Optional Math 8

Unit 3 Matrices

Specification Grid Table

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

No. of Questions – 1 1 – 2 6 10
Weight – 2 4 –

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

Objectives : At the end of the lesson
• Students are able to represents the numbers in matrix form.
• Students are able to know the concept of order of matrix and types of

matrices.
• Students are able to operate the matrix addition, subtraction, transpose

and multiplication by scalar.
• Students are able to know the properties of addition matrices.

Materials Required:
• Chart paper.
• List of price chart of goods of a market.
• List of properties of matrix addition
• List of types of matrices

PRIME Opt. Maths Book - VIII 47

3.1 Matrices:

The cost per kg of vegetables in Kalimati vegetable market in a particular

day is as follows. Cauli�lower Cabbage
Potato

Shop A 15 25 20

Shop B 20 30 15

Shop C 18 24 12

Above cost of the vegetables caSSSSSSSSRT112n580be r322e054pre112s250eWWWWWWWWXVnted using square bracket as,

Here, Rows represents the shops and column represents the type of
vegetables. This type of presentation of numbers is called a matrix.

Matrices The rectangular array of the numbers in different
rows and columns which are taken into a square
bracket or round bracket is called a matrix.

• A matrix is denoted by capital letter A, B, C, ...
•
• The members used in a matrix are called elements which are
•
• denoted by small letters a, b, c, d, e, f, ........................
•
The horizontal arrangement of the number is a matrix are called
48
rows.

The vertical arrangement of the number in a matrix are called

columns.

The no. of rows (m) and no. of columns (n) can be written as

m × n which is called the order of the matrix. We read m × n as m

by n.

An element of a matrix is represented by the no. of row and no. of

columns. It lies as lower subscript e.g. the element lies in 2nd row

and 3TSRSSSSSSSaaar132d111 coaaalu132222mnaaais132333dWWXVWWWWWW enoted by a23.
A=

PRIME Opt. Maths Book - VIII

Here, a11 = element in �irst row and �irst column.
a23 = element in second row and third column.

Order of the matrix A = 3 by 3 written as 3 × 3.

Types of matrices
1. Row matrix

The matrix having only one row is called row matrix.

Ex : A = [a11 a12 a13]1 × 3 B = [2 4]1 × 2

2. Column matrix:

The matrix having only one column is called column matrix.
B = SSSSRTSSSSSSS1532XWVWWWWWWWWWW4×1
Ex : A = RTSSSSSSSSaaa132111XWWWWVWWWW3×1

3. Null matrix: Matrices

The matrix having all the elements zero is called null matrix. It is

denoted by ‘O’.

Ex : O = <00 0 00F2×3
0

4. Rectangular matrix:
The matrix having unequal number of rows and columns is called
rectangular matrix.

Ex : A = <12 3 –42F2×3
–3

5. Square matrix:

The matrix having equal number of rows and columns is called square

matrix. B = <13 24F2×2 C = STSSSSSSRS174 2 369WXWWVWWWWW3×3
Ex : A = [2]1×1 5
8

PRIME Opt. Maths Book - VIII 49

6. Diagonal matrix
The square matrix having the diagonal elements from left top to right

bottom are non-zero and remaining elements zero is called diagonal
matrix.

a0
Ex: A = < F

0b
2×2

7. Scalar matrix:
The diagonal matrix having all non-zero diagonal elements are equal
iEsxc:aAlle=dSRSSSSSSSTs00aca00alar00amWWVXWWWWWW3×a3trix.

8. Identity matrix (Unit matrix):

The square matrix having diagonal elements from left top to right

bottom all one and remaining elements zero is called identity matrix.

Matrices It is denoted by ‘I’. SSSRSTSSSS100 100XWWWWVWWWW3×3

10 0
Eg : I = < F I = 1
9. 0 1 0
2×2

Equality of the matrices

Any two matrices having same order and same corresponding

elements are called equal matrices.

Ex : A = <2 3F , B= <2 3F
4 1 4 1

Here, A = B

10. General element of a matrix:

The element of the matrix is denoted by aij as the general element
where i is no. of rows and j is no of columns.

Ex : If aij = 2i – j
Then, a11 = 2 × 1 – 1 = 1
a12 = 2 × 1 – 2 = 0
a21 = 2 × 2 – 1 = 3
a22 = 2 × 2 – 2 = 2
<a11 a12F 1 0
\ 2 × 2 matrix is, A = a21 a22 = < F
3
2

50 PRIME Opt. Maths Book - VIII

Worked out Examples

1. The cost of fruits in a fruits shop in different days is given below

represent the informations in a matrix with appropirate meaning.

Apple Orange Banana

Day 1 150 80 75

Day 2 120 100 70

Solution :

The cost of fruits given in table of two days are represented by matix

as follows where,

rows → represents the days

columns → represents the type of fruits

A= <115200 80 7705F
100

2. If A = <–32 2 –51F i) Find the order of matrix A.
4 ii) Find the elements a12, a21, & a23.

Solution, Matrices

i) No. of rows = 2

No. of columns = 3

∴ order of matrix A is 2 by 3 (2 × 3).

ii) a12 = element in 1st row 2nd column = 2
a21 = element in 2nd row 1st column = –2
a23 = element in 2nd row 3rd column = –1

3. If A = B, where A = =x + 2 13G and B = <6 3 13F , �ind the value of x & y.
y – 2y

Solution : A = =x + 2 13G, B = <6 3 13F
y – 2y

Here, matrix A & B are equal.

By equating the corresponding elements

x+2=3 and y = 6 – 2y

or, x = 3 – 2 y + 2y = 6

or, x = 1 3y = 6

∴x=1 y= 6
3
∴ y=2 ∴ x = 1, y = 2

PRIME Opt. Maths Book - VIII 51

Exercise 3.1

1. Write down the following informations in matrix form with appropriate

meaning. Also write down the order of the matrices.

i) The number student in a class are given below.

1st column 2nd column

1st row 4 3

2nd row 5 2

3rd row 6 4

4th row 3 5

ii) The cost of clothes in three shops are given below.

Shirt Pants Vest

1st shop 500 700 300

2nd shop 450 600 350

3rd shop 400 550 250

Matrices

iii) The production of crops in Jhapa district in different years in

metric tonne is given below.

Rice Wheat Maize others

2072 BS 300 150 100 150

2073 BS 350 170 120 130

2074 BS 475 180 200 300

2. Write down the types of the matrices from the followings. Also write

diiv)o )w nSSSSSSSTRS<13t022he02154oFWXVWWWWWWW r der of the mviia)) t ricSSSSSRTSSS<e0000as.0000bF 00cWWWWWWWXWV viiii)) SSSRSSTSSSTRSSSSSSSS100–141100520100132WVXWWWWWWW XVWWWWWWWW
viii) SSSSRTSSSS324WWWWWWWVWX ix) 63 –2 1@
2 3 1
vii) <–2 2 F

–2

52 PRIME Opt. Maths Book - VIII

3. Answer the following questions from the matrix RTSSSSSSSS132 3 –151WWWXVWWWWW
i) Element in �irst row and second column. 0
ii) The element a22. 2

iii) a31 + a23 + a33
iv) Element aij where i = 3 & j = 2.
v) Write down the order of matrix.

