Equation, Inequality and Graph
The present age of the son = (x + 4) years = (4 + 4) years = 8 years.
The present age of the father = (8x + 4) years = (8 × 4 + 4) years = 36 years
So, the present age of the father and the son are 36 years and 8 years respectively.
EXERCISE 12.2
General Section
1. Let’s say and write the value of x as quickly as possible.
a) x + 5 = 9, x = .......................... b) x – 5 = 3, x = ..........................
= ..........................
c) 3x = 12, x = .......................... d) x = 5, x
4 = ..........................
e) 2 = 13, x = .......................... f) x – 1 = 2, x
x 3
2. Let’s say and write the answers as quickly as possible.
a) The sum of x and 5 is 12, then x = ......................
b) The difference of y and 4 is 5, then y = ......................
c) The difference of 15 and x is 5, then x = ......................
d) The product of 7 and x is 21, then x = ......................
e) The quotient of x divided by 4 is 6, then x = ......................
f) The quotient of 18 divided by x is 3, then x = ......................
3. Let x be the unknown number. Let’s say and write the required equation as
quickly as possible. Then, find the value of x.
a) The sum of two numbers is 12 and one of them is 4.
The equation is ............................................................. then x = .........................
b) The difference of two numbers is 3 and the greater one is 9.
The equation is ............................................................. then x = .........................
c) The difference of two numbers is 6 and the smaller one is 10.
The equation is ............................................................. then x = .........................
d) The product of two number is 30 and one of them is 5.
The equation is ............................................................. then x = .........................
e) The quotient of dividing a number by 3 is 7.
The equation is ............................................................. then x = .........................
f) The quotient of dividing 24 by a number is 6.
The equation is ............................................................. then x = .........................
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 149 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
Creative Section - A b) 7y + 3 = 17 c) 5a – 1 = a + 11
4. Solve: a
2
a) 5x – 2 = 23
d) 4x – 7 = 2x – 5 e) 3(x – 2) = 2 (x – 1) f) a + =3
g) x – x =1 h) p – 1 – p – 2 = 1 i) 2(x – 1) – 3(x – 2)= 1
3 4 2 3 3 4
y + 1 3 x + 2 1 2 3
j) y – 1 = 4 k) x + 3 = 2 l) x = x + 1
m) 1 + x 1 1 = x 2 n) y 2 1+ y 3 1 = 5 o) y – 1 = 2y – 3
x + –1 – + y y + 1 2y + 3
5. a) The sum of two numbers is 45. If one of them is 27, find the other number.
b) The difference of two numbers is 6. If the smaller one is 15, find the greater
number.
c) The difference of two numbers is 10. If the greater one is 32, find the smaller
number.
6. a) The sum of three consecutive odd numbers is 51. Find the numbers.
b) The sum of three consecutive even numbers is 78. Find the numbers.
7. a) If the sum of two numbers is 37 and the greater number exceeds the smaller by
5, find the numbers.
b) If the sum of two numbers is 56 and the smaller one is 8 less than the bigger one,
find the numbers.
c) If the sum of two numbers is 44 and their difference is 6, find the numbers.
d) A sum of Rs 135 is divided into two parts. If the greater part exceeds the
smaller by Rs 25, find the parts of the sum.
e) A sum of Rs 190 is divided into two parts. If the smaller part is Rs 5 less than half
of the greater part, find the parts of the sum.
f) A sum of Rs 140 is divided into two parts. If the greater part is Rs 20 more
than the double of smaller part, find the parts of the sum.
g) A sum of Rs 105 is divided into two parts. If two times the greater part is 15 less
than three times the smaller part, find the parts of the sum.
Creative Section - B
hAemdaenposspietsntdhse51repmaratinoifnhgissuinmcoomf Reso8n4h0i0s i13ncpoamret .on
8. a) life insurance and food. If
in a bank, find his
1
b) Mrs. Chaudhary had some money. She spent 6 part on her children’s education
1
and 4 part to run her family. If she deposited the remaining sum of Rs 9275 in a
bank, how much money did she have ?
9. a) Divide Rs 4800 in the ratio of 2 : 3.
b) If the angles of a triangle are in the ration of 3 : 4 : 5, find them.
c) If the angles of a quadrilateral are in the ratio of 1 : 2 : 3 : 4, find them.
10. a) The length of a rectangular ground is 18 m longer than its breadth and its
perimeter is 124 m. Find its length and breadth.
Vedanta Excel in Mathematics - Book 8 150 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
b) A rectangular field is three times longer than its breadth and its perimeter is
200 m. Find its length and breadth.
c) The length and breadth of rectangular hall are in the ratio 6 : 5. If its perimeter
is 66 m , find its length and breadth.
11. a) Let the C.P be Rs x. If a man gains 20% profit by selling an article for Rs 816, find
C.P.
b)Let the C.P. be Rs x. If a shopkeeper makes 12% loss by selling a watch for Rs 484,
find C.P.
c) Let the M.P. be Rs x. If you buy a radio for Rs 1350 at 25% discount, find M.P.
12. a) Ram is 9 years older than Hari. Three years ago, he was two times as old as Hari
was. Find their present age.
b) Father is 30 years older than his son. After five years, he will be three times as
old as his son will be. Find their present age.
c) A mother is three times as old as her daughter. Five years ago, she was five times
as old as her daughter was. Find their present age.
d) Seven years ago, father was seven times as old as his son was. Three years
hence, he will be three times as old as his son. Find their present age.
e) Two years hence a boy will be two times as old as a girl. Before four years, he was
three times as old as the girl was. Find their present age.
It’s your time - Project work
13. a) Let’s make linear equations in your own choice and solve each of them to get the
following value of variables: (i) x = 2 (ii) x = 5 (iii) a = –1 (iv) y = –3
b)Let’s write the total number of students and the number of girls in your class. Let
the number of boys be x. Then, make an equation and find x.
c) Let’s write the total number of students and the number of boys in your class. Let
the number of girls be y. Then make an equation to find y.
12.7 Linear equation with two variables
Let’s consider an equation x + y = 7
This equation contains two variables x and y. Each variable has the power 1. Such equa-
tions are called the first degree equations and they are known as linear equations with
two variables.
Y
12.8 Graph of linear equation (–2,7) (0,5)
Consider a linear equation x + y = 5. X (1,4)
It can be written as y = 5 – x. It gives us the rule for + Y (3,2)
taking values of x and using them to calculate values 5
for y. For example, =
X' O X
When x = 0, y = 5 – 0 = 5, When x = 1, y = 5 – 1 = 4
When x = 3, y = 5 – 3 = 2, When x = – 2, y = 5 – (– 2) = 7 Y'
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 151 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
Now, the results can be shown in a table given below.
x 0 1 3 –2
y=5–x 5 4 2 7
The values of x and y can be written in the form of coordinates, such as (x, y). Here, the
values of (x, y) are (0, 5), (1, 4), (3, 2), and (– 2, 7). If these coordinates are plotted on
graph and the points are joined, a straight line is obtained.
Thus, a linear equation always gives a straight line.
12.9 Simultaneous equations
Let’s consider a linear equation, y = x + 3
It is a linear equation on two variables x and y.
The equation has as many pairs of solutions as we wish to find.
The table given below shows a few pairs of solutions of the equation y = x + 3.
x 0 1 3 6 –1 –3 –5
y 3 4 6 9 2 0 –2
Thus, (0, 3), (1, 4), (3, 6), (6, 9), (– 1, 2), (– 3, 0), (– 5, – 2), ... are a few pairs of solutions
that satisfy the equation y = x + 3.
Again, let’s consider another linear equation, y = 4x – 6.
A few pairs of solutions of this equation are shown in the table below.
x012 3 – 1 – 2 –3
y –6 –2 2 6 – 10 – 14 – 18
Thus, (0, – 6), (1, – 2), (2, 2), (3, 6), (– 1, – 10), (– 2, – 14), (– 3, – 18)... are a few pairs of
solutions that satisfy the equation y = 4x – 6.
If you observe the pairs of solutions of both equations, you will find that a pair (3, 6)
is common to both equations. Here. (3, 6) satisfies both equations and it is the solution
of both equations. Such pair of equations that have only one pair of solution are called
simultaneous equations.
12.10 Methods of solving simultaneous equations
There are various methods to solve simultaneous equations. Here, we discuss about
three methods:
(i) Graphical method
In this method, we find a few pairs of solutions of each of two given equations in
two separate tables. The pairs of solutions of each equation are plotted in a graph
and by joining them two separate straight lines are obtained. The coordinates of the
point of intersection of two straight lines are the solutions of the given equations.
Vedanta Excel in Mathematics - Book 8 152 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
Worked-out examples
Example 1: From the given graph, find the solutions of the two equations
represented by the straight lines. Y
l2
Solution:
In the graph, the coordinates of the point of intersection of two (2,1)
X
ostfrtahigehetqluinaetisoln1,s arnepdrle2saernete(2d, 1). So, (2, 1) is the common solution
by l1 and l2.
X' O
l2
? x = 2 and y = 1.
Example 2: Solve graphically x + y = – 1 and 2x – y = – 11. Y'
Solution: (-4,3)
Here, x 0 2 –2 Y
x + y = –1
or, y = –1 – x y –1 –3 1 2 – = –11
The points (0, – 1), (2, – 3) and (– 2, 1) are plotted on the = –1
+
graph and they are joined to get a straight line. X' OX
Again, Y'
2x – y = – 11 x –3 –4 –5
or, 2x + 11 = y y531
or, y = 2x + 11
The points (– 3, 5), (– 4, 3) and (– 5, 1) are plotted on the
graph and they are joined to get another straight line.
In the graph, the coordinates of the point of intersection of two straight lines represented
by the equations is (– 4, 3).
? (– 4, 3) is the common solution of the equations.
So, x = – 4 and y = 3.
EXERCISE 12.3
General Section - Classwork
1. Let’s say and write the solutions to each pair of simultaneous equations represented
by straight line graphs
a) Y b) Y c) Y
X' O X X' O X X' O X
Y' Y' Y'
x = ........, y = .......... x = ........, y = .......... x = ........, y = ..........
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 153 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph e) Y f) Y
d) Y
X' O X X' O X X' O X
Y' Y' Y'
x = ........, y = .......... x = ........, y = .......... x = ........, y = ..........
