Math 8

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

8

Author

Shyam Datta Adhikari

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

8

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. 2076 (2019 AD)
B.S. 2078 (2021 AD) (Completely revised)

Contributors
Man Bahadur Tamang
Shuva Kumar Shrestha

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 humble 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 learning of process.

I have endeavoured 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 Megh Raj Adhikari, Madan Kumar Shrestha,
Shanta Kumar Sen (Tamang), Man Bahadur Tamang, Laxmi Gautam,
Sunil Kumar Chaudhary, Ram Prasad Sapkota, Saroj Neupane and
Yadav Siwakoti for their invaluable suggestions and contributions.

My gratitude goes to Managing Director Mr. Harish Chandra Bista for
his invaluable support and cooperation in getting this series published
in this shape.

At the end, constructive and practical suggestions of all kinds for further
improvement of the book will be appreciated and incorporated in the
course of revision.

Shyam Datta Adhikari
Author

March 2019

Contents

Geometry

Unit : 1 Angle................................................................................ 2

1.1 Some Special Pairs of Angles................................ 2

1.2 Angles Made by a Transversal with the lines.... 8

Unit : 2 Triangle, Quadrilateral and Polygon ........................ 21

2.1 Triangles.................................................................. 21

2.2 Types of quadrilateral........................................... 32

2.3 Construction of Rectangle..................................... 38

2.4 Construction of Regular Polygon........................ 40

Unit : 3 Congruency, Similarity, Circle, Solids and Their Nets 45

3.1 Congruent Figures................................................. 45

3.2 Similar Figures........................................................ 51

3.3 Circles....................................................................... 56

3.4 Solids and Their Nets............................................. 63

Coordinates

Unit : 4 Co-ordinates.................................................................... 66

4.1 Pythagoras Theorem.............................................. 66

4.2 Distance Formula................................................... 71

Mensuration

Unit : 5 Perimeter, Area and Volume....................................... 79

5.1 Perimeter and Area of Plane Figures.................. 79

5.2 Volume of Cube and Cuboid............................... 84

Transformation

Unit : 6 Transformation............................................................... 92

6.1 Reflection.................................................................. 92

6.2 Rotation.................................................................... 97

6.3 Translation or Displacement................................ 102

Unit : 7 Bearing and Scale Drawing......................................... 104

7.1 Bearing..................................................................... 104

7.2 Scale Drawing......................................................... 108

Contents

Sets

Unit : 8 Sets................................................................................. 114

8.1 Review................................................................... 114

8.2 Operations on Sets.............................................. 118

8.3 Cardinality of Two Sets...................................... 128

Arithmetic

Unit : 9 Number System of Different Bases............................. 136

9.1 Binary and Quinary Number System.................. 136

9.2 Addition and Subtraction of Binary Numbers........... 144

9.3 Addition and Subtraction of Quinary Numbers....... 147

Unit : 10 Integers........................................................................... 150

10.1 Integers.................................................................. 150

10.2 Simplification of Integers................................... 153

Unit : 11 Rational and Irrational Numbers............................ 156

11.1 Rational Numbers................................................ 156

11.2 Operations on Irrational Numbers................... 160

11.3 Scientific Notation of Numbers........................ 164

Unit : 12 Ratio and Proportion.................................................. 167

12.1 Ratio....................................................................... 167

12.2 Proportion............................................................. 172

12.3 Types of Proportion............................................ 175

Unit : 13 Percentage.................................................................... 180

Unit : 14 Unitary Method........................................................... 186

14.1 Unitary Method.................................................. 186

14.2 Time and Work.................................................. 191

Unit : 15 Profit and Loss ............................................................ 193

15.1 Profit and Loss..................................................... 193

15.2 Calculation of C.P. or S.P., profit and loss........ 198

15.3 Discount and VAT............................................... 201

Unit : 16 Simple Interest............................................................ 207

Contents

Statistics

Unit : 17 Statistics........................................................................ 215

17.1 Collection of Data................................................ 215
217
17.2 Cumulative Frequency Distribution................. 220
221
17.3 Line Graph or Time Series Graph..................... 224

17.4 Pie-chart or Pie-diagram....................................

17.5 Measures of Central Tendency..........................

Algebra

Unit : 18 Some Special Product Formulae............................... 239

Unit : 19 Factorisation.................................................................. 251

19.1 Factorisation......................................................... 251

19.2 Factorisation of Difference of Two squares..... 253

19.3 Factorisation of the Trinomial of the

form a2 ± 2ab +b2................................................... 257

19.4 Factorisation of the Trinomial of the

form ax2 + bx + c.................................................. 258

Unit : 20 H.C.F. and L.C.M......................................................... 263

20.1 Highest Common Factor (HCF)........................ 263

20.2 Lowest Common Multiple (LCM).................... 266

Unit : 21 Rational Expressions................................................... 270

21.1 Rational Expressions-Review............................ 270
21.2 Multiplication and Division of Rational Expressions 272
21.3 Addition and Subtraction of Rational Expressions 275
Unit : 22 Indices............................................................................ 280

Unit : 23 Equation, Inequality and Graph............................... 284

23.1 Linear Equation in one Variable....................... 284
23.2 Applications of Linear Equation....................... 287
23.3 Graph of the Linear Equation............................ 290
23.4 Slope of a Straight Line....................................... 291
23.5 Intercepts of a Straight Line............................... 295
23.6 Simultaneous Equations..................................... 298
23.7 Quadratic Equations........................................... 304
23.8 Inequality (In-equation)...................................... 308
Model Question.......................................................... 313

Geometry

42Estimated Teaching Hours

Contents
• Angle
• Triangle, Quadrilateral and Polygon
• Circle
• Solids and their Nets

Expected Learning Outcomes
At the end of this unit, students will be able to develop the following
competencies:

• State the features of various special pair of angles and solve the prob-
lems related to them

• Identify alternate angles, corresponding angles and co-interior angles
and use these property to solve the problems related to them

• State the properties of triangle, various types of quadrilaterals and pol-
ygons and use these properties to solve the problems related to them

• Verify properties of parallelogram and rectangle experimentally
• Construct rectangle and regular polygons
• Make two triangles congruent using different axioms (AAS, ASA, SAS,

RHS, SSS)
• Make two triangles similar
• Find the circumference and area of a circle
• Draw the nets of prism, pyramid, cube, cuboid, etc.

Teaching Materials

• Compass, protractor, set square, models of geometrical shapes, etc.

Oasis School Mathematics-8 11

Unit

1 Angle

1.1 Some Special Pairs of Angles (Review):

Vertically opposite angles AD

In the given figure, two straight lines AB and CD intersect each
other at a point O. The angles AOC and BOD are vertically C O B
opposite angles; the angles AOD and BOC are also 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 intersect at a point, vertically opposite angles so formed are always equal
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.

A

D AD

A C D
O O O

B CB

C
B

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

To verify: ∠BOC = ∠AOD
∠ AOC = ∠BOD

2 Oasis School Mathematics-8

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: If two lines intersect at a point, vertically opposite angles so formed are al-
ways equal.

Adjacent angles: A

In the given figure, ∠AOB and ∠BOC have a common vertex B
C
O and a common arm OB. They lie on opposite sides of the O
common arm OB.

So, ∠AOB and ∠BOC are adjacent angles.

Two 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 on same side of straight
is 1800.

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

Now, ∠ACD and ∠BCD are a pair of adjacent angles on a same side of straight line.

AD D
A
D

CC

B AC B B

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

Oasis School Mathematics-8 33

To verify: ∠ACD + ∠BCD = 1800

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. ..................0 ..................0 ∠ACD + ∠BCD = 1800
iii. ..................0 ..................0 ∠ACD + ∠BCD = 1800

Conclusion: The sum of adjacent angles on the same side of straight line is 1800.

