Prime Optional Mathematics 6

Exercise : 4.6

1. Find the value of the followings:
i. Sin0° + Cos0° + Cos60° + Sin30°
ii. Tan45° + Cos0° + Sin90°
iii. Sin60° + Cot45° – Cos30°
iv. Tan60° +Tan45° – Cot30°
v. 2Sin30° + 2Cos30° – Tan60°

2. Find the value of the followings:

i. 2Sin²45° + 4Cos²45° – 3Tan²45°
4 3
ii. 3 Tan²60° + 2 Sec²30° – 4Sin²90°

iii. Cos²0° + Cos²30° – Sin²45°

iv. Sin60°.Cos30° + Cos60°.Sin30°

v. Cos30°.Cos60° – Sin30°.Sin60°

3. Prove that the followings :

i. Tan²30°.Tan²60°.Sin²30°.Sec²60° = 1

ii. 2Sin²45° + 4Sin²60° – Tan²45° = 3
Cot30° + 1
iii. Tan60°–1 = 2+ 3

iv. 1–Sin60° = 7–4 3
1 + Cos30°

v. Tan45°–Tan30° = 2– 3
1 + Tan45°.Tan30°

4. Prime more creative questions:
1 – Tan30°
a. i. Prove that : 1 + Cot60° = 2– 3

ii. Prove that : 1 + Cot60° = 1 + Cos30°
1 – Tan30° Sin30°

iii. Evaluate : 13 + 7 + 1 + 2 3 Sin²60°

iv. Prove that : 3Tan²30° + 4Sin²60° – 2Cos²45° = 3

v. Simplify : Sin²30° + Cos²60° – Sin²45°.

96 PRIME Opt. Maths Book - VI

b. Answer the followings from the right angled triangles.
i) Find the length of AB from the given diagram.
A

B 60° C
20 m
ii) Find the length of AB from the given diagram.
A

10 m

BC
iii) Find the length of side PQ from the given triangle.

P

Q 45° R
20m
iv) A ladder of length 20 m is taken against the wall as shown in
diagram. Find the height of the wall.
D

40m

F 30° E

v) A boy is flying a kite as shown in diagram where height of
the kite from the ground is 50m and angle made by string
with the ground is 30°. Find the length of the string.

A

50 m

B 30° C

PRIME Opt. Maths Book - VI 97

Answer

1. i) 2 ii) 3 iii) 1 1 iv) 1 v) 1
2. i) 0 4 iv) 1 v) 1
4.a. iii) 4 ii) 2 1 iii) 1
b. i) 2 iv) 20m v) 100m
20 3 v) –

ii) 20m iii) 20m

Trigonometry

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. What is trigonometry?
2. a. Convert 12°14’18” into seconds.
IfifnCdotshAe=va54lu,efionfdTSainn4A5. ° + 2Sin30° – Cos0°
b.
c.
3. a. Prove that : Sin4A – Cos4A = 1 – 2Cos²A.
b. Find the ratio of the angles 72° and one right angle.

4. Find the angles of a triangle in degrees where the angles are in
the ratio 1:2:3.

Unit Test - 2
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. What is the value of 90° in grades?
2. a. Prove that : Cosq.Tanq.Cosecq = 1

b. If CosA = 1, find the value of CosecA.
c. Find the value of Sin²45° + Cos²45° = 1
3. a. One angle of a right angle triangle is 40g, find the third angle in degrees.

b. Prove that 1 – Tan30° = 2– 3
1 + Cot60°

4. Prove that : 1 – Sin4 i = 1 + 2Tan²q.
Cos4 i

98 PRIME Opt. Maths Book - VI

5 Vector

Specification Grid Table TM Periods

K(1) U(2) A(4) HA(5) TQ

No. of Questions 1 – 1 – 25 4

Weight 1–4–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can identify the vector and scalar quantity.
• Students can find magnitude, direction and unit vectors.
• Students can operate vector addition and scalar multiplication.
• Students can use the vector in our daily life.
• Students can find the position vector in different situation.

Materials Required:

• Chart paper.
• Graph paper.
• Chart of list of types and application of vectors.
• List of the examples of vector and scalar.
• Geo-board.
• Graph board.

PRIME Opt. Maths Book - VI 99

5.1 Introduction

Some of the quantities can be measured but some can not be measured
out of different quantities in our surroundings. The quantities which
can be measured are called Physical quantities. The Physical quantities
are classified into two groups scalar and vector.
The Physical quantities which can be measured and has magnitude as
well as direction are called vector and that which has only magnitude
are taken as scalar quantities.

The Physical quantities which has only magnitude
is called scalar.
Examples: Distance, time, mass, area, speed,
density etc.

The Physical quantities which have magnitude as
well as direction are called vector.
Examples: Displacement, velocity, acceleration,
work, power etc.

Scalar quantities is represented by using a line segment only where a
vector is represented by using a directed line segment.

BB
AA

Scalar – AB Vector – AB

• Vector is represented by AB, a , PQ, b , etc.

100 PRIME Opt. Maths Book - VI

Vector quantities using co-ordinates.

Y
(x2, y2)
B

(x1, y1) C
A

OM NX

Here,

A(x1, x2) and B(x2, y2) are any two points where,

AC = MN = ON – OM = x2 –x1

BC = BN – CN = BN –AM = y2 –y1

Then,
Vector AB is de�ined by using two component for the directed line

segment AB . where,

x - component = AC = x2 – x1

y- component = BC = y2 – y1

` AB = e yx - component o = d x2 –x1 n
- component y2 –y1

Component of vector in graph:

The x- component and y- component can be expressed taking
horizontal and vertical displacement of the given vector respectively
as shown in diagram.