2. Let’s say and write the values of x and y, and complete the tables.
a) y = x + 1 b) y = x – 3
x 1 –1 x0 5
y4
0 y 04
c) y = 5 – x d) y = 4 + x
x15 x –3 1
y3
y0 2 0
Creative section
3. Copy and complete the table of values for x and y. Plot the coordinates separately
from each table. Find the solutions of each pair of simultaneous equations.
a) y = 7 – x 4 y=x–1 7
x24
x02
y y
b) y = 8 – x 5 1 y=x+2 5 –6
x0 2 3 x 7
y 2 6
6 y3
c) y = 2 – x
x y=x–6 –3
y –1 x1
d) y = 10 – x y
x
y9 y = 2x – 11 2
x1
y
4. Solve the following simultaneous equations graphically:
a) y = 4 – x and y = x – 2 b) y = x + 3 and y = 7 – x
c) y = 9 – x and y = x + 1 d) y = x – 4 and y = 10 – x
Vedanta Excel in Mathematics - Book 8 154 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
e) y = 2x + 2 and y = x – 1 Equation, Inequality and Graph
g) y = x + 4 and y = 2x – 1 f) y = x + 3 and y = 3x – 1
i) x + y = 5 and x – y = 1 h) y = 2x – 4 and y = x – 3
k) 2x – y = 5 and x – y = 1 j) x + y = 8 and x – y = 2
m) 2x – y = 0 and 3x + 2y = – 14 l) 3x + y = 7 and x = 2y
n) 4x + 5y = 9 and 3x = 18
It’s your time - Project work
5. Let’s make 5 pairs of simultaneous equations in your own choice and solve them
graphically to get the following values of variables.
(i) x = 2 (ii) x = 3 (iii) x = 1 (iv) x = –1 (v) x = 4
y=1 y=2 y=3 y = –2 y = –3
(ii) Elimination method
In elimination method, we add or subtract the given equations to eliminate one of
the variables by making their coefficient equal. Then, we solve the equation with one
variable and its solution is substitute in any simultaneous equation to get the value of
another variable.
Worked-out examples
Example 1: Solve x + y = 7 and x – y = 1.
Solution:
x + y = 7 .................... (i)
x – y = 1 .................... (ii)
Adding equations (i) and (ii) to eliminate y,
x+y=7
x–y =1 The coefficients of y are the same in both
equations and they are of opposite signs.
2x = 8 So, one equation is added to the other.
or, x = 8 =4
2
Substituting the value of x in equation (i), we get,
4+y =7
or, y = 7 – 4 = 3
? x = 4 and y = 3
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 155 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
Example 2: Solve 2x + y = 5 and x – 2y = 5.
Solution:
2x + y = 5 .................... (i)
................... (ii)
x – 2y = 5
Multiplying equation (i) by 2 and adding equations (i) and (ii), we get,
4x + 2y = 10 To eliminate y, the coefficients of y in
both equations should be the same. So,
x – 2y = 5 equation (i) is multiplied by 2 to make
the same coefficients of y.
5x = 15
15
or, x = 5 = 3
Substituting the value of x in equation (i), we get,
2u3+y =5
or, 6 + y = 5
or, y = 5 – 6 = – 1
? x = 3 and y = – 1.
(iii) Substitution method
In substitution method, a variable is expressed in terms of another variable from one
equation and it is substituted in second equation. Then, the second equation containing
one variable is solved and the value of the variable is substituted in any one equation to
find the value of another variable.
Example 3: Solve x + y = 4 and 2x + 3y = 6.
Solution:
x + y = 4 .................... (i)
2x + 3y = 6 .................... (ii)
From equation (i),
y =4–x .................... (iii) y is expressed in terms of x.
Substituting for y in equation (ii), we get,
2x + 3 (4 – x) = 6
or, 2x + 12 – 3x = 6
or, – x = 6 – 12
or, – x = – 6
or, x = 6
Now, substituting the value of x in equation (iii), we get,
y=4–6 =–2
? x = 6 and y = – 2.
Vedanta Excel in Mathematics - Book 8 156 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
12.11 Application of simultaneous equations
We can use simultaneous equations to find unknown quantities asked in a verbal
problem. From the two unknown quantities which are asked to find in a word problem,
we make a pair of simultaneous equations under the two given conditions. By solving
the equations, we find the unknown quantities.
Worked-out examples
Example 1: The sum of two numbers is 27 and their difference is 3. Find the
Solution: numbers.
Let the greater number be x and the smaller number be y.
From the first given condition,
x + y = 27
or, y = 27 – x .................... (i)
From the second given condition,
x–y =3 .................... (ii)
Substituting the value of y from equation (i) in equation (ii), we get,
x – (27 – x) = 3
or, x – 27 + x = 3
or, 2x = 3 + 27
or, 2x = 30
or, x = 30 =15
2
Substituting the value of x in equation (i), we get,
y = 27 – 15 = 12
Hence, the required numbers are 15 and 12.
Example 2: The difference of two numbers is 4. If six times the smaller number is
equal to five times the bigger one, find the numbers.
Solution:
Let the smaller number be x and the bigger one be y.
From the first condition, When the bigger number is y, the smaller
y–x =4 number is x and the difference is 4, then
y–x=4
or, y = x + 4 .................... (i)
From the second condition,
6x = 5y .................... (ii) 6 times the smaller number is 6x
5 times the bigger number is 5y
Substituting the value of y from
equation (i) in equation (ii), we get,
6x = 5 (x + 4)
or, 6x = 5x + 20
or, 6x – 5x = 20 I can mentally check the
or, x = 20 answers !
Substituting the value of x in equation (i), we get, 24 – 20 = 4
y = 20 + 4 = 24 6 × 20 = 120 and 5 × 24 = 120
So, the required numbers are 20 and 24.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 157 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
Example 3: The total cost of 3 exercise books and 4 pens is Rs 230. If the cost of
2 exercise books is same as the cost of 5 pens, find the cost of 1
exercise book and 1 pen.
Solution:
Let the cost of 1 exercise book be Rs x and that of 1 pen is Rs y .
From the first condition, The cost of 1 exercise books = Rs x
3x + 4y = 230 The cost of 3 exercise books = Rs 3x
or, 4y = 230 – 3x The cost of 1 pen = Rs y
or, y = 230 – 3x .................... (i) The cost of 4 pens = Rs 4y
4
From the second condition,
2x = 5y .................... (ii)
Substituting the value of y from equation (i) in equation (ii), we get,
2x = 5(230 – 3x)
4
or, 8x = 1150 – 15x
or, 8x + 15x = 1150
or, 23x = 1150
1150
or, x = 25 = 50
Substituting the value of x in equation (ii), we get,
2 × 50 = 5y
or, 5y = 100
or, y = 100 = 20
5
Hence, the cost of 1 exercise book is Rs 50 and that of 1 pen is Rs 20.
Example 4: The sum of the age of a father and his son is 42 years. Six years ago,
the father was 5 times as old as his son was. Find their present age.
Solution:
Let, the present age of the father be x years and that of the son is y years.
From the first condition,
x + y = 42
or, y = 42 – x .................... (i) The present age of the father is x years.
6 years ago, his age was (x – 6) years.
From the second condition, The present age of the son is y years.
x – 6 = 5 (y – 6) 6 years ago, age of the son was (y – 6)
years.
or, x – 6 = 5y – 30
or, x – 5y = – 30 + 6
or, x – 5y = – 24 .................... (ii)
Substituting the value of y from equation (i) in equation (ii), we get,
x – 5 (42 – x) = – 24
or, x – 210 + 5x = – 24
Vedanta Excel in Mathematics - Book 8 158 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
or, 6x = 210 – 24
or, 6x = 186
or, x = 186 =31
6
Substituting the value of x in equation (i), we get,
y = 42 – 31 = 11
So, the present age of the father is 31 years and that of the son is 11 years.
Example 5: The length of a rectangular ground is two times its breadth. If the
perimeter of the ground is 96 m, find the length and breadth of the
ground.
Solution:
Let, the length of the ground be x m and the breadth be y m.
From the first condition, From the second condition,
length = 2 × breadth Perimeter = 96 m
x = 2y .................... (i) 2 (l + b) = 96
96
(x + y) = 2
x + y = 48 .................... (ii)
Substituting the value of x from equation (i) in equation (ii), we get,
2y + y = 48
or, 3y = 48
or, y = 48 =16
3
Substituting the value of y in equation (i), we get,
x = 2 × 16 = 32
Hence, the length of the ground is 32 m and the breadth is 16 m.
EXERCISE 12.4
General Section - Classwork
1. Let’s say and write the values of the variables as quickly as possible.
a) In x + y = 5, if x = 3, then y = .......... b) In x – y = 4, if x = 1, then y = .........
c) In 2x + y = 9, if y = 1, then x = ....... d) In 3x – 2y = 4, if y = 1, then x = .....
e) If (x1, 2) satisfies the equation x + y = 6, then x1 = ..............
f) If (3, y1) satisfies the equation x – y = –2, then y1 = ..............
2. Let’s say and tick the correct pair of solutions of the given pair of equations.
a) x + y = 4 and x – y = 2 (i) (2, 2) (ii) (1, 3) (iii) (3, 1)
b) x + 2y = 5 and 2x + y = 4 (i) (1, 2) (ii) (2, 1) (iii) (2, 2)
c) 2x – y = 6, and x + y = 6 (i) (3, 0) (ii) (4, 2) (iii) (2, 4)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 159 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
3. Let’s say and write ‘True’ or ‘False’ whether the given ordered pair is a solution of
the given pair of equations.
a) x + y = 7, x – y = 3 o (5, 2) .......... b) x + y = 10, x – y = 4 o (6, 4) ..........
c) y = x – 2, y = 2x + 2 o (–4, 6) .......... d) y = 2x + 1, y = x – 1 o (–2, –3) ..........
Creative Section - A
4. Solve the following simultaneous equations by elimination method.
a) x + y = 3 and x – y = 1 b) x + y = 5 and x – y = 1
c) x + y = 5 and x – y = 7 d) x + y = 6 and x – y = 4
e) x + y = 8 and x – y = 10 f) x + y = 9 and x – y = 3
g) x + y = 7 and 2x + y = 10 h) x – y = – 3 and y – 2x = 1
i) 2x – 3y = 1 and x + 2y = 18 j) 3x + 4y = 16 and x + 3y = 7
5. Solve the following simultaneous equations by substitution method.
a) y = x – 2 and x + y = 4 b) y = x – 2 and x + y = 6
c) y = x + 2 and x + y = 8 d) y = x + 6 and x + y = 14
e) x – y = 5 and y = 11 – x f) 2x – y = 3 and y = 9 – x
g) x + 2y = 19 and x + y = 13 h) 2x – y = 2 and x – y = – 3
i) 3x + 2y = 5 and x – y = – 10 j) 2x + 5y = 29 and x + y = 7
6. a) The sum of two numbers is 18 and their difference is 4. Find the numbers.
b) The sum of two numbers is 39. If the greater number is 5 more than the smaller
one, find the numbers.
c) The sum of two numbers is 54. If the smaller number is 6 less than the greater
one, find the numbers.
d) The difference of two numbers is 6. If three times the smaller number is equal
to two times the greater one, find the numbers.