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

E D
A
• In the given figure, ACB is a straight line,

Hence, ∠ACE + ∠ECD + ∠BCD = 1800 A C B

Complementary angles

In the given figure,

( i) ∠ABC = 60° and ∠ABD = 30°.

∴ ∠ABC + ∠ABD = 60° + 30° = 90°

∠ ABC and ∠ABD are complementary angles. ∠ABC is 300
600
the complement of ∠ABD and ∠ABD is also called the complement of ∠ABC. C
Hence, two angles are complementary angles if their sum is 90°. B

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 a 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

4 Oasis School Mathematics-8

Worked Out Examples

Example: 1

Find the value of x from the given figure.

Solution: (x+10)0
(3x–6)0
In the given figure,
(2x–4)0
(x + 10)0+ (3x – 6)0 + (2x – 4)0 = 1800 [Angles on a straight line]

or, x + 100 + 3x – 60 + 2x – 40 = 180°

or, 6x = 1800
or,
∴ x = 1800
6

x = 300

Example: 2

Find the value of x in the given figure.

Solution:

In the given figure, 1040 900
x
x + 650 + 250 + 900 + 1040 = 3600 [Angles around a point] 250
650

or, x + 2840 = 3600

or, x = 3600 – 2840

or, x = 760

Example: 3

If (2x + 30)0 and (2x – 50)0 form a linear pair, find the value of x.

Solution:

Hence, (2x + 30)0 + (2x – 50)0 = 1800 [Sum of the linear pair is 1800]

or, 4x – 20 = 1800

or, 4x = 2000
or, x = 2000
∴ 4
x = 500

Oasis School Mathematics-8 55

Exercise 1.1

1. Find the value of x. (b) C (c) D C
(a) C

x 1350 x 750 800
D D x 500
A B A B A O B

(d) C D (e) F (f) D
G
A
4x 5x 200 8x 3x x C
A 8x 6x B 3x O B
D E
O O

(g) D C (h) ST

R

600 600 200
4x0 4x0 500
B O B P x Q
O

2. Find the value of x, y and z in the given figures.

(a) D (b) P (c) BC
B
A x 400 R x A z 600 D
C 800 400 x
O
y yO
z S

Q FE

(d) M 700 Q (e) S (3x+30)0 Q
300 Y
P (2x+10)0 P
O
N

3. Without measuring, find the value of x in the given figures.

P (c) X

(a) A B (b)

800 4x W 900 800 Y
Ox 3x O
R x O 1000

5x Z
Q

(d) A (e) A

B (5x–20)0 D
O
3x0 (5x+20)0 1040 E
2x O D
5x0 B 250
750

C

C

6 Oasis School Mathematics-8

4. (a) Write the complement of: (i) 65° (ii) 58°
(b) Write the supplement of: (i) 100° (ii) 115°
5. (a) If (3x + 11)° and (2x + 4)° are complementary angles, find the value of x.
(b) If (3x + 15)° and (2x + 5)° are supplementary angles, find the value of x.

6. (a) In the given figure, ∠AOB and ∠BOC are complementary angles, find the
value of x.

A
B

(3x–4)0 C
(3x+10)0

O

(b) In the given figure, ∠ABC = 90°, find x and the unknown angles.

A
x

(x+20)° y
((27xx+-11)1°)° C
B
A
(c) Find the value of a, b and x in the given figure: B

a

F 5x–10 b
3x+4
a



E C
D

Answer
1. (a) 450 (b) 1050 (c) 500 (d) 100 (e) 200 (f) 150 (g) 150 (h) 500
2. (a) 400 (b) x = 1000 , y = 800 , z = 1000
(c) x = 400, y = 600, z = 800 (d) x = 300 (e) x = 400
3. (a) 2800 (b) 300 (c) 900 (d) 200 (e) 330
4. (a) (i) 250 (ii) 320 (b) (i) 800 (ii) 650 5. (a) 150 (b) 320
6. (a) 140 (b) 80, 280, 170, 450 ( c) x = 120, a = 400, b = 1000

Oasis School Mathematics-8 77

1.2 Angles made by a Transversal with the Lines

Transversal: In the given figure, PQ cuts two or more than two lines. PQ is the

transversal. P P

P

QQ Q

A straight line, which cuts two or more straight lines (parallel or non–parallel) at
different points is called a transversal. In each of the above figures, PQ is a transversal
line.

Angles formed by two lines and their transversal line P

In the given figure, AB and CD are any two straight lines and R B

transversal PQ cuts AB and CD at points R and S respectively. A ab
cd

It makes four angles with each of the lines AB and CD. These

angles are marked a, b, c, d, e, f, g, h respectively. C ef

In the figure, angles a, b, h, g are exterior angles whereas d, g hS D

c, e, f are interior angles. Q Fig. (i)

Alternate angles:



In all of the above figures, a pair of black marked angles are interior, non–adjacent
and lie on the alternate sides of transversal.

Such pair of angles are alternate angles.

When a transversal cuts a pair of straight lines, two non–adjacent interior angles
that lie on the alternate side of transversal are called alternate angles.

Again,

If a transversal cuts a pair of parallel lines then alternate E

angles so formed are equal. A M B
ab
In the given figure, AB//CD, so a = d and b = c. cd
C N D

F

8 Oasis School Mathematics-8

Conversely, if a transversal cuts two straight lines BD
such that a pair of alternate angles are equal, then x
the two lines are parallel. In the given figure if x = y,
then AB//CD are parallel. y
AC
Experiment III

If a transversal cuts two parallel lines, then the alternate angles so formed are equal.

Draw three pairs of parallel lines AB and CD.
Draw a transversal EF which cuts AB and CD at G and H respectively.
Now, ∠AGH and ∠GHD are a pair of alternate angles.

Similarly,
∠BGH and ∠ GHC are also a pair of alternate angles.

E E B AC
A GB D E
G
G
AH
CH D C H
F
F F
Fig. (ii)
BD

Fig. (i) Fig. (iii)

To verify: ∠AGH = ∠GHD

∠BGH = ∠GHC

Observations:

Measure each pair of alternate angles and tabulate the values as below.

Figure ∠AGH ∠GHD ∠BGH ∠GHC Remarks
i. ..................0 ..................0 ..................0 ..................0 ∠AGH = ∠GHD,
∠BGH = ∠GHC
ii. ..................0 ..................0 ..................0 ..................0 ∠AGH = ∠GHD,
∠BGH = ∠GHC
iii ..................0 ..................0 ..................0 ..................0 ∠AGH = ∠GHD,
∠BGH = ∠GHC

Conclusion: Hence, if a transversal cuts a pair of parallel lines, alternate angles so formed
are equal.

Oasis School Mathematics-8 99

Worked Out Examples

Example: 1

From the adjoining figure, calculate the values of x, y and z. 500
x
Solution:
y
x = 500 [Vertically opposite angles]
z

Here, y = x = 50° [Alternate angles]

Lastly, y + z = 180° [Linear pair]

or 50° + z = 180°

∴ z = 180° – 50° = 130°,

Hence, x = 50°, y = 50° and z = 130°

Example: 2

In the given figure, AB||CD, ∠ABE=30°, ∠CDE=40°, find the value of x.

Solution: A B
Here, AB || CD, ∠ABE = 30°, ∠CDE = 40°
∠BED = x° 300
Through E draw PQ || AB
P Ex Q

C 400 D

i.e. PQ || AB || CD

Here, ∠BEQ = ∠ABE = 30° [Alternate angles]

Again, ∠DEQ = ∠CDE = 40° [Alternate angles]

Hence, ∠BED = x° = ∠BEQ + ∠DEQ = 30° + 40°= 70° [Whole parts axiom]

Example: 3 B
A 300
In the given figure, find the two pairs of parallel lines.
Solution: C 300 300 D
Here, ∠ABC = ∠BCD = 300 [Given] E
then AB//CD. [Being alternate angles equal]
Again,
∠BCD = ∠CDE = 300 [Given]
Then CB //ED. [Being alternate angles equal]
Hence, AB//CD and CB//ED are two pairs of parallel lines.