P
B

M
AD

Q

N C

PRIME Opt. Maths Book - VI 101

Here,

The directed line segment which represent the vector which their

components can be expressed as follows.
3 –2
AB = ` 2 j , CD = ` 4 j

PQ = ` 3 j, MN = ` –2 j
–4 –4

Note:

Horizontal displacement $ Right side = +ve

(x-component) Left side = –ve

Vertical displacement $ Upward = +ve

(y-component) Downward = –ve

Types of Vector:

Column Vector:

The Vector quantity where the components are taken in a column is

called column Vector.
-
i.e. AB = c x - component m
y component

Row Vector :
The Vector quantity where the components are taken in a row is called
row vector.

i.e. AB = (x- component y- component)

Null Vector :

The Vector having both the quantities zero is called null Vector.
0
i.e. a = a 0 k

Equal vectors:
Any two vectors are said to be equal if both the components of the
vectors are same. i.e. The vectors having same direction as well as
equal magnitude are called equal vectors.

102 PRIME Opt. Maths Book - VI

Examples: B
i. 4 units Q

A
4 units

P

Here, AB = PQ

ii. AB = ` 3 j and PQ = ` 3 j
4 4

Here, AB = PQ

Addition of vectors:

The addition of any two vectors having same direction is the

combination of the vectors which is obtained by adding their
corresponding components.
x1 x2
i.e. a =a y1 k and b = a y2 k

Then,

a+b =a x1 k+a x2 k = a x1 + x2 k
Examples: y1 y2 y1 + y2

If a = ` 1 j and b = ` 3 j
2 4
Then,

a+b = ` 1 j + ` 3 j = e1 + 3 o = ` 4 j
2 4 2 + 4 6

`a + b = ` 4 j
6

Worked out Examples

1. Show the vector a = d 3n and b = d–2 n with
–4 5
directed line segment in graph paper.

Solution: b

The given vectors are:

a = d 3n and b = d–2 n a
–4 5

PRIME Opt. Maths Book - VI 103

2. If A(1, 2) and C(–3, 5) are any two points, find the vector AB .

Solution:

The given points are:
A(1, 2) = ((xx21
B(7, 5) = , y1)
, y2)

Then,

AB = dx2 – x1 n
y2 – y1

= d7 – 1n
5 – 2

= d6 n
3

` AB = d6n
3

3. If P(3, 1), Q(5. 4), R(4, 3) & S(6, 6) are the four points, prove that
PQ = RS .

Solution:

The points for PQ are,
P(3, 1) = ((xx12
Q(5, 4) = , y1)
, y2)

Then,

PQ = dx2 – x1 n
y2 – y1

= d5 –3n
4 –1

= d2 n
3

Again, the points for RS . are,
R(4, 3) = ((xx21
S(6, 6) = , y1)
, y2)

Then,

104 PRIME Opt. Maths Book - VI

RS = d x2 – x1 n
y2 – y1

= d6 – 4n
6 – 3

= d2 n
3

` PQ = RS proved.

Exercise : 5.1

1. i. What is vector quantity? Write down four examples of it.
ii. What is scales quantity? Write down four examples of it.
iii. Write down the difference between the following straight
lines.
B B

AA
iv. What do you mean by column vector?
v. What do you mean by null vector?

2. Find the vectors represented by the following directed line
segments.
BM

i. AB A C v. RS
QN
D
ii. CD P

RS
iii. PQ iv. MN

PRIME Opt. Maths Book - VI 105

3. Show the following vectors in graph paper using directed line
segment.

i. AB = d2 n ii. CD = d–3n iii. PQ = d 2n
4 4 –5

iv. RS = d–4 n v. MN = d3 n
–3 0

4. Find vectors AB and CD from the followings.
i. A(1, 5) and B(4, 10)
ii. A(–3, 2) and B(–5, 6)
iii. C(4, 2) and D(6, 0)
iv. C(1, 4) and D(4, 2)
v. A(–2, –1) and B(–4, –5)

5. Prime more creative questions:
i. If A(1, 3), B(4, 4), C(3, 2) and D(6, 3), find the vectors AB
and CD .
ii. Write down the relation of vectors AB and CD of Q. No. 1
iii. If P(1, 5), Q(3, 8), R(2, 6) and S(4, 9) are any four points.
Prove that PQ = RS .

iv. Find AB where A(1, –3) and B(4, 1) are the any two points.
Also find BA .

v. Find AB from the adjoining diagram.
A

3 units 4 units B
C

106 PRIME Opt. Maths Book - VI

Answer

1. Show to your teacher.

2. i) d3 n ii) d–4 n iii) d–3 n iv) d5 n v) d5 n
4 –4 2 –4 0

3. Do your self.

4. i) d3 n ii) d–2 n iii) d2n iv) d3 n v) d–2 n
5 4 –2 –2 –4

5. i) d3 n, d3 n ii) AB = CD
1 1

iv) d3 n , d–3 n v) d4n
4 –4 –3

PRIME Opt. Maths Book - VI 107

5.2 Vector operations

Magnitude of a vector:

The length of the directed line segment used in the vector is called

magnitude of the vector. It is also called the modulus of the vector.
-
For the vector AB = c x - component m
y component

Magnitude of AB ,
AB = ^x - componenth2 + ^y - componenth2

For the vector a = ` x j
y
a = x2 + y2

For the vector of a = ` 3 j
4
Magnitude of a is

|a | = x2 + y2

= 32 + 42

= 25

= 5 units.

The modulus of a vector which represents the
length of the directed line segment is called
magnitude of the vector.

i.e. |a | = x2 + y2

Unit vector:

The vector having magnitude one is called unit vector.
i.e. If AB = 1 unit, then the vector AB is called unit vector.

