SOLUTIONS
PULLOUT
Worksheets
Term 2 (October to March)
Mathematics
Class
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1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-1
Section - A
1.
1
x – 6y = 5
2.
4y = ax + 5
Here,
x = 2 and y = 3
So,
4(3) = a(2) + 5

12 = 2a + 5
½

2a = 7  a = 7/2.
½
Section - B
3.
Given,

2x – 5y = 7
5y = 2x – 7
2x  7

y =
5
2(–2) – 7
12

Now, when x = – 2, then y =
–2
5
5
Hence, the point (– 3, – 2) does not lie on the given line.
1
1
Section - C
4.
...(i)
Given equation is (p + 1) x – (2p + 3)y – 1 = 0
If x = 2, y = 3 are the solutions of the equation (i), then
(p + 1)2 – (2p + 3)3 – 1 = 0
2p + 2 – 6p – 9 – 1 = 0
–4p–8 = 0
2
p = – 2.
Put the value of p in the eqn. (i), then
–x + y – 1 = 0
or
5.
Given equation is
4x + 3y = 12
Put x = 0, then
0 + 3y = 12

1
x – y + 1 = 0.
y
y = 4
1½
(0, 4)
Hence point on y-axis is (0, 4).
Now put y = 0, then

4x + 0 = 12
Hence point on x-axis is (3, 0).
Section - D
6.
Given,

1½
x = 3
O
x
(3, 0)
x = 3y – 4
3y = x + 4
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
P-1

x4
3
y =
x
2
5
8
y
2
3
4
1
y
7
1
6
5
(11, 5)
4
3
(8, 4)
(5, 3)
2
(–1, 1)
1
(2, 2)
1
x'
x
–2
–1
–1
1
2
3
4
5
6
7
8
9
10
11
–2
y'
From the graph, it is clear that :
(i) when x = – 1, then y = 1,
(ii) when y = 5, then x = 11.
P-2
M A T H E M A T I C S --
1
1
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-2
Section - A
1.
The general form of a linear equation in two variable is ax + by + c = 0; where a, b, c are real
numbers and both a, b  0.
1
Section - B
2.
Putting x = 2k – 3, y = k + 2 in 2x + 3y – 15 = 0, we get
2(2k – 3) + 3(k + 2) + 15 = 0

4k – 6 + 3k + 6 + 15 = 0
15

k =
.
7
½
½
1
Section - C
3.
Given,
y = 2x
1
Putting y = 2x in the equation x + y = 6, we get
x + 2x = 6  3x = 6  x = 2

1
y = 2×2=4

4.
1
Required point is (2, 4).
Given,
3x + 2y = 1
1  3x
= y
2

(i) Put x = 1, then
y =
1  3(1) 2

 1
2
2
1
y =
1  3(3) 1  9

 4
2
2
1
(ii) Put x = 3, then
(iii) Put x = 5, then
1  3(5) 1  15

 7
2
2
Hence, three different solutions for the equation 3x + 2y = 1 are
1
y =
1
3
5
y
– 1
– 4
– 7
360
Section - D
5.
Graph of equation,
y = 90 x,
where x be time and y be distance.
From the graph,
Distance (in km)
x
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
y = 90 x
270
Q
180
90
0
(2, 180)
P (½, 45)
1
2 3 4
Time (in hours)
2
P-3
1
hour = 45 km
2
(ii) Distance travelled in 2 hours = 180 km.
2x + y = 8

y = – 2x + 8
1
(i) Distance travelled in
6.
x
– 1
0
1
y
10
8
6
1
2
From the graph it is clear that line intersects x-axis at (4, 0) and y-axis at (0, 8).
1
y
(–1, 10) 10
8
(1, 6)
6
1
4
2
x'
(4, 0)
–1
0
1
2
3
4
5
6
7
8
x
y'
P-4
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-3
Section - A
1.
1
Infinite.
2.
px – 2y = 10
At
x = 1 and y = –1
½
p(1) – 2 (–1) = 10

