum of complementary angles is always 900.
• Sum of supplementary angles is always 1800. • Angle around the point is 3600.

8 Oasis School Mathematics – 7

Worked Out Examples

Example: 1

If 2x° and (x – 15)° are pair of complementary angles, find them.
Solution:

Here, 2x° + (x–15)° = 90° (sum of complementary angles)
or, 2x° + x° = 90° + 15°

or, 3x° = 105°

or, x° = 105°
or, 3

x° = 35°

∴ The angles are 2x° = 2×35° = 70° and (x–15)° = (35–15)° = 20°

Example: 2

The pair of supplementary angles are in the ratio 5:4. Find them.
Solution:

Here, Let two angles in the ratio 5:4 be 5x and 4x.

Now, 5x + 4x = 180° [∵ sum of supplementary angles)

or, 9x = 180°

or, x = 180° = 20°
9

∴ The angles are 5x = 5 × 20° = 100° and 4x = 4 × 20° = 80°.

Example: 3 Z

In the given figure, two straight line segments TZ X a b Y
and XY intersect at O. If ∠XOZ = 35°, find the size of 350 T
marked angles a, b and c.
cO
Solution:

Here, TZ and XY are straight line segments that intersect at O.

So, ∠XOZ + ∠ZOY = 180° (∵ By linear pair axiom.)

or, 35 + a = 180°.

or, a = 180° – 35° = 145°

Again, ∠YOT = ∠ZOX (∵ Vertically opposite angles)

or, b = 35° and

∠XOT = ∠ZOY (∵ Vertically opposite angles)

or, c = 145°

Oasis School Mathematics – 7 9

Example: 4 130º
y
From the figure, find the value of y. y O 160º
Solution:

Here, y, y, 130° and 160° are sizes of angles around a point.
So, their sum must be 360° so we have,

Now, y + y + 130° + 160° = 360°

or, 2y + 2900 = 3600

or, 2y = 3600 – 290°

or, 2y = 70°

or, y = 70°
2

= 35°

Exercise 1.3

1. (a) If 'a' and 'b' are vertically opposite angles, what is their relation?
(b) If 'a' and 'b' are complementary angles, what is their sum?
(c) If 'p' and 'q' are supplementary angles, what is their sum?
(d) Are 45° and 60° supplementary angles? Justify your answer .

(e) Are 72° and 18° complementary angles? Justify your answer.

2. Find the complement of the following angles:

(a) 25° (b) 80° (c) 56° (d) 5°

3. Find the supplement of the following angles:

(a) 110° (b) 60° (c) 50° (d) 105°

4. (a) Does the angle 165° have complement? Justify your answer.

(b) Does the angle 182° have supplement? Justify your answer.

5. Two lines AB and CD intersect each other at point O. A D
B
O

(i) Name two pairs of vertically opposite angles. C
(ii) Are ∠AOC = ∠BOD ? Why?

6. (a) (2x – 5)° and (3x + 5)° are the pair of supplementary angles, find their sizes.

(b) (4x – 10)° and (5x + 10)° are supplementary angles. Find the size of each angle.

(c) A pair of supplementary angles are in the ratio 5: 4, find the angles.

7. (a) (3x – 10)° and 2x° are the pair of complementary angles. Find the value of 'x'.

(b) (a + b)° and (a–b)° are the complementary angles. Find the value of a°.

10 Oasis School Mathematics – 7

(c) A pair of complementary angles are in the ratio 2 : 3. Find them.

8. Find the value of x.

(a) (b) (c) 140º

x 130º x 55º x

9. Find the value of x, y and z from the given figure.

(a) (b) (c) 30º

y 110º y z
zx
x 95º yx
z

10. Find the value of 'x' from the given figures.

(a) (b) xº (c) xº
xº
x° 2xº3xº 50º
35º

11. Find the value of 'x' and 'b' from the given figures.

(a) (b)

2x –20º x (3b + 4)º (b + 6)º
x
(b + 5)º (b + 9)º

12. Find the value of 'p' and 'x' from the given figures.

(a) (b) (c)

4p p 2p x 4x
2p 35° 3x 2x

8p 7p 150° 85° 3x
6x

Answer 

1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher. 4. Consult your teacher.

5. Consult your teacher 6. (a) 67° and 113° (b) 70° and 110° (c) 100° and 80°

7. (a) 50° and 40° (b) 45° (c) 36° and 54° 8. (a) 50° (b) 125° (c) 40°

9. (a) x = 95°, y = 85°, z = 85° (b) x = 110°, y = 70°, z = 70° (c) x = 30°, y = 150°, z = 150°

10. (a) x = 55° (b) x = 15° (c) x = 20° 11. (a) 50° (b) 26° 12. (a) 15° (b) 90° (c) 20°

Oasis School Mathematics – 7 11

1.4 Transversal and Parallel Lines

Interior and exterior angles b a d
c
In the given figure, c, d, e and f are interior angles:
ef
Again, a, b, g and h are exterior angles. gh

From the above examples it is clear that if a transversal
cuts straight lines, angles within two straight-line

segments are called interior angles and the angles outside

the straight-line segments are called exterior angles. E

A G B
H D
In the given figure, line segment EF intersects two line
segments AB and CD at two distinct points G and H, C
so EF is a transversal line.

F
When a line intersects two or more than two lines at distinct points then the line is

called transversal line.

Corresponding angles

Look at the diagrams carefully and identify the patterns.



In all diagrams the marked angles are on the same sides of the transversal lines one
interior and one exterior non-adjacent so they are called corresponding angles.

In the adjoining diagram, 1 and 5, 2 and 6, 3 and 7, 12
4 and 8 are called pairs of corresponding angles. 34
Furthermore, if the line AB and CD are parallel
(AB||CD), then the corresponding angles are equal. 56
78

Conversely, if a pair of corresponding angles are equal A 2B
then the lines are parallel. For example, in the figure C 1D
AB||CD so ∠1 = ∠2

A a = 69º B
b = 69º D
In the adjoining figure, a = b so AB||CD.