10 Oasis School Mathematics-8

Exercise 1.2 a b
c d
1. From the given figure, (i) name four interior angles ef
and (ii) name four exterior angles. hg

2. Copy the following figures in your copy and shade the alternate angles of the
given angle.

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

3. Identify in which of the following figures alternate angles are equal.
(a) (b) (c)




4. Find the value of unknown angles in the given figures.

(a) (b) (c) (d) 1200

700 1100 300 x zy
x0 x
y

(e) y0 (f) (g)
x0
250 400 x y
580 x

300

5. Find a pair of parallel lines in the given figures and also justify your answer.

(a) P M O (c) A B
500 500
Q (b)

560

R 500 S 560 C 500 D
N
P

Answer

1. Consult your teacher. 2. Consult your teacher. 3. Consult your teacher.
4. (a) 700 (b) 1100 (c) 300 (d) x = 1200 , y = 1200, z = 1200 (e) x = 580, y = 1220
(f) 550 (g) x = 400, y = 400 5. Consult your teacher.

Oasis School Mathematics-8 1111

1.3 Corresponding angles





In all of the above figures, let's find the nature of black marked angles.

• each pair are non adjecent angles

• one is interior and another is exterior E
• both lie on the same side of transversal

These pair of angles are corresponding angles. ba B
cd D
A

Again, if a transversal cuts two parallel lines, corresponding

angles so formed are equal.

In the given figure, C fe

hg

a and e are corresponding angles. F
b and f are corresponding angles.

c and h are corresponding angles.

d and g are corresponding angles.

Since AB||CD,

a = e, b = f, c = h and d = g.

In the given figure, if x = y then AB||CD.

A B
x D

C
y

E

In the above figure x and y are a pair of corresponding angles and they are equal.

Experiment IV

If a transversal intersects two parallel lines, corresponding angles so formed are
equal.

Draw two pairs of parallel lines AB and CD.

Also draw a transversal EF which meets AB and CD at G and H respectively.

Now, ∠AGE and ∠GHC are a pair of corresponding angles.

Similarly ∠AGH and ∠CHF, ∠EGB and ∠GHD, ∠BGH and ∠DHF are other pairs
of corresponding angles.

12 Oasis School Mathematics-8

E E A C
A GB B E

GD G

CH D A H H
C
F F BD F
Fig (ii)
Fig (i) Fig (iii)

To verify : ∠AGE = ∠GHC
∠AGH = ∠CHF
∠EGB =∠GHD and ∠BGH = ∠DHF

Observations:
Measure each angles and tabulate their values as below.

Figure ∠AGH ∠CHF ∠GHD ∠EGB Remarks

i. ................0 ................0 ................0 ................0 ∠AGH = ∠FHC,

∠GHD = ∠EGB

ii. ................0 ................0 ................0 ................0 ∠AGH = ∠FHC,

∠GHD = ∠EGB

iii ................0 ................0 ................0 ................0 ∠AGH = ∠FHC,

∠GHD = ∠EGB

Conclusion: Hence, if a transversal intersects two parallel lines, then the
corresponding angles so formed are equal.

Worked Out Examples

Example: 1 A ba B
cd D
From the given figure, find the four pairs of corresponding angles.
fe
Solution: C hg

In the given figure, four pairs of corresponding angles are

a and e, b and f, c and h, & d and g.

Example: 2 E

Identify whether AB and CD are parallel or not.. A I 70º B

Solution: C J 70º D
In the given figure,

∠EIB = ∠IJD = 70° F

Oasis School Mathematics-8 1133

Since two corresponding angles are equal, then AB||CD. E

Example: 3 A
P
In the adjoining figure, BA||DC, find the value of x. B 3x C

Solution: 1300
Here, BA || DC, ∠BOD = 130°
Through O, draw OP || BA || DC. O

D 2x
F

Now, ∠BOP = ∠ABE = 3x [Corresponding angles being BA || OP]

∠DOP = ∠CDF = 2x [Corresponding angles being DC || OP]

Now, ∠BOD = ∠BOP + ∠DOP [Whole parts axiom]

or, 130° = 3x + 2x

or, 130° = 5x

∴ x = 1350° = 26°

Hence, the value of x is 26°.

Example: 4

State, with reasons, whether AB is parallel to CD or not. AC

Solution:

Here, mark the angles a, b and c as shown in the adjoining diagram. a=1200 b c = 600

Here, a = 120°, c = 60°. BD

Now, b + c = 180° [Being linear pair]

or b + 60° = 180°

∴ b = 180° – 60° = 120°…....(i)

Again, a = 120°.......(ii)

From (i) and (ii), we get a = b

Hence, AB || CD [∵ a = b, being corresponding angles equal]

Exercise 1.3

1. Copy the following figures in your copy and mark the angle corresponding to the
marked one.

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

14 Oasis School Mathematics-8

2. In the given figure, write the corresponding P
angles of the following angles. A RB

(i) ∠PRB (ii) ∠BRS CS D
(iii) ∠PRA (iv) ∠ARS

Q

3. Identify whether corresponding angles formed in the given figure are equal or

not.

(a) (b) (c)

Q
600 N
4. (a) Identify whether PQ is parallel to MN or not? Justify
600
your answer. P

EA M
(b) Identify whether AB is parallel to CD or not. 130º F
C

Justify your answer. 120º G

BD

5. Find the values of unknown angles in each of the following.

(a) (b) (c) x (d) y0
220
x0 800 x 500 320
x0
250
y

(e) (f) (g) (h) 58º
500 420 x
4x0 x

b 600 y 2x 62º

a



Answer
1. Consult your teacher.
2. Consult your teacher.
3. (a) not equal (b) equal (c) equal
4. (a) parallel (b) not parallel
5. (a) x = 250, (b) x = 800, y = 800 (c) x = 280 (d) x = 320, y = 320
(e) a = 500, b = 500 (f) x = 150, y = 600 (g) x = 420 (h) x = 1200

Oasis School Mathematics-8 1155

1.4 Co–interior angles:

Let's find the nature of black marked angles in all of the figures below:

• Each pair are non–adjacent angles.
• Both angles are interior angles.
• Both lie on the same side of the transversal.

These pair of angles are co-interior angles.
Again,

If a transversal cuts two parallel lines, sum of co-interior angles so formed is 1800.

In the given figure, E

d and e are co-interior angles. A ba B
c and f are co-interior angles. cd

Since AB||CD

d + e = 180° fe D

C hg

c + f = 180° F

Conversely, if the sum of co-interior angles is 1800, then two lines are parallel.

Note: b and h, a and g are two pairs of co-exterior angles

Experiment IV

If a transversal intersects two parallel lines, then the sum of two co-interior angles
so formed is 180°.

Draw two pairs of parallel lines AB and CD.
Also draw a transversal EF which meets AB and CD at G and H respectively.
Now, ∠AGH and ∠GHC are a pair of co-interior angles.
Similarly, ∠BGH and ∠GHD are a pair of co-interior angles.

E E A C
A GB B E

GD G

CH D A H H
C
F F BD F
Fig (ii)
Fig (i) Fig (iii)

To verify: ∠AGH + ∠GHC = 180°
∠BGH + ∠GHD = 180°

16 Oasis School Mathematics-8

Observations:
Measure the angles and tabulate the values as below:

Figure ∠AGH ∠GHC ∠BGH ∠GHD ∠AGH+∠GHC ∠BGH+∠GHD Remarks

i. ............0 ............0 ............0 ............0 ............0 ............0 ∠AGH+ ∠GHC=1800
ii. ............0 ............0 ............0 ............0 ............0 ............0 ∠BGH+ ∠GHD=1800
iii ............0 ............0 ............0 ............0 ............0 ............0 ∠AGH+ ∠GHC=1800
∠BGH+ ∠GHD=1800
∠AGH+ ∠GHC=1800
∠BGH+ ∠GHD=1800

Conclusion: If a transversal intersects two parallel lines then the sum of two
co-interior angles so formed is 1800.

