Prime Math 6 - Final rabi_final

Prime

Mathematics
Series

6

Raj Kumar Mathema

Dirgha Raj Mishra Bhakta Bahadur Bholan

Uma Raj Acharya Yam Bahadur Poudel

Naryan Prasad Shrestha Bindu Kumar Shrestha

Prime

Mathematics
Series

6

Raj Kumar Mathema

Dirgha Raj Mishra Bhakta Bahadur Bholan

Uma Raj Acharya Yam Bahadur Poudel

Naryan Prasad Shrestha Bindu Kumar Shrestha

Prime

Mathematics
Series

Approved by 6
Government of Nepal, Ministry of Education, Science and
Technology Curriculum Development Centre Sanothimi,

Bhaktapur as and additional learning materials.

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

Editors

Anil Kumar Jha
Dhurba Narayan Chaudhary

Hari Krishna Shrestha

Language Editor
Mrs. Tara Pradhan

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

Printing history
Frist Edition 2074 B.S.
Revised Edition 2077 B.S.

Layout and design
Pragya Desktop Team

© Publisher

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

ISBN : 978-9937-9170-2-5
Printed in Nepal

Pragya Books Published by
and
Pragya Books and Distributors Pvt. Ltd.
Distributors Pvt. Ltd.
Kathmandu, Nepal
E-mail : pragyabooks100@gmail.com

Preface

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

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

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

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

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

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

CONTENTS

Unit Topic Page

1. GEOMETRY 1-45
Lines and Angles 2
Angles in pairs 25
Triangles 30
Quadrilaterals and Polygon 36
Construction of Regular Polygons 42
2. CO-ORDINATES 46-51
3. MENSURATION 52-73
Perimeter of plane figures 53
Area of plane figures 60
Volume of solids 69
4. TRANSFORMATION, SYMMETRY, DESIGN OF POLYGONS
AND TESSELATIONS 74-89
Transformation 75
Symmetry, Design of Polygons and Tesselation 80
5. SETS 90-99
6. ARITHMETIC 100-133
Whole Numbers 101
Integers 125
Rational Numbers 131
7. FRACTIONS AND DECIMALS 134-179
Fraction 135
Decimals 160
8. RATIO, PROPORTION & PERCENTAGE 180 - 191
Ratio, Proportions and Percentage 181
9. PROFIT & LOSS 192-199
10. UNITARY METHOD 200 - 203
Unitary Method 201
11. SIMPLE INTEREST 204-209
Simple Interest 205
12. STATISTICS 210-217
Statistics 211
Bar Graph 214
13. ALGEBRA 218 - 259
Algebric Expressions 221
Equation & Inequality 242
ANSWERS 260
PRIME Questions for More Practice 275
Model Questions 283

Objectives Estimated periods – 31

At the end of this unit, the students will be able to:
identify intersecting and parallel lines.

identify and construct parallel and perpendicular lines using set squares and compass.
classify angles and construct angles using protractor and compass.

evaluate the measure of unknown angles using relations of pair of angles.

classify and construct triangles.
find the sum of angles of polygons and regular polygons.

construct regular polygons.
identify faces, edges and vertices of solids and construct the nets and models of special

solids. P

Q
AB

0123456789

6543

Teaching Materials Activities

Models showing parallel and perpendicular It is better to:
lines and size of angles, geoboard, paper discuss about intersecting, parallel,
models of triangles and polygons, models perpendicular lines and concurrent lines
of solid objects, geometry box. and demonstrate the construction of
parallel and perpendicular lines.
discuss about angles and their types and
demonstrate the construction of angles
using protractor and compass.
discuses with experiments, relation of
pair of angles, interior angles of triangles
and polygons.
demonstrate the nets and models of
special solid objects.

Lines and Angles Estimated periods – 10

Intersecting and parallel lines: The lines which meet each other and go across are
called intersecting lines.

(i) A D Two lines AB and CD intersect at point
O.

O

CB

(ii) A B Two lines AB and CD are not intersecting
but on producing towards the ends B

D and D, meet at certain point.

C B Two lines AB and CD are not actually
(iii) A intersecting but on producing CD
towards the end D, meets the line AB
D at a certain point.

C

Such lines are intersecting lines or non parallel lines. The point where two lines
meet is called the point of intersection. In the figure (i), O is the point of intersection
of the lines AB and CD.

Note: Two straight lines never meet at more than one point.
Observe the following:

If more than two lines are involved, different line pairs intersect at different points.
But if three or more than three lines have a common point of intersection, they are
said to be concurrent.

2 Prime Mathematics Book - 6

A E Perpendiculars dropped from A
F
the vertices to the opposite E

O sides of a triangle are F O

B C concurrent. B D C
D

Medians of a triangle are concurrent.

Perpendicular lines

If two straight lines intersect each other making a right angle (angle of 90o), they

are said to be perpendicular to each other.

(i) A (ii) A D

C OD C B
(iv) A B
B
(iii) C

E

B D
A
In each figure, AB^CD. C

Parallel lines:

If two lines are drawn such that every point of one is equidistant from the other
line, the two lines are said to be parallel to each other. Parallel lines never meet
each other on extending to any extension to either side.

A P P1 P2 B

d dd D
C Q Q1 Q2

Prime Mathematics Book - 6 3

Here, in the figure, consider the points P, P1, P2, etc on the line AB and draw PQ,
P1Q1, P2Q2 perpendiculars to the line CD. We get PQ = P1Q1 = P2Q2 = d. So, the lines
AB and CD are parallel. AB parallel to CD is written symbolically as AB//CD.

Note: Pairs of paralled lines should be with arrows. B
AD

C

Constructing perpendicular and parallel lines
To construct a line perpendicular to given line through a given point using set squares.