C



12 Oasis School Mathematics – 7

Alternate angles:
Look at the given diagrams and identify the patterns.

In the given diagrams, the black marked angles are of the same patterns. All
are interior, each pairs lie on opposite sides of the transversal line, all pairs are
non-adjacent. So, these pairs are called alternate angles.

In the figure (i) below c and f, d and e are alternate angles. Furthermore, if the

line segments are parallel then alternate angles are equal. In the given figure (ii),

AB||CD so ∠a = ∠b.

A ab B Aa B A a = 720 B
cd

C ef DC b DC b = 720 D
gh

figure (i) figure (ii) figure (iii)

Conversely, if the pair of alternate angles are equal then the lines must be parallel.
In the given figure (iii) a= b = 720. Hence AB//CD.

Co–interior angles:
Look at the given diagrams and identify the patterns.

In the given diagrams, all black marked angles are on the ab
same side of the transversal line and all are interior, so cd
they are called co–interior angles. ef
gh
In the adjoining figure, c and e are co–interior angles.
Similarly, d and f are also co–interior angles. a
b
P Q
S
Moreover when a pair of parallel lines are intersected by
transversal, then the sum of co–interior angles so formed
is two right angles (180°). In the adjoining diagram R
PQ||RS, hence a + b = 1800

Conversely, if the sum of co-interior angle is 1800, then A 100° B
80º D
the pair of lines must be parallel. In the adjoining figure,

the sum of 100° and 80° is 1800. So, AB||CD C

Oasis School Mathematics – 7 13

Remember !

When a transversal intersects a pair of lines that are parallel to each other, then
• the alternate angles are equal to each other
• the corresponding angles are equal to each other
• the sum of co–interior angles is 1800.
Conversely,
When two lines are intersected by a transversal.
• if the alternate angles so formed are equal, then the lines are parallel.
• if the corresponding angles so formed are equal, then the lines are parallel.
• if the sum of co–interior angles is 180°, then the lines are parallel.

Worked Out Examples

Example: 1 ab
cd
In the given figure, find the value of 'a', 'b', 'c','d', 'e',
'g' and 'h'. e 70°
Solution: hg

Here, given two lines are parallel, so the alternate angles so formed
are equal, the corresponding angles so formed are equal and the sum
of two co-interior angle is 180°.

Now, b = 70° [Corresponding angles]

h = 70° [Vertically opposite angles]

c = 70° [Alternate angles]

Again, g + 70° = 180° [Linear pair]

g = 180° – 70°

∴ e = 110°

Again, g = e [Vertically opposite angles]

110° = e

∴ e = 110°

Again, e = a [Corresponding angles]

a = 110°

Again, g = d [Corresponding angles]

110° = d

∴ d = 110°

14 Oasis School Mathematics – 7

Example: 2 A 120º B
D C
Find the sizes of unknown angles in the following diagram. x
E 150º
Solution: Here, AB||DC, ∠EAB = 120°,
∠EDC = 150º, ∠AED = x°.

To find the value of x, draw a line EF parallel to AB and CD. A

∠DEF + ∠EDC = 180° (∵ Sum of co–interior angles) 120º B
or C
∠DEF + 150° = 180° D

or, ∠DEF = 180° – 150° = 30° xº 150º

EF

Again, ∠BAE + ∠AEF = 180° (∵ Sum of co–interior angles on parallel lines)

or 120° + ∠AEF = 180°

or, ∠AED + ∠DEF = 60°

or, x + 300 = 600
or, x = 60° – 30° = 30°.

Exercise 1.4 ℓ1 ab
cd
1. (a) Name four exterior angles from the given figure.
(b) Name four interior angles from the given figure. ℓ2 ef
gh

2. Name the four pairs of corresponding angles from each of the following figures.

Identify whether they are equal or not. Give reason.

(b) hg AD
(a) a b cd (c)

cd ab EG H F
ef ef
gh BC

3. Name the two pairs of alternate angles in each of the following given figures.

Identify whether they are equal or not. Give reason. xy
zt
(a) ab (b) q s (c)
r pq
p u sr
v
cd t w

Oasis School Mathematics – 7 15

4. Name the pairs of co–interior angles from each of the following figures. Identify

whether their sum is 180° or not.
ab
(a) cd (b) (c)

ef ab eh ad
gh cd f g bc

5. Study the given figure and identify whether the given statements are true or false.

(i) a and e are corresponding angles

(ii) a = e

(iii) b and e are co-interior angles. ab ef
(iv) b = e cd gh
(v) d = g
(vi) b + e = 180º i

(vii) b and g are alternate angles

(viii) d = e

(ix) f and i are corresponding angles.

(x) e + f = 180°

6. From the following figures, find the unknown angles.

x 80º z
(a)
(b) x (c) 65º

82º x
y
y

(d) 60º (e) (f) 90º x
x (i)
110º x
y 105°
(g) x x ya
z
120º (h) 20 + xº

110º

16 Oasis School Mathematics – 7

7. Which of the following pairs of lines are parallel? Write with reasons.

(a) 65º (b) (c) (d)

120º 40º a

65º 110º 140º a+1

(e) (f) (g)

122º 160º

8. Find the sizes of unknown angles from the following figures.

(a) E (b) E (c) 70º x 5 0 º (d) A E
G
A G BA G B
xº B (7xº+ 10º) y D
y zb
(x + 20)º

C 3x H H (x – 20)º a (3xº+10º) z
DC D
C H

FF F

(e) (f) 40º 50º (g) 70º b (h) 130º
d 140º
yx x y

120° p qr yz x a

65º

(i) (j) 120º (k) 3 5º (l)