Conversely: If a transversal cuts two other straight lines such that the sum of two interior

angles on the same side of the transversal is two right angles, then the two lines are

parallel.

Consider the following figure. AC

In the given figure, ∠ABD + ∠BDC = 180° 95° E
[Sum of co-interior angles is 180°, then AB||CD.] D
85°

B

Worked Out Examples

Example : 1 ef
ab
From the given figure, find two pairs of co-interior angles.
dc
Solution: hg
In the given figure, two pairs of co-interior angles are a
E
and d, b and c.

Example : 2

In the given figure, identify whether AB||CD or not.

Solution: AG 120º B
Here, ∠BGH = 120°, ∠GHD = 60°.

∠BGH + ∠GHD = 120° + 60° = 180° C 60º D
Since the sum of co-interior angles is 180°, AB||CD. H

Example: 3 F

In the given figure, AB || DE. Find the value of ∠BCD. D E
Here, AB || DE, ∠ABC = 130°, ∠CDE = 110°
Draw a line XY through C parallel to AB and DE. 1100

AB

1300

Now, ∠ABC + ∠BCX = 180° X CY

Oasis School Mathematics-8 1177

or 130° + ∠BCX = 180°(sum of co-interior angles being AB||XY)

∴ ∠BCX = 180° – 130° = 50°

Again, ∠CDE + ∠DCY = 180° (Sum of co-interior angles being DE//CY)

or 110° + ∠DCY = 180°

∴ ∠DCY = 180° – 110° = 70°

Lastly, ∠BCD + ∠BCX + ∠DCY = 180° (Sum of angles on same side of a straight line)

or ∠BCD + 50° + 70° = 180°

or ∠BCD + 120° = 180°

∴ ∠BCD = 180° – 120° = 60°

∴ ∠BCD = 60°

Example: 4

In the given figure, AB||CD, if ∠A = 120º, find the values of x, y and z.

Solution: A zB

1200 y

Here, ∠A = 120° x C
∠A + ∠D = 180°
D

[Co-interior angles]

or, 120° + x = 180°

x = 180° – 120° = 60°

Again, x + y = 180° [Co-interior angles]

or, 60° + y = 180°

or, y = 1800 – 600 = 1200

Again, y + z = 1800 [Co-interior angles]

or, 1200 + z = 1800

or, z = 1800 – 1200

z = 600

∴ x = 600, y = 1200, z = 60°

Exercise 1.4

1. Copy the following figures in your exercise book and shade the angle co-interior
to the marked one.



18 Oasis School Mathematics-8

E

2. In the given figure, write the co-interior angle of AG B
(i) ∠AGH (ii) ∠GHD. CH D

3. State whether AB is parallel to CD or not, give reasons. F

(a) A C (b) A C

P 300 1500 1500

Q P R 500 S Q

BD BD

4. In each of the following figures, find the values of the unknown angles.

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

1200 1100 500 z 1200 c
a x y
b ab

Answer

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

3. (a) parallel (b) not parallel 4. (a) a = 600, b = 1200 (b) x = 700

(c) y = 500, z = 1300 (d) a = 600, b = 120, c = 600

Miscellaneous Exercise

1. (a) Study the given figure and answer the questions given below. A E B
(i) Which is the alternate angle of c?
(ii) Which is the corresponding angle of c? ba
cd

(iii) Which is the co-interior angle of c? C fe D
(iv) Which is the co-interior angle of d? gh
(v) Which is the alternate angle of f?
F

(vi) Which is the corresponding angle of e?

(vii) Is a = e? Why?

(viii) Is d + e = 1800? Why?

(b) Study the given figure and answer the questions given below. E

(i) Which is the alternate angle of ∠AGH? AG B
(ii) Which is the corresponding angle of ∠AGH? CH D
(iii) Which is the co-interior angle of AGH?

(iv) Is ∠AGH = ∠GHD? Why? F

(v) Is ∠AGH = ∠GHC? Why?

(vi) Is ∠AGH + ∠GHC = 1800? Why?

(vii) Is ∠AGH = ∠CHF? Why?

Oasis School Mathematics-8 1199

2. Find the unknown angles from the given figures.

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

600 800 (a+50)0 1200
(2a+30)0 xy
3x x
y x


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

a x 320 x
1200
y 1220 700
p 1020
q

3. Find the values of the unknown angles in each of the following figures.

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

700 x0 600 y 500 y
1200 x z

y z 1300 x 1200
x
z

4. Evaluate each of the unknown angles from the following figures.

(a) 320 (b) (c)
3x0 1200

600 1500 1600
a0 2x0 x

(d) 1080 (e) (f) x

x0 1050 x0 220 1620
1120 250

(g)
1000

1100 x0

Answer
1. (a) Consult your teacher. (b) Consult your teacher.
2. (a) x = 200 (b) x = 800, y =800 (c) a = 200 (d) x = 1200, y = 600
(e) a = 1200, p = 600, q = 1200 (f) x = 1020, y = 1020 (g) x = 900 (h) x = 1100
3. (a) x = 1100 (b) x = 600, y = 1200 z= 1200 (c) x = 700, y = 500, z = 600 (d) x=500, y = 600,

z = 700 4. (a) a = 280, (b) x = 300, (c) x = 1000 (d) x = 1400, (e) x = 1300, (f) x = 400, (g) x = 300

20 Oasis School Mathematics-8

Unit Triangle, Quadrilateral

2 and Polygon

2.1 Triangle A

A triangle is a closed plane figure bounded by three line seg- b
ments. It is denoted by a Greek letter ∆(Delta). A triangle has c

three sides and three angles. BC

a

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

Types of Triangles A3cm 2cm

I. According to sides 3cm

a. Scalene triangle : In the given figure, AB = 2cm, B 4cm C
BC = 4cm and AC = 3cm. If all the three sides of a triangle A
are different in length, then it is a scalene triangle.

In the figure alongside,
i.e. AB ≠ BC ≠ AC. ∆ABC is a scalene triangle

b. Isosceles triangle: In the figure alongside, AB = AC = 3cm 3cm 2cm

∆ABC is an isosceles triangle. If any two sides of a triangle are B 4cm C

equal, then it is an isosceles triangle.

2cm A

c. Equilateral triangle: In the figure, AB = BC = AC. ∆ABC is an 2cm C
equilateral triangle. If all the sides of a triangle are equal, then
it is an equilateral triangle.

B

Note: Each angle of an equilateral triangle is always 60°.

II. According to angles A

a. Right angled triangle: In ∆ABC, ∠B = 90°. ∆ABC is a right angled

triangle. If one angle of a triangle is 900, then it is a right angled

triangle. C
C
B

b. Obtuse angled triangle: In ∆ABC, ∠B is 1200 which is obtuse. A
So ∆ABC is an obtuse angled triangle. If one angle of a triangle

is more than 900 and less than 1800, then it is an obtuse angled 1200
triangle. B

Oasis School Mathematics-8 2211

c. Acute angled triangle: In ∆ABC, ∠A = 550 ∠B = 600 and A

∠C = 650, all three angles are less than 900. i.e. all are acute. ∆ABC is 550

an acute angled triangle. Hence, all the angles of a triangle are less B 600 650

than 90°, then it is called an acute angled triangle. C

Note:

Type of triangle Number of acute Number of Number of obtuse
Right angled triangle angles right angles angles

2 1 0

Acute angled triangle 3 0 0

Obtuse angled triangle 2 0 1

Experimental verification of sides and angles of triangles

Experiment : 1

Sum of all interior angles of a triangle is 180°.

Given: Draw three triangles ABC of different size as shown in the figure.