½
p = 10 – 2 = 8.
Section - B
3.
Put x = 4 and y = 3, then
3y – 2x = 3(3) – 2(4) = 1
So, (4, 3) is the solution of equation.
x = 2 2 and y = 3 2 , then
Again put
1
3y – 2x = 3(3 2 ) – 2(2 2 ) = 5 2  1
1
So, (2 2 , 3 2 ) is not solution of given equation.
Section - C
4.
Y
9
8
7
2
y=
3x
6
5
4
3
2
1
–2
–1
0
1
2
3
4
5
6
X
–1
–2
–3
–4
–5
–6
When
x
–1
–2
y
3
–6
1
Section - D
5.
Given
2x – 3y – 12 = 0  y =
2 x  12
3
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1½
P-5
x
6
9
3
y
0
2
– 2
y
5
4
3
2x – 3y –12 = 0
2
1
x'
–2 –1 0
–1
•
(9, 2)
(6, 0)
1
2
3
4
•
5
6
7
8
9
x
10
1½
•
(3, –2)
–2
–3
y'
Section - D
6.
Given, 3x + 4 = 5x + 8  5x – 3x = 4 – 8  2x = – 4  x = –2
1
(i) Representation on number line
x = –2
–4
–3
–2
–1
0
1
2
3
1
4
(ii) Representation on cartesian plane
y
4
x = –2
3
2
1
x'
–4
–3
–2
–1 0
–1 1
2
2
3
4
X
–2
–3
–4
y'
P-6
M A T H E M A T I C S --
IX
T E R M -- 2
On representation of solution on cartesian plane, we get a straight line parallel to y-axis and
this line possess infinitely many solutions.
7.
Given,

3x + 15 = 0
x =–5
P
–5
O
–4
–3
–2
–1
0
1
2
3
(i) Equation in one variable (Number line). A point P at a distance of 5 units to left of O on the
number line.
1
(ii) Equation in two variables (Cartesian plane).
A line AB parallel to y-axis at a distance of 5 units to the left of y-axis.
1
y
A
2
x'
–6
–5
–4
–3
–2
–1
B
1
0
2
3
x
y'
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
P-7
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-4
Section - A
1.
Here,
x = 0 and y = 2
So,
2x + 3y = k
or
2.
1
k = 2(0) + 3(2) = 6.
1
The equation of a line passing through origin is 0.x + 0.y = 0.
Section - B
3.
1
2x + 3y + k = 0
If x = 2 and y = 1 is the solution of the linear equation 2x + 3y + k = 0, then
2(2) + 3(1) + k = 0
k = – 7.
1
Section - C
4.
Given,
7x – 3y = 15
15  3 y
x =
7
x = 0;
7(0) – 3y = 15
0 – 3y = 15

At y-axis



½
½
½
½
15
=–5
3
Given line intersects the y-axis at y = – 5.
½
y =
x
+2
3
x
– 2x
3
x – 6x
– 5x
x
5.




½
= 2x – 3
=
=
=
=
–3–2
–5×3
– 15
3
1
Plot x = 3 on the cartesian plane.
On cartesian plane x = 3 is a line parallel to y-axis.
1
y-axis
x=3
3
2
1
–2
–1
1
1
2
x-axis
2
P-8
M A T H E M A T I C S --
IX
T E R M -- 2
Section - D
6.
2x – y = 4
y = 2x – 4

x
0
2
1
y
– 4
0
– 2
1
x+y = 2
y = 2–x

x
0
2
1
y
2
0
1
1
y
4
2x–
y=
4
3
(0, 2)
A
2
(1, 1)
1
x'
–4
–3
–2
–1
1
O
2
C
–1
(2,0)
–2
(1, –2)
–3
3
4
x
1
x+y=2
(10, –4)
B
–4
y'
From the graph  ABC is the required triangle and its vertices are A(0, 2), B (0, – 4) and C
(2, 0).
1
Value Based Question
7.
Since total number of voters who do not cast their votes = 40%
Hence total number of voters who cast their votes = 60%
Let the total number of voters are x and number of voters cast their votes is y.
Then according to question,
60
x
100
y = 60
y = 120
y = 180
...(1) 1
y =
Now, when x = 100, then
when x = 200, then
when x = 300, then
Thus, we have the following table :
x
100
200
300
y
60
120
180
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1
P-9
By plotting the Points (100, 60), (200, 120) and (300, 180) on the graph and by joining them, we get
the graph of equation (1) as shown in fig. From the graph we see that :
(i) If 300 voters cast their votes, then total number of votes = 500.
(ii) If the total number of voters are 1000, then number of votes cast = 600.
½
(iii) Everyone should cast his vote to elect an honest candidate.
½
y
1200
1100
1000
900
no of
voters
cast
800
(1200, 720)
700
(1000, 600)
600
y=
60 x
100
500
400
300
200
100
1
(300, 180)
(200, 120)
(100, 60)
x
O 100 200 300 400 500 600 700 800 900 1000 1100 1200
Total number of voters
P-10
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-5
Section - A
1.
Here,
5y = 2
2
y =
.
5
So, clearly,
1
Section - B
2.
Given, (– 1, – 5) lies on the graph of 3x = ay + 7