The vector having magnitude one is called unit vector.
1
Unit vector of a = a .a .

108 PRIME Opt. Maths Book - VI

Example: 3
4
If a = ` j , �ind the unit vector of a .

Solution:

a = ` 3 j
4

|a | = x2 + y2

= ^3h2 + ^4h2

= 25

= 5 units.

Then, unit vector of a

= 1 ^a h
a

= 1 ` 3 j
5 4
KKKKKJLKK OOOONOOO
= 3 P
5
4
5

Worked out Examples

1. If a =e 5 o , Find the magnitude of a .
11

Solution:

a =e 5 o
11

We have,
Magnitude of a is,
|a | = x2 + y2

= ^ 5h2 + ^ 11h2

= 5 + 11

= 16
= 4 units

PRIME Opt. Maths Book - VI 109

2. If A(1, 4) and B(5, 7) are any two points, find the magnitude of AB .

Solution:

The two points are:
A(1, 4) = ((xx21
B(5, 7) = , y1)
, y2)

Then,

AB = d x2 – x1 n
y2 – y1

= d5 – 1n
7 – 4

= d4 n
3

Magnitude of AB is,
; AB ; = x2 + y2

= 42 + 32

= 25
= 5 units.

3. If magnitude of a = ` 4 j is 5 units find the value of ‘m’.
m
Solution:

a = ` 4 j
m
|a | = 5 units
We have,
|a | = x2 + y2
or, 5 = 42 + m2

squaring on both sides,
or, (5)2 = ^ 16 + m2h2
or, 25 = 6 + m2
or, 25 – 16 = m2
or, 9 = m2
\ m=3

110 PRIME Opt. Maths Book - VI

4. Find the unit vector of a = a 8 k.
Solution 6
8
: a = a 6 k

Then,

|a | = x2 + y2 = 82 + 62 = 10 units.

Also,

unit vector of a is,
a = 1 ^ah

a

= 1 a 8 k
10 6

= c 8/10 m
6/10

= c 4/5 m
3/5

5. .IfAals=o`fi32njdatnhde mb a=gan54itkudareeoafnay+twbo. vectors, find the value of a + b
Solution :

Here,

a = ` 3 j
2

b = a 5 k
4

Then

a +b = ` 3 j + a 5 k
2 4

= a 3 + 5 k
2 + 4

= a 8 k
6
Again, magnitude a + b is,
| a + b | = x2 + y2
= 82 + 62
= 100
= 10 units

PRIME Opt. Maths Book - VI 111

Exercise : 5.2

1. Find the magnitude of the vectors form the followings:

i. a = d3 n , ii. b = d6n iii. p =d 5 n
4 8 12

iv. q =e 3o v. q=d 2 n
1 5

2. Find the unit vector of the followings.

i. a =d 1n ii. b = d4 n iii. c = d6 n
3 3 8

iv. p = d12n v. q =e 5o
5 2

3. Add the following vectors:

i. a = d1 n and b = d3 n ii. c = d4n and d = d1 n
2 4 3 –1

iii. p = d2 n and q = d1 n iv. a = d–2 n and b = d 1n
3 2 3 –2

v. b = d3 n and c = d 1n
4 –3

4. Subtract the followings.

i. a = a 1 k and b = a 2 k ii. a = a 6 k and b = ` 1 j
0 3 8 4

iii. p = ` 4 j and q = a –2 k iv. q = ` 1 j and r =a 2 k
3 3 1 –3

v. b = ` 0 j and c = ` 4 j
4 2

5. Prime more creative questions:
i. Find the magnitude of AB where A(1, 3) and B(4, 7) are any
two points.

ii. If a = d1 n and b = d5 n , find the magnitude of a +b.
2 6

112 PRIME Opt. Maths Book - VI

iii. If magnitude of a = dxn is 5 units, find the value of x.
4

iv. Prove that a = d1 n is the unit vector of a .
0

v. Prove that a m= LJKKKKKKKe34an55 OOOPNOOOObiys the unit vector of a .
vi. What do you magnitude of a vector?
vii. What do you mean by unit vector?

1. i) 5 units Answer iii) 13 units
iv) 2 units
ii) 10 units
v) 3 units

2. i) KKKLKKKKJ 1 OOOOONOPO ii) KKKKJLKKK 4 OOOOOOPNO iii) KJLKKKKKK 3 POOONOOOO iv) KKKKKKLKJ 12 NOPOOOOOO v) KJLKKKKKK 5 OOOOOOPNO
3. i) 2 5 5 13
3 3 4 5 3
5 5 13 2
2 3

d4n ii) d5 n iii) d3 n iv) d–1 n v) d4n
6 2 5 1 1

4. i) d–1 n ii) d5 n iii) d6n iv) d–1 n v) d–4 n
–3 4 0 4 2

5. i) 5 units ii) 10 units iii) 3

PRIME Opt. Maths Book - VI 113

Vector

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. What do you mean by vector quality?
2. a. Find vector AB from the two points A(3, –2) and B(5, 1).

b. If a = d3 n , find the magnitude of vector a .
4

c. If a = dx1n and b = d x2 n , find a + b .
y1 y2

3. a. If magnitude of vector a = d6 n is 10 units, find the value of m.
m

b. If A(1, –2), B(3, 4) and C(5, 10) are the three points, prove that
AB = BC .

4. If a = d3n and b = d5 n , find the value of a +b. Also find the
1 5

magnitude of vector a + b .

114 PRIME Opt. Maths Book - VI

6 Transformation

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
–1 2 7 10
No. of Questions – 1 –5

Weight –2

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can define the transformation.
• Students can identify the transformations.
• Students can find the coordinate of image of a point under

transformations.
• Students can find the image of a point using co-ordinate.
• Students can plot the object and image in graph.