A. When the point is on the line. N
Let’s consruct a line perpendicular of AB at M.
Steps:
i. Adjust the edge of scale on the line.

ii. Adjust one of the perpendicular edge
of the set square on the scale.

iii. Slide (run) the set square along the

scale such that we can draw a line MB
though the given point on the line A

along the other perpendicular edge 0123456789
of the set square . 6543

iv. Draw MN. Thus, MN^AB at M.

B. When the point is outside the line.
Let’s draw a line perpendicular to AB from an external point P.

Steps:

i. Adjust the edge of scale on the line. P

ii. Adjust one of the perpendicular edges of
the set square on the scale.

iii. Slide (run) the set square along the

scale such that we can draw line Q

through the given point P along the A 1 2 3 4 5 67 8 B
other perpendicular edge of the set
0 9
square.
6543

vi. Draw PQ. Thus, PQ^AB from P.

4 Prime Mathematics Book - 6

To construct a line perpendicular to the given line through a given point using
compass:

A. When the point is on the line: AB be X
the line and we need to draw a line
perpendicular to AB through the point
P on the line.

Steps: A M P NB
Y
i. Put the needle of the compass at the
point P and mark two points M and
N equidistant from P on the line
(taking arc).

ii. Extending the compass more than half of MN by putting the needle of
compass at M and N, take two points to either side of AB, equidistant
from M and N say X and Y (taking arcs).

iii. Join X and Y.

Thus, XY^AB is drawn through P.

B. When the point is outside the line: AB be the given line and we need to draw
a line perpendicular to AB through P which is outside of AB.

Steps:

i. Put the needle at P and extend the P
compass such that we can take an
arc cutting AB at two points M and
N.

ii. Extend the compass more than half Q NB

of MN and take a point X equidistant A M
from M and N, next side of AB to P.

iii. Join P and X which cuts AB at Q, thus X
PQ^AB is drawn.

Prime Mathematics Book - 6 5

Bisector and perpendicular bisector of a line segment:

A line XY cuts another line segment A X X
AB at M such that AM = MB, then
XY is bisector of AB. B M
M Y
If a line XY cuts the line segment AB at M B
such that AM = MB and XY^AB, then XY is A B
perpendicular bisector of AB.
Y

To construct perpendicular bisector of a line segment. X
Steps:
Given line segment AB.

1. Extend the compass more than half of the line A M
segment AB. Y

2. From A and B (putting the needle at A and B) draw
arcs to either side of AB to cut at X and Y.

3. Join X and Y. We get ÐXMA = Ð XMB = 900 and AM= BM. Thus XY is the
perpendicular bisector of AB.

Note: Any line passing through the midpoint of the given line segment is the bisector. There
are numerous bisectors but there exists only one perpendicular bisector.

To construct a line parallel to the given line through the given point: A.
Using set square: Let’s draw a line parallel to AB through the point P.

Steps: B

1. Adjust one edge of a set square to 0
the given line AB.
61
2. Set the ruler (scale) to one of the
other edge of the set square. 2 P

3. Slide (run) the set square along the 5 3
scale such that the point P lies on 4
the edge which is adjusted to the
line. 5

A 46
7

3 8
9

4. Remove the scale keeping the set square fixed. Then draw line through
P. Thus, the line parallel to AB through P is drawn.

6 Prime Mathematics Book - 6

B. Using compass: Let’s draw line parallel to AB through P.

Steps: Y
B
1. From P, draw PQ to cut AB at Q.
Q
2. With the help of compass, take P

ÐPQA = ÐQPY as alternate angles. X

3. Extend the arm XY. Thus XY//AB is A
drawn through P.

Exercise 1.1

1. Write the pairs of intersecting line segments in each Figure.

(i) U (ii) C D
R

TS BA

(iii) X (iv) A F
Y C B
E D

Z

2. Write the pairs of parallel lines in each of the following figures.

(i) C D (ii) N

MO

BA K
LP

Prime Mathematics Book - 6 7

3. Write whether the following line pairs are parallel or perpendicular:

(a) A (b) A C (c)
P
Q

B C BD R S
(d) M (f) F
(e) X Z

ON P H
E

WY G

4. The corner of the given blackboard are marked as A, B, C and D. Write the
pairs of all the parallel and perpendicular edges.
AB

DC

5. Name the parallel line segments of the following figures.

(a) A B (b) E F

D C H G
(c) M N (d) P S
R
P OQ

8 Prime Mathematics Book - 6

6. Write the pairs of perpendicular line segments of the following figures.

(a) A D A
(b)

B CB C
(c) W Z (d) P

XY QS R
7. (i) If AB//CD and CD//EF,
(ii) If PQ AB and RS AB,
are AB//EF ? what can you say about PQ
EF and RS ?
CD
AB PR

(iii) If AB//CD and AB MN, AQ SB
how do CD and MN relate ?
(iv) The line segment MN is
AC perpendicular to AB and CD.
Write whether AB and CD
are parallel or not ?

AC

MB M N
DN
D
B
Prime Mathematics Book - 6 9

Exercise 1.2

1. Copy the following figures approximately and construct a line parallel to
each of them through the given points using scale and set square.

(a) A B (b) C

Q

(c) E P D
G
(d)
R

S
FH

2. Copy the following figures approximately in your copy and construct a line
segment perpendicular to the given line from the given external points
using scale and set square.

(a) X (b) P

QR
AB

(c) M (d) W

P X
Y
N
Prime Mathematics Book - 6
10

3. Copy the following figures approximately in your copy and construct a line

segment perpendicular to each of them through the given point on the line

using set square. P

(a) (b)

M N X
A Q

(c) Z (d) S

M
YT

4. Copy the following figures approximately in your copy and construct a line
segment perpendicular to each of them through the given point on the line
by using compass.
Q
(a) (b)

M N X
A

R

5. Copy the following figures approximately in your copy and construct a line

segment perpendicular to each of them through the given external point

by using compass. M

(a) P (b)

X

A BN

Prime Mathematics Book - 6 11

X
(c) (d) P

Z Q
R

Y

6. (a) Draw a line segment AB of length 7cm and draw a perpendicular bisector
of it.