4. If the general element of a matrix is aij = 3i – 2j, what will be the matrices
of following order.

i) 2 × 2 matrix ii) 2 × 3 matrix iii) 3 × 3 matrix

iv) 3 × 2 matrix v) 1 × 3 matrix

5. Find the value of ‘x’ and ‘y’ from the following equal matrices.
i) A = <53 x6–1F & B = <2y3+ 1 76F

ii) A = =2x2+ 3 y 5 1G , B = <7 – 2x 5 5F Matrices
+ 2 3y –

iii) P= =2x3+ y 2 7yG, B = <73 2 7 3F
5 5 2y –
M = RTSSSSSSSS124 x–7+2yVWWWWWWWXW , N = SSSTSSSSRS124 –772XWVWWWWWWW
iv) 3 3
x–y 1
5
5

v) A= <2x5+ y 17F= <57 3x – 2yF
1

Answer

1. Show to your teacher. 2. Show to your teacher.

3. i) 3 ii) 0 iii) 9 3RTSSSSSSSS174 iv) 230–3WWWWWWXWVWiii) v) 3 × 3
4 i) <14 –21F ii) <14 iii) –1 v) 61 –1 –3@
5. i) x = 8, y = 2 –1 0–3F 1, y = 2 x = 2, y = 3
2 5

ii) x =

iv) x = 4, y = 3 v) x = 3, y = 1

PRIME Opt. Maths Book - VIII 53

3.2 Operation on matrices

The simpli�ication of two or more matrices in a single matrix by using
any kind of mathematical operations indicates the operation on matrices.
Which are addition, subtraction, multiplication with scalar, transpose etc.
Here we are discussing some of them in grade VIII.

3.2.1.Addition of matrices
Marks obtained by three students Sita, Pranav and Pranisha in two monthly
test examinations in optional maths are as follows of two months.

Ashadh 1st 2nd Shrawan 1st 2nd

Sita 60 75 Sita 70 75

Pranav 85 90 Pranav 95 80

Pranisha 95 85 Pranisha 80 95

Total marks obtained by them in two tests in two months as,

1st 2nd

Matrices Sita 60 + 70 = 130 75 + 75 = 150

Pranav 85 + 95 = 180 90 + 80 + 170

Pranisha 95 + 80 = 175 85 + 95 = 180

TSSTSSRSSSS689h055is 78i9n550fWWXWWWWVWWor+mSSSSSSSTRSa789t005ion789505cVWWWXWWWWWan=bSSRTSSSSSSe689055ex+++p798r005ess789e550d+++in798550mWWWWWWWWVXa=triSSSRTSSSSSx111738fo500rm111758a000sWWWWWVXWWW,

The sum of any two matrices having same order
is called a new single matrix obtained by adding
the corresponding elements of the matrices. The
single matrix so formed has the order same as the
given matrices.

54 PRIME Opt. Maths Book - VIII

eg. If A = <–21 53F, B = <–34 –21F

Then,
A + B = <–21 53F + <–34 –21F

= =–21+–34 3 + 12G
5 –

= <–55 54F

3.2. 2. Difference of the matrices.

The difference of any two matrices having same Matrices
order is called a new single matrix obtained by
subtracting the corresponding elements of the
matrices. The single matrix so formed has the
order same as the given matrices.

eg. If P = <52 3 12F, Q = <–21 1 53F
–3 1

Then,

P–Q = <52 3 12F – <–21 1 53F
–3 1

= =52 + 1 3–1 1 – 53G
– 2 –3–2 2 –

= <33 2 ––14F
–4

PRIME Opt. Maths Book - VIII 55

3.2.3.Transpose of the matrix.

Let us taking an example where cost of apple of a shop of three days

are given.

1st shop 2nd shop

Sunday 150 140

Monday 160 145

Tuesday 170 150

It can be written in such a way by changing the information of rows

and column as,

Sunday Monday Tuesday

1st Shop 150 160 170

2nd Shop 140 145 150

Matrices SAu=chSSSSTSRSSS111e765x000am111p544l500eWWWWWWWWXVs can be written in matrix form as,
After changing row and columns = <115400 117500F
160
145

It is called transpose of the matrix A and denoted by AT.

The new matrix obtained by interchanging the rows
and columns of a matrix is called transpose of the
matrix.
• Transpose of A is denoted by AT or A` or A.

Note : Order of the transpose matrix is different than matrix A for the

rectangular matrix but same for the square matrix.

eg. If A = <32 4 95F then.
6

AT = SSTSSSSSRS534 629WVWWWXWWWW

56 PRIME Opt. Maths Book - VIII

3.2.4. Multiplication of a matrix with a scalar.

Let us consider the cost of apple given above becomes double in the next
week which can be written as

2(SSSRTSSSSS111o756ld000co111s454t050)WWWWWXWWWV =(SSSSTRSSSSn333e402w000 c322o089st000)WWWWXWWWVW

The new matrix formed by multiplying each element of

a given matrix by a given scalar quantity is called the

multiplication of a matrix with the scalar.

a b ka kb
i.e. If A = < F ; then, KA = < F
c d kc kd

Eg. If A = <13 ––21F, find 3A. Matrices

1 –2 1 –2 3 –6
Solution : A = < F 3A = 3 < F = < F
3 –1 3 –1 9 –3

Worked out Examples

1. If A = <32 –11F and B = <43 –12F find 3A – 2B.

Solution :

21 3 –2
A = < F B = < F
3 –1 4 1

Then,

2 1 3 –2
3A – 2B = 3 < F – 2 < F
3 –1 4 1

6 3 6 –4
= <9 F – < F
–3 8 2

= =69 – 6 3 + 4G
– 8 –3 –2

07
= <1 F
–5

PRIME Opt. Maths Book - VIII 57

2. If A = <–11 32F , B = <32 –12F , prove that (A + B)T = BT + AT.

Solution :

A+B= <1 2F + <3 –2F = <4 0F
–1 3 2 1 1 4

4 0T 4 1
\ L.H.S. = (A + B)T = < F = < F

14 04

3 –2 T 1 2 T 3 2 1 –1 4 1
\ R.H.S. = BT + AT = < F + < F = < F + < F = < F
2 1 –1 3 –2 1 2 3 0 4

\ L.H.S. = R.H.S. proved.
3. If A + B = <43 14F and A – B = <––52 32F, find the matrices A and B.

Solution :

A+B= <4 1F ................ (i)
3 4

Matrices –2 3
A – B = < F ................ (ii)
–5 2

Adding (i) and (ii), we get,

41
A + B = < F
3 4

A–B= <–2 3F
–5 2

24
2A = < F
–2 6

\ 12
A= < F
–1 3

Putting the value of ‘A’ in equation (i)

B = <4 1F – <1 2F = <3 –1F
3 4 –1 3 4 1

1 2 3 –1
\ A = < F & B = < F
–1 3 4 1

58 PRIME Opt. Maths Book - VIII

Exercise 3.2

1. Add the following matrices.

1 2 3 –1 3 –2 27
i) A= < F and B = < 1 F ii) M = < 4 F, N = < F
–1 3 4 –1 6 1

–2 3 2 –3 3 5 –2 –3
iii) P= <4 F , Q = < 1 F iv) A= < F , B = < F
–1 –4 –2 7 5 –3

421 3 –1 2
v) M= < F , N = < F
3 1 5 –2 1 0

2. Subtract the matrices given below.

1 2 3 –1 3 –2 27
i) A= < F and B = < 1 F ii) M = < 4 F, N = < F
–1 3 4 –1 6 1

–2 3 2 –3 3 5 –2 –3
iii) P= <4 F , Q = < 1 F iv) A= < F , B = < F
–1 –4 –2 7 5 –3

421 3 –1 2 Matrices
v) M= < F , N = < F
3 1 5 –2 1 0

3 24 –1 , find the following operations.
3. If A = < F , B = < F
1 –2 2
1

i) 2A + B ii) 3A – B
iii) A + 3B iv) 3A – 2B
v) 4A – 3B

1 –2 3 –6
4. i) If M = < F and N = < 9 F prove that 3M – N is a null matrix.
2 3 6