50º 120º
x
xº x xº yº
40º 30º y 30º



Answer
1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher. 4. Consult your teacher.
5. Consult your teacher. 6. (a) x = 82° (b) x = 80°, y = 80° (c) x = 65°, y = 65°, z = 65°
(d) x = 60°, y = 60° (e) x = 70° (f) x = 90° (g) x=1200 ­ (h) x = 900 (i) x = 750, a = 750, y = 1050, z = 1050
7. Consult your teacher. 8. (a) x = 45° (b) x = 90° (c) x = 60°, y = 70°, z = 50°, a = 110°, b = 130° (d) x=16°,

y = 58°, z = 1220 (e) x = 65°, y = 60°, p = 60°, q = 65°, r = 115° (f) x = 40°, y = 130°, z = 50° (g) x = 110°,
b = 110°, d = 70° , a = 110° (h) y = 90° (i) x = 90°, (j) x = 900 (k) x = 35°, y = 35° (l) x = 30°, y = 150°

Oasis School Mathematics – 7 17

Objective Questions

Choose the correct alternatives. PR
1. Adjacent angle of ∠QOR is

(i) ∠POR (ii) ∠POQ (iii) ∠QOR itself Q

2. Vertically opposite angle of ∠AOC is A O
D

(i) ∠AOD (ii) ∠BOD (iii) ∠BOC O

C B

3. In the given figure, which of the following statement is not true? A D
B
(i) ∠AOD = ∠BOC (ii) ∠AOC = ∠BOD (iii) ∠AOC = ∠AOD O
B
C D

4. If the sum of two angles is 1800, then the angles are called Q
S
(i) complementary angles

(ii) supplementary angles

(iii) vertically opposite angles.

5. Sum of the angles around a point is

(i) 900 (ii) 3600 (iii) 1800.

6. Supplement of 800 is

(i) 800 (ii) 1000 (iii) 1800

7. Which of the following statement is not true?

(i) Vertically opposite angles are equal.

(ii) Complementary angles are equal.

(iii) Sum of supplementary angles is 1800. P

8. Alternate angles of ∠AQS is A
C
(i) ∠QSD (ii) ∠AQP (iii) ∠PQB Q

9. Which one of the following is not true? E S
R

(i) ∠EGB = ∠AGH (ii) ∠AGH = ∠GHD A G B
(iii) ∠AGH = ∠GHC H
C D
T
10. Corresponding angle of ∠TVQ is F
PV
(i) ∠VWS (ii) ∠VWR (iii) ∠SWV RW

U

18 Oasis School Mathematics – 7

Unit Triangles, Quadrilaterals

2 and Circles

2.1 Triangle (Review) A

A triangle is a closed plane figure bounded by three line c b

segments. It is denoted by a Greek letter ∆(Delta). A triangle has

three sides and three angles. BC

a

Note: The sides opposite to vertex A, B, C are denoted by a, b, c respectively.

Properties of triangle

Activity – 1 A
• Draw a triangle with the help of scale and pencil.
• Take the measurement of all angles.
• Find the sum of all angles.

∠A ∠B ∠C ∠A + ∠B + ∠C
......° ......° ......° ......°

BC

What is the conclusion of above activity?

Is the sum of all angles of a triangle 180°? If yes, repeat the activity for other triangles
and check whether it is true or false. From the above activity, you will find that the sum
of three angles of a triangle is 180°.

Activity – 2

Draw a triangle ABC with the help of pencil and scale. Take the measurement of AB,

BC and CA. A

AB BC AC AB + BC AC+AB AB + BC

Check whether the following relations are true or false. B C
AB + BC > AC, AB + AC > BC
BC + AC > AB

Oasis School Mathematics – 7 19

Draw another triangle and repeat the same activity again and draw the conclusion.
From this activity, you will conclude that sum of any two sides of a triangle is

longer than the third side.

i.e. In the triangle ABC A
(i) AB + BC > AC

(ii) AB + AC > BC

(iii) BC + AC > AB BC

Activity – 3

Draw a triangle having three unequal sides, let us say AB>AC>BC.
• Take the measurement of sizes of ∠A, ∠B and ∠C and compare Athem.
∠A ∠B ∠C

You will get ∠C >∠B >∠A. B C

Repeat the above activity by drawing other different
types of triangles.

Conclusion: From the above activity, you conclude that angle opposite to the longest
side is the greatest and the angle opposite to the shortest side is the smallest.

Converse of the above statement is also true i.e. side opposite to the greatest angle
is the longest and side opposite to the smallest angle is the shortest.

Activity – 4

• Draw a triangle ABC and produce side AB up to D. C

• Take the measurement of ∠C, ∠A and ∠CBD by protractor.

• Find the sum of ∠C and ∠A.

∠A ∠C ∠CBD ∠A + ∠C DB A

Check whether the sum is equal to ∠CBD or not.

Repeat the above activity by producing any other side.
You will get the sum of any two interior angles of a triangle equal to exterior

non-adjacent angle.

Conclusion: When a side of a triangle is produced then the exterior angle is equal
to the sum of non–adjacent interior angles.

20 Oasis School Mathematics – 7

Types of triangles based on length of sides (Review)

There are three types of triangles according to the length of sides, they are as follow.

(i) Equilateral triangle

A triangle whose all sides equal is called an equilateral triangle. A
Properties:
• All sides are equal. 3.2cm 3.2cm
• All angles are equal.
• Each angle measures 60°.

In the figure, AB = BC = AC = 3.2 cm. So, ∆ABC is an BC

equilateral triangle so, ∠A = ∠B = ∠C = 60°. 3.2cm

∴ Each angle of an equilateral triangle is 60°.

(ii) Isosceles triangle X
Any two sides of a triangle are equal, then the triangle is isosceles.

Properties: YZ
• Two sides are equal.
• Angles opposite to equal sides are equal.

In the given figure, sides XY = XZ. Angle opposite to XY is ∠Z and angle opposite

to XZ is ∠Y. So, ∠Y = ∠Z.

∴ Base angles of an isosceles triangle are equal.

(iii) Scalene triangle 2.5 cm P
The triangle having none of its side equal is called scalene tri-
3.2cm

angle. In the given figure, PQ < PR < QR, so the triangle PQR Q 5 cm R
is scalene.