3 (– 1) = a(– 5) + 7

– 3 = – 5a + 7

5a = 7 + 3
10

a =
5

a = 2.
1
1
Section - C
3.
4.
1
Let larger number is x, then 5 times of larger number = 5x and smaller number is y.
 Quotient = 2 and remainder = 9
So according to question,
5x = 2y + 9

5x – 2y – 9 = 0.
Given,
1
1
y
x +y = 3
On the y-axis put x = 0, then
(0, 3)
B
y = 3
Hence, on the y-axis coordinate of B = (0, 3)
x+y=3
On the x-axis put y = 0, then
1
x = 3
Hence, on the x-axis coordinate of A = (3, 0)
Hence, a triangle whose sides are x = 0, y = 0
and
A
0 (0,0)
x + y = 3.
Again, the vertices of triangle are A (3, 0), B (0, 3) and O (0, 0).
(3, 0)
x
1
1
Section - D
5.
The fare for 1st km is Rs. 8
Let the total distance to be covered is x km
Fare for (x – 1) kilometre at the rate of Rs. 5 per km is 5 (x – 1)

Total fare y = 5 (x – 1) + 8

y = 5x – 5 + 8

y = 5x + 3
x
0
– 1
– 2
y
3
– 2
– 7
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
2
P-11
y
5
4
(0, 3)
3
2
1
x'
–5 –4
–3
–2
–1
O
X
1
–1
2
3
4
5
6
2
–2
(–1, –2)
–3
–4
–5
–6
(–2, –7)
–7
y'
Value Based Question
6.
P-12
(i) According to question, the equation of line is
x+y = 0
The line passes through a point whose y-coordinate is – 19·5

y = – 19·5
Put the value of y in equation (i), we get
x – 19·5 = 0
x = 19·5
So the required coordinates are (19·5, – 19·5).
(ii) According to question, the equation of line is
x = y
The line passes through point whose x-coordinate is 20.5

x = 20.5
Put this value of x in equation (ii), we get
y = 20.5
So the required coordinates are (20·5, 20·5).
(iii) The coordinates of all the points lying on a line satisfy the equation of the line.
(iv) Obedience and respect for elders or parents.
M A T H E M A T I C S --
IX
...(i)
1
...(ii)
1
1
1

T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-6
Section - A
1.
1
It cuts y-axis at –4.
3x =
2.


2 x+1

3 2 x = 1

x =
1
3 2



2
3 2
3
3 2

32

3 2

1
Section - B
3.
The equations of two lines on same plane which intersecting at point (2, 3) are
2
x + y = 5 and y – x = 1.
Section - C
4.
Writing in standard form
3x + 2y – 12 = 0
½
3x – 12 = 0
1
x = 4
½
On x-axis, y = 0 

Point on the x-axis = (4, 0).
On y-axis, x = 0 
2y – 12 = 0
½
y = 6

5.
½
Point on the y-axis = (0, 6).
According to question,
1
10x = 4y + 18

y
5x – 2y – 9 = 0.
Section - D
6.
6
Given,
2
x – y = 2
3
5

2x – 3y = 6
3

4
2
2x = 3y + 6

x =
1
(6, 2)
1
3y  6
2
x'
x
0
3
6
y
– 2
0
2
0
–1
(3, 0)
1
–2 (0, –2)
y'
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
2
3
4
5
6
7
x
2
P-13
7.
(i) When the line cuts x-axis, then put y = 0
i.e.,
2x

x
Hence point is (3, 0).
(ii) When the line cuts y-axis then put x = 0
i.e.,
3y + 6

y
Hence point is (0, – 2).
According to question
x + y = 10 y = 10 – x
= 6
= 3
1
= 0
= –2
1
x
0
2
3
4
5
y
10
8
7
6
5
...(1)
1
2x – y = 5  y = 2x – 5
Also,
...(2)
x
0
2
5
y
–5
–1
5
1
Plot these points on graph paper
y
(2,8)
8
(3,7)
7
(4,6)
6
5
2x – 7 = 5
(5,5)
4
2
3
2
x + y = 10
1
x'
–8 –7 –6 –5 –4 –3 –2 –1 0
–1 1
–2
x
2
3 4
(2, –1)
5
6 7
8
–3
–4
(0, –5)
–5
–6
–7
–8
y'
From graph it is clear that point of intersection is (5, 5).
P-14
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-7
Section - A
1.
It will intersect at y = 2.
1
2.
The solution of the equation is (1, – 1).
1
Section - B
3.
Equation :
4x + 3y = 12
For intersection with x-axis  y = 0