Materials Required:
• Chart paper.
• Graph board.
• Sample of transformations.
• List of the formula of transformations.
• Geo-board
• Chart paper.
• Shape of objects

PRIME Opt. Maths Book - VI 115

Introduction of transformation

Let us see,
i)

Here, the image of a car is seen into a mirror of the car, where
• The car is coming from back is the object.
• The car into the mirror is called the image.
ii)

Here,
Patterns of triangles is made by transferring from one place to another
place one after another.
Above examples show that one object can be transferred from one
place to another place with same shape & size or different size which
is called transformation.

116 PRIME Opt. Maths Book - VI

The process of changing the position or
size of an object under any geometrical
conditions is called transformation.

By using transformation, patterns of objects can be drawn which is
useful to fill the pictures in clothes, walls and in any kind of objects to
make them very attractive.

Here in grade VI we discuss only two types of transformations
i) Reflection iii) Translation

6.1 Reflection P A‘
B‘
The image of a person AB is A‘B‘ which is A Q
formed at equal distance from the mirror m
line m as the distance of object where AP =
PA‘ or BQ = QB‘. The image A‘B‘ so formed
is laterally inverted as the image formed B
in the looking glass but size is same as the
object.

Here, AB is called the object
m is called reflection axis (mirror line)
A‘B‘ is called the image.

The transformation of an object from one place to
another place by a mirror line is called reflection.
• Object and image so formed are always

congruent in reflection.

PRIME Opt. Maths Book - VI 117

Reflection using co-ordinates axes.

i) Reflection about x-axis (y = 0).

Y

4

3 M(2, 3)

2

1

X‘ P X
-4 -3 -2 -1 O 1 2 3 4
-1

-2

-3 M’(2, –3)

-4

Y‘
Draw MP⊥OX and produce MP to M‘ such that MP = PM‘
Here, M(2, 3) is an object.

x-axis (XX‘) is the reflection axis.
M‘(2, –3) is the image.

It shows that: P(x, y) → P‘(x, –y)

ii) Reflection about y-axis (X = 0)

Y

P‘(–3, 4) M4 P(3, 4)

3

2

1

X’ -4 -3 -2 -1 O 1 2 3 4 X
-1

-2

-3

-4

Y’

118 PRIME Opt. Maths Book - VI

Draw PM⊥OY & produce PM to MP‘ to make PM = MP‘.
Here, P(3, 4) is an object.
Y -axis (YY‘) is the reflection axis.
P‘(–3, 4) is the image.
It shows that : P(x, y) → P‘(–x, y

Worked out Examples

1. Draw the image of ∆ABC under the given mirror line.

A

B
C

Solution: A

B
M

C

P A`
N C`

B`

Here.
AM, BN and CP are the perpendicular drawn on the mirror line.

Also,
AM = MA`
BN = NB`
CP = PC`

Then,
the ∆A`B`C` is the image of ∆ABC after reflection.

PRIME Opt. Maths Book - VI 119

2. Find the image of a point P (3, 2) under reflection about y - axis.
Solution:
Under reflection about y -axis,
P (x, y) → P` (–x, y)
P (3, 2) → P` (–3, 2)

3. Find the image of ∆ABC having vertices A (–A, 3), B ( 2, 5) and C
(4, 0) under reflection on x-axis. Also plot the object and image in
graph.
Solution:
under reflection about x- axis,
P (x, y) → P` (x, –y)
A (–1, 3) → A` (–1, –3)
B (2, 5) → B` (2, –5)
C (4, 0) → C` (4, 0)

Y

B(2, 5)

A(–1, 3)

X’ O C(4, 0) X

C`(4, 0)

A`(–1, –3)

B`(2, –5)

Y’

120 PRIME Opt. Maths Book - VI

Exercise : 1.1

1. Draw the image of given triangles under reflection on mirror line
M.
i. ii. L
A M

B C K
M M

iii. iv.
L
S
M M
v. PK

QM

R

U M
T
121
SV

PRIME Opt. Maths Book - VI

2. Find the image of following points under the given reflection axis.
i. A (2, 3); under x - axis
ii. p (–3, 5); under y - axis
iii. B (–4, –5); under x - axis
iv. Q (–5, 2); under the line y = 0
v. F(3, –4); under the line x = 0

3. Find the image of the triangles under the given reflection axis.
Also plot the object and image in graph.
i. Having vertices A (2, 3), B (4, 6) and C (7, 1) under x- axis.
ii. Having vertices P (4, 2), Q (1, 5) and R (3, –3) under y- axis.
iii. Having vertices A (–3, 0), B (–6, 4) and C (0, 3) under x-axis.
iv. Having vertices P (–6, 1), Q (–3, 5) and R (–5, –4) under the
line x = 0.
v. Having vertices D(3, –2), E(5, 3) and F(–2, 5) under the line
y = 0.