Creative Section - B
7. a) The total number of students in a class is 36. If the number of girls is 6 more
than the number of boys, find the number of boys and girls.
b) The total cost of a box and a pen is Rs 100. If the cost of a box is same as the cost
8 . a) of 3 pens, find the cost of a box and a pen.
The sum of the age of a father and his son is 40 years. If the father is 28 years
older than his son, find their age.
b) The sum of the age of two sisters is 20 years. If one of them is 4 years younger
than other, find their age.
c) Father is three times as old as his son. If the difference of their age is 24 years,
find their age.
d) The sum of the age of a father and his son is 48 years. Four years ago, the father
was 9 times as old as his son was. Find their present age.
Vedanta Excel in Mathematics - Book 8 160 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
e) A father is 30 years older than his son. After six years he will be 3 times as old
as his son will be. Find their present age.
f) If twice the son’s age in years is added to his father’s age, the sum is 70. But if
twice the father’s age is added to the son’s age, the sum is 95. Find the present
age of the father and the son.
9. a) The length of a rectangular garden is two times its breadth. If the perimeter of
the garden is 72 m, find the length and breadth of the garden.
b) The perimeter of a rectangular field is 120 m. If the field is 10 m longer than its
breadth, find the length and breadth of the field.
It’s your time - Project work
10. a) Make the pair of simultaneous equations and solve them to get the following
values of variables:
(i) x = 2 (ii) x = 5 (iii) x = –3 (iv) x = –1
y=3 y = –1 y=6 y = –2
b) Let’s find the unit cost of each of the following pairs of items in your local
market. Consider the unit cost of one item as Rs x and another item as Rs y.
Then, make simultaneous equations with different number of items in each
case. Solve the equations and find the values of x and y.
(i) 1 pencil and 1 eraser (ii) 1 pen and 1 exercise book
(iii) 1 kg of potatoes and 1 kg of tomatoes (iv) 1 kg of sugar and 1 kg of rice
12.12 Quadratic equation – Introduction
Let’s consider an equation x + 5 = 0.
Here, the highest power of the variable x is 1. So, it is called a first degree equation of
one variable. A first degree equation is also called a linear equation.
On the other hand, let’s consider another equation x2 + 7x +12 = 0.
In this equation, the highest power of the variable is 2. So, it is called a second degree
equation. A second degree equation of one variable is called a quadratic equation.
The standard form of a quadratic equation is ax2 + bx + c = 0, where a, b, c R and
a ≠ 0.
12.13 Solution of quadratic equations
A quadratic equation is a second degree equation. Therefore, we obtain two solutions
(or roots) of the variable from a given quadratic equation. There are many methods of
solving quadratic equations. We discuss only about the factorisation method here.
12.14 Solving quadratic equations by factorisation method
In this method the quadratic equation of the form ax2 + bx + c = 0 is factorised and
expressed as the product of two linear factors. Then, each linear factor is separately
solved to get the required solutions of the equation.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 161 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
Worked-out examples
Example 1: Solve (x – 2) (x + 3) = 0.
Solution:
Here, (x – 2) (x + 3) = 0 I understood !
If a × b = 0, either a should
Either x – 2 = 0 .................... (i) be zero or b should be zero.
Or, x + 3 = 0 .................... (ii)
Solving equation (i),
x–2 =0
or, x = 2
Solving equation (ii),
x+3 =0
or, x = – 3
Hence, x = 2 and – 3.
Example 2: Solve x2 – 16 = 0
Solution: I’ve remembered !
a2 – b2 = (a + b) (a – b)
Here, x2 – 16 = 0 So, x2 – 42 = (x + 4) (x – 4)
or, x2 – 42 = 0 Alternative process
x2 – 16 = 0
or, (x + 4) (x – 4) = 0 or, x2 = 16
or, x2 = (r 4)2
Either, x + 4 = 0 .................... (i) removing square from both sides
of the equation, we get
or, x – 4 = 0 .................... (ii) or, x = r 4
Solving equation (i),
x+4 =0
or, x = – 4
Solving equation (ii),
x–4 =0
or, x = 4
Hence, x = – 4 and 3 or r 4.
Example 3: Solve x2 + x – 6 = 0
Solution:
Here, x2 + x – 6 = 0
or, x2 + (3 – 2) x – 6 = 0
or, x2 + 3x – 2x – 6 = 0
or, x (x + 3) – 2 (x + 3) = 0
Vedanta Excel in Mathematics - Book 8 162 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
or, (x + 3) (x – 2) = 0 .................... (i) Checking the answer
Either, x+3 =0 x2 + x – 6= 0, when x = –3,
(–3)2 + (–3) – 6 = 0
or, x – 2 = 0 .................... (ii)
9–3–6 =0
Solving equation (i), Solving equation (ii), 0=0
x+3 =0 x–2 =0 When x = 2
22 + 2 – 6 = 0
or, x = – 3 or, x = 2 4+2–6 =0
Hence, x = – 3 or 2. 0 =0
Hence, –3 and 2 are correct
solutions.
Example 4: Solve 2x2 – 9x – 18 = 0
Solution: Checking the answer
Here, 2x2 – 9x – 18 = 0 When x = 6
or, 2x2 – (12 – 3)x – 18 = 0 2 × 62 – 9 × 6 – 18=0
or, 72 – 54 – 18 = 0
or, 2x2 – 12x + 3x – 18 = 0 or, 18 – 18 = 0
or, 2x (x – 6) + 3 (x – 6) = 0 or, 0 = 0
or, (x – 6) (2x + 3) = 0 When x = – 3
2
Either, (x – 6) = 0 .................... (i) 3 3
2 × (– 2 )2 – 9 × (– 2 ) – 18 = 0
or, (2x + 3) = 0 .................... (ii)
Solving equation (i), Solving equation (ii), or, 9 + 27 – 18 = 0
2 2
x–6=0 2x + 3 = 0 or, 18 – 18 = 0
or, x = 6
or, 2x = – 3 0=0
3
Hence, x = 6 and – 3 or, x =– 2 Hence, 6 and – 3 are correct
2 solutions. 2
Example 5: If the sum of two numbers is 9 and the product is 20, find the numbers.
Solution:
Let, one of the numbers be x.
Then, the other number is (9 – x) The sum of two numbers is 9 and one of the
The product of these two numbers is 20. numbers is x. So, other number will be 9 – x.
? x (9 – x) = 20
or, 9x – x2 = 20
or, x2 – 9x + 20 = 0
or, x2 – (5 + 4)x + 20 = 0
or, x2 – 5x – 4x + 20 = 0
or, x (x – 5) – 4 (x – 5) = 0
or, (x – 5) (x – 4) = 0
Either, x – 5 = 0 .................... (i)
or, x – 4 = 0 .................... (ii)
Solving equation (i), Solving equation (ii), Checking the answer
Sum of 5 and 4 = 5 + 4 = 9
x–5 =0 x–4 =0
Product of 5 and 4= 5 × 4 = 20
or, x = 5 or, x = 4
Hence, the required numbers are 5 and 4.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 163 Vedanta Excel in Mathematics - Book 8
Equation, Inequality and Graph
EXERCISE 12.5
General Section - Classwork
1. Let’s say and write the values of variables as quickly as possible.
a) x2 = 1, x = .............. b) y2 = 4, y = ................ c) p2 = 9, p = ...............
d) a2 = 16, a = ........... e) x2 = 25, x = ............. f) y2 = 81, y = .............
g) x2 = 5, x = ........... h) m2 = 6, m = ............... i) x2 = 7, x = ...............
2. Let’s say and write the values of variables as quickly as possible.
a) (x + 2) (x – 2) = 0, x = ................ b) (y – 1) (y – 3) = 0, y = ........., .........
c) (a + 2) (a – 3) = 0, a = .........., ......... d) (x + 4) (x + 7) = 0, x = ........., ..........
Creative Section - A
3. Let’s solve the following equations:
a) (x – 2) (x – 3) = 0 b) (x – 1) (x – 2) = 0 c) (x + 3) (x – 4) = 0
d) (x – 4) (x + 5) = 0 e) (x – 9) (x – 3) = 0 f) (x + 2) (x + 11) = 0
g) x (x + 3) = 0 h) x (x – 8) = 0 i) 2x (x – 7) = 0
4. Let’s solve the following equations:
a) x2 – 1 = 0 b) x2 – 4= 0 c) x2 – 9 = 0 d) x2 – 16 = 0
e) x2 – 25 = 0 f) x2 – 36 = 0 g) x2 – 49 = 0 h) x2 – 64 = 0
i) x2 – 81 = 0 j) x2 – 100 = 0 k) 4x2 – 9 = 0 l) 16x2 – 25 = 0
m) 49 – 36x2 = 0 n) 100 – 81x2 = 0 o) 9x2 – 64 = 0 p) 64x2 – 121 = 0
5. Let’s solve the following equations:
a) x2 – 3x + 2 = 0 b) x2 – 5x + 6 = 0 c) x2 – 5x + 4 = 0
f) x2 – 4x + 4 = 0
d) x2 – 7x + 12 = 0 e) x2 – 12x + 36 = 0 i) x2 + 7x + 12 = 0
l) x2 + 14x + 45 = 0
g) x2 – 9x + 18 = 0 h) x2 + 5x + 6 = 0 o) x2 – 3x = 10
j) x2 + 9x + 14 = 0 k) x2 + 13x + 42 = 0
m) x2 – 9x – 70 = 0 n) x2 – 10x – 39 = 0
p) x2 + 6x = 40 q) x2 + 10x = 75 r) x2 + 3x = 28
6. Let’s solve the following equations:
a) 2x2 – x = 0 b) 3x2 – x = 0 c) 4x2 – x = 0
d) 2x2 – 3x = 0 e) 3x2 – 4x = 0 f) 4x2 – 5x = 0
g) 2x2 + 3x – 1 = 0 h) 2x2 – 3x + 1 = 0 i) 2x2 – x – 21 = 0
j) 2x2 – x – 6 = 0 k) 3x2 – 2x – 8 = 0 l) 3x2 – 10x + 8 = 0
m) 3x2 – 5x – 2= 0 n) 3x2 + 10x + 3 = 0 o) 4x2 + 12x + 9 = 0
p) 5x2 – 32x + 12 = 0 q) 6x2 + 17x + 12 = 0 r) 6x2 – x – 2 = 0
Vedanta Excel in Mathematics - Book 8 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Equation, Inequality and Graph
Creative Section - B
7. a) The sum of two numbers is 12 and their product is 32. Find the numbers.
b) The sum of two numbers is 15 and their product is 54. Find the numbers.
c) The difference of two numbers is 4 and their product is 60. Find the numbers.
d) The difference of two numbers is 5 and their product is 84. Find the numbers.
e) In a two digit number, the product of the digits is 18 and their sum is 9. Find the
number.
f) The sum of digits of two digit number is 7 and their product is 12. Find the
number.
g) A number exceeds than another number by 3 and their product is 40. Find the
numbers.
h) Divide 12 into two parts such that the product of the two parts is 35.
i) If four times a number is added to the square of the number, the result is 21.