32 54
ii) If A = < F and B = < F , prove that 2A – B is an identity
–1 –2 –2 –5

matrix.

x 3 =4 + x 2 G, C = 8 5 and A + B = C, find the
iii) If A = < F , B = –1 y–2 < F
2 1 1
1

value of ‘x’ and ‘y’ .

iv) 1 2 F , find AT.
If A = <
3 –2

PRIME Opt. Maths Book - VIII 59

v) 3 21 –3 , find (A + B)T.
If A = < F and B = < F
1 –1 34

5. PRIME more creative questions.

32 15
i) If A = < F and B = < F , prove that (A + B)T = BT + AT.
1 –2 –4 1

3 26 –1 , find the matrix B.
ii) If A = < F and (A + B)T = < F
–1 4 2
1

iii) 4 36 5 , find B and AT.
If A = < F and A + B = < F
21 43

6 5 2 1 , find the matrices A & B.
iv) If A + B = < F , A – B = < F
4 3 0
–1

4 7 2 –3 , find the matrices P and Q.
v) If P + Q = < F & P – Q = < F
–3 –1 5
–3

Matrices 6. Project work
Collect the price of potato and tomato from local shop of two days in

the month of Baishakh and Jestha. Present the price of such month in
matrix form. Also find the difference of the cost using matrix.

60 PRIME Opt. Maths Book - VIII

Answer

4 1 ii) <55 55F iii) <00 00F
1. i) < F
3
4

12 713
iv) <3 F v) <1 2 5F
4

–2 3 1 –9 –4 6
2. i) < F ii) <–7 3 F iii) < 8 –2F
–5 2

58 1 3 –1
iv) <–7 F v) <5 0 5 F
10

10 3 57 15 –1
3. i) < 4 F ii) <1 –7F iii) < 7 1 F
–3

18 0 11
iv) <–1 F v) <–2 –11F
–8

4. i) ii) iii) x = 2, y = 2 Matrices

iv) <12 3F v) <–41 34F
–2

30
5. i) < F ii)
0 –3

iii) B = <22 22F , AT = 4 2
< F
3
1

iv) A = <4 3F , B = <22 22F
2 1

32 15
v) P = < F , Q = < F
1 –2 –4 1

PRIME Opt. Maths Book - VIII 61

Matrix

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 order of a matrix?

2 31 –2 , find the matrix 3A – 2B
2. a. If A = < F and B = < F
–1 4 3
1

2x–y 44 2z , find the value of x and y.
b. If < 3y F = < F
5 6
5

c. If A = SSSSSSSTRS–023 3 –144WWWXVWWWWW , find the value of a23 – a32. Also write down the
–2
–1

Matrices order of the matrix A.

3. a. If aij = 3i – 2j, find the matrix of order 2 × 2. Also find its transpose.
2 3 3 –1
b. If A = < F and B = < F , prove that (A + B)T = BT + AT.
1 –2 21

5 2 –1 4 find the matrices A and B.
4. If A + B = < F and A – B = < F
3 –1 –1
–3

62 PRIME Opt. Maths Book - VIII

Unit 4 Co-ordinate Geometry

Specification Grid Table

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

No. of Questions – 1 1 – 2 6 10
Weight – 2 4 –

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

Objectives : At the end of the lesson
• Students are able to know concept of co-ordinate system.
• Students can plot the points in graph.
• Students can find distance between any two points.
• Students can use section formula and mid-point formula to find the section

point.
• Students can find the centroid of a triangle.

Materials Required:
• Graph board
• Graph paper
• Geo-board
• Sheet of paper
• Scissors
• Chart of sign used in quadrents

PRIME Opt. Maths Book - VIII 63

4.1 Coordinate system Y

As shown in fig. let a person starts B r P(x, y)
moving from ‘O’ to reach the point P. He O
q y
can reach P in a straight way OP = r units x AX

making an angle q with the horizontal

line (OX). The position of the person at

P is denote as (r, q). On the other hand,
the person can move first OA = x units
along OX direction and then AP = y units

along OY direction where OX^OY. Then

the position of P is expressed as (x, y).

Co-ordinate Geometry In particular,
OA = 4 equal parts = 4 units = run
AP = OB = 3 equal parts = 3 units = rise

Y

4

B3 P(4, 3)

2

1

X’ -4 -3 -2 -1 O 1 2 3 4 A X
-1

-2

-3

-4

Y’
Then the point P is written as (4, 3).

The horizontal distance travelled is called run and the
vertical distance travelled is called rise for a point.
Run and Rise may be taken from origin in standard
position.

64 PRIME Opt. Maths Book - VIII

Rectangular Coordinate axes: Y
The horizontal line XOX‘ and 4
vertical line YOY‘ are intersected 3
at point ‘O’ perpendicularly
which are called x- axis and

y-axis respectively and the 2

intersecting point ‘O’ is called 1

the origin point. X’ -4 -3 -2 -1 01 2 3 4 X
Here, -1
-2
• In the line OX, only +ve -3
-4
numbers are counted

in horizontal axis.
• In the line OX‘, only –ve

numbers are counted Y’

in horizontal axis. Co-ordinate Geometry
• In the line OY, only +ve numbers are counted in vertical axis.
• In the line OY‘, only –ve numbers are counted in vertical axis.
• The plane is divided by the lines XOX‘ & YOY‘ into four equal parts

as below.

Plane YOX " 1st quadrant.
Plane YOX‘ " 2nd quadrant.
Plane Y‘OX‘ " 3rd quadrant.
Plane XOY‘ " 4th quadrant.

A point in the quadrant. Y
Let us consider a point
P(3, 4) in first quadrant 4 P(3, 4)

where PM^OX. 3

Here, 2

OM = 3 units 1
= X-coordinate X‘
= abscissa -4 -3 -2 -1 O 1 2 3 4 X
-1 M
PM = 4 units -2

= Y-coordinate -3

= ordinate -4

The two values abscissa Y‘

and ordinate can be written

PRIME Opt. Maths Book - VIII 65

as (3, 4) which is called co-ordinate. It is written in the form of
P(x, y) in general in the co-ordiante system.

Sign in quadrants; Y

(–, +) (+, +)
X’
X
O
(–, –) (+, –)

Co-ordinate Geometry Y’

The coordinate of the points according to sign in different quadrants

can be written as for a point P(3, 4).

(+, +) → 1st quadrant = P(3, 4) Y
(–, +) → 2nd quadrant = P‘(–3, 4)
(–, –) → 3rd quadrant = P‘‘(–3, –4)
(+, –) → 4th quadrant = P‘‘‘(3, –4)

P‘(–3, 4) P(3, 4)

X’ X
O

P‘‘(–3, –4) P‘‘‘(3, –4)

Y’
66 PRIME Opt. Maths Book - VIII

Distance between any two points.

Distance between the points A(x1, y1) and B(x2, y2).

Y B (x2, y2)

d

(x1, y1) C
A

OM NX

Let, ‘d’ be the distance between the points A(x1, y1) and B(x2, y2)
Draw, AM⊥OX, BN⊥OX and AC⊥BN.

Then, OM = x1, AM = y1 Co-ordinate Geometry
ON = x2, BN = y2
AC = MN = ON – OM = x2 – x1
BC = BN – CN = BN – AM = y2 – y1
AB = d

In right angled ∆ABC,
h2 = p2 + b2

or, AB2 = AC2 + BC2
or, d2 = (x2 – x1)2 + (y2 – y1)2

∴ d = (x2 – x1) + (y2 – y1)2
It is the distance formula,

Some important informations:
• Origin point is O(0, 0)
• A point on x – axis is (x, 0)
• A point on y – axis is (0, y)
• To prove equilateral triangle. All sides should be equal.
• To prove right angled triangle.
h2 = p2 + b2
• To prove isosceles triangle.
Any two sides should be equal.