Answer

1. Show to your teacher.
2. i. (2, 3) ii. (3, 5) iii. (–4, 5) iv. (–5, –2) v. (–3, –4)
3. i. A`(2, –3), B`(4, –6), C`(7, –1); graph

ii. P`(–4, 2), Q`(–1, 5), R`(–3, –3); graph
iii. A`(–3, 0), B`(–6, –4), C`(0, –3); graph
iv. P`(6, 1), Q`(3, 5), R`(5, –4); graph
v. D`(3, 2), E`(5, –3), F`(–2, –5); graph

122 PRIME Opt. Maths Book - VI

6.2 Translation P’

P

Q’ B R’

QA

R

Here, vector AB is the magnitude & direction of translation.
DPQR is an object.
PP‘ = QQ‘ = RR‘ = AB
PP‘ ' QQ‘ ' RR‘ ' AB
DP‘Q‘R‘ is the image of DPQR

The transformation of an object from one place
to another place according to the magnitude and
direction of the given vector is called translation.

i) Translation using co-ordinate:

Y

A’(7, 5)
A(4, 3)

X’ O B’(6, 1) C’(9, 1)
X

B(3, -1) C(6, -1)

Y’

PRIME Opt. Maths Book - VI 123

Here, A(4, 3) is translated to A’(7, 5)
B(3, –1) is translated to B’(6, 1)
C(6, –1) is translated to C’(9, 1)

i.e. All the points are translated with constant number 3 for
x-component and 2 for y - component.

i.e. A (4, 3) → A‘(4 + 3, 3 + 2) = A‘(7, 5)

i.e. Translation vector is T = <3F
2

i.e. under translation T = <aF
b

P(x, y) → P‘(x + a, y + b)

ii) Translation using vector:
Let us consider a translation vector is AB where A(1, 2) and B(3,
5) are any two points.
Then, according to the concept of column vector AB
AB = <x2 – x1F = <3 – 1F = <2F
y2 – y1 5 – 2 3

\ Translation vector T = AB = <2F
3

Worked out Examples

1. Draw the image of ∆PQR under translation on given vector.
P

Qv
R

124 PRIME Opt. Maths Book - VI

Solution: P

P`

Q v
Q`

R R`
Here,PP`, QQ` and RR` are drawn parallel to the vector line V
where, PP` = QQ` = RR` = V
\ ∆PQR is the image of ∆PQR.

2. Find the image of a point P (2, –3) under a vector T =:32 D
Solution:
under translation about T = :32 D
then,
P (x, y) → P` (x + a, y + b)
→ P` (x + 2, y + 3)
P (2, –3) → P` (2 + 2, –3 + 3)
→ P` (4, 0)

3. Find the image of ∆ABC having vertices A (1, 2), B (3, 5) and C (6,

1) under in the translation about T = <–2F . Also plot the object
and image in graph. 4

Solution: 8-42

Under translation about T = B
P (x, y) → P` (x + a, y + b)
→ P` (x –2, y + 4)
A (1, 2) → A` (1 – 2, 2 + 4) = A1 (–1, 6)
B (3, 5) → B` (3 –2, 5 + 4) = B1 (1, 9)
C (6, 1) → C` (6 –2, 1 + 4) = C1 = (4, 5)

PRIME Opt. Maths Book - VI 125

Y

B`

A` B C`

A

C

X’ O X

Y’

Exercise : 6.2

1. Draw the image of the triangles given below under translation on
given vector ‘v’.
P

i. A ii.
V Q
R
BC
V

iii. A D iv. P
B V
Q R
C V

126 PRIME Opt. Maths Book - VI

v. T

SU
V

RQ

2. Find the image of the points given below under the given vector
‘T’.
i. A (3, –5) ; under T = 812 B

ii. P (–2, 1) ; under T = 832 B

iii. M (3, –2) ; under T = 8–41 B

iv. B (–4, –2) ; under T = :26 D

v. Q(4, 5); under T = <–5F
–2

3. Find the image of the triangles given below under the translation
‘T’. Also plot the object and image in graph.
i. Having vertices A (1, 2), B (3, –2) and C (4, 5) under T =:32 D
ii. Having vertices P (0, 1), Q (3, 5) and R (1, –4) under T = 834 B
iii. Having vertices A (–3, 2), B (3, 6) and C (4, –3) under T = AB

iv. Having vertices P (5, 2), Q(3, 5) and R(7, –4) under T = PQ
v. Having vertices P(–3, –4), Q(0, 3) & R(5, 0) under a translation

vector T = <–4F .
3

PRIME Opt. Maths Book - VI 127

Answer v. (–1, 3)

1. Show to your teacher.

2. i. (4, –3) ii. (1, 3) iii. (2, 2) iv. (–2, 4)

3. i. A`(3, 5), B`(5, 1), C`(6, 8); graph.
ii. P`(3, 5), Q`(6, 9), R`(4, 0); graph.
iii. A`(–5, 5), B`(1, 9), C`(2, 0); graph.
iv. A`(1, 0), B`(–1, 3), C`(3, –6); graph.
v. P`(–7, –1), Q`(–4, 6), R`(1, 3); graph.

Transformation

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Write down the image of a point P(a, b) under reflection about x-axis.
2. a. What is transformation of a point? Write down the importance

of transformation.
b. Find the image of a point A(3, –2) under reflection about y-axis.

c. Find the image of a point P(–2, 5) uwnidtherattrraannslsalatitoionnveTct=or<baTF=, fi<n13dF
3. a. If A(3, 2) is translated to A’(4, 5)
the value of a and b.
b. Find the image of a triangle having vertices A(3, 1), B(5, –2)
and C(6, 4) under reflection about y = 0. Also plot the object
and image in graph.
4. Find the image of DPQR having vertices P(–2, 1), Q(2, 4) and

R(0, –3) under a translation vector T = <4F . Also plot the object
and image in graph. 5

128 PRIME Opt. Maths Book - VI

7 Statistics

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
–1 2 76
No. of Questions – 1 –5

Weight –2

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can find the values of central tendency mean, median &

mode.
• Students can find the partition values like median.
• Students can find the measures of dispersion range.
• Students can tabulate the collected data.
• Students can present the data in diagrams.