(b) Draw a line segment of length 9cm and locate its mid point.
7. (a) Draw a line segment AB of length 7cm. Draw the perpendicular line

segments AD and BC of length 5cm each through A and B respectively.
Join C and D. Name the figure ABCD so formed.
(b) Draw a line segment WX of length 6cm. Draw a line segment WZ and XY
of length 6cm each through the points W and X respectively. Join Y
and Z. Name the figure WXYZ so formed.
8. AB is a line segment and P is a point not in AB.
(a) How many lines can be drawn from P parallel to AB?
(b) How many lines perpendicular to AB from P can be drawn ?

P

AB

12 Prime Mathematics Book - 6

9. In the given figure, BC AB, MN is perpendicular bisector of AB and
NO BC. Measure AN,CN, BO and CO.

(a) Does NO bisect BC ? C

(b) Is N mid point of AC ?

8 cm N
O

B M A
6 cm

Prime Mathematics Book - 6 13

Angles : O B A
Angle
When two lines/rays imanate from a point, the divergence
between the two rays is called an angle. Here in the
figure, two rays OA and OB imanate from the point O
and ÐAOB is formed where the rays OA and OB are called
arms and the point O is called the vertex. ÐAOB can
also be writen as ÐBOA with the vertex O at the middle.

BB

O BA O A OA

O AO AO BA
B B

Let two rays OA and OB coincide each other and let the ray OB revolve

about O. Angle formed in a complete revolution of the ray OB is supposed to be 360o.

[Read as 360 degrees]. B

When the rreevvoolluvitniognl,inOeBO^BOcAoamnpdlethteesm14eoafsuthree OA
complete

of ÐAOB = 90o.

We use a protector to measure the angle. A protractor is generally a transparent,
and semi-circular device which can measure angles up to 180o. But we can also get
a circular protractor which can measure any angle.

0o-180o protractor

0o-360o protractor

14 Prime Mathematics Book - 6

To measure the given angle using protractor:

Process:
- Given ÐABC.

- Place the centre O of the protractor at the vertex B of the angle and a base
line (zero line) coinciding with the arm BC (or AB).

- Read the angle indicated by the arm BA (or BC).

- Here, ÐABC = 40o.
A
0o

B 40o C A

180o 180o O 0o C
B
C
Construction of an angle of given magnitude using protractor. 0o
Process: B

Let’s draw an angle of 30o; 180o O C
A A B
- Draw a line segment AB.
30o
- Place the centre of the protractor at A
and base line (zero line) on AB.

- Mark angle 30o on the protractor (starting
from Oo shown by AB) at C.

- Remove the protractor and join A
and C.

Thus ÐBAC = 30o is constructed.

Classification of angles:
Angles are classified according to the size of the angle compared with angle of
magnitude 90o.

1. Right angle: An angle of magnitude 90o is called a right angle.

Prime Mathematics Book - 6 15

C C
ÐABC = 90o
90o
A 90o A
0 B

B

2. Acute angle: An angle which is less than 90o is called an acute angle.
C
70o C
ÐABC = 70o < 90o

B A B 70o A

3. Obtuse angle: An angle which is greater than 90o (right angle) and less than
180o (two right angle) is called an obtuse angle.

C 120o C
90o

180o 0o A 120o A
B
B

ÐABC = 120o > 90o
ÐABC = 120o < 180o

4. Straight angle: An angle which is exactly 180o (two right angles) is called a
straight angle. Two arms of such angle form a straight line.

180o B 0o A C B A
C
ÐABC = 180o
16
Prime Mathematics Book - 6

5. Reflex angle: An angle which is greater than 180o (two right angles) is called
a reflex angle.

180o B 0o A B A
C ÐABC > 180o

C

To measure a reflex angle: Generally we do not get 0o - 360o protractor. To measure
a reflex angle using usual 0o - 180o protractor, measure obtuse or acute part of the
angle and subtract it from 360o.

B B A
A 120o

CC

ÐABC = 120o \ Reflex ÐABC = 360o - ÐABC = 360o - 120o = 240o

B B A
A 60o

CC

ÐABC = 60o \ Reflex ÐABC = 360o - ÐABC = 360o - 60o = 300o

Note: Ð is used to denote acute or right or straight or obtuse angle and ‘reflexÐ` is used
to denote reflex angle.

Prime Mathematics Book - 6 17

Exercise 1.3

1. Measure the following angles and write their name, size and type as in (a).

(a) A (b) W (c) Q P

BC X Y
ÐABC = 50o, an acute angle. R

(d) (e) S (f)
O Q

Z YX
P

RT

2. Under your estimation (without actual measurement of the angles) classify
the following angles.

(a) C (b) D (c) M L

BA N

(d) P Q E F M
R (e) (f) L
Y
X ZK
(c) 180o (d) 90o
3. Classify the following angles: (g) 345o (h) 360o

(a) 25o (b) 150o

(e) 89o (f) 200o

4. Construct the following angles using protractor.

(a) 30o (b) 45o (c) 60o (d) 75o

(e) 90o (f) 120o (g) 135o

18 Prime Mathematics Book - 6

5. From the given figure write the acute angle, C
obtuse angle and straight angles.

B O A

6. Measure the indicated angles of the following figures.

(a) A (b) D

B C E F
(c) A (d) P

Q O
R

DC B B A
(e) P S (f) DE

Q R C

Prime Mathematics Book - 6 19

Construction of angles by using set squares and compass
We can draw the following angles using set squares.