Find the number
j) The square of a number is four more than three times the number. Find the
number.
8. a) The difference of the age of two sisters is 4 years and the product of their age is
45. Find the age of the two sisters.
b) The sum of the age of two children is 13 years and the product of their age is 36.
Find their age.
9. a) A room is 4 m longer than its breadth. If the area of the floor of the room is
96 sq. m, find the length and breadth of the room.
b) The length of a room is 3 m longer than its breadth. If the area of its floor is
70 sq. m, find the length and breadth of the room.
It’s your time - Project work
10. a) Let’s write the quadratic equations in the form of x2 – a2 = 0, where a2 is a perfect
square number. Then, solve them to get the following solutions:
(i) x = ± 1 (ii) x = ± 2 (iii) x = ± 3 (iv) x = ± 4 (v) x = ± 5
Let’s check the equations replacing x by the given solutions.
b) Let’s write any 3 quadratic equations of the for x2 – a2 = 0, where a2 is a square
number. Solve each equation.
c) Let’s write any 3 quadratic equations of the form ax2 – b = 0, where a and b are
square numbers. Solve each equation.
d) Let’s write any 4 quadratic equations of the following forms.
(i) x2 + ax + b = 0 (ii) x2 + ax – b = 0
(iii) x2 – ax – b = 0 (iv) x2 – ax + b = 0
Solve each equations and find the solutions.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 165 Vedanta Excel in Mathematics - Book 8
Unit Transformation
13
13.1 Transformation –Review
The rule under which the position or size (or both) of an object (or a geometrical figure)
may be changed is known as transformations:
There are four fundamental transformation.
(i) Reflection (ii) Rotation
(iii) Translation (or Displacement) (iv) Enlargement
Here, we shall discuss about reflection, rotation and displacement.
Let's study the following illustrations and investigate the ideas about transformation.
Reflection of 'ABC Rotation of of 'PQR
A Q'
B P R' XM' ’ R
M C
P'
B' M’
XQ
P' C'
O
A'
P
'ABC is shifted from the position
'PQR is shifted from the position X
P to the position P' reflecting by the
axis of reflection MM'. to the position X' due to the rotation
It is the transformation of 'ABC due through 90° about O in anticlockwise
direction.
to reflection and 'A'B'C' is the image
It is the transformation of 'PQR due
of 'ABC.
to rotation and 'P'Q'R' is the image of
'PQR.
13.2 Reflection
The reflection of a geometrical figure means the formation of the image of the figure
after reflecting about the line of reflection. The line of reflection is also called the axis
of reflection. In the given figure, 'ABC is reflected by the axis of reflection MM’ to form
the image 'A'B'C'.
Properties of reflection C B
M
(i) The geometrical figure and its image are at equal A
distance from the axis of reflection. C'
A' M'
(ii) The areas of the geometrical figure and its image are
equal. B'
(iii) The appearance of the image of a figure is opposite to
the figure.
Vedanta Excel in Mathematics - Book 8 166 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
Now, let’s learn to draw the image of a geometrical figure reflected about an axis of
reflection.
In the figure, ABC is a triangle. MM' is the axis of reflection. A C
B
(i) Draw perpendiculars AP, BQ and CR from each vertex of
'ABC on the axis of reflection.
(ii) Produce AP, BQ, and CR. MP Q R M'
(iii) Measure the length of AP by using compasses and cut C'
A' B'
off PA' = PA. Similarly cut off QB' = BQ and RC' = CR.
(iv) Join A', B' and C' by using a ruler.
Thus, 'A'B'C' is the image of 'ABC.
13.3 Reflection of geometrical figure using coordinates
Let’s learn to find the coordinates of the images of geometrical figures formed due to
the reflection about x-axis and y-axis. Here, x-axis and y-axis are the axes of reflection.
(i) x-axis as the axis of reflection
Study the following illustrations and learn to find the coordinates of the image of a
point in different quadrants when x-axis is the axis of reflection.
The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
A(3,5)
A(-4,3) A' (-2,4) A' (3,4)
X' O X X' O X X' O X X' O X
A' (3,-5) A' (-4,-3) A(-2,- 4) A(3,–4)
Y' Y' Y' Y'
A (3, 5) o A' (3, –5) A (–4, 3)o A' (–4, –3) A (–2, –4)o A' (–2, 4) A (3, – 4)o A' (3, 4)
? P (x, y) o P' (x, –y) ? P(–x, y)o P' (–x, –y) ? P(–x, –y)o P' (–x, y) ?P (x, – y)o P' (x, y)
Only the sign of Only the sign of Only the sign of Only the sign of
y-coordinate is y-coordinate is y-coordinate is y-coordinate is
changed changed changed changed
From the above illustrations, it is clear that when a figure is reflected about x-axis, the
coordinate of the image of vertex remains the same and the sign of y-coordinate of the
image is changed.
Therefore, P(x, y) o P'(x, –y)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 167 Vedanta Excel in Mathematics - Book 8
Transformation
Worked-out examples
Example 1: P(3, 6), Q (– 2, 4) and R(5, 1) are the vertices of ∆PQR. Find the coordinates
of its image when it is reflected about x-axis. Also, draw the graphs of
this transformation. Y P(3,6)
Solution: Q(-2,4)
P (3, 6), Q (– 2, 4), and R (5, 1) are the vertices of 'PQR. When
'PQR is reflected about x-axis, X' R(5,1)
Y
P (3, 6) o P' (3, – 6) O
Q (– 2, 4) o Q' (– 2, – 4) R' (5,-1)
R (5, 1) o R' (5, –1) Q' (-2,-4) P' (3,-6)
Y'
? P' (3, – 6), Q' (– 2, – 4) and R' (5, – 1) are the vertices of the
image of 'PQR.
(ii) y-axis as the axis of reflection
Study the following illustrations and learn to find the coordinates of the image of point
in different quadrants when y-axis is the axis of reflection.
The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
A'(-4,3) A(4,3) A(-3,4) A'(3,4)
X' O X
X' O X X' O X X' O X
A(-5,-4) A'(5,-4) A'(-5,-2) A(5,-2)
Y' Y' Y' Y'
A(4, 3) o A'(– 4, 3) A(– 3, 4) o A'(3, 4) A(–5, –4) o A'(5, –4) A(5, –2)o A'(–5, –2)
? P(x, y) o P'(– x, y) ? P(– x, y) o P'(x, y) ? P(–x,–y)o P'(x, –y) ? P(x,–y)o P'(–x, –y)
Only the sign of Only the sign of Only the sign of Only the sign of
x-coordinate is x-coordinate is x-coordinate is x-coordinate is
changed changed changed changed
From the above illustrations, it is clear that when a figure is reflected about y-axis, the
coordinates of the image of vertex remains the same and the sign of x-coordinate of the
image is changed.
Therefore, P(x, y) o P'(–x, y)
Example 2: D (0, – 2), E (6, – 4), and F (2, 5) are the vertices of ∆DEF. Find the
coordinates of its image when it is reflected about y-axis. Also, draw
the graphs of this transformation.
Vedanta Excel in Mathematics - Book 8 168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
Solution: Y
D (0, – 2), E (6, – 4) and F (2, 5) are the vertices of 'DEF. F'(-2,5) F(2,5)
When 'DEF is reflected about y-axis.
X' O D(0,-2) X
D (0, – 2) o D' (0, – 2)
E (6, – 4) o E' (– 6, – 4) E'(-6,-4) D'(0,-2)
F (2, 5) o F' (– 2, 5) E(6,-4)
? D' (0, – 2), E' (– 6, – 4) and F' (– 2, 5) are the vertices of the
Y'
image of 'DEF.
EXERCISE 13.1
General Section
1. The dotted lines are the axes of reflection. Let's draw the image of the following
figures by using set-squares and ruler.
a) A b) P c) H G
QE
BC R F
2. Let's say and write the coordinates of images as quickly as possible.
a) X-axis is the axis of reflection.
P (x, y) o .................... P (–x, y) o ......................... P (– x, – y) o ...................
P (x, y) o .................... A (–3, 4) o ........................ B (3, – 4) o .....................
C (1, 5) o ................... D (– 1, – 5) o .................... E (0, – 2) o .....................
b) Y-axis is the axis of reflection.
P (x, y) o ..................... P (– x, y) o ................... P (–x, – y) o ...................
P (x, – y) o .................. A (2, – 7) o ................... B (8, 3) o .......................
C (– 4, – 5) o ............... D (– 1, 5) o .................... E (3, 0) o .......................
Creative Section - A
3. Copy the following figures in your own graph papers and draw their images under
the reflection about x-axis. Also, write the coordinates of the vertices of images.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 169 Vedanta Excel in Mathematics - Book 8
Transformation
a) Y b) Y c) Y
FQ
A C X X’ E
X’ D EP RO K FO RX
Q
B X X’
O
LM H G S
P
Y’ Y’ X’
4. Copy the following figures in your own graph papers and draw their images under
the reflection about y-axis. Also, write the coordinates of the vertices of images.
a) Y b) Y c) Y
A
AF D
X X’
X’ B CE GK C X
O O B R
X X’
OS
QR M PQ
L
Y’ Y’ Y’
5. Plot the following points in graph papers and draw triangles joining the points in
order. Draw the image of each triangle under the reflection about x-axis. Write the
coordinates of the vertices of images.
a) P (3, 6), Q (5, 9), R (4, 2) b) A (– 2, 4), B (3, 3), C (– 5, 7)
c) K (– 1, – 4), L (– 6, – 3), M (– 2, 4) d) D (3, – 5), E (2, 6), F (4, – 1)
6. Plot the following points in graph papers and draw triangles joining the points in
order. Draw the image of each triangle under the reflection about y-axis. Write the
coordinates of the vertices of images.
a) A (1, 4), B (4, 7), C (6, 3) b) P (–2, 3), Q (–4, –6), R (–5, 5)
c) E (3, -7), F (5, –2), G (0, –3) d) X (1, 4), Y (4, –5), Z (2, 6)
7. a) A (0, –4), B (– 3, 0) and C (2, 5) are the vertices of 'ABC. Find the coordinates of
its image under the reflection about x-axis.
b) P (– 7, – 8), Q (4, 6) and R (– 6, 2) are the vertices of 'PQR. Find the coordinates
of its image under the reflection about y-axis.
Creative Section- B
8. a) W (4, 2), X (– 3, 6) and Y (– 1, – 4) are the vertices of 'WXY. Find the coordinates
of the vertices of 'W'X'Y' under the reflection about x-axis. Also, find the
coordinates of 'W''X''Y'' when 'W'X'Y' is reflected about y-axis.
b) R (7, – 2), S (– 4, – 5) and T (– 6, 3) are then vertices of 'RST. Find the coordinates
of the vertices of 'R'S'T' under the reflection about y-axis. Also, find the
coordinates of 'R''S''T'' when 'R'S'T' is reflected about x-axis.