Materials Required:
• Chart paper.
• Sample of data collection.
• Sample of tabulation of data.
• List of formula used in statistics.
• Graph paper.
• Scissors
• Geo-board

PRIME Opt. Maths Book - VI 129

Measure of central tendency

The measurement of single value of the observations like marks
obtained by the students, age of the students, number of boys and
number of girls of different classes etc can be calculated as the average
value. It can be measured by using different ways in statistics which
is called the measure of central tendency. It can be calculated by using
three different ways called arithmetic mean, median and mode.

7.1 Arithmetic mean (Average)

The average value of the observations can be calculated by taking

the sum of the observations and by dividing with the number of
observations.

i. For individual observations:
NObos. eorfvoabtsioenrvsa=tixo1n, sx2=, xN3,..............................xn
Sum of the observations = x∑1x+ x2 + x3 +....................... xn
=

\ Mean(x) = 6x
N

ii. For desecrate observations:
PFOrrbeosqdeuurevcnat ttailnoynd=ssfu=1,mxf21,,fx3==2, ,..∑fx.1.3.xf.,.x1.....+........f...2...x.....2.....+.,..f.f.n3..x.,3x+n ...................... + fnxn

Sum of frequencies (N) = ∑f
Then,

\ Mean(x) = 6fx
N

130 PRIME Opt. Maths Book - VI

Worked out Examples

1. Find the arithmetic mean of the marks obtained by 10 students

of grade VI as 14, 16, 18, 20, 12, 15, 5, 10, 13, 17.

Solution:

The marks obtained by the students in ascending order are:

5, 10, 12, 13, 14, 15, 16, 17, 18, 20

No. of student (N) = 10.

Arithmetic mean,
Rf
x = N

= 5 + 10 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 20
10

= 140
10

= 14

∴ Mean (x) = 14

2. If arithmetic mean of 3, 10, m, 12 and 14 is 10 �ind the value of ‘m’

Solution:

The given observations in ascending order are:

3, 10, m, 12, 14

No. of observations (N) = 5

Arithmetic mean (x) = 10

We have,
X
(x) = N

or, 10 = 3 + 10 + m + 12 + 14
5

or, 10 × 5 = 39 + m

or, 50 – 39 = m

or, 11 = m

∴ m = 11

PRIME Opt. Maths Book - VI 131

3. Find the arithmetic mean of the age of the students given below.
Age 8 10 12 14 16

f 37442

Solution:

Age (x) f f×x

83 24

10 7 70

12 4 48

14 4 56

16 2 32

We have, N= 20 ∑fx = 230

Arithmefic mean,
/fx
x = N

= 230
20

= 11.5
\ Arithmetic mean of age = 11.5 years.

Exercise : 7.1

1. Answer the following questions:
i. What is central tendency?
ii. What do you mean by arithmetic mean?
iii. Name the measures of central tendency.
iv. Write down the formula to calculate arithmetic mean of
individual observations.
v. Write down the formula to calculate arithmetic mean of
discrete observations.

2. Find the arithmetic mean of the followings:
i. 3, 5, 6, 7, 9 ii. 12, 18, 24, 30, 36

132 PRIME Opt. Maths Book - VI

iii. 18, 22, 26, 30, 14, 10 iv. 18, 23, 28, 33, 13, 8, 3
v. 4, 8, 12, 16, 32, 28, 24, 20

3. Find the followings:
i. If the mean of 2, 7, p, 10 and 12 is 8, find the value of p.
ii. If the mean of 12, 18, 24, m, and 36 is 24, find the value of m.
iii. If arithmetic mean of a, 14, 18, 22, 26 and 30 is 20, find the
value of ‘a’.
iv. If arithmetic mean of 2x, x + 3, x + 6, x + 9, x + 12 and x + 15
is 18, find the value of x.
v. If arithmetic mean of m, m + 2, m + 4, m + 6 and m + 8 is 14,
find the value of m.

4. Find the arithmetic mean from the following:

i. Weight 10 12 20 25 35
5
f 3 10 15 7
45
ii. Age 5 15 25 35 5
f 7 18 20 10
20
iii. Marks 8 10 15 4
f 5 88
9
iv. Height 5 678 4
f 4 8 14 10
80
v. x 10 30 50 70 10
f 7 8 10 15

Answer

1. Show to your teacher. iii. 20 iv. 18 v. 18
2. i. 6 ii. 24 iv. 9 v. 10
3. i. 9 ii. 30 iii. 10 iv. 7.05 v. 53.2
4. i. 20 ii. 23 iii. 12.8
133
PRIME Opt. Maths Book - VI

7.2 Median

The single value of the observations arranged in ascending order
which divides the whole observations into two equal half is called
median.

Median

i. Individual observations:
• Arrange the observations in ascending order.
N+ th
• Find the size of a 2 1 item.
• k

Corresponding observation is the median

ii. Discrete observations:
• Arrange the observations in frequency distribution table.
• Find cumulative frequency column.
+ th
• Find size of a N 2 1 item.
• k

Corresponding observation of just greater than a N + 1 th
2
k

item in c.f. column is the median.

Worked out Examples

1. Find the median of the observations 8, 12, 16, 28, 24, 32, and 20.

Solution:

The given observations in ascending order are:
8, 12, 16, 20, 24, 28, 32.
No. of observations (N) = 7

Then,

Median = Size of a N + 1 th item
2
k

= Size of a 7 + 1 th item.
2
k

= Size of 4th item
= Corresponding observation is 20
` Median = 20

134 PRIME Opt. Maths Book - VI

2. Find median marks obtained by 8 students of class VI in optional
math as 12, 18, 8, 20, 10, 16.

Solution:

The obtained marks in ascending order are
8, 10, 12, 16, 18, 20.
No. of observations (N) = 6

Then,

Median (Md) = size of a N+ 1 th item
2
k

= Size of a 6 + 1 th item
2
k

= Size of 3.5th item

= 3rd + 4th
2

= 12 + 16
2

= 14

3. If median of the observations taken in order 8, 12, 16 , x + 15, 24,
28 and 32 is 20, find the value of x.

Solution:

The observations taken in order are
8, 12, 16, x + 15, 24, 28, 32
No. of observations (N) = 7
MWeedhiaavne(,md) = 20
+ th
Median = size of a N 2 1 item
k

or, 20 = Size of a 7 + 1 th item
2
k

or, 20 = size of 4th item
or, 20 = x + 15
or, 20 – 15 = x
or, 5 = x
` x = 5.