4x = 12.
or
x = 3
 Co-ordinates are (3, 0)
For intersection with y-axis 
x = 0

3y = 12
or
y = 4.
 Co-ordinates are (0, 4).
½
½
½
½
Section - C
4.
The equation is
y = 9x – 7
A(1, 2) :
2 = 9(1) – 7
2 = 2;
B(–1, –16) :
1
True
1
True
1
–16 = 9(–1) – 7 = – 9 – 7
= –16;
C(0, –7) :
–7 = 9(0) – 7
= 0 – 7 = –7;
5.
True
Equation :
3x + 4y = 7
x
1
– 3
y
1
4
Equation :
y
1
3x–2y=1
3x – 2y = 1
x
1
– 1
y
1
– 2
(1,1)
From graph point of intersection is (1, 1).
3x
+4
x
y=
7
1
Section - D
6.
Let,
Age of Amit = x years
Age of Akhil = y years
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1
P-15
(i)

According to question the linear equation for above situation is
x + y = 25
y = 25 – x
x
0
10
15
y
25
15
10
1
y
25 (0, 25)
20
(10, 15)
15
11
(15, 10)
10
5
0
5
10 14 15
20
25
x
1
(ii) From the graph, when Amit’s age = 14 years, then Akhil’s age = 11 years.
P-16
M A T H E M A T I C S --
IX
1
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-8
Section - A
1.
The equation is
2.
The graph of x = –1 is parallel to y-axis.
1
x + y = 0.
1
Section - B
3.
On x-axis y co-ordinate is zero. So put y = 0 in 3x – 2y = 6, we get
1
3x – 0 = 6

x =
6
=2
3
½
 3x – 2y = 6 meets x-axis at (2, 0).
½
Section - C
4.
Given,

y 3 = 8x +
8x – y 3 +
3
3 = 0
Putting x = 0, y = – 1, we get
3 +
3  0
1
 (0, – 1) is not the solution of given equation.
Putting x =

3 , y = 9, we get
8 3 –9 3 +
3 = 0, which is correct
 ( 3 , 9) is a solution of given equation.
5.
Given,
1
2x – 3y = 12
 x  12
.
3
When line cuts x-axis,
y = 0
 x  12

0 =
3
x = 6
When line cuts y-axis, put x = 0, then
y = –4
 Point of intersection with x-axis and y-axis are (6, 0) and (0, – 4).

1
y =
1
1
1
Section - D
6.
Equation,
3x – 5y – 15 = 0
Substituting y = 0 in the equation 3x – 5 × 0 – 15 = 0  x = 5
Hence the point on x-axis is (5, 0).
When x = 0 in the equation, we get
(3 × 0) – 5y – 15 = 0  y = –3
 point of y-axis is (0, – 3).
Plotting A (5, 0) and B(0, – 3), we get the triangle AOB.
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1
P-17
y
5
4
3
2
1
x'
–4
–3 –2
A (5, 0)
–1 O
–1
B (0, –3)
1
–2
3x
–3
2
–5
3
y–
4
=
15
x
5
2
Fig.
0
–4
y'
Since,
7.
OA = 5 units and OB = 3 units
1
1

Area of triangle AOB =
× OA × OB =
× 5 × 3 = 7.5 sq. unit.
2
2
Let F be the force applied and a be the acc. produced, therefore
F  a

F = ka, where k is constant
Replace a by x and F by y.
y = k(x)

y = xk
Here,
k = 5

y = 5x
x
0
1
– 1
y
0
5
– 5
1
1
1
y
(1, 5)
5
4
3
2
x'
1
1
–5 –4 –3 –2 –1 O
2 3 4
5
x
–1
–2
–3
–4
–5
(– 1, – 5)
y'
2
P-18
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-9
Section - A
1.
 9 
The solution in two variables is   , m  .
 2 
½+½
Section - B
2.
If point (3, 4) lies on 3y = ax + 7, then