AA A

B C B C BC
Fig. (i) Fig. (ii) Fig. (iii)

To verify: ∠A + ∠B + ∠C = 180°

Observations:

Measure ∠A, ∠B and ∠C with the help of a protractor and tabulate the values as
follows .

Figure ∠A ∠B ∠C Remarks
°........................... °........................... ∠A + ∠B + ∠C = 1800
(i) °........................... °........................... °........................... ∠A + ∠B + ∠C = 1800
°........................... °........................... ∠A + ∠B + ∠C = 1800
(ii) °...........................

(iii) °...........................

Conclusion: The sum of all interior angles of a triangle is 180°.

Experiment : 2

If two sides of a triangle are equal, then the angles opposite to them are also
equal.

or, the base angles of an isosceles triangle are equal.

22 Oasis School Mathematics-8

Draw three isosceles triangles ABC in which AB = AC with different measurements

as shown in the figures below.

A
AA

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

To verify : ∠B = ∠C.

Observations:
Measure ∠B and ∠C with the help of a protractor and tabulate them as follows:

Figure ∠B ∠C Remarks
(i) °........................... °........................... ∠B = ∠C
(ii) °........................... °........................... ∠B = ∠C
(iii) °........................... °........................... ∠B = ∠C

Conclusion: The base angle of an isosceles triangle are equal.

Experiment : 3

Each of the base angles of an isosceles right angled triangle is 45°.

Draw three right angled triangle ABC in which AB = BC and ∠B = 900 with different

shapes and sizes as shown in figures below. A

AA

B CC B CB
Fig. (iii)
Fig. (i) Fig. (ii)

To verify : ∠A = ∠C = 45°
Observations:
Measure ∠A and ∠C with the help of a protractor and tabulate them as follows:

Figure ∠A ∠C Remarks
°........................... ∠A = ∠C = 450
i. °........................... °........................... ∠A = ∠C = 450
°........................... ∠A = ∠C = 450
ii. °...........................

iii. °...........................

Conclusion : Each of the base angles of an isosceles right triangle is 45°.

Oasis School Mathematics-8 2233

Experiment : 4

Each angle of an equilateral triangle is 600.
Draw three equilateral triangles ABC of different measurements.

A
AA

B C BC

Fig. ( i i) BC
Fig. (iii)
Fig. (i)

To verify : ∠A = ∠B = ∠C = 600

Observation:
Measure ∠A, ∠B and ∠C of each figure and tabulate their values as follows:

Figure ∠A ∠B ∠C Remarks
i. ................0 ................0 ................0 ∠A = ∠B = ∠C = 600
ii. ................0 ................0 ................0 ∠A = ∠B = ∠C = 600
iii. ................0 ................0 ................0 ∠A = ∠B = ∠C = 600

Conclusion: Hence, each angle of an equilateral triangle is 600.

Altitude and Median of a Triangle A

Altitude: In the given ∆ABC, AD ⊥ BC, BE ⊥ AC and CF ⊥ AB FE
AD, BE and CF are altitudes of ∆ABC.

Perpendicular drawn from any vertex to its opposite side is called B D C

the altitude of a triangle.

Note: • A triangle has three altitudes.
• All three altitudes intersect at the same point.

Median: In the given triangle ABC, D, E and F are the mid points A

of BC, AC and AB respectively. AD, BE and CF are its medians. FE

The straight line joining a vertex to the mid–point of oppo-

site side is the median of a triangle. BD C

Note:
• A triangle has three medians.

• The three medians always intersect each other at the same point.

24 Oasis School Mathematics-8

Experimental Verification on Medians and Altitudes

Experiment : 1

The line joining the vertex to the mid–point of the base of an isosceles triangle is

perpendicular to the base.

Draw three isosceles triangles ABC, where AB = AC.

Find the mid–point of BC (D) and join AD.

A A
A

BD C B DC BD C
Fig. (iii)
Fig. (i) Fig. (ii)

To verify: AD⊥BC i.e. ∠ADB = ∠ADC = 900.

Observations: Measure ∠ADB and ∠ADC from each of the figures and tabulate the
values as below:

Figure ∠ADB ∠ADC Remarks
i. ..............0 ..............0 AD ⊥ BC
ii. ..............0 ..............0 AD ⊥ BC
iii. ..............0 ..............0 AD ⊥ BC

Conclusion: The line joining the vertex to the mid–point of the base of an
isosceles triangle is perpendicular to the base i.e. median of an isosceles triangle is
perpendicular to the base. Its converse is also necessary to verify experimentally.

Worked Out Examples

Example: 1 A

In the given figure, find the value of each angle.

Solution: 4x
Here, ∠A = 4x, ∠B = 5x and ∠C = 6x

∠A + ∠B + ∠C = 1800 [Sum of three angles of a triangle is 1800] B 5x 6x C
or, 4x + 5x + 6x = 1800

or, 15x = 1800
or, x = 180° = 120
15

Oasis School Mathematics-8 2255

∴ ∠A = 4x = 4×120 = 480
∠B = 5x = 5×120 = 600
∠C = 6x = 6 ×120 = 720

Example: 2 A
70°
In the given figure, ABC is a triangle in which
AB = AC, ∠A = 70°, find ∠B. BC

Solution: Here,

Given: In ∆ABC, AB = AC, ∠A = 70°

Let, ∠B = ∠C = x [Base angles of an isosceles triangle]

Now, ∠A +∠B +∠C = 180° [Sum of all interior angles of a triangle]

or, 70° + x + x = 180°

or, 70° + 2x = 180°

or, 2x = 180 – 700
or,
∴ ∠B = ∠C = 550. x = 110° = 55°
2

Example: 3 B

In the given figure, ∆ABD is an equilateral triangle and ∆BCD

is an isosceles triangle in which BC = CD. Find ∠CBA. C 800 A

Solution: Here,

∆ABD is an equilateral triangle D
∠ABD = 60° [Each angle of an equilateral triangle is 60°]

Again, ∠CDB = ∠CBD[ Being, BC = CD]

Now, ∠BCD + ∠CDB + ∠CBD = 180° [Sum of all interior angles of a triangle]

or 80° + ∠CBD + ∠CBD = 180°

or, 2∠CBD = 180° – 80°

∠CBD = 100° = 50°
∴ 2

∠CBA = ∠CBD + ∠ABD [Whole part axiom]

= 50° + 60° = 110°

26 Oasis School Mathematics-8

Example: 4 A

From the given figure, find the value of each angle. x

Solution: ∠C = ∠B = 2x B 2x C

[In ∆ABC, base angle of an isosceles triangle]


Now, ∠A + ∠B + ∠C = 180° [Sum of all interior angles of a triangle]

or x + 2x + 2x = 180°

or 5x = 180°

or x = 1850° = 36°

Hence, vertical angle = ∠A = x° = 36° and

Each base angle = ∠B = ∠C = 2x° = 2 × 36° = 72°.

Example: 5 A

In the given figure, AB = AC. If ∠BAC = 400, find the value of x. 400

Solution: x

Here, AB = AC and ∠BAC = 400. B C D

Here, ∠ABC = ∠ACB [ Base angles of an isosceles triangle]

Now, ∠ABC + ∠ACB + ∠BAC = 1800 [Sum of all interior angles of a triangle]

or ∠ACB + ∠ACB + 40° = 1800 [ ∵∠ABC = ∠ACB]

or 2∠ACB = 1800 – 400
or
or ∠ACB = 140°
2

∠ACB = 700

Again, ∠ACB+∠ACD = 1800 [ ∵ Being linear pair]

Again, 700 + ∠ACD = 1800

or, ∠ACD = 1800 – 700
∴ x = 1100

Oasis School Mathematics-8 2277

Exercise 2.1

1. Answer the following questions.

(a) How many acute angles does an acute angled triangle have?

(b) How many obtuse angles does an obtuse angled triangle have?