PRIME Opt. Maths Book - VI 135

4. Find the median of the observations: 50
3
Marks 15 25 35 45
5
f 348

Solution:

Marks (X) f c.f.
15
25 3 3
35
45 4 3+4=7
50 8 7 + 8 =15
5 15 + 5 = 20
20 + 3 = 23
3
N = 23

Median = size of a N + 1 th item
2
k

= Size of a 23 + 1 th item
2
k

= Size of 12th item
= 15 is just greater than 12 in c.f.
= Corresponding observation is 35
` Median = 35

Exercise 7.2

1. Answer the following questions:
i. What do you mean by median value?
ii. What steps should be done to find the median of individual
observations?
iii. What steps should be done to find the median of discrete
observations
iv. What is the median of the observations a, b, and c?
v. What is the median of the observations a and b?

136 PRIME Opt. Maths Book - VI

2. Find median from the followings:
i. 7, 9, 11, 15, 17 ii. 28, 32, 16, 12, 24, 8, 20
iii. 28, 26, 24, 22, 20, 36, 32, 34, 30
iv. 6, 10, 14, 18, 22, 24
v. 34, 28, 22, 16, 10, 40, 46, 52
vi. 30, 24, 36, 66, 48, 60, 54, 42, 78, 72

3. Find the unknowns from the followings:
i. If median of 10, 12, x + 4, 16, 18 which are taken in order is
14, find the value of x.
ii. If median of the observation taken in order 7, 10, m + 10,
m + 12, 19 and 22 is 15, find the value of m.
iii. If median of the observation taken in order 8, 12, P + 4,
P + 8, p + 12, 28 and 32 is 20, find the value of p.
iv. If median of the observations taken in order 6, k + 2, k + 6,
k + 10, 22 and 24 is 16, find the value of ‘k’.
v. If median of the observation taken in order m + 1, 2m,
2m + 4, 3m + 3, 4m + 2 and 5m – 1 is 16, find the value of m.

4. Find the median from the followings:
i. x 10 20 30 40 50
f 25864

ii. Marks 5 15 25 35 45
f 5
7 12 6 3

iii. Weight 30 10 50 70 90
f 7 5 8 14 9

iv. Age 30 24 18 42 36

f 15 18 14 15 12

v. x 100 200 300 400 500 600
f 4 5 8 10 5 4

PRIME Opt. Maths Book - VI 137

Answer

1. i. Show to your teacher. ii. Show to your teacher.
a+b
iii. Show to your teacher. iv. b v. 2
2. i. 11 ii. 20 iii. 28 iv. 16
v. 31 vi. 51
3. i. 10 ii. 4 iii. 12 iv. 8 v. 5
4. i. 30 ii. 25 iii. 70 iv. 30 v. 400

138 PRIME Opt. Maths Book - VI

7.3 Mode

Let us discuss the shoes numbers produced by a factory in a day are
26, 26, 30, 30, 30, 32, 32, 36, 36, 36, 36, 36, 38, 37, 38, 40, 40, 42, 43,
44. Out of 20 pairs of these shoes in a day, 36 number is manufactured
more then the others accorded to the demand of consumers in the
locality. It is taken as the model number.

The single value represents the observations which is the most
repeated or the most frequent observation is called mode.

• The most repeated observation in individual
observations is mode.

• The most frequent observation in the discrete
observations is mode

7.4 Range

In the above example highest number of shoes is 42 and lowest number
is 26. The difference of the number of shoes where the consumers
used is 42 –26 = 16. It mean there are the consumers of 16 varieties
according to using shoes.

It is the measure of dispersion of the observations which is the
difference between highest and lowest observations of the data is
called range.

i.e Range = Highest observation – Lowest observation
R=H–L
H – L
Also, Coefficient of range = H + L

PRIME Opt. Maths Book - VI 139

Worked out Examples

1. Find mode of the observations
12, 13, 12, 14, 14, 15, 16, 16, 15, 15, 14, 14.
Solution:

The observations in ascending order are:
12, 12, 13, 14, 14, 14, 14, 15, 15, 15, 16, 16
Here,
Mode = The most repeated observation
Here, 14 is repeated most (4 times)
` Mode = 14

2. Find mode from the observations:
Weight 40 45 50 55 60 65
f 4 7 13 8 9 5
Solution:

The given frequency table is,

Weight 40 45 50 55 60 65
f 4 7 13 8 9 5

Here,
Mode = The most frequent observation.

= 50 (Highest frequency is 13)
` Mode = 50.