1
3 × 4 = 3a + 7

3a = 12 – 7 = 5  a =
5
.
3
1
Section - C
3.
4.
(a)


(b)

(c)

When y = 3, then
2x + 3
2x
x
2(4) + y
y
2 +y
When x = 4, then
When x = 1, then
one more solution is (1, 5).
Cost of a pen
Number of pen bought

total cost

16x

16x – y
=
=
=
=
=
=
7
4
2.
7
7 – 8 =– 1
7y=5
=
=
=
=
=
` 16
x
16x
y
0
1
1
1
Y
8
7
6
5
4
3
2
X'
1
–4
–3
–2
–1
1
2
3
4
X
1
–1
–2
–3
–4
–5
16x – y = 20
–6
–7
–8
Y'
Cost of 6 pen = 16 × 6 = ` 96.
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1
P-19
Section - D
5.
Given,

x +y = 7
y = 7–x
x
5
7
4
y
2
0
3
1
y
5
4
(4,3)
3
(5,2)
2
1
(7,0)
x'
–1
0
1
2
3
4
5
–1
6
7
8
9
(8,–1)
x
2
–2
y'
From graph it is clear that (8, – 1) lies on the line AB
Hence (8, – 1) is a solution of the given equation.
1
100
P-20
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-10
Section - A
1.
The highest degree of x in linear equation is 1.
1
Section - B
2.
½
½
½
½
Substituting x = – 1 and y = – 1 in 9kx + 12ky = 63, we get
9k (– 1) + 12k (– 1) = 63
– 9k – 12k = 63
–21k = 63  k = – 3.
Section - C
3.
4.
On putting (3, 4) in the equation of the line
3y =
3(4) =
12 =

3k =
kx + 7
k (3) + 7
3k + 7
5
5
k =
3
The equation becomes
9y = 5x + 21
Two more solutions are (– 6, – 1) and (3, 4).
Given,
Equation is
3x – 5y – 15 = 0

1
1
y
4
3
5y = 3x – 15
3 x  15
y =
5
3
y =
x–3
5
Put x = 0, then y = – 3


2
1
x'
–5 –4
–3 –2
Put x = –5, then y = –6
B
C
x
0
5
–5
y
–3
0
–6
–1
1
–1
2
3
4
5
B (5, 0)
x
–2
Put x = 5, then y = 0
A
1
–5
C(
, –6
)3
5
x–
y
5
–1
=0
–3
A (0, –3)
1
–4
–5
–6
y'
To plot these points on graph paper join these points.
The graph of the line intersects x-axis at (5, 0) and y-axis at (0, –3).
1
1
Section - D
5.
Total students in the class = y
Let the boys in the class = x, then equation between the students and the boys is
4
x
y =
3
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1
P-21
Now,
x
0
30
60
y
0
40
80
To draw a graph between students and boys is
y
80
70
60
students
(60, 80)
50
40
(30, 40)
2
30
20
10
0
x
10 20 30 40 50 60 70 80
boys
6.
1
From the graph it is clear that there are 30 boys in a class of 40 students.
Let Sita contribute = Rs. x and Gita contribute = Rs. y
According to question,
x + y = 200
y = 200 – x
x
0
200
100
y
200
0
100
1
1
y
(0, 200)
(100, 100)
2
(200, 0)
P-22
x
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-11
Section - A
1.
The equation of a line parallel to y-axis is
x = a.
1
Section - B
2.
700 + 25x = y
1½
25x – y + 700 = 0.
1½
According to question,
Section - C
3.
3x = y + 3; three solutions are x = 1, y = 0; x = 2, y = 3 and x = 0, y = – 3.
4
3
1
(2, 3)
2
1
–1
–2
–3
4.
1
(1, 0)
1
2
3
4
(0,– 3)
From graph it is clear that line meet x-axis at (1, 0) and y-axis at (0, – 3).
(i) Given,
3x + 2y = k
put x = 2, y = 1, then
3(2) + 2(1) = k  k = 8.
(ii) Given,
2x – ky = 6
put x = 2, y = 1, then
2(2) – k (1) = 6
4 – k = 6  k = 4 – 6 = –2.
x y