(c) How many right angles does a right angled triangle have?

(d) In ∆ABC, if AB = 5cm, BC = 6cm, AC = 4.5cm, what type of triangle is ∆ABC on
the basis of its sides?

(e) In ∆PQR, PQ = 6.5cm, QR = 6.5 cm, PR = 6cm, which type of triangle is this?

(f) In ∆XYZ, XY = YZ = XZ = 6.8cm, which type of triangle is ∆XYZ?

2. Find the value of x and y in each of the given figures.

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

x 600 6x 3x

7x 2x Z E 2x F
Y
B 550 650 C Q 2x 3x R

(e) A (f) S (g) G (h)
W
x Py 1200 D y
x

1200 1100 EQ 420 620 R E x 1120 600 x 1300 Z
DB C F HX Y

3. Find the values of unknown angles from the given figures. P

(a) A (b) X (c) D (d)

560 500 x

Bx CYx ZE y 600 F 1500 x yR
SQ
(e) A (f) P

x

Q R
yw xS
D z
B C

4. Find the value of x from the given triangles.

(a) A (b) P (c) W
Xx
600

x cm 4 cm Z

B 600 600 C Q x 600

RS Y

28 Oasis School Mathematics-8

5. Find the values of x, y, z from the figures below.

A (b) S

(a)

300 R 550
x0
y
C 500 D
B 400 P z x 650 T
300

Q

6. Evaluate x, y, z from the following figures.

(a) D (b) P R (c) M P (d) P S
G x R
350 750 550 y
xz y z 500
250 T
H Q S
y
E 600 yF N 650 Q Qx
xz

O

7. Find the value of x, y from the given figures.

(a) A (b) Q

x y

S 300 P
Cx
B 500 M
R


D

8. (a) In ∆ABC, AB = AC and AE||CB; E A
prove that ∠DAE = ∠EAB.

BC

(b) In the given figure, AB//CD. GI is the bisector of ∠BGH E

and HI is the bisector of ∠GHD. Prove that ∠GIH = 900. A G I B

 CH D

F

Answer
1. Consult your teacher. 2. (a) x = 600 (b) x = 240 (c) x = 120 (d) x = 180

(e) x = 500 (f) x = 760, y = 1040 (g) x = 520 (h) x = 500, y = 700
3. (a) x = 620 (b) x = 650 (c) x = 600, y = 600 (d) x = 300, y = 300 (e) w=600, x = 600,

y = 1200, z = 600, (f) x = 1350

4. (a) x = 4cm (b) x =1200 (c) x = 1200 5. (a) x=1200 (b) x = 600, y = 300, z = 1200

6. (a) x = 600, y = 850, z = 850 (b) x = 750, y = 800, z= 800 (c) x = 600, y = 550, z = 650

(d) x = 650, y = 650 7. (a) x = 400, (b) x =600, y = 600

Oasis School Mathematics-8 2299

2.2 Types of Quadrilateral

a. Parallelogram D C
B
In the given figure, where AB = DC, AB||DC, AD = BC, AD||BC,

then ABCD is a parallelogram. Hence, a parallelogram is a

quadrilateral whose opposite sides are equal and parallel. A

Note: Opposite angles of a parallelogram are equal. In a parallelogram ABCD,
∠A = ∠C and ∠B =∠D.

b. Rectangle C
B
In the given figure, ABCD is a rectangle where, AB = DC, D
AB||DC, AD= BC, AD||BC, ∠A = ∠B = ∠C =∠D = 90°.

A rectangle is a quadrilateral whose opposite sides are equal A
and parallel and each angle is right angle.

Note: Every rectangle is a parallelogram. D
A
c. Square C

Here, ABCD is a square. B
Where AB = BC = CD = AD and ∠A = ∠B = ∠C = ∠D = 90°.
A square is a rectangle whose all four sides are equal.

Note: Every square is a rectangle.

d. Rhombus D C
B
Here, ABCD is a rhombus. Where, AB = BC = CD = AD and

AB||DC, AD ||BC. A
A rhombus is a parallelogram whose all sides are equal.

Note: Every square is a rhombus. Every rhombus is a parallelogram.

e. Trapezium A B

In the given figure, ABCD is a trapezium, where AB||DC. C
A
A trapezium is a quadrilateral in which a pair of opposite sides D
are parallel. D

f. Kite B

In the given figure, ABCD is a kite, where AB = AD and BC = DC.

A kite is a quadrilateral in which two pairs of adjacent sides are equal.

C

30 Oasis School Mathematics-8

Experimental Verification on Parallelogram, Rectangle and Square

Experiment : 1

The opposite sides of a parallelogram are equal.

Draw three parallelograms ABCD where AB || CD and AD || BC with different

sizes with the help of set square and ruler.

A DA DA D

BC B CB C
Fig. (i) Fig. (ii)
Fig. (iii)

To verify : AB = CD and AD = BC.

Observation: Measure AB, CD, AD, BC and tabulate their values as follows:

Figure AB CD AD BC Results
i. ..........cm AB = CD and AD = BC
ii. ..........cm ..........cm ..........cm ..........cm AB = CD and AD = BC
iii. ..........cm AB = CD and AD = BC
..........cm ..........cm ..........cm

..........cm ..........cm ..........cm

Conclusion: Hence, the opposite sides of a parallelogram are equal.

Experiment : 2

The opposite angles of a parallelogram are equal.

Draw three parallelograms ABCD, where AB || CD and AD || BC with different

sizes with the help of set square and ruler.

A DA DA D

BC B CB C
Fig. (i) Fig. (ii)
Fig. (iii)

To verify : ∠A = ∠C and ∠D = ∠B.

Observation: Measure ∠A, ∠B, ∠C, ∠D and tabulate their values as follows:

Figure ∠A ∠B ∠C ∠D Results
i. ..........cm ∠A = ∠C and ∠D = ∠B
ii. ..........cm ..........cm ..........cm ..........cm ∠A = ∠C and ∠D = ∠B
iii. ..........cm ∠A = ∠C and ∠D = ∠B
..........cm ..........cm ..........cm

..........cm ..........cm ..........cm

Conclusion : Hence, the opposite angles of a parallelogram are equal.

Oasis School Mathematics-8 3311

Experiment : 3

The diagonals of a parallelogram bisect each other.

Draw three parallelograms ABCD of different sizes by using set square and ruler. In
each figure, draw diagonals AC and BD which intersect each other at O.

A DA DA D
O O O

B C B C B C
(i) (ii) (iii)


To verify: AO = OC and BO = OD

Observation: Measure AO, OC, BO, OD and tabulate them as follows

Figure AO OC BO OD Results
i. ....cm ...cm ...cm ...cm AO = OC and BO = OD
ii. ....cm ...cm ...cm ...cm AO = OC and BO = OD
iii. ....cm ...cm ...cm ...cm AO = OC and BO = OD

Conclusion : Hence, the diagonals of a parallelogram bisect each other.

Experiment : 4

Diagonals of a rectangle are equal in length.

Given:
• Draw three rectangles ABCD of different size.
• Draw the diagonals AC and BD in each rectangle.

A BA A B
B

D CD CD C
(i)
(ii) (iii)

To verify : AC = BD

Observations: Measure the lengths of AC and BD from each rectangle and tabulate
their values as below:

Figure AC BD Remarks

i. ....cm ...cm AC = BD

ii. ....cm ...cm AC = BD

iii. ....cm ...cm AC = BD

Conclusion: Hence, the diagonals of a rectangle are equal.

32 Oasis School Mathematics-8

Experiment : 5

Diagonals of a square bisect each other at right angles.

Given:

• Draw three squares ABCD of different size.
• Join the diagonals AC and BD of each square which meet at point O.

AB AB A B

O O O

DC DC DC
(ii) (iii)
(i)

To verify: AO = OC, DO = OB, AC ⊥ BD.