Types of triangle based on size of angle

There are three types of triangles according to the sizes of angles. A

(a) Acute angled triangle 59º

In the figure, ∠A = 59° (acute) ∠B = 63° (acute) and ∠C = 58°

(acute). So ∆ABC is an acute angled triangle. B 63º 58º C

Hence, a triangle whose all angles acute is called an acute angled triangle. A

(b) Right angled triangle: 38º

In the given figure, ∠B = 90° so ∆ABC is a right angled

triangle. If one angle of a triangle is a right angle, the other C 52º 90º
two are acute and their sum must be 90°. B

Hence, a triangle having one angle 900 is called a right angled triangle.

Oasis School Mathematics – 7 21

(c) Obtuse angled triangle: P

In the figure, ∠Q = 118° (obtuse angle). Therefore, ∆PQR is an 1180 R
obtuse angled triangle.
Q

Remember !

• The sum of angles of a triangle is 180°.

• Sum of any two sides of a triangle is greater than the third side.

• Side opposite to the greatest angle is the largest and the side opposite to the smallest
angle is the shortest.

• Angle opposite to the longest side of a triangle is the greatest and the angle opposite
to the shortest side is the smallest one.

• The exterior angle of a triangle is equal to the sum of two non adjacent interior angles.

Worked Out Examples

Example: 1 A

Find the value of x from the given figure. 520
Solution: Here,
B 670 xC
In the given figure,

∠A + ∠B + ∠C = 1800 [∵ sum of three angles of a triangle is 1800]

or, 52° + 67° + x° = 180°

or, 119° + x = 180°

or, x = 180 – 119°

∴ x = 61°.

Example: 2

Identify whether 4 cm, 5 cm and 10 cm are the sides of triangle or not?
Solution: Here,

Three sides of a triangle are 4cm, 5 cm and 10 cm.

Now, sum of first 2 sides is 4 cm + 5 cm = 9 cm.

Which is less than third side.

Since, the sum of two sides of a triangle must be greater than the third side, 4 cm,
5 cm and 10 cm are not the sides of a triangle.

22 Oasis School Mathematics – 7

Example: 3 A C350

Identify the longest and the shortest side in the given figure. 600
Solution: Here,
∠A = 60°, ∠B = 85° and ∠C = 35° 850
∠B is the greatest angle. Opposite side to ∠B is AC.
So, AC is the longest side. B
Again, ∠C is the smallest angle. Opposite side to ∠C is AB.
So, AB is the shortest side.

Example: 4

Find the value of x, from the given figure.

A

50º

B xº C

110º

D

In the given figure, ∠ BAC + ∠ACB = ∠CBD
[∵ An exterior angle is equal to the sum of two non-adjacent interior angles]

or, 50° + x = 110°

or, x = 110° – 50°

∴ x = 60°

Example: 5 A

Find the value of x, from the given figure.

xº

70º C

B

In the given figure, AB = AC [Given]

∠ABC = ∠ACB = 70° [Base angles of an isosceles triangle]

or, ∠ABC + ∠ACB + ∠BAC = 180°

or, 700 + 700 + x = 180

or, x = 180° – 140°

∴ x = 40°

Oasis School Mathematics – 7 23

Example: 6 E

In the given figure, AB // CD, find the value of a.

A G B
x
x

C H yy aT

D

Solution: F

In the given figure, AB||CD and EF intersects AB and CD at G and H, GT and TH
bisect ∠BGH and ∠GHD. So,

∠BGT = ∠TGH = x, ∠GHT = ∠THD = y, ∠GTH = a.

Now,

(i) ∠BGH + ∠GHD = 180° (Sum of co–interior angles on parallel lines)

or, 2x + 2y = 180°

or, 2(x + y) = 180°
or,
or, x + y = 180°
2

x + y = 90°

(ii) In ∆ TGH,

∠GTH + ∠GHT + ∠HGT = 180° (Sum of angles in a triangle)

or, a + x + y = 180°

or, a + 90 = 180°

or, a = 180° – 90°

or, a = 90°

Example: 7

(2x + 10)°, (x – 10)° and (3x – 30)° are the angles of a triangle, find them.
Solution:

Here, (2x + 10)°, (x – 10)° and (3x – 30)° are the angles of a triangle so,

(2x + 10)° + (x – 10)° + (3x – 30)° = 180° (Sum of angles of a triangle) A

or, 2x + x + 3x + 10 – 10 – 30° = 180°. (2x+10)0

or, 6x = 180° + 30° B (x–10)0 (3x–30)0 C
or, x = 210°

6

24 Oasis School Mathematics – 7

∴ x = 35°

∴ The angles are (2x + 10)° = (2 × 35 + 10)°

= (70 + 10)°

= 80°

(x – 10)° = (35 – 10)°

= 25°

(3x – 30)° = (3 × 35 – 30)°

= (105 – 30)°

= 75°

Example: 8

If the angles of a triangle are in the ratio of 4:7:4, find the size of all angles.
Solution:

Here, the angles are in the ratio 4:7:4. So let us suppose the angles are 4x, 7x, 4x then their
sum is 180°.

i.e. 4x + 7x + 4x = 180°.

or, 15x = 180°
or, 180°

x = 15
= 12º

∴ The angles are 4 × 12°, 7 × 12°, 4 × 12°,

or, 48°, 84°, 48°.

Exercise 2.1

1. Find the sizes of unknown angles from the given figures:

(a) A (b) P (c) X

600 600

550 x0 B Q 3x0 x0 R 2x0 x0 Z
2
C (e) Y
G
(d) D
8x
2x0

3x0 x0 F 6x 4x I
Oasis School Mathematics – 7
E H

25

2. Which of the following are the sides of a triangle and why?

(a) 3 cm, 3 cm, 4 cm, (b) 8 cm, 5 cm, 2 cm,

(c) 6 cm, 6cm, 1 cm (d) 8 cm, 4 cm, 3 cm.