3. Find the range and its coefficient of range of 12, 15, 18, 20, 24, 28,
38.

Solution:

The given observations are:
12, 15, 18, 20, 24, 28, 38
Highest observation (H) = 38
Lowest observation (L) = 12

Then, =H–L
Range (R) = 38 – 12
= 26

140 PRIME Opt. Maths Book - VI

Coefficient of range = H–L
H+L

= 38 – 12
38 + 12

= 26
50

= 0.52

4. Ifrangeandlowestobservationofadataare30and10respectively,
find the highest observation.
Solution:
Lowest observation (L) = 10
Range = 30
Highest observation (H) = ?
We have,
Range = H – L

or, 30 = H – 10
or, 30 + 10 = 40

` H = 40

Exercise 7.3

1. Find the mode of the observations:
i. 10, 15, 20, 20, 25, 15, 20, 30, 25, 20.
ii. 8, 12, 14, 16, 12, 16, 14, 16, 8, 18, 18, 20, 16, 20.
iii. x
12 16 20 24 28 32
5
f 2 6 8 13 4
40
iv. Marks 15 20 25 30 35 10
f 8 12 15 20 16
5.6
v. Height 4.4 4.6 4.8 5.2 5.4 8
f 3 7 9 12 10

PRIME Opt. Maths Book - VI 141

2. Find the range and its coefficient from the followings:
i. 10, 15, 20, 25, 30, 35, 40.
ii. 20, 30, 40, 50, 60
iii. 15, 25, 35, 45, 55, 65.
iv. x
20 30 40 50 60
7
f 6 8 12 9

v. x 15 25 35 45 55 65
f 268531

3. Answer the followings:
i. If heighest observation and range of a data are 60 and 40
respectively, find the lowest observation.
ii. If range and lowest observation of a data are 20 and 12
respectively, find the highest observation.
iii. If highest marks obtained by the students of class vi is 40
and lowest marks is 10 out of full marks 40, find the range
and coefficient of range of the marks.
iv. If large number of people need shoes of 40 number out of so
many shoes. What is the mode of shoes.
v. What do you mean by range ? Write down it’s formula.
vi. What do you mean by mode of the observations?

Answer

1. i. 20 ii. 16 iii. 24 iv. 30 v. 5.2
2. i. 30, 0.6 ii. 40, 0.5 iii. 50, 0.625
iv. 40,0.5 v. 50, 0.625
3. i. 20 ii. 32 iii. 30, 0.6 iv. 40

142 PRIME Opt. Maths Book - VI

Statistics

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:
1. Write down the calculating formula of arithmetic mean for

discrete observations.
2. a. Find the mean of 10, 12, 14, 16, 18.

b. Find the median of 18, 14, 10, 22, 26.
c. Find the mode of the observations.

12, 11, 8, 12, 11, 12, 8, 15, 15, 14, 12, 14.
3. a. Find the range and coefficient of range of the observations 12,

16, 20, 24, 28
b. If median of the observations 8, 12, x + 6, 2x, 24 and 28 is 18,

find the value of x.
4. Find the arithmetic mean of the observations.

Class 0-10 10-20 20-30 30-40 40-50
f 23852

PRIME Opt. Maths Book - VI 143

Proposed Syllabus with Grid for

First Terminal Examination

S.N. Contains Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Order Pairs 11 1 – 37 5
ii. Cartesian Product

2 Matrices All 111 – 37 5

3 Co-ordinate i. Distance Formula 1 1 1 1 4 12 4

Geometry

4 Trigonometry Exercise 4.1 1 3 3 1 8 24 12
Exercise 4.2
Exercise 4.3
Exercise 4.4

Total Questions 4 6 6 2 18

Total Marks 4 12 24 10 50 26

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

Model Question Set for First Terminal Examination

Group ‘A’ [4 × 1 = 4]
1. a. What do you mean by ordered pairs ?

b. Find the transpose of the matrix <2 –3F
1 4

2. a. Write down the co-ordinate of the points an x-axis and y- axis.
b. Write down the relation of Cosθ in terms of Sinθ.

Group ‘B’ [6 × 2 = 12]
3. a. If (x + 1, y – 2) = (3, 1) are equal ordered pairs, find x and y.

b. If A = <1 2F and B = <2 1F , find A+B.
–1 3 3 –1

4. a. Find the distance between the points A(3, –2) and B (7, 1).
b. Convert 20° 15’ 10’’ into seconds.

5. a. Find the sum of Sin²θ + 2Sinθ.cosθ – 3Cos²θ and Sin²θ – 3Sinθ. Cosθ +
2Cos²θ.

b. Find Sinθ from the given right angled triangle.
A

4cm

B C
3cm

144 PRIME Opt. Maths Book - VI

Group ‘C’ [6 × 4 = 24]
6. If A = {a, b, c} B = {p, q}, Find the Cartesian product A×B and B×A.

7. If A = <2 1F and B = <1 3F , Find 2A – B.
3 –2 –1 2

8. Prove that the points (4, 3) and (–3, 4) are equidistant from the origin.
9. One angle of a triangle is 60° and the other is 80g. Find the third angle in

degrees.
10. Prove that : Sec4θ – Tan4θ = 1 + 2Tan²θ.
3
11. If Sinθ = 5 , find the value of Cosθ and Tanθ.

Group ‘D’ [2 × 5 = 10]
12. Prove that the points A (3, 2), B (3, –3) and C (8, –3) are the vertices’s.
1 – Cosi
13. Prove that : 1 + Cosi = (Cosecθ – Cotθ)2

Proposed Syllabus with Grid for

Second Terminal Examination

S.N. Contains Topics K-1 U-2 A-4 HA-5 TQ TM Periods

1 Algebra i. Polynomial 111 – 37 6
ii. Surd

2 Matrices – 111 – 37 –

3 Co-ordinate Remaining All –11 – 26 4

Geometry

4 Trigonometry Remaining All 1 2 3 – 6 17 12

5 Transformation Reflection 1– – 1 26 6

6 Statistics Arithmetic Mean –1 – 1 27 6

First Term Review 6

Total Questions 4 6 6 2 18

Total Marks 4 12 24 10 50 40

K = Knowledge, U = Understanding, A = Application, HA = Higher ability

PRIME Opt. Maths Book - VI 145