30o 90o 45o 90o
90o 45o 45o, 90o 30o

60o 90o + 30o = 120o
30o, 60o, 90o

90o
45o

45o 90o + 45o = 135o
60o 30o
60o + 45o = 105o 45o
45o + 30o = 75o
90o
60o 90o 90o

90o + 60o = 150o 90o + 90o = 180o

These angles are known as standard angles. Only standard angles and their halves,
1 1
4 ths, 8 ths etc. can be constructed by using compass.

20 Prime Mathematics Book - 6

To bisect a given angle using compass B A
M
Process:
P
- Given angle ABC.
C
- Taking the vertex B as centre, take an arc of N
suitable radius to cut the arms AB and BC at
M and N respectively.

- Again taking M and N as centre and radius more than MN draw arcs to cut at
P within the interior of the ÐABC.

- Join B and P. We get ÐABP = Ð PBC

Thus, BP bisects the ÐABC.

Note: A general angle cannot be trisected by using Euclidian tools. It is an unsolved problem
in mathematics.

Construction of angle 60o using compass. B
Process: N
Let’s construct ÐAOB = 60o.
i. Draw a line segment OA. O MA
ii. Taking O as centre, draw a semi circle

(or a large arc) to cut OA at M.

iii. Taking M as centre and radius same as in step II, take an arc to cut the semi
circle at N.

iv. Join O and N and produce to B.
Thus, ÐAOB = 60o is constructed.

Note: Since, OM = ON = MN = radius (r) of the B
circle, OMN is an equilateral triangle. N
So, ÐAOB is 60o.
r
r A

O rM

Prime Mathematics Book - 6 21

Construction of angle 30o. B C
Let’s draw ÐAOC = 30o. N
i. Construct ÐAOB = 60o
O MA
(No need to draw OB, mark N only, such
that OM = MN).
ii. Draw bisector OC of ÐAOB = 60o.
Thus, ÐAOC = 30o is constructed.

Construction of angle 120o. CP A
M
Let’s draw ÐABC = 120o.
N
i. Draw a line segment BA.
BL
ii. Taking B as centre and with suitable radius,
draw a semi circle (or an arc) to cut BA
at L.

iii. Taking L as centre and radius same as in
step (ii), take an arc to cut the semi circle
at M so that ÐABP = 60o (mark M only no
need to draw BP)

iv. Taking M as centre and same radius, take an arc to cut the semi circle at N.
v. Join B and N and produce to C.

Thus, ÐABC = 120o is constructed.

Construction of angle 90o. R C P
N B M
Let’s construct ÐABC = 90o.
L A
i. Construct ÐABP = 60o and ÐABR = 120o
(only mark M and N on the semi circle
corresponding to 60o and 120o, no need to
draw BP and BR)

22 Prime Mathematics Book - 6

ii. Draw bisector BC of angle PBR or MBN.
Thus, ÐABC = ÐABP + ÐPBC = 60o + 30o = 90o is constructed.

Construction of angle 45o. Q R P
Lets draw ÐABC = 45o. B C

i. Construct ÐABR = 90o (dotted line are not N M
necessary to draw only mark on the circle)
LA
ii. Draw bisector BC of ÐABR.

Thus, ÐABC = 1 ÐABR = 1 × 90o = 45o is constructed.
2 2

Construction of angle 75o.

Given figure shows the steps to construct ÐABC = 75o.

i. What is the size of ÐABL ? N C
ii. What is the size of ÐLBN ? M L

iii. Draw bisector BC of ÐLBN. What is RS Q
the size of ÐLBC ?

iv. ÐABL + ÐLBC = ?, ÐABC = ? B PA

[Dotted lines are not necessary to draw, only mark Q, R and S on the semi
circle]

Construction of angle 105o C N
M
Given figure shows the steps to construct L
ÐABC = 105o.
B A
i. What is the size of ÐNBM ?

ii. Draw bisector BC of ÐNBM. What is the
size of ÐNBC ?

iii. ÐABN + ÐNBC = ?

iv. ÐABC = ?

Prime Mathematics Book - 6 23

Exercise 1.4

1. Construct the following angles by using protractor and bisect them. Measure
the size of each half.

(a) 60o (b) 30o (c) 80o (d) 90o

(e) 100o (f) 140o (g) 120o (h) 150o

2. Construct the following angles using set squares:

(a) 90o (b) 30o (c) 45o (d) 60o (e) 75o

(f) 105o (g) 120o (h) 135o (i) 150o

3. Write the size of the angles under the following constructions.

(a) (b) P (c) Z

C

B AN O Y
(d) Y (e) (f)
B

WX R Q
O AP

4. Construct the following angles using compass.

(a) 60o (b) 30o (c) 120o (d) 90o (e) 45o
(f) 75o (g) 105o (h) 150o (i) 135o (j) 165o

5. (a) Draw a line segment AB = 6cm. At A, draw ÐBAD = 90o and at B, draw
ÐABC = 90o. Take AD = BC = 4cm and join CD. Name the figure ABCD so
obtained (use compass to draw the angles)

(b) Draw a line segment PQ = 5cm. At P, draw ÐQPS = 90o and at Q, draw
ÐPQR = 90o. Take PS = QR = 5cm and join RS. Name the figure so obtained.

(c) Draw a line segment AB = 5cm. At A, draw ÐBAC = 60o and at B, draw
ÐABC = 60o using compass and produce the arms to cut at C. Measure
the lengths of side AC and BC.

24 Prime Mathematics Book - 6

Angles in pairs Estimated periods – 3

Adjacent angles: B
C
CC BB
B
OA

O AO AO AC

In each of these figures, ÐAOB and ÐBOC have a common vertex (O) and a common
arm OB, other arms lying to either side of the common side. Such pairs of angles
are called adjacent angles.