3. Name the greatest and smallest angles from the given figures.

(a) A (b) P (c) X

6 cm 3.2 cm
3 cm 2.5 cm 5 cm
3 cm

B 3.5 cm C Q 5 cm RY 4 cm Z

4. Name the longest and smallest sides from the given figures.

(a) A (b) P (c) L

600 2x0 3x0

580 620 C Q x0 R 2x0 x0 N

B M P
M 2x
5. Find the value of x from the given figures.

(a) A (b) P (c) N (d)

600 590 500

x0 500 480 R X x0 600 xO

DB C Q x0 1200 Y N

SZ

6. Find the value of x from the given figures.

(a) A (b) D (c) G
500 750
2x0

Bx x FH x0 I

CE

7. Find the sizes of unknown angles. P

(a) A (b) x (c) My x N
Q 600 P
x R
O 400 z
y 1300 D 500 y

B C Q S

(d) P (e) W Z (f) A C B

620 x xz
y

Sx yT y 500 450 400 E
z0 R z
480 X D
Y
Q

26 Oasis School Mathematics – 7

R

(g) x (h) A F B

P

y T 650 Ex

320 z C G D

Q S

8. Answer the following questions:
(a) Is it possible to draw a triangle having angles 45°, 55° and 90°? Give reason.

(b) Is it possible to draw a triangle having sides 7 cm, 3 cm, and 2.5 cm? Give reason.

(c) In ∆ABC, ∠A = 60°, ∠B = 80° and ∠C = 40°, arrange the sides of ∆ABC in
ascending order.

(d) In ∆ABC, AB = 8cm, BC = 6.5 cm and AC = 7 cm, arrange the angles in descending
order.

9. (a) Three interior angles of a triangle are in the ratio 1:2:3. Find the angles in degree.
(b) Three interior angles of a triangle are (3x + 10)0, (2x – 30)° and 3x°. Find the value of each angle.

(c) In an isosceles triangle, the vertical angle is 50°. Find the value of base angle.

Answer

1. (a) 65° (b) 48°, 72° (c) 30°, 60°(d) 30°, 60°, 90° (e) 40°, 60°, 80° 2. Consult your teacher.
3. Consult your teacher 4. Consult your teacher

5. (a) 110° (b) 107° (c) 70° (d) 60° 6. (a) 65° (b) 30° (c) 36°

7. (a) x = 80°, y = 50° (b) x = 65°, y = 115°

(c) x = 40°, y = 20°, z = 20° (d) x = 48°, y = 70°, z = 70° (e) x = 50°, y = 65°, z= 65°

(f) x = 45°, z = 40°, y = 95° (g) x = 32°, z = 83°, y = 83° (h) 90°
8. Consult your teacher. 9. (a) 30°, 60°, 90° (b) 85°, 20°, 75° (c) 65°

Oasis School Mathematics – 7 27

2.2 Construction of Triangles Rough Sketch

Type: I A

If three sides are given BC
Example : Construct a triangle having 4 cm
AB = 6 cm, BC = 4 cm and AC = 5 cm.
6 cm

6 cm
A 5 cm
B 4 cm
5 cm
Steps:

• Draw a line segment and take radius of 4 cm
on the compass and make BC = 4 cm.

• Take radius of 6 cm on your compass and cut
by an arc of 6cm from B.

C • Take radius of 5 cm and cut by an arc of 5cm
from C to get A.

• Join AB and AC using ruler.

∆ABC is the required triangle.

Type: II

If two sides and angle between them are given: Rough Sketch
Example : Construct a ∆ ABC in which
AB = 6 cm, BC = 5 cm, ∠ABC = 45°. A

AP 6 cm

B 45° C
5 cm

6 cm Steps:
• Draw a line segment BC of length 5 cm.
B 45° 5 cm • Draw an angle of 450 at B and draw a

straight line BP.
• Cut AB = 6 cm from B.
C • Join A and C using ruler.
∆ABC is a required triangle.

28 Oasis School Mathematics – 7

Type: III Rough Sketch

If a side and angles at both ends are R
given:
Example : Construct a triangle PQR in which P 60º 45° Q
side PQ = 5.7 cm, ∠PA= 60°, ∠Q = 45°.
5.7 cm
B

R

Steps:

• Draw a line segment
PQ = 5.7 cm.

• Construct an angle of

P 60º 45° 60° and 45° at P and Q
Q respectively.

5.7 cm • Draw PA and QB so that
they intersect at R.

∆PQR is the required
triangle.

Type: IV

If the hypotenuse and a side of a Rough Sketch
triangle are given:
X
Example : Construct a right angled triangle XYZ in
which hypotenuse XY = 7 cm and side YZ = 4 cm.

7 cm X

7 cmYZ
4 cm

YZ Steps:
4 cm • Draw a line segment YZ = 4 cm.
• Draw an angle of 90° at vertex Z.
• From Y cut YX = 7 cm, then join XY.
∆XYZ is the required triangle.

Oasis School Mathematics – 7 29

Exercise 2.2

1. Construct a triangle PQR from the following data.
(a) PQ = 5 cm, QR = 6 cm, PR = 8 cm
(b) PQ = 6.5 cm, QR = 5.1 cm, PR = 4.8 cm
(c) PQ = QR = PR = 6 cm
(d) PQ = QR = 5 cm, PR = 4 cm
2. Construct a triangle ABC from the following data.
(a) AB = 5 cm, BC = 6 cm, ∠B = 60°.
(b) AB = 4.5 cm, AC = 5.6 cm, ∠A = 75°.
(c) AB = BC = 6 cm, ∠B = 30°.
(d) CB = 5.2 cm, CA = 4.3 cm, ∠C = 90°.
3. Construct a triangle XYZ from the following data.
(a) YZ = 4 cm, ∠Y = 60°, ∠Z = 30°.
(b) XY = 5.2 cm, ∠X = 120°, ∠Y = 30°
(c) XZ = 6.1 cm, ∠X = 75°, ∠Z = 45°.
4. Construct a right angled triangle ABC from the following data.
(a) Hypotenuse AC = 6 cm and AB = 3.5 cm
(b) Hypotenuse BC = 4.5 cm and AC = 3.2 cm.