It's your time - Project work
9. Let's draw a triangle in a squared graph paper. Write the coordinates of its vertices.
Then reflect it about X-axis and Y-axis. Write the coordinates of its image.
Vedanta Excel in Mathematics - Book 8 170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
13.4 Rotation
Study the following illustrations and investigate the idea of rotation of a point.
Rotation of a point P through 90° in anti- Rotation of a point P through 90° in
clockwise direction about the centre of
clockwise direction about the centre of
rotation O. P'
rotation O. O P
90°
90° P P'
O Here, P' is the image of P.
Here, P' is the image of P. Rotation of a point P through 180° in
clockwise direction about the centre of
Rotation of a point P through 180° in anti- rotation O.
clockwise direction about the centre of
rotation O. P' O P
180° P 180°
P' O Here, P' is the image of P.
Here, P' is the image of P.
To rotate a geometrical figure, following three conditions are required:
(i) Centre of rotation (ii) Angle of rotation (iii) Direction of rotation
We can rotate a figure in two directions.
(i) Anti-clockwise direction (Positive direction)
(ii) Clock-wise direction (Negative direction)
In anti-clockwise direction, we rotate a figure in opposite direction of the rotation of
second-hand of a clock. In clockwise direction, we rotate a figure in the same direction
of the rotation of second-hand of a clock.
Now, let’s learn to draw the image of a figure when it is rotated through the given angle
in the given direction about the given centre of rotation.
Rotation through 60° in anti-clockwise direction
(i) Join each vertex of the figure to the centre of rotation with dotted lines.
(ii) On each dotted line, draw 60q at O with the help of B'
protractor in anti-clockwise direction.
(iii) With the help of compasses, cut off OA' = OA, C'
OB' = OB and OC' = OC.
+60°
(iv) Join A', B', and C'. O A' C B
Thus, 'A'B'C'is the image of 'ABC formed due to the
A
rotation through 60q in anti-clockwise direction
about the centre O.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 171 Vedanta Excel in Mathematics - Book 8
Transformation
Rotation through 60° in clockwise direction O C
In this case, follow the steps as mentioned above. A B
–60°
At O, draw 60q with the help of protractor in clockwise A' C'
direction. In the adjoining diagram, 'A'B'C' is the image of
'ABC formed due to the rotation through 60q in clockwise
direction about the centre O. Similarly, we can rotate B'
the given geometrical figure through any given angle in
anti-clockwise or clockwise direction about the given centre of rotation.
13.5 Rotation of geometrical figures using coordinates
In this case, we shall discuss about the rotation of figures through some special angles
such as 90q and 180q in anti-clockwise and clockwise directions about the centre at
origin.
The rotation through 90q is also called a quarter-turn and the rotation through 180q is
called a half-turn.
Rotation through 90° in anti-clockwise about the centre at origin
The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
A(-3,4)
A'(-5,2) A(2,5) X' O A'(5,1)
X' O X' O A(-4,-2) A'(2,-4) X' O
A'(-4,-3) Y'
Y' Y' A(1,-5)
A(2, 5) o A'(–5, 2) A(–4, –2) o A'(2, –4)
? P(x, y) o P'(–y, x) A(–3, 4) o A'(–4, –3) ? P(–x, –y) o P'(y, – x) Y'
? P(–x, y) o P'(–y, –x) A(1, –5) o A'(5, 1)
? P(x, –y) o P'(y, x)
Thus, when a point is rotated through 90q in anti-clockwise direction (+ 90q) about the
origin as the centre of rotation, the x and y-coordinates are exchanged by making the
sign of y-coordinate just opposite.
i.e., P (x, y) o P' (– y, x)
Vedanta Excel in Mathematics - Book 8 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
Rotation through 90° in clockwise about the centre at origin
The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
A(3,2)
A(-4,5) A'(5,4) A'(-4,2)
X' O X' O X' O X' A'(-3,-1) O
A'(2,-3) A(-2,-4) A(1,-3)
Y' Y' Y' Y'
A(3, 2) o A'(2, –3) A(–4, 5) o A'(5, 4) A(–2, –4) o A'(–4, 2) A(1, –3) o A'(–3, –1)
? P(x, y) o P'(y, –x) ? P(–x, y) o P'(y, x) ?P(–x, –y) o P'(–y, x) ? P(x, –y) o P'(–y, –x)
Thus, when a point is rotated through 90q in clockwise direction (– 90q) about the origin
as the centre of rotation, the x and y-coordinates are exchanged by making the sign of
x-coordinate just opposite.
i.e. P (x, y) o P'(y, – x)
Rotation through 180° in anti-clockwise and clockwise about origin
When a point is rotated through 180q in anti-clockwise or in clockwise direction
about origin as the centre of rotation, the coordinates of the image are same. Study the
following illustration:
The point A is in The point A is in The point A is in The point A is in
the 1st quadrant. the 2nd quadrant. the 3rd quadrant. the 4th quadrant.
Y Y Y Y
A (-2, 4)
A(5,3) A (3,3) A'(-4, 3)
X' O X' O X' O X' O
A'(-3,-3)
A'(-5,-3) A'(2,-4) A (4, -3)
Y' Y'
A(–2, 4) o A'(2, –4) Y' Y'
A(5, 3) o A'(–5, –3) A(4, –3) o A'(–4, 3)
? P(x, y) o P'(–x, –y) ? P(–x, y) o P(x, –y) A(–3, –3) o A'(3, 3) ? P(x, –y) o P'(–x, y)
? P (–x, –y) o P(x, y)
Thus, when a point is rotated through 180q in anti-clockwise (– 180q) or in clockwise
(+ 180q) direction, about the origin as the centre of rotation, the x and y-coordinates of
the image remain the same just by changing their signs.
i.e. P (x, y) o P' (– x, – y)
Worked-out examples
Example 1: A (3, 1), B (4, – 2) and C (5, 3) are the vertices of ∆ABC. Find the coordinates
of its image under the rotation through 90° in anti-clockwise direction
about origin. Also, draw the graphs of this transformation.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 173 Vedanta Excel in Mathematics - Book 8
Transformation
Solution: Y
A (3, 1), B (4, – 2), and C (5, 3) are the vertices of 'ABC. When C'(-3,5) B'(2,4)C(5,3)
it is rotated through + 90q about origin,
A'(-1,3) A(3,1) X
A (3, 1) o A' (– 1, 3) X'
B (4, – 2) o B' (2, 4)
C (5, 3) o C' (– 3, 5) B(4,-2)
? A'(– 1, 3), B'(2, 4), and C'(– 3, 5) are the vertices of the image Y'
of 'ABC.
Example 2: P (2, 3), Q (–2, 6), and R (– 4,2) are the vertices of ∆PQR. Find the
coordinates of its image under the rotation through 180° in clockwise
direction about origin. Also, draw the graphs of this transformation.
Solution: Y
P (2, 3), Q (– 2,6), and R (– 4, 2) are the vertices of 'PQR. Q(-2,6)
When 'PQR is rotated – 180° about origin. X' R(-4,2) O P(2,3)
P (2, 3) o P' (– 2, –3) R'(4,-2)X
Q (– 2, 6) o Q' (2, – 6)
R (–4, 2) o R' (4 –2) P'(-2,-3)
? P' (–2, – 3), Q' (2,– 6), and R' (4, – 2) are the vertices of the
z
Y' Q'(2,-6)
image of 'PQR.
Example 3: D(1, 4), E (–3, –5), and F (2, –3) are the vertices of ∆DEF. Find the
coordinates of the vertices of ∆D'E'F' under the rotation through –90°
about origin. Also, find the coordinates of ∆D''E''F'' when ∆D'E'F' is
rotated through +180° about origin.
Solution:
D (1, 4), E (– 3, – 5), and F (2, –3) are the vertices of 'DEF. When it is rotated
through –90° about origin.
D (1, 4) o D' (4, – 1)
E (– 3, 5) o E' (5, 3)
F (2, – 3) o F' (– 3,–2)
? D' (4, – 1), E' (5, 3), and F' (–3, – 2) are the vertices of 'D'E'F'
Again when 'D'E'F' is rotated through +180° about origin.
D' (4, –1) o D'' (– 4, 1)
E' ( 5, 3) o E'' (– 5, – 3)
F' (– 3, – 2) o F'' (3, 2)
? D'' (– 4, 1), E'' (–5, – 3) and F'' (3, 2) are the vertices of of 'D''E''F''
Vedanta Excel in Mathematics - Book 8 174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
EXERCISE 13.2
General Section - Classwork
1. Let's say and write the coordinates of images as quickly as possible.
a) Rotation through 90° in anticlockwise direction
P (x, y) o ................. P (– x, y) o ................. P (–x, –y) o ................
P (x, –y) o ................. A (–2, 3) o ................. B (2, –5) o .................
C (–4, –1) o ................. D (6, 7) o ................. E (0, –4) o .................
b) Rotation through 90° in clockwise direction.
P (x, y) o ................. P (– x, y) o ................. P (–x, –y) o ................
P (x, –y) o ................. A (–2, –1)o ................. B (3, –5) o .................
C (2, 3) o ................. D (–4, 5) o ................. E (–6, 0) o .................
c) Rotation through 180° in anticlockwise or clockwise direction
P (x, y) o ................. P (– x, y) o ................. P (–x, –y) o ................
P (x, –y) o ................. A (–6, 5) o ................. B (3, –9) o .................
C (1, 3) o ................. D (–4, –2)o ................. E (0, 4) o .................
Creative Section - A
2. Draw the images of the following figures rotating through the given angles in
anti-clockwise and clockwise direction about the given centre of rotation.
a) A C b) E c) P R
O F G O
B O Q
Rotation through 45q Rotation through 60q Rotation through 90q
d) S e) A
f) OE
PR C M
QO BD
Rotation through 120qO PR
Rotation through 75q Rotation through 180q
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 175 Vedanta Excel in Mathematics - Book 8
Transformation
3. Copy the following figures on your own graph papers. Draw their images rotating
through 90q in (i) anti-clockwise (ii) clockwise directions about the centre at origin.
Also, write the coordinates of the vertices of images.
a) Y b) c)
Y
B Y E
A
F
EF
X’ C X X’ D OK X X’ AO G X
Q PO L
R BC
M
Y’ Y’ Y’
4. Copy the following figures on your own graph papers. Draw their images rotating
through 180q in anti-clockwise direction. Also, write the coordinates of the vertices of
images. Y b) Y c) Y
a) PB DE X X’ FE
A
Q RC X X’ KF OA GX
X’ O O C
LM B
Y’ Y’ Y’
5. a) P (– 4, 2), Q (3, 7), and R (– 1, – 6) are the vertices of 'PQR. Find the coordinates
of its image under the rotation through 90q in
(i) anti-clockwise (ii) clockwise direction about origin.
b) A (– 5, – 8), B (2, – 3), and C (– 2, 9) are the vertices of 'ABC. Find the coordinates
of its image under the rotation through 180q in anti-clockwise direction about
origin.