PRIME Opt. Maths Book - VIII 67

Co-ordinate Geometry • To prove equilateral triangle.
All sides should be equal.

• To prove rhombus.
All 4 sides should be equal.

• To prove square.
All 4 sides should be equal and
h2 = p2 + b2 [diagonal is taken as hypotenuse] [or diagonals are equal]

• To prove parallelogram.
Opposite sides should be equal. (Mid points of diagonal must be
same.)

• To prove rectangle.
Opposite sides should be equal and h2 = p2 + b2 for a diagonal h.
(or diagonals are equal)

Worked out Examples

1. Plot the points A(1, 3) and B(4, 7) in graph paper and �ind the distance
between the points A and B by drawing perpendicular lines to the axes
using run & rise.
Solution: The given points A(1, 3) and B(4, 7) in graph as.
Y

B

AC

X’ X
OM N

Y’
Here, AC = MN = ON – OM = 4 – 1 = 3 = run

BC = BN – CN = BN – AM = 7 – 3 = 4 = rise
∴ AB = ^runh2 + ^riseh2

68 PRIME Opt. Maths Book - VIII

= 32 + 42

= 25
= 5 units

2. Find the distance between the points (2, –3) and (5, 1). Co-ordinate Geometry
Solution:
The given points are:
A(2, –3) = (x1, y1)
B(5, 1) = (x2, y2)
Using distance formula,
d = (x2 – x1)2 + (y2 – y1)2
d(AB) = (5 – 2)2 + (1 + 3)2
= 32 + 42
= 5 units.

3. Prove that the line joining the points A(3, 4), B(7, 7) and C(11, 10) are
collinear points.
Solution :
The given points are A(3, 4), B(7, 7) & C(11, 10)
Using distance formula,
d = (x2 – x1)2 + (y2 – y1)2

d(AB) = (7– 3)2 + (7 – 4)2 = 5 units

d(BC) = (11 – 7)2 + (10 – 7)2 = 5 units

d(AC) = (11 – 3)2 + (10 – 4)2 = 10 units
Here,
AC = AB + BC
or, 10 = 5 + 5
∴ 10 = 10
Hence, They are collinear points.

4. Prove that the points A(–2, 3), B(–2, –4), A(–2, 3) D(5, 3)
C(5, –4)
C(5, –4) & D(5, 3) are the vertices of a square.

Solution :

The given points are

A(–2, 3), B(–2, –4), C(5, –4), D(5, 3)

Using distance formula

d = (x2 – x1)2 + (y2 – y1)2 B(–2, –4)

PRIME Opt. Maths Book - VIII 69

d(AB) = (–2 + 2)2 + (–4 –3)2 = 7units

d(BC) = (5 + 2)2 + (–4 + 4)2 = 7units

d(CD) = (5 + 5)2 + (3 + 4)2 = 7units

d(AD) = (5 + 2)2 + (3 –3)2 = 7units

d(AC) = (5 + 2)2 + (–4 –3)2 = 7 2 units
Here, AB = BC = CD = DA
Also, AB2 + BC2 = AC2
or, 72 + 72 = (7 2 )2
or, 98 = 98
Hence, They are the vertices of a square.

Co-ordinate Geometry 5. Find the co-ordiante of a point on Y (5, 4)
x-axis which is 5 units distance O
from a point (5, 4). d = 5units X
Solution : Let, the point on x-axis (x, 0)
be A(x, 0)
The given point is B(5, 4)
Using distance formula,
d2 = (x2 – x1)2 + (y2 – y1)2
or (5)2 = (x – 5)2 + (0 – 4)2
or, 25 = x2 – 10x + 25 + 16
or, x2 – 10x + 16 = 0
or, x2 – (8 + 2)x + 16 = 0
or, x2 – 8x – 2x + 16 = 0
or, x(x – 8) – 2(x – 8) = 0
or, (x – 8) (x – 2) = 0

Either, or

x–8=0 x–2=0

∴x=8 ∴x=2

∴ The required point is (8, 0) or (2, 0)

70 PRIME Opt. Maths Book - VIII

Exercise 4.1

1. Plot the following points in graph and join to each other.
i) (1, 2), (3, 6) and (7, 0) ii) (1, 1), (3, –3) and (–2, –2)
iii) (–1, 5), (–7, –1) and (–3, 1) iv) (2, –3), (1, 7) and (–2, 1)
v) (–1, –1), (3, –5), (–2, –6) and (–5, 2)

2. Plot the following points in the same graph and join each points with
a straight line one after another continuously.

(5, 1), (6, 2), (4, 5), (–4, 5), (–5, 2), (–7, 4), (–6, 0), (–7, –3), (–5, –2),
(–4, 4), (4, –4), (6, –1), (5, 1)

3. Find run and rise of the following pairs of points. Also find the distance

between the points using run and rise.

i) (3, 2) and (6, 6) ii) (1, –2) and (–5, 6) Co-ordinate Geometry

iii) (7, –9) and (3, –6) iv) (10, –1) and (6, 1)

v) (–5, –1) and (1, 3)

4. Find the distance between the following pairs of points.

i) A(2, 1) and B(5, 5) ii) M(3, –2) and N(9, 6)

iii) P(3, –1) and Q(–1, –4) iv) X(–6, –7) and Y(–3, –1)

v) A(a, –b) and B(b, a)



5. Find the distance between the following points.

i) M(a, 0) and N(0, b)

ii) P(p + q, p –q) and Q(p – q, p + q)

iii) A(–1, 1) and B( 3 , 3 )

iv) (–3, –3) and (–3 3 , 3 3 )

v) (3, 3 ) and (0, 2 3 )

6. Solve that following problems.

i) Prove that the points (3, –4), (5, 0) and (–4, –3) are equidistance

from the origin.

ii) Prove that AB = BC where the points are A(–1, 5), B(5, –3) and

C(11, –11).

iii) P(–1, 3), Q(7, –3) and R(4, 1) are the three points. Prove that QR

= 1 PQ.
2

PRIME Opt. Maths Book - VIII 71

Co-ordinate Geometry iv) A(3, 4), B(7, 7) and C(11, 10) are the collinear points, prove that,
B is the mid-point of AC.

v) Prove that P(–2, 0) is equidistance from the points A(2, 3) and
B(–5, 4).

7. Solve the following problems.
i) Prove that the vertices A(4, 8), B(3, –1) and C(–5, –7) are of a

scalene triangle.
ii) Prove that the points P(3, 5), Q(–2, 0) and R(8, 0) are the vertices

of an isosceles triangle.
iii) Prove that the points (–3, –3), (–3 3 , 3 3 ) and (3, 3) are the

vertices of an equilateral triangle.
iv) Prove that the points (–4, 2), (1, 2) and (1, –1) are the vertices of

a right angled triangle.
v) Prove that the points (3, 2), (0, 5) and (–3, 2) are the vertices of a

right angled isosceles triangle.

8. Answer the following questions.
i) Prove that the points (1, –2), (3, 6), (5, 10) and (3, 2) are the

vertices of a parallelogram.
ii) Prove that the points (–2, 3), (–2, –4), (7, –4) and (7, 3) are the

vertices of a rectangle.
iii) Prove that the points (3, 2), (–5, 2), (–5, –6) and (3, –6) are the

vertices of a square.
iv) Prove that the points (–4, 3), (5, 4), (4, –5) and (–5, –6) are the

vertices of a rhombus
v) Prove that the points (7, 9) (3, –7) and (–3, 3) are the vertices of

a isosceles right angled triangle.