(iii) Given,
= 5k
4 3
2 1

put x = 2, y = 1, then
= 5k
4 3

6 + 4 = 60k
1

k =
.
6
1
1
1
Section - D
5.
5x – 2 = 3x – 8
2x = – 6
x = –3
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
1
P-23
(a) Point P (– 3, 0) represents the solution x = – 3 on the number line
P
–4
–3
–2
–1
0
1
2
3
1½
4
(b) Line AB represents the solution in the cartesian plane.
y
B
x = –3
3
2
1
x'
–4
–2
–1
0
1
2
3
4
1½
x
–1
A
P-24
–2
y'
M A T H E M A T I C S --
IX
T E R M -- 2
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-12
Section - A
1.
At x-axis, y = 0
so,
3x = 15

x = 5.
1
Section - B
2.
Equation is
2x – 3y = 12
or
½
3y = 2x – 12

y =
2 x  12
3
On x-axis, y = 0
2 x  12
= 0x=6
3
½
 At point (6, 0) the given line cuts the x-axis.
On y-axis,
x = 0
y =

2  0  12
3
y = –4
1
At point (0, – 4) the given line cuts the y-axis.
Section - C
3.
The line passing through (2, 14) is
2y = 14x
or,
1
y = 7x
1
Infinitely many lines are there.
The equation in the form ax + by + c = 0 is
1
7x – y + 0 = 0.
4.
Let,
the cost of a toy elephant = x, ball = y
3y = x  y =
x
3
LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
P-25
x
3
6
9
y
1
2
3
1½
y
4
(9, 3)
3
2
(6, 2)
1
x'
–2 –1 0 –1 1
y'
5.
Given,
(3, 1)
2
3
4
5
1½
6 7
8
9 10
–2
x+y=3
2x + 2y = 8
x
1
2
x
1
2
y
2
1
y
3
2
4
1
(1, 3)
(1, 2) (2, 2)
3
(2, 1)
2
2
1
–4
–3
–2
–1
–1 1
–2
2
3
4
x
5
+
y
=
3
2x
+
2y
=
8
–3
–4
1
The lines are parallel to each other.
6. Given equation is
3x + 4y = 6
(i) When it cuts x-axis, then put y = 0 i.e.,
3x = 6

x = 2
Hence point on x-axis is (2, 0).
(ii) When it cuts y-axis then put x = 0 i.e.,
P-26
M A T H E M A T I C S --
1
IX
T E R M -- 2
4y = 6 y = 3/2
3
Hence point on y-axis is  0, 
2


1
3x + 4y = 6

6  3x
4
y =
x
2
– 2
6
y
0
3
– 3
1
y
(–2, 3)
(2, 0)
x'
x
– 6 – 5 – 4 –3 –2 –1 0
–1
–2
–3
1
(16, –3)
–4
y'

LINEAR EQUATIONS IN TWO VARIABLES
L I N E A R E Q U A T I O N S I N O N E V A R I A B LE
P-27
1
LINEAR EQUATIONS IN TWO VARIABLES
WORKSHEET-13
Type A
1.
Coordinate of A can be obtained by substituting x = 0 in the equation of the given line, as the
point lies on the line. Hence,
Equation of line is

1
2x + 3y = 6
2.0 + 3.y = 6, as x = 0

3y = 6

y = 6/3

y = 2
So, A is (0, 2).
Similarly for point C, substitute y = 0, to get C (3, 0).
However, the points can also be found out simply by observation
Hence, D(6, –2) and B (–3, 4).
2.
This is the graph of linear equation.
1
3.
Yes, for x = 1 :
1
2x + 3y = 6

2.1 + 3y = 6

3y = 6 – 2
Also, for x = 2 :
4
3
2x + 3y = 6

2.2 + 3y = 6

4 + 3y = 6

3y = 2

y =
2
.
3
Hence two other solutions for the given equation are : (1, 4/3) and (2, 2/3).

y =
4.
There are infinite possible solutions of this equation.
1
5.
The triangle formed by the given line and the coordinate axes is a right triangle AOC.
1
Type B
3.
each.
4.
x-axis.
5.
Equation in two variables and degree 2.
6.
2(1 +
7.
0.x + 3y + 5 = 0.
P-28
5 ).
M A T H E M A T I C S --
IX
T E R M -- 2
2
QUADRILATERALS
WORKSHEET-14
Section - A
1.
The value of x is 95º.
1
2.
The value of Q – S = 0º.
1
Section - B
3.
In BAC and DCA
C
D
1
3
AC
BAC
=
=
=