Answer
Consult your teacher.

2.3 Quadrilateral

A quadrilateral is a closed plane figure bounded by four line segments.

Every quadrilateral has four sides and four angles.

30 Oasis School Mathematics – 7

Draw a quadrilateral ABCD as shown in figure with the help of a pencil and scale.

• Take the measurement of all angles ∠A, ∠B, ∠C and ∠D. A B

∠A ∠B ∠C ∠D ∠A + ∠B + ∠C + ∠D

D C
• Find their sum.

Repeat the activity again by drawing any type of quadrilateral.
Now, write the conclusion.

Sum of all angles of a quadrilateral is always 3600.

Some special types of quadrilateral

As we have already discussed, quadrilateral is a polygon having four sides.

Parallelogram, rectangle, square, rhombus, trapezium and kite have some special

properties, so they are called special types of quadrilateral.

I. Parallelogram: A quadrilateral is called a A D

parallelogram if its opposite sides are parallel. In

the figure, AB||CD and AD ||BC, so ABCD is a

parallelogram. B C

Properties:

• Opposite angles are equal. AD
• Opposite sides are equal.

• Diagonals bisect each other. O
In the given parallelogram, AB = CD, AD = BC BC

[Opposite sides of a parallelogram]

∠A = ∠C, ∠B = ∠D [Opposite angles of a parallelogram]

OA = OC, BO = OD [Diagonals of a parallelogram bisect each other]

II. Rectangle: It is a parallelogram having each angle A D
90°. In the figure ∠B = 90°, AB||DC and AD||BC C
so, ABCD is a rectangle.

Properties: B

• All angles of a rectangle are equal. (i.e; every angle is 900)

• Opposite sides and angles are equal.

• Diagonals are equal.

Oasis School Mathematics – 7 31

In the given rectangle ABCD, AD
∠A = ∠B = ∠C = ∠D = 90°
AD = BC, AB = DC and O
AC = BD BC
III. Rhombus: Rhombus is a quadrilateral having all
AD
sides equal. In the given figure, AB = BC = CD = AD. BC
So, ABCD is a rhombus.
Properties: AD
• All sides are equal.
• Opposite angles are equal. O
• Diagonals bisect at right angle. BC
In the figure, ABCD is a rhombus.
∠A = ∠C, ∠B = ∠D A D
AO = OC, BO = OD B C
∠AOD = 90°
IV. Square:
Square is a quadrilateral having all angles and all

sides equal.
Properties:
• All angles and sides of square are equal.
• Each angle is 90°.
• Diagonals are equal.
• Diagonals are bisected at right angle.
In the figure, ABCD is a square.
So, AB = BC = CD = AD
∠A = ∠B = ∠C = ∠D = 90°
∠AOD = 90°.
AO = OC, BO = OD, AC = BD.
V. Trapezium: Trapezium is a quadrilateral having

one pair of opposite sides parallel. In the figure,
ABCD is a trapezium because AD||BC.
Property: A pair of opposite sides are parallel.

32 Oasis School Mathematics – 7

VI. Kite: Quadrilateral having two pairs of adjacent A D
sides equal is a kite. In the figure, AB = BC and B S
AD = DC. So ABCD is a kite.
C
Properties: P
• Two pairs of adjacent sides are equal.
• Diagonals are perpendicular to each other. QO
• Long diagonal bisects the shorter one.
It is the combination of two isosceles triangles. R
In the given figure, PQRS is a kite.
So, PQ = PS, QR = SR, QO = OS.
PR and QS are perpendicular to each other.
∆PQS and ∆QRS are two isosceles triangles.

Summary
Quadrilateral

Parallelogram Trapezium Kite

Rhombus Square Rectangle

Worked Out Examples

Example: 1

From the figure, find the value of x.
Solution:

Here ABCD is a quadrilateral. We know that sum of angles in a quadrilateral is 360°

so, ∠A + ∠B + ∠C + ∠D = 360°. A D

or, 2x + x + x + 2x = 360° 2x 2x

or, 6x = 360°
or,
x = 360º = 60° Bx xC
6

Oasis School Mathematics – 7 33

Example: 2

If the angles of a quadrilateral are in the ratio 4:2:1:3, find the angles.
Solution:

Let the angles of a quadrilateral be 4k, 2k, k and 3k.

then, 4k + 2k + k + 3k = 360° (∵ sum of 4 angles of quadrilateral.)

or, 10 k = 360°

or, k = 360° = 36°
∴The angles are 10

4 k = 4 × 36° = 144°

2k = 2 × 36° = 72°

k = 1 × 36° = 36°

3k = 3 × 36° = 108°

Example: 3 Dz C
x
In the given figure, ABCD is a parallelogram. If ∠A = 125°, 1250
find the values of x, y and z. A yB
Solution:

Here, ABCD is a parallelogram.

Then, ∠BAD = ∠BCD [∵ opposite angles of a parallelogram]

125° = x

∴ x = 125°

Again, ∠ABC + ∠BAD = 180° [∵ Sum of co-interior angles]

y + 125­° = 180°

or, y = 180° – 125° ∴ y = 55°

Again, ∠ABC = ∠ADC [∵ opposite angles of a parallelogram]

or, y = z

or, 55° = z

∴ z = 55°

Example: 4

In the given figure, PQRS is a parallelogram. Find the P Q
value of x, y and z. (5x+5)° z

Solution: S (4x–5)° yR

Here, PQRS is a parallelogram.