Creative Section - B
6. a) K (1, 4), L (– 3, – 5), and M (5, – 2) are the vertices of 'KLM. Find the coordinates of
'K'L'M' under the rotation through –90q about origin. Also, find the coordinates
of 'K''L''M'' when 'K'L'M' is rotated through +180q about origin.
b) D (– 4, – 1), E (0, – 6), and F (5, 0) are the vertices of 'DEF. Find the coordinates of
'D'E'F' under the rotation through +90q about origin. Also, find the coordinates
of 'D''E''F'' under the reflection about x-axis.
It's your time - Project work!
7. Draw a triangle in separate squared graph paper. Write the coordinates of its vertices.
Rotate the triangle through (i) +90° (ii) –90° (iii) +180° (iv) –180°.
Then write the coordinates of the vertices of image in each case.
Vedanta Excel in Mathematics - Book 8 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
13.6 Displacement
It is the transformation of geometrical figures in which each vertex of a figure is displaced
by equal distance to the same direction. It is also called translation. Study the following
illustrations. S
PR
A'
A
B' C' Q
S’
B
C P’ R’
Q’
'ABC is displaced to 'A'B'C'
Here, AA' = BB' = CC' Quadrilateral PQRS is displaced to P'Q'R'S'
Also, AA' // BB' // CC' Here, PP' = QQ' = RR' = SS'
Also, PP' // QQ' // RR' // SS'
Thus, the displacement of a geometrical figure has magnitude as well as direction. So, it is
a vector quantity.
Now, let’s learn to displace a given geometrical figure.
13.7 Displacement of geometrical figures using coordinates
In the graph given below, A (2, 4), B (8, 2), and C (5, 9) are the vertices of 'ABC. When
'ABC is displaced by 7 units right and 3 units up, each of its vertices should be displaced
by 7 units right and 3 units up. It means the x-coordinate of each vertex is added by 7
and y-coordinate is added by 3. If A', B', and C' are the images of the vertices A, B and C
of 'ABC respectively, then Y
A (2, 4) o A' (2 + 7, 4 + 3) o A' (9, 7)
B (8, 2) o B' (8 + 7, 2 + 3) o B' (15, 5) C' (12, 12)
C (5, 9) o C' (5 + 7, 9 + 3) o C' (12, 12)
C (5, 9)
Thus, A' (9, 7), B' (15, 5), and C' (12, A' (9, 7) B' (15, 5)
12) are the vertices of the 'A'B'C'. This B (8, 2)
is the case of right and up direction of X' A (2, 4) X
displacement. O
Now, let's take the case of left and down Y'
direction of displacement.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 177 Vedanta Excel in Mathematics - Book 8
Transformation
In the given graph P (3, 7), Q (–4, 5), and R (1, 3) are the vertices of 'PQR. Let's displace
the triangle 6 units left and 5 units down. If P', Q', and R' be the images of the vertices P,
Q and R of 'PQR respectively, then, Y
P (3, 7) o P' (3 – 6, 7 – 5) o P' (–3, 2)
Q (–4, 5) o Q' (–4, – 6, 5 – 5) o Q' (–10, 0) Q (–4, 5) P (3, 7)
R (1, 3) o R' (1 – 6, 3 – 5) o R' (–5, –2) Q' R (1, 3) X
(–10, 0) P' (–3, 2)
Thus, in the case of left and down X'
displacement, the number of units of O
left displacement is subtracted from
x-coordinate of each vertex. Similarly, the R'(–5, –2)
number of units of down displacement
is subtracted from y-coordinate of each Y'
vertex.
In this way, if a be the number of units of right or left displacement and b be the number
of units of up or down displacement, then,
P (x, y) o P' (x + a, y + b) [Right and up displacement]
P (x, y) o P' (x – a, y – b) [Left and down displacement]
P (x, y) o P' (x + a, y – b) [Right and down displacement]
P (x, y) o P' (x – a, y + b) [Left and up displacement]
Worked-out examples A
Example 1: Displace the adjoining ∆ABC in the magnitude C
p
and direction of vector p (p ). B
A C
Solution: p
B
(i) From A, draw AA' // p and cut off AA' = p A’ C’
(ii) From B, draw BB' // p and the cut off BB' = p' B’
(iii) From C, draw CC' // p and cut off CC' = p' .
(iv) Join A', B' and C'
(v) 'A'B'C' is the required image of 'ABC under the
displacement of p .
Vedanta Excel in Mathematics - Book 8 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
Example 2: A (3, 6), B (–2, 5), and C (2, 8) are the vertices of 'ABC. Find the coordinates
of the image of the triangle in the graph due to the displacement of 5
units to the right and 4 units down. Also, check your answer by using
the right and down displacement rules.
Solution:
A (3, 6), B (–2, 5), and C (2, 8) are Y
the vertices of 'ABC. When it is C (2, 8)
displaced by 5 units right and 4 B A
(–2, 5) (3, 6)
units down, the coordinates of each C' (7, 4)
O B' A' (8, 2)
vertices of its image are A' (8, 2), (3, 1)
B' (3, 1) and C' (7, 4) X' X
Now, by applying the rules of right and
down displacement,
A (3, 6) o A' (3 + 5, 6 – 4) o A' (8, 2) Y'
B (–2, 5) o B' (–2 + 5, 5 – 4) o B' (3, 1)
C (2, 8) o C' (2 + 5, 8 – 4) o C' (7, 4)
Example 3: Find the coordinates of the image of the point P (–3, 5) when it is
displaced by 4 units left and 6 units up. Also find the coordinates of the
final image when it is again displaced by 6 units right and 9 units down.
Solution:
Here, P (–3, 5) is the given point
When it is displaced by 4 units left and 6 units up,
P (–3, 5) o P' (–3, – 4, 5 + 6) o P' (–7, 11)
? P' (–7, 11) is the coordinates of the image of the point P (–3, 5) due to the first
displacement.
Again, when P' (–7, 11) is displaced by 6 units right and 9 units down,
P' (–7, 11) o P" (–7 + 6., 11 – 9) o P" (–1, 2)
? P" (–1, 2) is the coordinates of the final image due to the two successive displacements.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 179 Vedanta Excel in Mathematics - Book 8
Transformation
EXERCISE 13.3
General Section
Let's say and write the answers as quickly as possible.
1. a) When a point P (x, y) is displaced by a units right and b units up, coordinates
of its image are ...............................................................
b) When a point A (2, 4) is displaced by 1 unit right and 3 units up, coordinates
of its image are ...............................................................
c) When a point P (x, y) is displaced by a units left and b units down, coordinates
of its image are ...............................................................
d) When a point A (3, 1) is displaced by 3 units left and 2 units down, coordinates
of its image are ...............................................................
e) When a point P (x, y) is displaced by a units right and b units down,
coordinates of its image are ...............................................................
f) When a point A (–2, 5) is displaced by 3 units right and 2 units down,
coordinates of its image are ...............................................................
g) When a point P (x, y) is displaced by a units left and b units up, coordinates
of its image are ...............................................................
h) When a point A (3, –4) is displaced by 5 units left and 6 units up, coordinates
of image are ...............................................................
Creative Section
2. Displace the following geometrical figures on the magnitude and direction of p .
a) b) c) E
P A
p
Q R p
d) P
M B C F G
R p e) A f) K p
E
R T T
E
p
p
I
I
Vedanta Excel in Mathematics - Book 8 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Transformation
3. Plot the following points in the graph paper. Find the coordinates of the image
of each point under the given displacements.
a) A (1, 3) [2 units right and 4 units up]
b) B (5, 2) [3 units left and 3 units down]
c) C (–2, 4) [5 units right and 2 units down]
d) D (–3, –5) [4 units left and 1 unit up]
4. a) P (3, 6), Q (6, 2) and R (5, 8) are the vertices of 'PQR. Draw the triangle in
the graph paper and find the coordinates of the image of its vertices when
they are displaced by 7 units right and 5 units up. Also check your answer
by using the displacement of right and up rules.
b) Plot the points A (2, 5), B (–2, 0), C (–8, 4), and D (–5, 7) in the graph paper
and join them. Displace the figure so formed by 4 units left and 6 units up.
Find the coordinates of the vertices of the image so formed. Also, check your
answer by using the displacement of left and up rules.
5. a) Find the units of displacement of the point (4, 2) to get the image of (7, 7).
b) What must be the units of displacement so that the image of a point (–3, 5)
can be (–8, 2)?
6. a) Find the coordinates of the image of the point M (–4, 7) when it is displaced
by 3 units right and 5 units down. Also, find the coordinates of the final
image when it is again displaced by 6 units left and 9 units up.
b) A (1, 4), B (3, –2), and C (–4, –5) are the vertices of 'ABC. If it is displaced
by 2 units right and 4 units down and then again 5 units left and 1 unit up,
find the coordinates of the final image of 'ABC due to the two successive
displacements.
It's your time - Project work!
7. a) Let's draw a triangle in a squared graph paper. Write the coordinates of it's
vertices. Displace the triangle under the different units of the following
displacement of your own. Then, write the coordinates of the vertices of its
image.
(i) Right and up (ii) Right and down (iii) Left and up (iv) Left and down
b) Let's draw a quadrilateral in a squared graph paper. Write the coordinates of
its vertices. Displace the quadrilateral under the following displacements with
your own units of displacements. Then, write the image of each vertex.
(i) Right and up (ii) Right and down (iii) Left and up (iv) Left and down
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 181 Vedanta Excel in Mathematics - Book 8
Unit Geometry: Angle
14
14.1 Different pairs of angles – Looking back
Classroom - Exercise
1. Let's say and write the pairs of adjacent angles, liner pairs, vertically opposite
angles, complementary angles, and supplementary angles from the following
figures.
a
db
x mn cp
y q
Adjacent Linear pairs Vertically Complementary Supplemen-
angles opposite angle
........and........ ........and........ angles tary angles
........and........ ........and........
........and........ ........and........ ........and........ ........and........
........and........ ........and........
........and........ ........and........
........and........
........and........
........and........
........and........
........and........
........and........
........and........
........and........
2. Let's say and write the pairs of alternate angles, corresponding angles and
co-interior angles.
Alternate angles Corresponding angles Co-interior angles ab
dc
........and........ ........and........ ........and........
ef
........and........ ........and........ ........and........ hg
........and........
........and........
(i) Adjacent angles C C
B
AOB and BOC are a pair of adjacent B
angles. They have a common vertex O and a O AO A
common arm OB.