2
4
(Alternate pair)
AC
(Common)
DCA, (By ASA)
2
1
4
1
3
1
A
B
AM  DC
4.
AN  BC
In quadrilatral AMCN,
A + M + C + N = 360°

A + C
50° + C

C
In parallelogram,
A
B
B
A
=
=
=
=
=
5 0°
180°
180°
130°
D
C = 130°
D = 180° – 130° = 50°.
N
C
M
1
1
Section - C
5.
DE || PR and DE =
1
PR (mid-point theorem)
2
P
EF || PQ and EF =
1
PQ
2
DF || QR and DF =
1
QR
2
As
D
Q
1
1
1
PQ =
QR =
PR
2
2
2
DE = EF = DF

  DEF is an equilateral triangle.
Given,
R
1½
1
½
PQ = QR = RS, PQR = 128°
(180  128)
1 = 2 =
2
52º
= 26º
2
PTQ = QRP = 26°.
=

E
PQ = QR = PR (PQR is an equilateral triangle)
So,
6.
F
Q U A D R I L A T E R A L S
1
P-29
PTS = 78°
ROS = 2 RTS = 2 × 26° = 52°.
1
1
Section - D
7.
8.
1½
1
½
As AC and BD bisect each other.
Therefore, AC and BD are the diagonals of the parallelogram
 ABCD is a parallelogram.
D
  A =  C and  B =  D
But, ABCD is a cyclic quadrilateral
  A +  C = 180° and  B +  D = 180°
 2 A = 180° and 2 B = 180°
A
  A = 90° and  B = 90°
  A =  B =  C =  D = 90°
 ABCD is a rectangle and diagonals AC and BD are diameters.
(i) Since ABCD is a square and DCE is an equilateral triangle



Similarly, we have
C
O
½
B
½
ADC = 90° and EDC = 60°
ADC + EDC = 90° + 60°
A
ADE = 150°
B
1
BCE = 150°
Thus in ADE and BCE, we have
9.
P-30
AD = BC
ADE = BCE = 150°
and
DE = CE
So by SAS congruence criterion, we have
ADE  BCE

AE = BE
(ii) In EAD, we have
AD = DE

EAD = AED = x (say)
Now,
ADE + AED + DAE = 180°

150° + x + x = 180°

2x = 180° – 150° = 30°

x = DAE = 15 °.
Given : Two triangles ABC and DEF, such that
AB = DE and AB || DE
Also
BC = EF and BC || EF
B
(a) In a quadrilateral ABED,
AB = DE and AB || DE
 one pair of opposite sides are equal and parallel.
 ABED is a parallelogram
 AD = BE and AD || BE.
(b) In quadrilateral BCFE,
BC = EF and BC || EF
 One pair of opposite sides are equal and parallel.
 BCFE is a parallelogram.
 CF = BE and EF || BE.
C
M A T H E M A T I C S --
D
1
E
Proved. 1
Proved. 1
A
C
D
E
Proved.
...(i) 1
F
Proved.
....(ii) 1
IX
T E R M -- 2
(c) From equations (i) and (ii), we get

AD =
 ACFD is a parallelogram

AC =
(d) In ABC and DEF,
AB =
BC =
and
AC =
So by S.S.S.
ABC 
Q U A D R I L A T E R A L S
CF and AD || CF
DF and AC || DF
DE (given)
EF (given)
DF (Proved above in part (c))
DEF.
Proved.
1
Proved.
1
P-31
2
QUADRILATERALS
WORKSHEET-15
Section - A
1.
The value of RQT is 55º.
1
2.
Square.
1
Section - B
APR = DRP (Alternate interior angles)
3.
1 = 2
or
But these are alt. int angles

t
SP || QR, SR || PQ
A
PQRS is a parallelogram
P
1
APR + BPR = 180°, (linear pair)
1
1
1
APR + BPR =
× 180°
2
2
2
1 + 3 = 90°
S
B
l
1
3
Q
2
C
R
m
D
SPQ = 90°
 PQRS is a rectangle.
1
Section - C
4.
(i)
AB || CD, AD || BC
 2 =  4,
D
1 = 3
3
1 = 2
But
(As diagonal AC bisects A)
1
3 = 4
4
2
1
A
 AC bisects  C.
C
B
 2 =  4 But  1 =  2
(ii)
1 = 4
1
AB = BC (sides opp. to equal angles)
 ABCD is a rhombus. In a parallelogram, if one pair of adjacent sides are equal, then it is
rhombus.
1
5.
Construction : Join AC.
D
In DAC, S and R are mid points of DA and DC.
By mid-point theorem : SR || AC and SR =
1
AC
2
1
AC
2
From (1) and (2), we get SR || PQ and SR = PQ
Hence, PQRS is a parallelogram.
Similarly, in BAC, PQ || AC and PQ =
P-32
...(1)
...(2)
M A T H E M A T I C S --
R
C
1
Q
S
A
P
1
B
1
IX
T E R M -- 2
Section - D
6.
A
B
E
F
H
D