9. Solve the following problems.
i) If distance between the points A(2, –3) and B(x, 1) is 5 units, find

the value of ‘x’.
ii) If distance between the points P(3, y) and Q(9, –2) is 10 units,

find the value of ‘y’.
iii) Find the co-ordiante of a point on M(x, 0) which is 13 units distant

from the point N(–3, 12).
iv) Find the co-ordiante of a point on P(0, y) which is 17 units distant

from the point Q(15, –5).

72 PRIME Opt. Maths Book - VIII

v) The distance between the points C(a, 2a) and D(4, 1) is 13 Co-ordinate Geometry
units, find the value of ‘a’.

PRIME more creative questions
10. i) Find the co-ordiante of a point on x-axis. Which is 5 units distance

from the point (1, 3).
ii) Find the co-ordinate of a point on y-axis which is 10 units distant

from the point (8, 4).
iii) Find the co-ordinate of a point on x-axis which is equidistance

from the points (5, 4) and (–2, 3).
iv) Find the co-ordinate of a point on y-axis which is equidistance

from the points (8, –3) and (–6, –5).
v) If a point P(x, y) is equidistance from the points A(1, 2) and

B(3, –2), prove that x – 8y – 8 =0.

11. i) Prove that the points (3, 3), (–3, –3) and (3 3 , –3 3 ) are the
vertices of an equilateral triangle.

ii) Prove that the points (3, 4), (7, 7) and (11, 10) are the collinear
points.

iii) Find the co-ordiante of a point (2a, a) which is 10 units distance
from the point (0, –5)

iv) Find the value of ‘a’ where distance between A(2a, 3) and B(8, a)
is 17 units.

v) If A(3, 4), B(3, –3) and C(6, –3) are the points, prove that
AB^BC.

PRIME Opt. Maths Book - VIII 73

Answer

1. Show to your subject teacher.

2. Show to your subject teacher.

3. i) 3, 4, 5, units ii) –6, 8, 10 units

iii) –4, 3, 5 units iv) –4, 2, 2 5 units

v) 6, 4, 2 13 units

4. i) 5 units ii) 10 units
iii) 5 units iv) 3 5 units
v) 2 (a2 + b2) units

5 i) a2 + b2 units ii) 2q 2 units
iii) 2 2 units iv) 6 2 units
v) 2 3 units
Co-ordinate Geometry
9. i) –1 or 5 ii) 6 or 10
iv) 3 or – 13
iii) 2 or – 8
ii) (0, –2) or (0, 10)
v) 2 or 2 iv) (0, 3)
5

10. i) (–3, 0) or (5, 0)
iii) 2, 0

11. i) (6, 3) or (–10, –5) iii) (6, 3) or (–10, –5)

iv) 2 or 28
5

74 PRIME Opt. Maths Book - VIII

4.2 Section Point

Let us consider a point P cuts a line segment AB at a point where the point

divides the line segment in two sectors AP and BP. Hence the point P is

called the section point of a line segment AB as shown in diagram. The

two sections AP and PB may be equal or

may be in a certain ratio. Here, we discuss B
the process of finding the coordinates of P

the point which divides a line segment in A
two parts (may be equal or unequal) in the

given ratio.

The point which cuts a line segment in equal sections is Co-ordinate Geometry
called mid-point.
The point which cuts a line segment in unequal sections
either externally or internally is called section point.

Centroid of a triangle

The intersecting point of medians of a triangle is called
the centroid of the triangle. It cuts the medians in the
ratio 2:1 from the vertex.

Here, A
In DABC,
P, Q and R are the mid-point of sides of DABC where
AP, BQ and CR are called medians.

The intersecting point of the medians AP, BQ and R Q
G
CR is ‘G’ which is called the centroid of DABC.
P
Here, AG : GP = 2 : 1 B C

BG : GQ = 2 : 1

CG : GR = 2 : 1

PRIME Opt. Maths Book - VIII 75

Co-ordiante of section point of a line segment.

i. The point cuts internally to a line segment joining the points

A(x1, y1) and B(x2, y2). Y
Let, a point C(x, y) cuts the

line joining the points B(x2, y2)
A(x1, y1) and B(x2, y2) C(x, y)m2 S
internally in the ratio m1:m2.

Draw the perpendiculars, A m1
AM^OX, BN^OX, (x1, y1) Q
CP^OX, AQ^CP,
CS^BN, O M P NX

Co-ordinate Geometry Then, OM = x1, AM = y1,
ON = x2, BN = y2, OP = x, CP = y
AQ = MP = OP – OM = x – x1.
CS = PN = ON – OP = x2 – x.
CQ = CP – QP = CP – AM = y – y1.
BS = BN – SN = BN – CP = y2 – y.
AC : CB = m1:m2

In DACQ and DCBS

i) \ Q = \ S → Both being 90°

ii) \ A = \ C → Being corresponding angles.

iii) \ C = \ B → Being remaining angles

\ DACQ ∼ DCBS → By AAA axiom.

\ AQ = CQ = AC
CS BS CB

or, x – x1 = y – y1 = m1
x2 – x y2 – y m2

Taking,

or, x – x1 = m1
x2 – x m2

or, m2x – m2x1 = m1x2 – m1x

or, x(m1 + m2) = m2x1 + m1x2

\ x= m1 x2 + m2 x1
m1 + m2

76 PRIME Opt. Maths Book - VIII

Again taking,

y – y1 = m1
y2 – y m2

or, m2y – m2y1 = m1y2 – m1y
or, y(m1 + m2) = m2y1 + m1y2

\ y= m1 y2 + m2 y1
m1 + m2

\ Co-ordiante of section point is + m2 y1
m1 y2 + m2
C(x, y) = ( m1 x2 + m2 x1 , m1 )
m1 + m2

It is the co-ordinate of internal division point.

ii) Mid-point of a line segment

If point ‘C’ be the mid-point of the line segment AB in the above

diagram, then m1 = m2. The section point so formed is called the mid-
point of the line segment AB, where,
m1 y2 + m2 y1 Co-ordinate Geometry
(x, y) = ( m1 x2 + m2 x1 , m1 + m2 )
m1 + m2

If m1 ==m( 2mfo1 (r2xmm1 +i1dx-p2)o,inmt 2 (y1 + y2) )
or, (x, y) 2m2

\ (x, y) = ( x1 + x2 , y1 + y2 )
2 2

It is the co-ordinate of the mid-point of a line segment.

iii. If the section point cuts a line segment externally.