Vedanta Excel in Mathematics - Book 8 182 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Angle
Linear pair C B
OA
If the sum of a pair of adjacent angles is 180q, they are
said to be linear pair.
In the figure alongside, AOB + BOC = 180q.
So, AOB andBOC are the linear pair.
(ii) Vertically opposite angles A O C
D B
In the adjoining figure, AOC and BOD are vertically C B
opposite angles formed by intersected line segments. They 90° A
have a common vertex and they are lying to the opposite
side of the common vertex. AOD and BOC are another O A
pair of vertically opposite angles. B
Vertically opposite angles are always equal. O
AOC = BOD and AOD = BOC.
(iii) Complementary angles
A pair of angles are said to be complementary if their sum is a
right angle (90q). In the figure, AOB and BOC are a pair of
complementary angles.
? AOB + BOC = 90q
Also, complement of AOB = 90q – BOC.
Complement of BOC = 90q – AOB.
(iv) Supplementary angles
A pair of angles are said to be supplementary if their
sum is two right angles (180q). In the figure, AOB and
BOC are a pair of supplementary angles.
? AOB + BOC = 180q. C
Also, the supplement of AOB = 180q – BOC
The supplement of BOC = 180q – AOB.
14.2 Experimental verifications of pair of angles formed by two
intersecting lines
Experiment 1: The sum of a pair of adjacent angles formed by two intersecting lines
is 180°.
Step 1: Draw three pairs of intersecting lines AB and CD intersecting at O.
A A D A D
CO O O
D
B CB C B
fig (i) fig (ii) fig (iii)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 183 Vedanta Excel in Mathematics - Book 8
Geometry: Angle
Step 2: Measure AOC and AOD (or AOC and BOC or BOC and BOD or AOD
and BOD) with the help of protractor and write the measurements in the table.
Fig. No. AOC AOD AOC + AOD Result
(i) AOC + AOD = 180°
(ii)
(iii)
Conclusion: The sum of pair of adjacent angles formed by two intersecting lines
is 180°.
[Note: The pair of adjacent angles whose sum is 180° are said to be the linear pair.]
Experiment 2: Each pair of vertically opposite angles formed by two intersecting lines
are equal.
Step 1: Draw three pairs of intersecting lines AB and CD intersecting at O.
A A D A D
CO O O
D
B CB C B
fig (i) fig (ii) fig (iii)
Step 2: Measure AOC and BOD, AOD and BOC with the help of protractor. Write
the measurements in the table.
Fig. No. AOC BOD AOD BOC Result
(i) AOC = BOD
(ii)
(iii) AOD = BOC
Conclusion: Each pair of vertically opposite angles formed by two intersecting lines
are equal.
Worked-out examples
Example 1: If x° and (x + 6)° are a pair of complementary angles, find them.
Solution:
Here, xq + (x + 6)q = 90q [The sum of a pair of complementary angles]
or, 2xq = 90q – 6q
8=24°42=q
or, = 42q and xq = 42q = 48q.
? xq (x + 6)q + 6q
Vedanta Excel in Mathematics - Book 8 184 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Angle
Example 2: If a pair of supplementary angles are in the ratio 4:5, find them.
Solution:
Let, the required supplementary angles be 4xq and 5xq.
Now, 4xq + 5xq = 180q [The sum of a pair of supplementary angles]
or, 9xq = 180q
or, xq = 1890°= 20q
? 4xq = 4 u 20q = 80q and 5xq = 5 u 20q = 100q.
Example 3 : Find the unknown sizes of angles.
a) B b) c)
D C
3x 3x x+80°
2x 5x°
Ox 4x+50°
A x+30° x B
AO
Solution:
a) x + 2x + 3x = AOB b) x + 3x + (x + 30°) = BOA
or, 6x = 90° or, 5x + 30° = 180°
or, x = 15° or, x = 30°
? x = 15°, ? x = 30°
2x = 2 × 15° = 30° 3x = 3 × 30° = 90°
3x = 3 × 15° = 45° x + 30° = 30° + 30° = 60°
c) (x + 80°) + 5x° + (4x + 50°) = 360°
or, 10x + 130° = 360°
or, x = 23°
? x + 80° = 23° + 80° = 103°
5x° = 5 × 23 = 115°
4x + 50° = 4 × 23° + 50° = 142°
EXERCISE 14.1
General Section- Classwork
1. Let's say and write the answers as quickly as possible. = .....................
a) If x° and 100° are a linear pair, then x° = .....................
b) If y° and 50° are vertically opposite angles, then y° = .....................
c) If a° and 30° are a pair of complementary angles, then a° = .....................
d) If p° and 120° are a pair of supplementary angles, then p°
Creative Section - A
2. a) If xq and 50q form a linear pair, find xq.
b) If 2xq and 3xq are adjacent angles in linear pair, find them.
c) If yq and 48q are a pair of complementary angles, find yq.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 185 Vedanta Excel in Mathematics - Book 8
Geometry: Angle
d) If 2pq and (p + 15)q are a pair of complementary angles, find them.
x°
e) If xq and 4 are a pair of supplementary angles find them.
f) If pair of complementary angles are in the ratio 4:11, find them.
g) If a pair of supplementary angles are in the ratio 7:5, find them.
3. Find the value of x in the each of figures given below.
a) b) c) d)
x° 58° x° 4x° 100° 3x° 2x°
125°
e) f) g) h)
5x° 4x° 3x°+50° 2x°
x° ) x )° ) x4 )°
2
x°
4. Find the unknown sizes of angles. c) d) 92°
4x° x° z° y°
a) b) x°
3x° 2x°
(x+20)° 2x° x° 2x° 3x°
x°
e) f) g) h) q° x°
x° 3x° 2x° x+80° 3q° x+25°
z° y° 2x–30° 5x° p°
4x+50°
x+35°
2x+30°
5. a) In the given figure, calculate the values of angles p and q, qp
where p = 21q.
b) In the figure, given alongside, if x = 3y, find the sizes of x
angles represented by a and b. ay
b
Creative Section -B x c
ab
6. a) In the adjoining figure, if a + b + c = 180q,
prove that x = b + c. az b
xy
b) In the figure given alongside, a = x and b = y.
Show that x + y + z = 180q.
Vedanta Excel in Mathematics - Book 8 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Angle
7. a) Draw a line segment AB. Mark a point O on AB and draw an angle BOC. Measure
BOC and AOC. Verify that BOC + AOC = 180°.
b) Draw two intersecting line segment AB and CD intersecting at O. Measure the
size of each pair of vertically opposite angles. Verify that each pair of vertically
opposite angles are equal.
14.3 Different pairs of angles made by a transversal with two
straight lines
In the adjoining figure, straight line segment EF intersects two E B
line segments AB and CD at the point G and H respectively. A 1G 2 D
Here, EF is called a transversal. 43
1, 2, 3, 4, 5, 6, 7 and 8 are the angles made by 5H 6
87
the transversal with two line segments. C F
Here, 1, 2, 7 and 8 are the exterior angles.
3, 4, 5 and 6 are the interior angles.
Also, 4 and 6, 3 and 5 are two pairs of alternate angles.
1 and 5, 2 and 6, 4 and 8, 3 and 7 are four pairs of corresponding angles.
4 and 5, 3 and 6 are two pairs of co-interior angles.
14.4 Relation between pairs of angles made by a transversal
with parallel lines
Two line segments are said to be parallel if they do not P RD
intersect each other when they are extended to either C
directions.
In the figure, AB and CD are two parallel lines. It is written A Q SB
as AB//CD.
The perpendicular distance between two parallel lines is always equal.
? PQ = RS.
Let's investigate the relation between the pair of alternate angles, pair of corresponding
angles and the pair of co-interior angles formed due to the intersection of two parallel
lines by a transversal.
Experiment 3: Each pair of alternate angles formed by a transversal with two parallel
lines are equal.
Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF at
G and H.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187 Vedanta Excel in Mathematics - Book 8
Geometry: Angle A C AC
F
E E
G BG H
A
H G
DF
H EB D
C BD
fig (iii)
F fig (ii)
fig (i)
Step 2: Measure AGH and GHD, BGH and GHC with the help of protractor. Write
the measurements in the table.
Fig. No. AGH GHD BGH GHC Results
(i) AGH = GHD
(ii)
(iii) BGH = GHC
Conclusion: Each pair of alternate angles formed by a transversal with two parallel
lines are equal. ab
G
Furthermore, in the given figure, AB // CD and EF B
intersects AB and CD at G and H respectively. Here, a A H D
cd
and d, b and c are two pairs of alternate exterior
F
angles. They are the exterior angles lying towards the
alternate sides of the transversal. The alternate exterior C
angles made by a transversal with parallel lines are
always equal.
? a = d and b = c.
Experiment 4: The sum of each pair of co-interior angles formed by a transversal with
two parallel lines is 180°.
Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF at
G and H. F E A C
G E
AC A
H B G
G CH DB H
F
EB D D F
fig (ii)
fig (i) fig (iii)
Vedanta Excel in Mathematics - Book 8 188 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Angle
Step 2: Measure AGH and CHG, BGH and DHG with the help of protractor. Write
the measurements in the table.
Fig. AGH CHG AGH + CHG BGH DHG BGH + DHG Result
No.
(i) AGH + CHG = 180°
(ii) BGH + DHG = 180°
(iii)
Conclusion: The sum of each pair of co-interior angles formed by a transversal with two
parallel lines is 180°.
Experiment 5: Each pair of corresponding angles formed by a transversal with two
parallel lines are equal.
Step 1: Draw three pairs of parallel lines AB and CD intersected by a transversal EF
at G and H.
AC AC E
AG
HF F B
H
G GH
E CD
B D EB D F
fig (i) fig (ii) fig (iii)
Step 2: Measure AGE and CHG, BGE and DHG, AGH and CHF, BGH and
DHF with the help of protractor. Write the measurements in the table.
Fig. AGE CHG BGE DHG AGH CHF BGH DHF Result
No.
(i) AGE = CHG
(ii) BGE = DHG
(iii) AGH = CHF
BGH = DHF
Conclusion: Each pair of corresponding angles formed by a transversal with two parallel
lines are equal.
The co-interior angles, alternate angles, and corresponding angles made by a transversal
with two parallel lines appears in the following shapes.
Co-interior angles
ab b b ab
ab a aa b
a and b are co-interior angles. So, a + b = 180q.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 189 Vedanta Excel in Mathematics - Book 8
Geometry: Angle
Alternate angles
a aa ba b
bb
a and b are alternate angles. So, a = b.
Corresponding angles
ab ba ab
b a ba ab
a and b are corresponding angles. So, a = b.
Worked-out examples
Example 1: Find the size of unknown angles in the following figures.