∠QPS + ∠PSR = 180° [Sum of co-interior angles is 180°]

or, 5x + 5 + 4x – 5 = 180°

or, 9x = 180°
or,
x = 180°
9

34 Oasis School Mathematics – 7

∴ x = 20°
Again, ∠QPS = y [Opposite angles of a parallelogram]
or, 5x + 5 = y
or,
or, y = (5 × 20 + 5)°
∴ y = (100 + 5)°
Again, y = 105°
or, ∠PQR = ∠PSR [Opposite angles of a parallelogram]
or, z = 4x – 5
or, z = (4 × 20 – 5)°
∴ z = (80 – 5)°
z = 75°

Exercise 2.3

1. Answer the following questions.
(a) What is the name of a quadrilateral whose opposite sides and opposite angles are equal?
(b) What is the name of a quadrilateral, whose a pair of opposite sides are parallel?
(c) What is the name of a parallelogram, whose adjacent sides are equal?

2. Fill in the blanks.

(a) If the diagonals of a quadrilateral bisect each other, it is a ………… .

(b) If the diagonals of a quadrilateral bisect each other at right angle, it is a …… .

(c) In a parallelogram, ………… sides and ………… angles are equal.

(d) In a rhombus, ………… sides and ………… angles are equal.

3. (a) Write one difference between a parallelogram and a rectangle.

(b) Write one difference between a rhombus and a square.

4. Find the unknown angles in the following figures.

(a) A 1280 B (b) P 530 2 x Q (c) C D
4a
120º 3a
2a a
D 48º xC S 105° 96° R F

E

Oasis School Mathematics – 7 35

5. (a) Four angles of a quadrilateral are 2x, 4x, 5x and 7x, find the angles in degree.

(b) Four angles of a quadrilateral are in the ratio 1 : 2 : 3 : 3, find the angles in
degree.

6. Find the value of x and y, from the given parallelogram.

(a) 3x+2 (b) 2x–3

2y+1 3y-2

2x+3 x+5

7. Find the value of x, from the given parallelogram.

(a) 3x+5 5x+1 (b)

x-2
3x-4

8. Find the unknown angles in the following figures.

(a) A z B (b) A y B (c) P 3x+20° yQ
120° x

S 3x–20°

y x D 70° z z
D C C R



(d) A B (e) W t X (f) 120° T
z
120° 2x P x Q
a

60° y x y 82° A Sz yR
D C Z Y

9. Find the unknown angles x0 and y0 in the following figures.

(a) 2x y (b) x + 40° y

x 60° x 80°

Answer

1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher.

4. (a) x = 64° (b) x =106° (c) a = 360, 2a = 720, 3a = 1080, 4a = 1440

5. (a) 40°, 80°, 100°, 140° (b) 40°, 80°, 120°, 120°

6. (a) x = 1, (b) x = 8, y = 3 , 7. (a) x = 2, (b) x = 1

8. (a) x = 120°, y = 60°, z = 60° (b) x = 1100, y = 700, z = 1100

(c) x = 30º, y = 70°, z = 110° (d) x = 30°, y = 120°

(e) x = 82°, y = 98°, z = 98°, t = 82° (f) x = 60°, y = 120°, z = 60°, a = 120°

9. (a) x = 60°, y = 120­° (b) x = 70­°, y = 100°

36 Oasis School Mathematics – 7

Activity

I. Sum of four angles of a quadrilateral is 360º.

a c a
b db

c

d

• Draw a quadrilateral as shown in the figure.
• Mark the angle by a, b, c and d as shown in the figure.
• Cut each angles along the dotted line.
• Arrange these angles at a point as shown in the figure.
From the figure, we see that the sum of four angles of a quadrilateral is 360º.

II. Formation of kite by paper folding AA
• Take a rectangular sheet of paper.
• Fold it as shown. C' C C
• Cut along ABC. BB
• Open it.

2.4 Construction of parallelogram

Type: I Rough Sketch

When two adjacent sides and angle between AD
them are given:
4 cm
Example : Construct a parallelogram ABCD in
which AB = 4 cm, BC = 6 cm and ∠ABC = 60°. B 60º C
6 cm
X

Think! A 6 cm Steps:
Opposite sides of
parallelogram are • Draw a line segment BC = 6
cm.
equal.
• Make∠XBC = 60° at the point
D B.

4 cm • From B cut BA = 4 cm, on BX.

4 cm • Opposite sides of parallelo-
gram are equal so cut off CD
B C = 4 cm and AD = 6 cm from C
6 cm and A to get D.

∴ ABCD is the required parallelogram. • Join AD and CD.

Oasis School Mathematics – 7 37

Type: II

When a side, a diagonal and angle between Rough Sketch
them are given :
P 4.9cm S

Example : Construct a parallelogram PQRS in

which QR = 4.9 cm, QS = 6.1 cm, ∠SQR = 30°. 6.1 cm

X Q 30º R
S 4.9cm
P 4.9 cm

Steps:

6.1 cm • Draw a line segment QR = 4.9 cm.
• At Q, draw an angle ∠XQR =30°.

Q 30º R • Cut off QS = 6.1 cm from Q on QX:
4.9 cm Opposite sides of a parallelogram
are equal so cut SP = 4.9 cm from S
∴ PQRS is the required parallelogram. and QP = RS from Q.

• Join PQ, PS and SR.

Construction of rhombus

A rhombus is also parallelogram. Its all four sides are equal in length. Process of
construction of rhombus is same as the construction of parallelogram.

Type: I Rough Sketch

When a side and an angle is given A 6.5 cm D
Example : Construct a rhombus ABCD where
6.5 cm
AB = 6.5cm and ∠B = 45°. 6.5 cm

A 6.5 cm D B 450 C
6.5 cm

6.5 cm Steps:

• Draw a line segment BC = 6.5cm.

6.5 cm • AT B, draw an angle of 450.
• From B, cut by an arc of 6.5cm to

get A.

B 45° • From A and C cut by an arc of
6.5cm to get D.
6.5 cm C
• Join AD and CD.

∴ ABCD is the required rhombus. Think!

All sides of rhombus are equal.

∴ Each side is equal to 6.5cm.