Let, a point C(x, y) cuts the line Y C(x, y)
joining the points A(x1, y1) and
B(x2, y2) externally in the ratio m1 B(x , y )
m1:m2.
2 2
Draw the perpendiculars,
m2 S

AM^OX, BN^OX and CQ^OX

AR^CQ and BS^CQ A R
(x1, y1) N QX
Then, OM
OM = x1, AM = y1, ON = x2
BN = y2, OQ = x, CQ = y
AR = MQ = OQ – OM = x – x1

PRIME Opt. Maths Book - VIII 77

BS = NQ = OQ – ON = x – x2
CR = CQ – RQ = CQ – AM = y – y1
CS = CQ – SQ = CQ – BN = y – y2
AC : BC = m1:m2

Now, In DARC and DBSC,

\ R = \ S → Both being 90°
Corresponding angles
\ A = \ B → Common angles
By AAA axiom.
\ C = \ C →

\ DARC ≅ DBSC →

\ AR = CR = AC
BS CS BC

or, x –x1 = y –y1 = m1
x – x2 y – y2 m2

Co-ordinate Geometry Taking,

1st and last ratios,

x –x1 = m1
x – x2 m2

or, m2x – m2x1 = m1x – m1x2
or, m1x2 – m2x1 = x(m1 – m2)

\ x= m1 x2 – m2 x1
m1 – m2

Again, taking,

2nd and last ratios

y –y1 = m1
y – y2 m2

or, m2y – m2y2 = m1y – m1y2
or, m1y2 – m2y1 = y(m1 – m2)

\ y= m1 x2 – m2 x1
m1 – m2

\ co-ordinate of section point externally is,

(x, y) = ( m1 x2 – m2 x1 , m1 y2 – m2 y1 )
m1 – m2 m1 – m2

78 PRIME Opt. Maths Book - VIII

iv. Co-ordinates of centroid of a triangle.
Let ‘G’ (x, y) be the centroid of a triangle having vertices A(x1, y1), B(x2,
y2) and C(x3, y3) where AP is a median and ‘G’ cuts the median AP in the
ratio 2:1.
A(x1, y1)

G(x, y)

B(x2, y2) P C(x3, y3)

Now,

using mid-point formula for BC, y2 + y3
x2 + x3 2
the co-ordiantes of P = ( 2 , ) Co-ordinate Geometry

Again, using section formula for AP [G cuts AP in the ratio 2:1.]

(x, y) = ( m1 x2 + m2 x1 , m1 y2 + mm2×22 y(1y2)
G(x, y) = m1 + m2 m1 +
+ + y3
=2×( x2 2 x3 ) + 1 × x1 2 ) + 1 × y1 G

2+1 , 2+1

∴ G(x, y) = ( x1 + x2 + x3 , y1 + y2 + y3 )
3 3

It is the co-ordinates of centroid of ∆ABC.

PRIME Opt. Maths Book - VIII 79

Worked out Examples

1. Find the co-ordinate of a point which divides the line joining the points

(3, 1) and (–2, 6) in the ratio 2: 3 internally.
23

A P(x, y) B

Solution:

The given points are,

A(3, 1) = (x1, y1)
B(–2, 6) = (x2 , y2)
Let, A point P cuts AB in the ratio 2:3 = m1 : m2.

By using section formula, m1 y2 + m2 y1
m1 x2 + m2 x1 m1 + m2
x = m1 + m2 , y =

Co-ordinate Geometry = 2 (–2) + 3 (3) = 2 (6) + 3 (1)
2+3 2+3

= –4 + 9 = 12 + 3
5 5

=1 =3

∴ The required point is (1, 3).

2. Find the co-ordinate of mid-point of a line joining the points P(3, 4)
and Q(5, – 6).

P(3, 4) M(x, y) Q(5, –6)

Solution:

The given points are,

P(3, 4) = (x1, y1)
Q(5, – 6) = (x2 , y2)
Let, A point M(x, y) cuts the line PQ at mid-point.

Using mid-point formula, y1 + y2
x1 + x2 2
x = 2 , y =

= 3+5 = 4 + (–6)
2 2
8 –2
= 4 = 2

= 4 = –1

∴ Mid-point of PQ is M(4, –1).

80 PRIME Opt. Maths Book - VIII

3. Find the co-ordinate of centroid of a triangle having vertices A(1, – 2),

B(3, 5) and C(5, –3). B(3, 5)
Solution:

The given vertices of DABC are,

A(1, –2) = (x1 , y1) RP
B(3, 5) = (x2 , y2) G
C(5, –3) = (x3, y3)
Let, G(x, y) be the centroid of DABC,

A(1, –2) Q C(5, –3)

Using centroid formula, y1 + y2 + y3
3
x = x1 + x2 + x3 y =
3

= 1+3+5 = –2 + 5 – 3
3 3

= 9 = 0
3 3

= 3 = 0 Co-ordinate Geometry

\ The required point is (3, 0).

4. If (2, 3) is the mid-point of line joining the points (a, –2) and (3, 0) find
the value of ‘a’ and ‘b’.

A(a, –2) M(2, 3) B(3, b)

Solution:

Let, M(2, 3) = (x, y) be the mid-point of line joining the points,

A(a, –2) = (x1 , y1)
B(3, b) = (x2 , y2)
Now,

By using mid-point formula, y1 + y2
2
x= x1 + x2 , y=
2

or, 2= a + 3 3 = –2 + b
2 2

or, a + 3 = 4 –2 + b = 6

\ a = 1 b = 8.

PRIME Opt. Maths Book - VIII 81

5. Prove that the vertices A(–1, 3), B(0, D(4, 0) C(5, –5)
B(0, –2)
–2), C(5, –5) and D(4, 0) are the vertices
S(a, b)
a parallelogram. R(7, 5)

Solution:

The given vertices are A(–1, 3) B(0, – 2),

C(5, –5) and D(4, 0). A(–1, 3)
Now,

Using mid-point formula for diagonal AC.
y1 + y2
x = x1 + x2 , y = 2
2

= –1 + 5 = 3–5
2 2

= 4 = –2
2 2

Co-ordinate Geometry = 2 = –1
\ Mid-point of diagonal AC is (2, –1)

Again,

For diagonal BD, y1 + y2
2
x = x1 + x2 , y =
2

= 0 + 4 = –2 + 0
2 2

= 2 = –1

\ Mid-point of diagonal BD is also (2, –1).

Hence, ABCD is a parallelogram.

6. The three vertices of a parallelogram P(3, 1)
are P(3, 1), Q(5, 7) and R(7, 5). Find the

co-ordinate of fourth vertex.

Solution:

The three vertices of a parallelogram

are P(3,1), Q(5, 7) & R(7, 5). Q(5, 7)
Let, fourth vertex be S(a, b).

Now,

82 PRIME Opt. Maths Book - VIII

Using mid-point formula for diagonal PR,
y1 + y2
x = x1 + x2 y = 2
2

= 3+7 = 1+5
2 2

= 10 = 6
2 2

= 5 = 3

Again,

The mid-point (5, 3) is also the mid-point of diagonal QS in a
parallelogram,

So, y1 + y2
2
x= x1 + x2 y =
2

or, 5= 5+a 3 = 7+b
2 2

or, 5 + a = 10 7 + b = 6 Co-ordinate Geometry

or, a = 5 b = –1

\ Fourth vertex is (5, –1).

Exercise 4.2

1. Find the Co-ordinate of a point which cuts the line joining the following
points in the ratio given below.

i) A(1, 1) and B(4, 7); ratio 2:1 internally.
ii) P(3, – 1) and Q(3, 8); ratio 1:2 internally.
iii) M( –1, 5) and N(4, 5); ratio 2:3 internally.
iv) A(2, –3) and B(2, 3); ratio 3:4 externally.
v) P(–3, –2) and Q(–3, 5); ratio 3:4 externally.
vi) M(–2, 3) and N(–4, 5); ratio 3:2 externally.

2. Find the co-ordinate of mid-point of a line joining the points given
below.

i) A(3, 2) and B(5, 0) of a line segment AB.
ii) P(5, 1) and Q(1, 7) of a line segment PQ.
iii) M(–2, 0) and N(–4, 6) of a line segment MN.
iv) A(–5, –1) and B(1, –7) of a line segment AB.
v) A(a, 3b) and B(3a, 7b) of a line segment AB.