Complementary angles:

If the sum of the two angles is 90o, they are said to be complementary to each other.

Here, in the given figure, A
ÐABC = 60o and ÐDEF = 30o D

B 60o CE 30o F

Since ÐABC + ÐDEF = 60o + 30o = 90o,

ÐABC and ÐDEF are complementary angles.

ÐABC is complementary of ÐDEF and ÐDEF is the complementary of ÐABC.

As, ÐABC + ÐDEF = 90o,

ÐABC+ complementary of ÐABC = 90o

\ Complementary of ÐABC = 90o - ÐABC.

Thus, complementary of an angle = 90o - the angle itself. C B

If adjacent angles are complementary, they are called
complementary adjacent angles.

Supplementary angles: If the sum of two angles is 180o, O A
they are called supplementary angles, or supplementary
to each other.

Prime Mathematics Book - 6 25

CD

120o 60o F
AB E

Here, ÐABC = 120o and ÐDEF = 60o

\ ÐABC + ÐDEF = 120o + 60o = 180o

So, ÐABC and ÐDEF are supplementary angles or supplementary to each other. ÐABC
is supplementary of ÐDEF.

As, ÐABC + ÐDEF = 180o

ÐABC+ supplementary of ÐABC = 180o

Supplementary of ÐABC = 180o - ÐABC.

\ Supplementary of ÐABC = 180o -the angle itself.

B

C 135o 45o A
O

If two adjacent angles are supplementary, they are called supplementary adjacent
angles or a linear pair.

Note: Complementary and supplementary angles may or may not be adjacent.

Vertically opposite angles: A O D
When two lines intersect at a point in the same plane, C B
the equal and opposite angles having common vertex
and no common arm are called vertically opposite
angles.

Here, two line segments AB and CD intersect at O.

ÐAOC = ÐBOD and ÐAOD = ÐBOC are two pairs of
vertically opposite angles.

26 Prime Mathematics Book - 6

In this case, four pairs of supplementary adjacent angles ( linear pairs) are also
formed.
ÐAOC + ÐAOD = 180o,
ÐAOC + ÐBOC = 180o,
ÐBOD + ÐAOD = 180o and
ÐBOD + ÐBOC = 180o

Experiment: A O D
On a sheet of paper, C B
draw AB and CD intersecting at O.
Cut along AO, OD and AD also along OB and BD.
Put ÐBOD over ÐAOC and ÐAOD over ÐBOC.
They exacltly match. What do you conclude?

Example 1. Find the size of complementary angle of 40o.
Solution: Given angle is 40o
Let x be the angle complementary to 40o, then
40o + x = 90o
or, x = 90o - 40o
\ x = 50o
Hence, complementary of 40o is 50o.

Example 2. Find the size of the angle supplementary of 60o.
Solution: Here,
Given angle = 60o = x (say).
Let the angle supplementary to x is y, then
x + y = 180o
or, 60o + y = 180o
\ y = 180o - 60o = 120o
Hence, the size of the angle supplementary to 60o is 120o.

Prime Mathematics Book - 6 27

Example 3: From the given figure, find the value of x. C
Solution:
Here, ÐAOB = 35o and ÐBOC = x are x
complementary to each other 35o B
\ x + 35o = 90o A
or, x = 90o - 35o O
\ x = 55o

Example 4: From the given figure, find the size of angle x.

Solution: Here, ÐAOB = 50o, ÐBOC = x and ÐCOD = 30o

ÐAOB + ÐBOC + ÐCOD = 180o (sum being straight angle)

or, 50o + x + 30o = 180o B
or, x + 80o = 180o
C

or, x = 180o - 80o = 100o D 30o x 50o A
\ x = 100o O

Example 5: From the figure given alongside, find the value of x, y and z.
Solution:
Here, ÐAOC = 50o A D

Being vertically opposite angles, x B
ÐBOD = ÐAOC 50o y
\ y = 50o
zO
C

Being supplementary adjacent angles (linear pair)

ÐAOD + ÐAOC = 180o

or, x + 50o = 180o

or, x = 180o - 50o

\ x = 130o

Again, being vertically opposite angles

z=x

z = 130o

\ x = 130o, y = 50o and z = 130o.

28 Prime Mathematics Book - 6

Exercise 1.5

1. Define. (b) Complementary adjacent angles.
(a) Adjacent angles. (c) Supplementary adjacent angles.
(c) Vertically opposite angles.

2. Identify pair of adjacent angles and vertically opposite angles from the

following figures. D (b) C B
(a) A

O B D OA
C E

3. Find the complementary angles of the following. (d) 81o

(a) 20o (b) 45o (c) 52 1o
2

4. Find the supplementary angles of the following.

(a) 25o (b) 90o (c) 81o (d) 110o (e) 1 3 5 o

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

(a) (b)

x 70o
38o X 50o
(d) 5x
(c)
3x x
x
4x (f) x
(e) y 55o
z
115o x

(g) (h) 4x
115o 55o y x
x Prime Mathematics Book - 6

29

Triangles Estimated periods – 4

A triangle is a plane figure bounded by three line segments. A

The figure given alongside is a triangle. The corner points A
C
A, B, C are the vertices and the line segments AB, BC,
A
and AC are its sides. ÐBAC, ÐABC and ÐACB are its three B MC C
D
angles. Triangle ABC is denoted as DABC.

- A triangle consists three angles, so it is named
triangle (tri-angle)

- The three angles are also called interior angles.

- The sum of interior angles of a triangle is 180o.

- If we produce any side, the supplementary adjacent B
angle to the interior angle a new angled is formed
which is called exterior angle. ÐACD is an exterior
angle.