Observation: Measure AO, OB, DO, OC and ∠AOB, ∠BOC, then tabulate their
values as below:

Figure AO OB DO OC ∠AOB ∠BOC Remarks

i. ....cm ...cm ...cm ...cm ......° ......° AO = OC, DO = OB,

AC ⊥ BD
ii. ....cm ...cm ...cm ...cm ......° ......° AO = OC, DO = OB,

AC ⊥ BD
iii. ....cm ...cm ...cm ...cm ......° ......° AO = OC, DO = OB,

AC ⊥ BD

Conclusion: Hence, the diagonals of a square bisect each other at right angle.

Remember !

• Opposite sides of a parallelogram are equal.
• Opposite angles of a parallelograms are equal.
• Diagonals of a parallelogram bisect each other.
• Every rectangle is a parallelogram.
• Every rhombus is a parallelogram.
• In a trapezium one pair of opposite sides are parallel.

Worked Out Examples

Example: 1 A D
C
1100

In the given figure, ABCD is a parallelogram. If ∠A = 1100, find

∠B, ∠C and ∠D. B

Oasis School Mathematics-8 3333

Solution:

ABCD is a parallelogram

∠A + ∠B = 180° [Sum of co-interior angles is 1800]

or 110° + ∠B = 180° [∵∠A = 110°]

∴ ∠B = 180° – 110° = 70°

Again, ∠C = ∠A = 110° [Opposite angles of a parallelogram are equal]

and ∠D = ∠B = 70°

Hence, ∠B = ∠D = 70°, ∠C = 110°

Example: 2

In the given figure, ABCD is a square. If AO = OD = 5cm and AB = 5 2cm, find (i) OB
(ii) OC (iii) ∠AOB (iv) AD (v) CD

Solution: A 5 2cm B
C
Here, 5cm

OD = OB [Diagonals of square bisect each other] 5cm O
D
∴ OB = 5 cm

Again, AO = OC [Diagonals of square bisect each other]

∴ OC = 5 cm

Again, ∠AOB = 90° [Diagonals of a square bisect each other at right angle]

Hence, AB = CD = AD [All sides of square are equal.]
∴ AD = CD = 5 2cm



Exercise 2.2

1. Fill in the blanks.
(a) The sum of all the interior angles of a quadrilateral is ……… .
(b) In a parallelogram, ……… sides are equal and ……… angles are equal.
(c) Value of each angle of a rectangle is ……… .
(d) If opposite sides are equal and each angle is 90°, then it is a ……… .
(e) In a square, ……… angles are equal and ……… sides are equal.
(f) In a rhombus, ……… angles are equal and ……… sides are equal.
(g) In a kite, ……… pair of adjacent sides are equal.
(h) In a trapezium ……… pair of opposite sides are parallel.

34 Oasis School Mathematics-8

2. Tick (√) or cross (×) in the box.

Features Parallelogram Rectangle Rhombus Square
√ √ √ √
Opposite sides are equal.
Opposite angles are equal.
All sides are equal.
All angles are equal.
Each angle is 90°.
Diagonals bisect to each other

3. Find the value of x, y in the given figures.

(a) (b) y (c) (d)
y
2x+1

5 cm x
2.6 cm
y+1
xx

2y-1

6 cm 3.2 cm 3x-2 4.6 cm
y
4. Find the size of x, y, z, etc. in the following figures.

(a) x (b) (c) 3x (d)
z x
600 z

600 y (2x+20)0 1200 2y z 800
2x
(e) (f)

3x-5

1500 4x+10

5. Find the values of x and y in the given figures.

(a) A B (b) P (c) P Q (d) A B
C
Q x

3x+1 7cm 12cm 2cm
2x+5 O 2x-1 O O
y xOy

D CS RS RD

6. Evaluate a, b, x, y, z in the given figures.

(a) (b) P Q (c) A xB
b 300

300 x y a
x 250 C
500
(d) A a D

x yD S x R
200 z
(e) z
B
C y 300 b

Oasis School Mathematics-8 3355

A B
C
7. In a quadrilateral ABCD, if ∠A = ∠C and ∠B = ∠D then
prove that ABCD is a parallelogram.

Answer D

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

3. (a) x = 5 cm, y = 6 cm (b) x = 2.6 cm, y = 3.2 cm

(c) x = 3 cm, y = 2 cm (d) x = 4.6 cm

4. (a) x= 600, y = 1200, z = 1200 (b) x = 200 (c) x = 400, y = 300, z = 600

(d) x =1000, y = 800, z = 800 (e) x = 150 (f) x = 250

5. (a) x = 4 cm (b) x = 4 cm (c) x = 6 cm, y = 6 cm (d) x = 2cm, y = 2 cm

6. (a) x = 300 (b) x = 450 (c) x = 250, y = 300, a = 1250, b = 1250

(d) x = 700, y = 200, z = 700, (e) x = 500, a = 300, b = 500, z = 1000, y = 1000

I. Activity:
To find the properties of rectangle by paper folding.

Material required:

• A4 size paper • coloured chart paper • glue • scissors.

Procedure:

• Take a rectangular sheet of paper and fold it as shown below a, b, c and d.

• Paste the paper (e) on the chart paper and give the name ABCD as in 'f'.

(a) (b) (c) (d) (e)

Fill up the given table after measuring it. (f) A D
C
∠A ∠B ∠C ∠D AB BC CD AD O

AC BD AO BO CO DO B


1. Are opposite sides equal?
2. Are opposite sides parallel?
3. Are all angles equal?
4. How many angles are right angles?
5. Are the diagonals equal?
6. Do the diagonals bisect each other?

Conclusion: ____________________________________

36 Oasis School Mathematics-8

II. Activity:
To find the properties of rhombus by paper folding.
Material required:
• A4 size paper • coloured chart paper • glue • scissors.
Procedure:
• Take a rectangular sheet of A4 size paper.
• Fold it as shown below.

• Fold it again to get four layers.

• Mark two points P and Q as shown in the figure.

• Fold along PQ and make a crease.

• Open it and you find a rhombus.

• Cut the rhombus and paste it on the chart paper.

• Give the name as shown in figure 'g'.
(a) (b) (c) (d) (e) (f)

Fill up the given table after measuring it. (g) A D

∠AOB ∠AOD ∠BOC ∠COD AB BC CD AD

O

AO BO CO DO BC


Answer the following questions.
1. Are all the four sides equal?
2. Are opposite angles equal?
3. Do the diagonals bisect each other?
4. Do the diagonals bisect each other at right angle?
5. Are all angles equal?
6. Are the diagonals equal in length?

Conclusion: _______________________________________


Oasis School Mathematics-8 3377

2.3 Construction of Rectangle

I. If a diagonal and angle between two diagonals are given

Example: 1

Construct a rectangle ABCD where diagonal AC = 6.8 cm and an angle between
AC and BD is 600.

Think ! Rough Figure

P • Diagonals of rectangle D 3.4cm C
D are equal. 600
3.4cm

• Diagonals of rectangle 3.4cm O 3.4cm B
bisect each other.
A

3.4cm

Steps

• Draw a line and cut it by an arc of
A C 6.8 cm such that AC = 6.8 cm.
3.4cm O 3.4cm

• Draw the perpendicular bisector of

AC to find O as the mid–point of AC.

3.4cm • At O, construct an angle POA = 60°

and produce it to Q.

• From O, cut by an arc of 3.4 cm to get
B points B and D.

Q • Join AB, BC, CD and DA by line
segments.

• ABCD is the required rectangle.

II. If a side and a diagonal are given:

Example

Construct a rectangle ABCD where AB = 6.4 cm and diagonal AC = 8 cm.

P Rough Figure

D C D 6.4cm C

8 cm
900

A 6.4cm B
8 cm

Think !

• Each angle of
rectangle is 900.

A 6.4cm B • Opposite sides of
rectangle are equal.