PRIME Opt. Maths Book - VIII 83

Co-ordinate Geometry 3. Find the co-ordinate of centroid of a triangle having vertices given
below.

i) A(1, 2), B(3, 6) and C(5, 1).
ii) P(3, 4) , Q(1, 1) and R(2, 1).
iii) K(2, –1) , L(4, –5) and M(3, 3).
iv) A(–3, –2), B(–4, 5) and C(–5, –3).
v) P(–1, 7) , Q(–5, –4) and R(–3, 3).

4. Prove that the following vertices are of a parallelogram.
i) P(1, 2), Q(3, 6), R(7, 8) and S(5, 4)
ii) A(3, 4), B(0,1), C(5, 6) and D(8, 9).
iii) K(3, 1), L(1, –4), M(7, –5) and N(9, 0).
iv) A(–3, 2), B(–5, 8), C(–7, 6) and D(–5, 0).
v) P(–1, –3) , Q(–7, –5), R(–11, –1) and S(–5, 1)

5. PRIME more creative questions:
i) If (3, 2) is the mid-point of line joining the points A(–2, b) and

B(a, 5), find the value of a and b.
ii) If A(m, n), B(2, 3) and C(6, 7), find the value of m and n where

AB = BC.
iii) If (1, 3) is the centroid of a triangle having vertices A(–2, 1),

B(p, 5) and C(3,q),find the value of p and q.
iv) The three vertices of a parallelogram are A(3, 1), B(5, 6) and

C(7, 3). Find the co-ordinate of fourth vertex D.
v) The three vertices of a parallelogram are P(2, 3), Q(1, –7) and

R(4, –11), find the co-ordinate of fourth vertex S.

6. Project work
Find the co-ordinate of the corners of your optional mathematics book

by putting in a sheet of graph paper. Verify it is a rectangle by calculate
distance of edges.

84 PRIME Opt. Maths Book - VIII

1. i. (3, 5) Answer iii. (1, 5)
iv. (2, –21) vi. (–8, 9)
2. i. (4, 1) ii. (3, 2) iii. (–3, 3)
iv. (–2, –4) v. (–3, 16)
3. i. (3, 3) ii. (3, 4) iii. (3, –1)
iv. (–4, 0) v. (2a, 5a)
5. i. a = 8, b = –1 ii. (2, 2) iii. p = 2, q = 3
iv. (5, –2) v. (–3, 2)
ii. m = –2, n = –1
v. (5, –1).

Co-ordinate Geometry

PRIME Opt. Maths Book - VIII 85

Co-ordinate Geometry Co-ordinate Geometry
Unit Test - 1

Time : 30 minutes
[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. Write down the co-ordinate of a point on x-axis and y-axis.
2. a. Find the distance between the points A(3, – 1) and B(7, 2).
b. Find the co-ordinate of mid-point of a line segment joining the
points A (3, –1) and B(–5, 7).
c. Find the section point which cuts a line segment joining the points
A(3, –2) and B(3, –5) in the ratio 1:2.
3. a. Prove that the vertices A(3, 5), B(–1, 2) and C(6, 1) are the vertices
of an isosceles triangle.
b. Prove that the points (–3, 2), (5, 4), (7, – 2)and (1, –4) are the
vertices of a parallelogram.
4. Find the co-ordinate of a point on y-axis which is 10 units distant from
the point (6, –5).

86 PRIME Opt. Maths Book - VIII

Unit 5 Trigonometry

Specification Grid Table

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

No. of Questions 1 1 2 – 4 11 20
Weight 1 2 8 –

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

Objectives : At the end of the lesson
• Students can solve the right angled triangle.
• Students are able to know the types of measurement of angles.
• Students can find the trigonometric ratios.
• Students can prove the trigonometric identities.
• Students can find the value of the ratio of the standard angles.
• Students can find height and distance.

Materials Required:
• Chart paper.
• Graph paper.
• Chart of list of formulae used in trigonometry.
• Chart of values of standard angles.
• Model of right angled triangle.

PRIME Opt. Maths Book - VIII 87

Trigonometry 5. Trigonometry

The word trigonometry is defined using three words; tri → three, gones
→ angles, metron → measurement. Tri + gones + metron = trigonometry
(Measurement of three angles of a triangle.) It is said that the origin of the
trigonometry is taken from the ancient Hindu Civilization as,

Tri – lq
gono – sf]0f
metry – dfk0f

The measurement of three angles of
a triangle is called trigonometry.

It is useful to find the angles of a right angled triangle and to find the
length of the sides of it. It is used by engineers to find the heights and
distance during constraction of structures which is useful to estimate the
constructing materials and cost. It is also useful for various purposes in
physics, mathematics, statistics as well as other scientific purposes.

5.1 Measurement of angles

Here, we discuss the important three ways of measuring an angle viz.
Sexagesimal measurement (British System), Centesimal measurement
(French System) and Circular measurement (Radian measurement).

i. Sexagesimal measurement:
In this system, the right angle is divided into 90 equal parts, called

degrees, one degree is divided into 60 equal parts, called minute and
one minute is divided into again 60 equal parts, called Second.
Thus,

One right angle = 90°
\ 1° = 60’
\ 1’ = 60’’

88 PRIME Opt. Maths Book - VIII

ii. Centesimal measurement:
In this system, one right angle is divided into 100 equal parts, called

grade, one grade is divided into 100 equal parts, called minute and
one minute is divided into 100 equal parts, called Seconds.
Thus,

One right angle = 100g
\ 1g = 100’
\ 1’ = 100’’

iii. Circular measurement:
As shown i$n figure, OA = radius = r
Length of AB = OA = r
Then the angle \AOB is fixed for all circles irrespective of the length

of the radius which is called one radian angle.

We denote it by 1c Trigonometry

i.e. \AOB = 1c B

radian angle (1c)

OA

$
Here, Radius OA = arc length AB
\ \ AOB = radian angle (1c).

1c angle : The central angle formed by an arc of
a circle equal to the radius of the circle is called
one radian angle.

Also we know C = 2pr (perimeter or circumference i.e. in circumference,
there are 2p number of arcs times radius.)

\ Central angle formed by circumference is 2pc.
But, central angle by circumference in degree is 360° or 400g.

PRIME Opt. Maths Book - VIII 89

∴ 2πc = 360° = 400g

πc = 180° = 200g 200 g
180° r
And, 1c = r =

Relationship table between the measurement of angles.

2rt angle = \ 180° = 200g = πc
rc
1 rt angle = 90° = 100g = 2

1° = a 10 g = ` r jc
9 180
k

1g = a 9 kc = ` r jc
10 200

1c = ` 180 jc = ` 200 jg
r r

Worked out Examples

Trigonometry 1. Convert into Seconds of 12°15’36’’.

Solution = (12 × 60 × 60 + 15 × 60 + 36)‘‘
12°15‘36‘‘ = (43200 + 900 + 36)‘‘
= (44136)‘‘

2. Convert into grades of 25 g35’’45’’

Solution = (25 + 35 + 45 )g
25g35‘‘45‘‘ 100 100 ×100

= (25 + 0.35 + 0.0045)g

= (25.3545)g

3. Convert into degrees of ` r jc .
125
Solution
kc
` r jc = a r × 180 [∴ 1c = 180g ]
125 125 r r

= a 180 kc
125

= a 36 kc
25

= 1.44°

90 PRIME Opt. Maths Book - VIII

4. Find the ratio of 40 g and a 3r c . A
10 70°
k 76°

Solution

1st angle = 40g
3r c
2nd angle = a 10 B C
k

= a 3r × 200 g [\ 1c = 200g ]
10 r r
k

= 60g

Ratio = 1st angle
2nd angle
40 g
= 60 g

= 2
3

= 2:3
5. If any two angles of a triangle are 70° and 76°, find the third angle in

grades. A Trigonometry
Solution :

Let, ABC be a triangle, 54°

Where,

\ A = 70° ` 3r c C
\ B = 76° 4
\C = ? j

B

We have,

\ A + \ B + \ C = 180° [\Sum of internal Angles of a D in indegre]

or, 70° + 76° + \ C = 180°

or, \ C = 180° – 146°

∴ \ C = 54°

∴ Third angle = (54 × 10 )g
9
= 60g

6. One angle of a triangle is 54° and the Second A
3r c 2x°
angle is a 8 , find the third angle in grades.
k 3x° 5x°

Solution :

Let, ABC be a triangle where,

\ A = 54° B C

= (54 × 10 )g = 60g
9

PRIME Opt. Maths Book - VIII 91

\ B = a 3r c
8
k

= a 3r × 200 g
8 r
k

= 75g

\ C = ?
We have,
\ A + \ B + \ C = 200g [\Sum of internal Angles of a D in grades]
or, 60g + 75g + C = 200g
or, C = 200g – 135g
∴ C = 65g
∴ The third angle is 65g.