- Generally, the side on which a triangle stands is

known as base and the perpendicular drawn from

the vertex to the base is called height (or altitude)

of the triangle. AM is the altitude (height) of the B

triangle ABC over base BC.

A special property of triangle:

Sum of the interior angles of a triangle is 180o.
Experiments:
1. Draw any triangle ABC in a chart paper.

A

ÐA + ÐB + ÐC = 180o
BAC
BC
- Cut the three corners of the triangle as shown and arrange them with
common vertex.
- The three angles form a straight angle.
Thus, sum of angles of a triangle is 180o.

30 Prime Mathematics Book - 6

2. Draw three triangles namely ABC of different shape and size using scale.

(i) A (ii) A (iii) A

B CB CB C
40o C
- Measure the angles of each triangles using protractor.

- Find the sum of the angles in each case.

Figure ÐA ÐB ÐC ÐA + ÐB + ÐC Remark

i.
ii.
iii.

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

Types of triangles: A
75o
A. According to angles, there are three types
of triangles 65o
B
i. Acute angled triangle:

In an acute angled triangle, all its three angles
are acute.

ii. Right angled triangle: In a right angled triangle, A
one of its three angles is a right angle (90o).

- A triangle can not have more than one right

angles. B C
- In a right angled triangle, two acute angles C

are complementary.

ÐABC = 90o \ ÐBAC + ÐACB = 90o A

iii. Obtuse angled triangle: In an obtuse angled 20 o

triangle, one of the three angles is obtuse

angle. B120o 40o

- A triangle can not have more than one obtuse angles.

Prime Mathematics Book - 6 31

B. According to sides, there are three types of triangles. A

i. Equilateral triangle: In an equilateral

triangle, all the three sides are equal in

length. An equilateral triangle is also known

as equi-angular triangle. B C
C
\ In an equilateral DABC, AB = BC = AC and ÐA = ÐB = ÐC = 60o.

ii. Isosceles triangle: In A A A
an isosceles triangle, CB CB

two sides are equal in

length. Here, in the

figure AB = BC B

(Naming each figure as DABC).

- An isosceles triangle may be acute angled or right angled or obtuse
angled.

- In an isosceles triangle, the angles opposite to equal sides are equal and
are called base angles and the angles between the equal sides is called
vertical angle. Here, in the given figures.

AB = BC and ÐBAC = ÐACB are base angles. A
\ ÐABC is vertical angle. 4cm

iii. Scalene triangle: In a scalene triangle, none of 3cm 5cm

the sides are equal. In the given DABC, AB ¹ BC ¹

AC. It is a scalene triangle. B C

- A scalene triangle may be acute angled or right angled or obtuse angled.

Example 1: Find the size of the unknown angle from A
Solution: the figure given alongside. x

Here, in DABC,

ÐA = x, ÐB = 65o, ÐC = 40o B 65o 40o C
We know,

ÐA + ÐB + ÐC = 180o (Sum of angles of triangles is 180o.)

or, x + 65o + 40o = 180o

32 Prime Mathematics Book - 6

or, x + 105o = 180o

or, x = 180o - 105o

\ x = 75o

Example 2: Find the value of x in the figure given A
Solution: alongside. 52o

Example 3: In the given triangle,
Solution:
ÐABC = 90o, ÐBAC = 52o and ÐACB = x B x C
C
Since, sum of acute angles of a right angled triangle is 90o.

ÐBAC + ÐACB = 90o A
or, 52o + x = 90o 2x
or, x = 90o - 52o
\ x = 38o Bx x

From the figure given alongside, find the
size of each angle of the triangle ABC.

Here, ÐBAC = 2x, ÐABC = x, ÐACB = x

\ ÐBAC + ÐABC + ÐACB = 180o (Sum of angles of triangle is 180o)

or, 2x + x + x = 180o

or, 4x = 180o

or, x = 180o = 45o
4
\ ÐBAC = 2x = 2 × 45o = 90o

and ÐABC = ÐACB = x = 45o

A

Example 4: Find the unknown angles in the given 80o
Solution:
figure. yx B
Here, ÐBAC = 80o and AB = AC. DC

\ ÐABC = ÐACB = x (Being base angles of isosceles triangle)

Now, ÐBAC + ÐABC + ÐACB = 180o

or, 80o + x + x = 180o (Sum of angles a of triangle is 180o)

Prime Mathematics Book - 6 33

or, 80o + 2x = 180o Again, x + y = 180o
or, 50o + y = 180o
or, 2x = 180o - 80o or, y = 180o - 50o
\ y = 130o
or, 2x = 100o \ x= 50o and y = 130o

or, x = 100o
2

\ x = 50o

Exercise 1.6

1. Define: (b) Right angled triangle
(a) Triangle

(c) Acute angled triangle (d) Obtuse angled triangle
(e) Equilateral triangle (f) Isosceles triangle

(g) Scalen triangle

2. Identify the types of the following triangles, where

(a) All the angles are equal ___________
(b) All the angles are acute ___________
(c) One angle is a right angle ___________
(d) Two sides are equal ___________
(e) None of the sides are equal ___________
(f) One angle is obtuse angle ___________
(g) All sides are equal ___________
(h) Two angles are equal ___________

3. In the given figure, how many A
triangles are there? Name them.
D F

BE C

4. Measure the sides of the following triangles and classify them.

(a) A (b) A (c) A

B C BC BC
Prime Mathematics Book - 6
34

5. Measure the angles of the following triangles and name them.

(a) A (b) A (c) A

B CB CB C
x
6. Find the size of the unknown angles in the given figures.

(a) (b) y (c)
60o

x 45o 40o y

(d) (e) y (f)
25o
xy 50o x x 40o y
(g) (h)
(i)
y x

xy x 75o 50o
55o

(j) x (k) (l) 3x
x yx 2x
40o x
63o 35o

7. (a) If one angle of a right angled triangle is 42o, find the size of other acute
angle.