38 Oasis School Mathematics – 7

Exercise 2.4

1. Construct a parallelogram PQRS from the following data.
(a) PQ = 5 cm, QR = 6 cm, ∠PQR = 60°.
(b) PQ = 6.2 cm, PS = 6.8 cm, ∠SPQ = 75°.
(c) QR = 4 cm, QS = 6 cm, ∠SQR = 60°.
(d) QR = 5 cm, RP = 7 cm, ∠PRQ = 45°.
(e) PQ = 7cm, QS = 6.5cm, ∠PQS = 600.

2. Construct a rhombus ABCD from the given information.
(a) AB = 6cm, ∠A = 60° (b) BC = 7.3cm, ∠C = 75°
(c) CD = 6.4cm, ∠C = 30° (d) AD = 5.6cm, ∠D = 120°
Answer
Consult your teacher.

2.5 Construction of rectangle Rough Sketch

DC

Type: I 3 cm

When two adjacent sides are given: A 6 cm B

Example : Construct a rectangle ABCD in which
AB = 6 cm and AD = 3 cm.

D Steps:
A 6 cm C • Draw a line segment AB of length 6 cm.

• Draw angle of 90° at vertex A.

• Cut AD = 3 cm from A.

• Opposite sides of a rectangle are equal

B so cut DC = 6 cm and BC = 3 cm to get
vertex C.

• Join DC and BC.

∴ ABCD is the required rectangle.

Think!
Each angle of rectangle is 900.

Oasis School Mathematics – 7 39

Construction of square Rough Sketch5.7cm

Type: I AD

When one side is given: BC
Example: Construct a square ABCD
where AB = 5.7 cm. 5.7cm

Steps:
A D • Draw a line segment BC of

length 5.7cm

• At B, draw an angle 900.

• From B, cut by an arc of 5.7 cm to
get A.

• From A and C cut by the arc of 5.7
cm to get point D.

B 6 cm C • Join AD and DC.

∴ ABCD is a required square. Think!

All sides of square are equal.
Each angle of square is 900.

Type: II Rough Sketch

When one diagonal is given: AD
Example: Draw a square ABCD in 5.2 cm

which one diagonal BD = 5.2 cm

YX BC
A

Steps:

• Draw a rough sketch.

B D • Draw a diagonal BD = 5.2 cm.
5.2 cm
• Diagonals bisect the angles of
C a square, construct ∠DBX =
∠BDY = 45°.
∴ ABCD is the required square.
• Mark the point of intersection as A.

• Take AB (or AD) as radius and
cut BC = DC = AB.

• Join BC and CD.

40 Oasis School Mathematics – 7

Exercise 2.5

1. Construct a rectangle ABCD in which,
(a) AB = 6 cm, BC = 5 cm

(b) AB = 5.2 cm, BC = 3.4 cm.

(c) CD = 6.8cm, AD = 5.8cm.

(d) AD = 6.4cm, CD = 7cm.

2. Construct a square ABCD in which,

(a) AB = 5.1 cm, (b) BC = 3.9 cm.

(c) AC = 6 cm. (d) BD = 8 cm.



Answer
Consult your teacher.

2.6 Polygons

Introduction:

A polygon is a closed plane figure bounded by three or more straight lines.

It is a polygon. It is not a polygon. It is a polygon.

Regular polygons:
Polygon having all sides and all angles equal are regular polygon.

These are regular polygon.

Oasis School Mathematics – 7 41

Interior angle of a regular polygons:


All of the above figures are regular polygons. The marked angles in each figure are
their interior angles. Interior angles of a regular polygon are equal.

Exterior angles of a regular polygon:

In each of the above figure marked angle is their exterior angle.
Hence, if one of the sides of a regular polygon is extended outside, the angle so

formed is called an exterior angle.

Sum of interior angles of a polygon

Draw all diagonals to form triangles through a single vertex of given polygon.



4 sides, 2 triangles 6 sides, 4 triangles

It is observed that the number of triangles formed is 2 less than the number of its
sides in polygon.
So, if number of sides of polygon is n, then the number of triangles formed will be (n – 2)
But the sum of angles of a triangle = 180°
∴ Sum of angles of (n – 2) triangles = (n – 2) × 180°

Hence, sum of interior angles of a polygon having n sides = (n – 2) × 180°
If the n-gon is regular, it has n-equal sides and n-equal angles.

∴ Each interior angle of polygon = (n – 2) × 180°
n

42 Oasis School Mathematics – 7

An exterior angle of a regular polygon

Let θ be the interior angle and α be the exterior angle of a regular polygon.

Now at each vertex of a polygon A

exterior angle + interior angle=180° [both the angles together B E
form a straight angle]

i.e. α + θ = 180° θα F
or CD
(n – 2) × 180°
or α + n = 180°

[∵ Interior angle (θ) = (n – 2) × 180°]
n

α = 180° – (n – 2) × 180°
n

or α = 180°n – 180°n + 360°
n

∴ α = 360°
n

Hence, the exterior angle of regular polygon = 360°
n

Note : If the sides of a polygon are produced in order, the sum of exterior angles

so formed is always 360°. θ
β

In the given figure, a + β + υ + δ = 3600. υ α

δ

Remember ! = (n – 2) × 180°

• Sum of the interior angles of a polygon = 360°
• Sum of the exterior angles of a polygon
• An interior angle of a regular polygon = (n – 2) × 180°
n
• An exterior angle of a regular polygon 360°
= n

Oasis School Mathematics – 7 43

The table below shows some polygons, number of sides and their names.

Num- Name Figure Number Name Figure
ber of of sides
sides

5 Pentagon 11 Undecagon

6 Hexagon 12 Dodecagon

7 Heptagon 13 13 – gon
8 Octagon 14 14 – gon
9 Nonagon 15 Quin –decagon

10 Decagon



Worked Out Examples

Example: 1
Using the formula, find the size of each interior angle and exterior angle in a regular
hexagon.
Solution:
Here, number of sides of hexagon (n) = 6

44 Oasis School Mathematics – 7