Trigonometry 7. Find the angles of a triangle in grades which are in the ratio 2:3:5.
Solution :
Let, ABC be a triangle and the angles are in the ratio 2:3:5 in grades.
Where,
\ A = 2xg
\ B = 3xg
\ C = 5xg
We have,
\ A + \ B + \ C = 200g
or, 2x + 3x + 5x = 200g
or, x = 20g

Then, the angles are,
\ A = 2 × 20g = 40g
\ B = 3 × 20g = 60g
\ C = 5 × 20 = 100g

92 PRIME Opt. Maths Book - VIII

Exercise 5.1

1. Express the following angles in sexagesimal Seconds.
i) 25° 26‘ 27‘‘ ii) 36° 25‘‘ iii) 45‘ 36‘‘
iv) 75° 50‘ 55‘‘ v) 42° 45‘

2. Express the following angles in centesimal Seconds.
i) 42g 35‘‘ 42‘‘ ii) 75g 65‘ 72‘‘ iii) 36g 38‘‘
iv) 55g 36‘ v) 25‘ 32‘‘

3. Express the following angles in degrees.
i) 22° 36‘ 42‘‘ ii) 72° 42‘ 54‘‘ iii) 50° 24‘‘
iv) 55° 45‘ vi) 65g 75‘ 42‘‘
v) 80g
45‘‘ viii) a 25r kc ix) ` 3r jc
vii) 42g 36‘
x) a 74r5 c

k

4. Convert the following angles in grades. Trigonometry
i) 85g 52‘ 45‘‘ ii) 27g 52‘ 46‘‘ iii) 44g 45‘‘
iv) 47g 27’ 35‘‘ vi) 66° 42‘
v) 72°
36‘‘ viii) a 130r kc ix) ` 4r0 jc
vii) 56° 24‘
x) a 22r5 c

k

5. Convert the following angles into radian.

i) 72° ii) 240° iii) 144°
vi) 350g
iv) 125g v) 400g

6. Answer the following problems:

i) Find the ratio of the angles 48° and 80g.
2r c
ii) Find the sum of the angles a 5 and 28° in degrees.
k

iii) Find the difference of the angles 90g and a 3r c in degree
10
k

measurement.

iv) Find the angles of a triangle in degrees which are in the ratio
1:2:3.

v) Find any two angles in degrees whose sum is 50° and difference
is 14°.

PRIME Opt. Maths Book - VIII 93

7. Find the angles of the triangle from the followings.

i) Find the angles of a triangle which are in the ratio 3:4:5 in degree

measurement.

ii) If two angles of a triangle are in the ratio 3:5 and the third angle
is 60°, find the angles in degrees.
iOIsfnt32ewaoonfaganlegrlioegfshatotafrnaiagtnlregi,alfeninigsdle53thaoreeftah4ir0ri°dghaantndagnl6eg8lien°
iii) and the second angle
iv) degrees.
respectively, find the

third angle in grades.

v) Find the angles of the right angled isosceles triangle in grades.

8. Solve the problems given below.

i) One angle of a triangle is 60g, Second angle is 36°. Find the third

angle in degrees.

ii) Two angles of a triangle are in the ratio 3:5 and the third angle is
40g, find the angles in degrees.
7r c
Trigonometry iii) One angle of a triangle is ` 20 , Second angle is 81°, find the
j

third angle in grades.

iv) iOsnae35arnkgcl,efionfdatthreiatnhgilrediasn29gleofina right angle and the Second angle
degrees.

v) The sum of any two angles of a triangle is 100° and the difference
is 20°, find the angles of the triangle.

9. PRIME more creative question:
i) Convert 70° 45‘ 30‘‘ into centesimal measurement.
ii) Convert 81g 15‘ 75‘‘ into sexagesimal measurement.
iii) Express the angle 82° 24‘ 42‘‘ into circular measurement. (p =

3.1416) kc

iv) The angles of a triangle are a 3x , a 2x g and ` rx jc , find the
5 3 75
k

angle in degrees.
v) The first angle of a triangle in grades, Second angle in degrees
and the third angle in radians are in the ratio 280 : 288 : p, find

the angles in degrees.

94 PRIME Opt. Maths Book - VIII

1. i) 91587” Answer iii) 2736”

iv) 273055” ii) 129625” iii) 360038”
v) 153900”
2. i) 423542” ii) 756572” iii) 50.0067°
v) 2532” vi) 59.17878°
iv) 553600” ii) 72.715° ix) 60°
v) 72°
3. i) 22.61167° viii) 72°

iv) 55.0125°

vii) 38.12805°

x) 28°

4. i) 85.5245g ii) 27.5246g iii) 44.0045g
v) 80g vi) 74.1111g
iv) 47.2735g viii) 60g ix) 5g

vii) 62.6778g

x) 16g
2r 4r c iii) a 45r kc
5. i) a 5 c ii) a 3
k
k

iv) a 58r c v) 2pc vi) ` 7r c Trigonometry
4
k j

6. i) 2:3 ii) 100° iii) 27°
iv) 30°, 60°, 90° v) 32° and 18° iii) 54°, 60°, 66°
7. i) 45°, 60°, 75° ii) 45°, 75°, 60° iii) 40g
iv) 80g v) 100g, 50g, 50g iii) 1.40345c
8. i) 90° ii) 54°, 90°, 36°
iv) 52° v) 40°, 60°, 80°
9. i) 78g62’03” ii) 73°2’30”
iv) 30°, 30°, 120° v) 63°, 72°, 45°

PRIME Opt. Maths Book - VIII 95

Trigonometry 5.2 Trigonometrical ratios

The triangle having an angle 90° is called right angled triangle where other
two angles remains as acute angles. The longest side of right angle triangle
is called hypotenuses and other two sides are taken as perpendicular and
base.

One of the acute angles in a right angled triangle is
considered as angle of reference. The side opposite to
right angle is called hypotenuse, the side opposite to
the angle of reference is called the perpendicular and
the remaining side is called the base. Side opposite
to reference angle is taken as perpendicular and
remaining side is taken as base.

P

Q qR
Here,
In a right angle triangle PQR,
\ Q = 90° (right angle)
\ R = q = reference angle
PR = hypotenuse (h) [opposite to right angle]
PQ = perpendicular (p)[opposite to reference angle]
QR = base (b) [remaining side or side which joins the right angle and the
angle of reference]

Also, h2 = p2 + b2 for the solution of sides.
i.e. PR2 = PQ2 + QR2

Ratio of perpendicular and hypotenuse is taken as Sine (is short Sin).
p PQ

i.e Sinq = h = PR

96 PRIME Opt. Maths Book - VIII