(b) If the sizes of two angles of a triangle are 88o and 42o, find the size of
the third angle.

(c) If the size of the vertical angle of an isosceles triangle is 100o, find the
size of the base angles.

(d) If the vertical angle of an isosceles triage is equal to the sum of the base
angle, find the size of each angle.

Prime Mathematics Book - 6 35

Quadrilaterals and Polygons Estimated periods – 5

A quadrilateral is a closed plane figure bounded by D C
four straight line segments. Given figure is a
quadrilateral ABCD where AB, BC, CD and AD are A B
four sides. A, B, C and D are the vertices and ÐBAD, D C
ÐABC, ÐBCD and ÐADC or simply ÐA, ÐB, ÐC and
ÐD respectively are the angles (interior angle). In
a quadrilateral or any polygon, the line joining
alternate vertices are called diagonals.

Note: A triangle doesn't have any diagonal. A B
A D
Types of quadrilaterals
On the basis of nature of sides and angles B C
quadrilaterals are classified as:

1. A rectangle: A rectangle is a quadrilateral having A D
opposite sides equal and each of four angles 90o. C

Given figure is a rectangle where opposite sides AB
= CD and AD = BC and ÐA = ÐB = ÐC= ÐD = 90o.
Opposite sides of a rectangle are also parallel and B
diagonals are equal. AB//DC, AD//BC and AC = BD.

2. A square: A square is a quadrilateral having all sides P Q

equal and each angle 90o. In the given figure, PQRS is R
Z
a square where the sides PQ = QR = RS
Y
= PS and ÐP = ÐQ = ÐR = ÐS = 90o S

3. A parallelogram: A parallelogram is a quadrilateral W

having opposite sides parallel. In the given figure,

WXYZ is a parallelogram where WX//ZY and WZ//XY.

X

Angles of a parallelogram may or may not be 90o each but opposite angles are
equal. Also, the opposite sides are equal.

36 Prime Mathematics Book - 6

4. A rhombus: A rhombus is a parallelogram with AD
all sides equal. In the given figure, ABCD is a
rhombus where AB = BC = CD = AD. B C
K L
5. A trapezium: A trapezium is a quadrilateral
having any two opposite sides parallel. In the N M
given figure, KLMN is a trapezium where KL//NM. A B

6. Trapezoid: A trapezoid is a quadrilateral in
which no sides are parallel.

Sum of angles of a quadrilateral: A DC
A. - Draw any quadrilateral B

ABCD in a chart paper. AB
- Cut along the edges. DC

- Cut the corners as shown. D C

- Arrange the corners about a point. They form a complete angle. So,

ÐA + ÐB + ÐC + ÐD = 360o.

Thus, the sum of angles of quadrilateral = 360o.

B. - Draw any quadrilateral ABCD Ap x yD
-
Divide it into two triangles by joining a
diagonal, say AC.

- Let ÐBAC = p, ÐABC = q, ÐACB = r and

ÐDAC = x, ÐADC = y, ÐACD = z, then Bq rzC

p + q + r = 180o (sum of Ðs of D)

and x + y + z = 180o (sum of Ðs of D)

Now, sum of Ðs of the quadrilateral ABCD

= ÐA + ÐB + ÐC + ÐD

=p+x+q+r+z+y

= (p + q + r) + (x + y + z) (just by grouping)

= 180o + 180o

Thus, = 360o of angles of a quadrilateral = 360o.
the sum

Prime Mathematics Book - 6 37

Polygons:

A polygon is a closed plane figure bounded by three or more line segments. A triangle
is a polygon with the least number of sides.

Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon

Polygons are named according to the number of sides.

In latin Angle formed by two
sides of a polygon
Mono = one towards its interior is
called interior angle.
Di = two

Tri = three interior exterior
angle angle
Tetra/quadri = four

Penta = five

Hexa = six If we produce a side of a polygon, the angle
formed by extended part of the side and other
Hepta = seven side to the outside of the polygon is called exterior
angle.
Octa = eight
If all the sides and angles of a polygon are equal,
Nona = nine it is known as a regular polygon.

Deca = ten

Mono deca = eleven

Dodeca = twelve

Quindeca = fifteen

Icosa = twenty

Sum of angles (interior angles) of a polygon: A rC
p uD
We have already observed that sum of angles of triangle, Bq
Ap s rt C
p + q + r = 180o = 1 × 180o
Dividing into two triangles, sum of angles of a quadrilateral q
ABCD is B
ÐA + ÐB + ÐC + ÐD = p + s + q + r + t + u

= (p + q + r) + (s + t + u)
= 180o + 180o = 360o = 2 × 180o

38 Prime Mathematics Book - 6

Consider a pentagon ABCDE. Ey x vD
Drawing two non intersecting diagonals AC and AD, we
su C
can divide it into three triangles, DABC, DACD and DADE w t r
Now, ÐA + ÐB + ÐC + ÐD + ÐE q B
A
=q+t+w+r+s+u+v+x+y

= (q + r + s) + (t + u + v) + (w + x + y)

= 180o + 180o + 180o (Sum of angles of triangle is 180o)

= 540o = 3 × 180o

Polygon No. of sides Figure No. of triangles Sum of interior General formula
Triangle formed (n) angles (n × 180o)

3 1 1 × 180o = 180o (3 - 2) 180o

Quadrilateral 4 2 2 × 180o = 360o (4 - 2) 180o

Pentagon 5 3 3 × 180o = 540o (5 - 2) 180o

Hexagon 6 4 4 × 180o = 720o (6 - 2) 180o

n - gon n n-2 (n - 2) × 180o (n - 2) × 180o

Thus, the sum of angles of a polygon with n-side is (n-2) 180o, where n is the number
of sides or angles.