38 Oasis School Mathematics-8

Steps

• Draw a line and cut it by an arc of 6.4 cm such that AB = 6.4 cm.
• At B draw an ∠ABP = 90°.
• Take an arc of 8 cm and from A cut BP at C such that AC = 8 cm.
• From C draw an arc of 6.4 cm and take the arc equal to BC and from A cut

the previous arc at point D.
• Join CD and AD.
• ABCD is the required rectangle.

III. If a side and angle between that side and a diagonal is given.

Example:

Construct a rectangle ABCD where BC = 6.2 cm, ∠DBC = 300.

R Q
P
A
D
Rough Figure

A D

300 900
B 6.2 cm C

300 6.2 cm C
B

Steps
• Draw a line and cut it by an arc of 6.2 cm. such that BC = 6.2 cm.
• At B draw an angle ∠PBC = 30°.
• At C draw an angle ∠QCB = 90°.
• Point of intersection of BP and CQ is the point D.
• Again at B draw ∠RBC = 90°.
• Take an arc equal to DC and from B, cut at point A.
• Join AD.
• ABCD is the required rectangle.


Oasis School Mathematics-8 3399

Exercise 2.3

1. a. Construct a rectangle ABCD where AC = 6.2 cm and angle between AC and BD
is 30°.
b. Construct a rectangle PQRS where PR = 7 cm, ∠PRQ = 45°.
c. Construct a rectangle EFGH where EG = 5.8 cm, ∠EGH = 60°.

2. a. Construct a rectangle ABCD where AB = 6.5 cm and diagonal BD = 7 cm.
b. Construct a rectangle PQRS where PQ = 5.6 cm and diagonal PR = 6 cm.
c. Construct a rectangle EFGH where HE = 5.8 cm and diagonal EG = 6.6 cm.

3. a. Construct a rectangle ABCD where AB = 6 cm, ∠CAB = 30°.
b. Construct a rectangle PQRS where PQ = 5.8 cm and ∠PQS = 45°.
c. Construct a rectangle EFGH where EF = 5.6 cm and ∠GEF = 60°.

2.4 Construction of Regular Polygon

Introduction:

Triangle Quadrilateral Pentagon

A closed figure formed by three or more than three line segments is a polygon.

Regular Polygon

Let's observe the following figures.

600

600 600 900

In all of the above figures, sides of each polygon are equal; angles of each polygon

are also equal. Such polygons are called regular polygons. Hence, if all the sides and

angles of a polygon are equal such a polygon is called a regular polygon.

Interior angle of regular polygons A

Given figure is a regular pentagon. Its angles are A, B, C, D B E
and E. The value of each angle is same, such angles are interior
angles.

Interior angle of a regular polygon can be obtained by the C D
relation,

Interior angle = (n – 2) × 1800 where n = number of sides.
n

40 Oasis School Mathematics-8

Exterior angle of regular polygons

Given figure is a regular pentagon. A side DE is produced AF
to E. Then, ∠AEF is an exterior angle.

Sum of all exterior angles of a polygon = 3600 BE

Hence,

An exterior angle of a regular polygon = 3600 . CD
n

Remember !

• Sum of all the interior angles of a polygon = (n – 2) 180°

• Sum of the exterior angles of a polygon = 360°.

• An interior angle of a regular polygon = (n – 2)×180°
n

• An exterior angle of a regular polygon = 3600
n


Construction of Regular Pentagon

Example: 1

Construct a regular pentagon having a side 4.5 cm.

Number of sides of regular pentagon (n) = 5

We have, = (n – 2) × 180°
Interior angle of regular polygon n

Then, interior angle of regular pentagon = (5 – 2) × 180° = 108°
5

D

1080 4.5 cm Steps:

4.5 cm • Draw a line segment and cut it by
an arc of 4.5 cm, such that AB = 4.5
cm.

E C • At B, draw an angle of 108° (with

1080 the help of a protractor).

• From B cut by an arc of 4.5 cm at C.

• At C draw an angle of 108°.

4.5 cm 4.5 cm • From C cut by an arc of 4.5 cm at D.

• Again, draw an angle of 108° at D.

1080 • From D, cut DE = 4.5 cm.
• Join A and E.
A 4.5 cm B • ABCDE is a regular pentagon.

Oasis School Mathematics-8 4411

Example: 2

Draw a circle of radius 3.5 cm and construct a regular pentagon inside it.

Number of sides of regular pentagon = 5

An exterior angle of regular pentagon = 3600 = 3600 = 72°
n 5

B

Steps:

• Draw a circle having radius 3.5
C cm taking O as the centre.

• Take any point A on the circum-
ference of the circle and join OA.

720 3.5 cm A • At O, draw ∠AOB = 72° [with the
help of a protractor].
O

• Take an arc equal to AB and from
B cut at C, from C cut at D, from D

D cut at E.

• Join AB, BC, CD, DE and AE.

E Hence, ABCDE is the required
regular pentagon.

Example: 3

Construct a regular hexagon having a side 4 cm.

Number of sides of regular hexagon (n) = 6

An interior angle of regular polygon = (n – 2) × 180°
n

= (6 – 2) × 180° = 120°
6
E
4 cm D Steps:

4 cm 4 cm 4 cm • Draw a line segment, cut it by an arc
4 cm of 4 cm such that AB = 4 cm.

• At B, draw an angle of 120° and cut
BC = 4 cm

F C • Similarly, at C draw an angle of 120°
and cut CD = 4 cm.

• Similarly, at D draw an angle of 120°
and cut DE = 4 cm.

A 4 cm B • At E draw and angle of 1200 and cut
EF = 4 cm.

• Join AF.

• Hence, BCDEF is a required regular
hexagon.

42 Oasis School Mathematics-8

Alternative Method D Steps

E • Draw a line segment AB of 4 cm.

4 cm • Take an arc equal to AB and cut from
A and B to get point O.
4 cm
4 cm • Taking O as the centre and OA as
C radius, draw a circle.
4 cm
4 cm • Take an arc equal to AB and from B
4 cm cut at C, from C cut at D and so on.

4 cm • Join BC, CD, DE, EF and FA.
FO
• Hence, ABCDEF is a regular
A 4 cm B hexagon.

Note: This hexagon can be drawn by constructing an exterior angle at the centre of this
circle as in example 2.

Example: 4

Construct a circle having radius 4.2 cm and also construct a regular octagon in it.

C

Steps:

G E • Draw a circle taking radius
B OA = 4.2 cm.

F • Produce AO to B then AB is a
diameter.

4.2 cm A • Draw the perpendicular bisesctor
of AB which meets the circle at C
O and D.

• Draw the bisectors OE and OG of
∠AOC and ∠BOC and produce

H them to F and H respectively.

D • Join AE, EC, CG, GB, BF, FD, DH
and HA.

• Hence, AECGBFDH is a required
regular octagon.

Oasis School Mathematics-8 4433

Exercise 2.4

1. Find the interior angle and construct the following pentagon.

a. having a side 5.4 cm

b. having a side 6 cm

c. having a side 6.4 cm

2. a. Draw a circle having radius 4.8 cm and construct a regular pentagon inside it.

b. Draw a circle having radius 5 cm and construct a regular pentagon inside it.

c. Draw a circle having radius 4.5 cm and draw a regular hexagon in it.

3. Find the interior angle and construct a regular hexagon having:

a. a side 5.2 cm. b. a side 6 cm.

4. a. Construct a regular hexagon having each side 5 cm inside the circle.

b. Construct a regular hexagon having each side 5.4 cm inside the circle.

5. Find the interior angle of following octagon and use a protractor to construct the
following regular octagon.

a. a regular octagon having each side 4.5 cm.

b. a regular octagon having each side 5 cm.

6. a. Draw a circle having radius 5 cm and construct a regular octagon inside it.
b. Draw a circle having radius 5.4 cm and construct a regular octagon inside it.



44 Oasis School Mathematics-8