(i) 70° z 60° (ii) 40°
80°
ax yb x
Solution: 160°
i) x = 70q [Being alternate angles]
y = 60q [Being alternate angles]
70q + z + 60q = 180q [Being the sum a straight angle]
or, z = 180q – 130q = 50q
a + 70q = 180q [Being the sum of co-interior angles]
or, a = 180q – 70q = 110q
b + 60q = 180q [Being the sum of co-interior anlges]
or, b = 180q – 60q = 120q
(ii) EF and GH parallel to AB and CD are drawn. A B
E F
a = 40q [Being corresponding angles] G a H
b = 160q [Being corresponding angles] c
c = 80q – a = 80q – 40q = 40q C d x D
c + d = 180q [Being the sum of co-interior angles]
b
or, 40q + d = 180q 160°
or, d = 180q – 40q = 140q
b + d + x = 360q [Being the sum a complete turn]
or, 160q + 140q + x = 360q
or, x = 360q – 300q = 60q
Vedanta Excel in Mathematics - Book 8 190 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Angle
EXERCISE 14.2 d c
General Section - Classwork bx
1. Let's say and write the answers as quickly as possible.
a) Corresponding angle of a is ...................
b) Alternate angle of x is ....................... a
c) a and d are a pair of ................................... angles.
d) The sum of b and c is ................ .
e) If b = 120°, then x = ........, c = .......... a = .......... d = ............
2. If two non-paralled lines are intersected by a transversal,
a) Are the alternate angles so formed equal ? ....................
b) Are the corresponding angle so formed equal ? ....................
c) Is the sum of a pair of co-interior angle 180° ? ....................
Creative Section A
3. Let's find the unknown sizes of angles.
a) b) c) d)
z y
105° x
y x x ba
z 80° 130° y xz
y 100°
e) f) g) h)
ba c b 115° y 112° f bc
a x x a 120° ad
78°
zy b 105°
e
85°
4. From the figures given below, find the value of x.
a) b) c) 2x d)
(3x+10)° (5x+12)° (6x–13)° (x+45)° (x–20)° (80–x)°
(4x–10)°
e) f) g) h) (x+40)°
(2x–15)° x° (2x–30)°
3x°
2x° (x+15)° 3x°
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 191 Vedanta Excel in Mathematics - Book 8
Geometry: Angle
5. Find the unknown sizes of angles.
a) b) c) y d)
110° x
z 130° x 96°
z ay
88° y x yx
z
e) f) g) h)
20° x
zy x y 60° x
x 120° 60° 70° z 80°
y
110° y ab
i) j) k) l)
58° 45°
50° x 152° 36°
zy 60° a x x
b 35°
28°
m) n) o) p)
x 40° 162°
135° 38° 44°
y 30° x 88°
72°
150° x
x
155°
6. Find the unknown sizes of angles in the following figures.
a) y z b) 110° c) a x d) 55° w
x 80° d a y z
c 30°
b 40° b y
x 75°
7. a) In the adjoining figure, PQ//MR NMR= 150q P Q
and QNM = 40q, calculate the value of x. x
M R
40° 150°
N
Vedanta Excel in Mathematics - Book 8 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Angle
A C E F
x°
b) In the given figure AB//DC and BE//DF. Find the y°
measures of x°, y° and z°.
40° z°
Creative Section -B BD
8. a) In the figure alongside, show that wx ab
zy dc
(i) q = d (ii) f = z
pq ef
(iii) x = h (iv) a = r sr hg
(v) p = c (vi) b = s
b) In the adjoining figure show that x + y + z = 180°. x
y za
b
c) In the figure alongside, show that yc b
a + b + c + d = 4 right angles.
d ax
A 50° B
9. a) In the adjoining figure, AB//CD. Show that, AC//BD.
70° 60° D
C
P T
100°
b) In the given figure, show that, PQ//TR.
Q 45° R 35° S
10. Draw two parallel line segments. Intersect them with a transversal.
a) Measure each pair of corresponding angles and verify that they are equal.
b) Measure each pair of alternate angles and verify that they are equal
c) Measure each pair of co-interior angles and verify that the sum of each pair is
180°.
It's your time - Project work
11. Let's put your ruler on a sheet of paper and draw two parallel lines through it's two
opposite edges along length. Draw a transversal to intersect the parallel lines at
two district points. Then measure eight angles so formed. Explore the relationships
between corresponding angles, alternate angles and co-interior angles.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 8
Unit Geometry: Triangle
15
15.1 Triangle - Looking back
Classroom - Exercise
1. Let's select the appropriate types of triangles and write in the blank spaces.
Scalene triangle, isosceles triangle, equilateral triangle, acute- angled triangle,
obtuse-angled triangle, right-angled triangle
a) 5 cm b) c)
4 cm 80° 90°
5 cm 52° 48°
d) e) f) 3 cm
4 cm
4 cm 3 cm 3 cm
140°
5 cm
2. a) If x°, y°, and z° are the angles of a triangle, x° + y° + z° = ...............
b) If a°, b°, and c° are the angles of a triangle and a° + b° = 100°, c° = ...............
c) If p°, q°, and r° are the angles of an equilateral triangle, p° = q° = r° = ...............
d) If m° and n° are the base angles of an isosceles triangle and m° = 50°,
then n° = .................
Vedanta Excel in Mathematics - Book 8 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Triangle
15.2 Experimental verification of properties of triangles
Experiment 6: Experimental verification of sum of the angles of a triangle.
Step 1: Draw three triangles ABC with different measurements.
AA A
B C B Fig (ii) C B C
Fig (i) Fig (iii)
Step 2: Measure the angles A, B and C with the help of protractor and write the measurements
in the table.
Fig. No. BAC ABC ACB Result
(i) BAC + ABC + ACB = 180q
(ii) BAC + ABC + ACB = 180q
(iii) BAC + ABC + ACB = 180q
Conclusion: The sum of the angles of a triangle is 180q (2 right angles).
Theoretical proof P AQ
Given: In 'ABC, ABC, BCA, and BAC are the
angles of the triangle.
To prove: ABC + BCA + BAC = 180q. BC
Construction: PQ//BC is drawn through A.
Proof:
(i) PAB = ABC [Being alternate angles] proved
(ii) QAC = ACB [Being alternate angles]
(iii) PAB + BAC + QAC = 180q [The sum is a straight angle]
(iv) ABC + BAC + ACB = 180q [From (i), (ii) and (iii)]
Experiment 7: Experimental verification of base angles of an isosceles triangle.
Step 1: Draw three straight line segments AB of different lengths.
Step 2: From A and B, draw two arcs towards the same sides with the equal radius
intersecting at C by using pencil compasses.
Step 3: Join C, A and C, B. Now, ABC is an isosceles triangle in each figure.
A
CA
C
AB BC Fig (iii) B
Fig (i) Fig (ii)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 8
Geometry: Triangle
Step 4: Measure the size of A and B in each triangle with the help of protractor. Write the
measurement in the table.
Fig. No. BAC ABC Result
(i) BAC = ABC
(ii) BAC = ABC
(iii) BAC = ABC
Conclusion: The base angles of an isosceles triangle are equal.
Experiment 8: Experimental verification of base angles of an isosceles right angled
triangle.
Step 1: Draw three straight line segments AB of different lengths.
Step 2: Draw ABC = 90q at B with the help of protractor in each figure.
Step 3: From B, cut off BC on BP such that BC = AB in each figure.
Step 4: Join A, C. Now, ABC is an isosceles right angled triangle right angled at B in
each figure. P P
C C
BA
A B BA C
Fig (i) Fig (ii) P Fig (iii)
Step 5: Measure the sizes of A and C in each figure with the help of protractor. Write
the measurement in the table.
Fig. No. BAC ACB Result
(i) BAC = ACB =45°
(ii) BAC = ACB =45°
(iii) BAC = ACB =45°
Conclusion: Each base angle of an isosceles right angled triangle is 45°.
Experiment 9: The line drawn from the vertex of an isosceles triangle to join the
mid-point of the base is perpendicular to the base.
Step 1: Draw three isosceles triangles ABC of different measurements.
A AA
BP C B P C BP C
fig (i) fig (ii)
fig (iii)
Vedanta Excel in Mathematics - Book 8 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Geometry: Triangle
Step 2: Find the mid-point P of the base BC of each triangle and join A, P.
Step 3: Measure APB and APC with the help of a protractor and write the
measurements in the table.
Fig. No. APB APC Result
(i) APB = APC = 90°
(ii)
(iii)
Conclusion: The line drawn from the vertex of an isosceles triangle to join the mid-point
of the base is perpendicular to the base.
Experiment 10: The interior angles of an equilateral triangle are equal and each of 60°.
Step 1: Draw three equilateral triangles ABC of different measurements.
C
A
AC
A
B C B B
fig (i) fig (ii) fig (iii)
Step 2: Measure A, B and C with the help of a protractor and write the measurements
in the table.
Fig. BAC ABC ACB Result
No. BAC = ABC = ACB = 60°
(i)
(ii)
(iii)
Conclusion: The interior angles of an equilateral triangle are equal and each of 60°.
Experiment 11: Experimental verification of exterior angle and its opposite interior
angles of a triangle.
Step 1: Draw three triangles ABC with different measurements.
Step 2: Produce the side AB to D in each figure.
C A CA
C
A BD B B
Fig (i) D Fig (ii) D Fig (iii)
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 8
Geometry: Triangle
Step 3: Measure the sizes of CBD, BAC and ACB in each figure. Write the
measurement in the table.
Fig. No. CBD BAC ACB Result
(i) CBD = BAC + ACB
(ii) CBD = BAC + ACB
(iii) CBD = BAC + ACB
Conclusion: The exterior angle of a triangle is equal to the sum of two opposite interior
angles.
Worked-out examples
Example 1: Find the unknown sizes of angles in the following figures.
(i) (ii) P (iii) A E
A 70° 30° C
B 100°
35° b a
D 20°
B x 95° C Qa bR
O
Solution:
(i) x + 95q + 35q = 180q [Being the sum of the angles of 'ABC]
or,
or, x + 140q = 180q
x = 180q – 140q = 40q
(ii) a = b [Being the base angles of isosceles triangle PQR]
a + b + 70q
or, = 180q [Being the sum of the angles of 'PQR]
or, a + a
or, 2a = 180q – 70q
?
a = 110q
110
a = 2 = 55q
= b = 55q
(iii) a + 100q + 30q = 180q [Being the sum of the angles of 'ABC]
or, a + 130q
or, a = 180q
Now, b + 20q
or, b + 20q = 180q – 130q = 50q
or, b
= a [Being the exterior and opposite interior
= 50q angles of 'BDO]
= 50q – 20q = 30q
Vedanta Excel in Mathematics - Book 8 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Vedanta Excel in Mathematics Book - 8 Final (2078)
Related Publications
Discover the best professional documents and content resources in AnyFlip Document Base.
Search