Sum of Ðs of n-gon = (n - 2) × 180o.

In a regular n-gon, there are n-equal interior angles.

\ Size of an angles of a regular polygon = (n-2)180o
n

Note: Geometrical proof of the formula will be shown in higher classes.

Prime Mathematics Book - 6 39

Example 1: Find the size of the unknown angle A 160o xD
Solution: of the quadrilateral given along side.

Here, ÐA = 160o, ÐB = 60o, ÐC = 50o B 60o 50o C
and ÐD = x
(Sum of Ðs of a
We have ÐA + ÐB + ÐC + ÐD = 360o
quadrilatreal is 360o.)
or, 160o + 60o + 50o + x = 360o

or, 270o + x = 360o

or, x = 360o - 270o \ x = 90o

Example 2: Find the size of the unknown angles z w
Solution: of the parallelogram given in the y x 80o
figure.

Here, x + 80o = 180o (Being linear pair)

or, x = 180o - 80o

\ x = 100o

z = x = 100o (Opposite angles of a parallelogram are equal)

Also, y = w = k (say) (Opposite angle of parallelogram)

Now, x + y + z + w = 360o (Sum of angles of quadrilateral = 360o)

or, 100o + k + 100o + k = 360o

or, 200o + 2k = 360o

or, 2k = 360o - 200o

or, 2k = 160o

\ k = 80o

\ y = w = k = 80o

\ x = z = 100o, y = w = 80o

Example 3: Find the sum of the angles of a regular hexagon.
Solution: No. of the sides of a hexagon (n) = 6
We have,
40
Sum of angles of a n-gon = (n - 2) × 180o
Sum of angles of a 6-gon = (6 - 2) × 180o

= 4 × 180o = 720o

Prime Mathematics Book - 6

Example 4: Find the size of the unknown angle of the D
Solution: given pentagon. 110o
In a pentagon, number of side (n) = 5 E x 142o C
\ Sum of Ðs of 5 - gon = (n - 2) × 180o
A 100o 108o
= (5 - 2) × 180o B
= 3 × 180o = 540o
Now, A + B + C + D + E = 540o
or, 100o + 108o + 142o + 110o + x = 540o
or, 460o + x = 540o
or, x = 540o - 460o
\ x = 80o

Exercise 1.7

1. Define:

(a) Quadrilateral (b) Rectangle
(c) Square (d) Parallelogram
(e) Rhombus (f) Trapezium
(g) Trapezoid (h) Polygon

2. Find the sum of the angles of the following polygons.

(a) A quadrilateral (b) A pentagon
(c) A hexagon (d) A heptagon
(e) An octagon (f) A decagon
(g) A dodecagon (h) A quindecagon

3. Find the size of the unknown angles of the following polygons.

(a) 120o 90o (b) 3x x (c) 100o

132o 120o

x 100o 2x 4x x 108o
(f) y
(d) 150o 80o (e) 100o
110o 140o 80o y 110o

120o x 130o x 60o 80o x 70o
Prime Mathematics Book - 6
41

Construction of regular polygons Estimated periods – 3

Construction of equilateral triangle C

Let’s construct an equilateral triangle of side 6cm. 60o 60o
A 6cm B
A. On the basis of angles.

Step - Draw a line segment AB = 6cm

- Since, each angle of an equilateral
triangle is 60o, using protractor (or
compass) draw ÐBAC = 60o and ÐABC
= 60o and produce the arms to intersect
at C.

Thus, the required equilateral triangle
ABC is constructed.

B. On the basis of sides. C

- Draw line segment AB = 6cm

- Taking A and B as centre and radius 6cm, take 6cm 6cm
arcs to cut at C.

- Join A and C, B and C.

Thus, the required triangle ABC is constructed. A 6cm B

Construction of square DC
To construct a square of side 5cm A 5cm B
- Draw a line segment AB = 5cm
- At A, draw ÐBAD = 90o and take AD = 5cm
- Taking B and D as centre and redius equal to AB

take arcs to cut at C.

- Construct BC and DC.
Thus, square ABCD is constructed.

42 Prime Mathematics Book - 6

Construction of a regular pentagon

To construct a regular pentagon of side 4cm

We know that size of an interior angle of a regular pentagon = (n - 2) × 180o
n
=(5 - 2) × 180o
5

= 108o

- Draw a line segment AB = 4cm D

- At B, draw ÐABC = 108o using protractor and take 108o
BC = 4cm.
E 108o C

- At C, draw ÐBCD = 108o, towards same side and
take CD = 4cm.

- At D, draw ÐCDE = 108o and take DE = 4cm. 108o
- Join A and E. A 4cm B

Thus, the required regular pentagon ABCDE is constructed.

Exercise 1.8

1. Construct a equilateral triangle of side:

(a) 4cm (b) 4.5cm

(c) 5cm (d) 6cm

2. Construct a square of side: (b) 4.8cm
(a) 4cm (d) 5.4cm
(c) 5cm

3. Construct a regular pentagon of side:

(a) 3.8cm (b) 4cm

(c) 5cm (d) 5.5cm

Prime Mathematics Book - 6 43

Unit Revision Test I

1. Copy the figures approximately in your copy draw the following and using

compass draw.

(a) A line perpendicular (b) A line perpendicular to CD through
the point P.
to AB through P P
A D

P
C

B

2. Measure the following angles and write their name, size and type.

(a) P (b) M

QR K

L

3. Construct an angle of 45o using protector and bisect it.

4. Construct the following angles using compass.

(a) 30o (b) 75o

5. Find the angles

(a) Complementary to 20o (b) Supplementary to 110o

6. From the figure given alongside, find x
the size of angles x and y. y 50o

7. From the figure given alongside, find 35o
the size of angles x and y.

xy

44 Prime Mathematics Book - 6