C
 21 A + 21 B 
1
1
CGB = 180° –  C + D 
2
2
FEH = AED = 180° –
1
FGH =
1
FEH + FGH

7.
G
360 –
Adding,
FEH + FGH = 180°
 EFGH is cyclic quadrilateral.
Const. : Join SP, PQ, QR, RS and AC
Proof : In DAC,
In ABC,
1
(A + CB + LC + CD)
2
1
1
Proved.
RS || AC and RS =
1
AC (mid-point theorem)
2
...(i) 1
PQ || AC and PQ =
1
AC (mid-point theorem)
2
...(ii) 1
From (i) and (ii), we get
D
R
C
1
RS || PQ and RS = PQ
8.
 PQRS is a parallelogram.
S
O
Since diagonals of parallelogram bisect each other.
 PR and QS bisect each other.
A
P
In ACE,
43° + 62° + d = 180°
d = 180° – (43° + 62°)
= 180° – 105° = 75°
a + d = 180° (opp. angles of cyclic quad. are supp.)
a + 75° = 180°
C
a = 80° – 75° = 105°
In ABF,
62° + 105° + b = 180°
b = 180° – (62° + 105°)
b = 180° – 167° = 13°
DEF = 180° – 75° = 105°
B
In DEF,
105° + 130° + c = 180°
a
D
118° + c = 180°
c
c = 180° – 118° = 62°
d
a = 105°, b = 13°, c = 62°, d = 75°.
A
E
Q U A D R I L A T E R A L S
Q
B
1
1
1
1
b
F
1
P-33
2
QUADRILATERALS
WORKSHEET-16
Section - A
1.
2.
1
mQ = 80º.
The value of S is 175º.
1
Section - B
3.
Since opposite angles of parallelogram are equal

3x – 2 = 63 – 2x

x = 13°
1
Angles of parallelogram :
(39 – 2)°, (180 – 37)°, (63 – 26)°, (180 – 37)°
1
i.e., 37°, 143°, 37°, 143°.
Section - C
4.
 1 = 2 (AE is angle bisector)
Here
 1 = 3 (alternate angles as AD || BC)
  3 =  2
BE = AB (sides opposite to equal angles)
But
Hence
But
BE =
AB =
and
5.
3
1
A
1
AD.
2
Given, PQRS is a parallelogram.

P + Q = 180° (adjacent angles)


1
=

E
B
180
 90
2
1
PTQ = 180  (P  Q)
2
= 180° – 90°
PTQ = 90°.
1
1
P  Q
2
2

2
1
Proved. 1
AB =

C
D
1
BC, (E is mid-point of BC)
2
1
BC
2
BC = AD (opposite sides of parallelogram)

1

1
1
Section - D
6.
Join B and E.
ADEB is a cyclic quadrilateral

P-34
ADE = EBC,
(ext. angle of a cyclic quad. = int. opp. angle)
M A T H E M A T I C S --
IX
...(1) 1
T E R M -- 2
Similarly FECB is a cyclic quadrilateral,

EBC + CFE = 180°
From (i),
ADE + CFE = 180°
But these are interior angles for the lines AD and CF.
Interior angles are supplementary.

AD || CF.
(opp. S are supp.)
1
1
Proved. 1
7.
D
R
C
S
A
Q
P
B
Const. : Join PQ, QR, RS, SP and AC.
Proof : In ABC,
P is mid-point of AB
Q is mid-point of BC (Given)

1
AC (Mid-point theorem)
2
1
SR || AC and SR =
AC
C
2
PQ || AC and PQ =
Similarly in ADC,
From the above results
1
AC
C
2
Hence PQRS is a parallelogram as pair of opposite sides are equal and parallel.
 PR and QS bisect each other (diagonals of parallelogram PQRS).
PQ || SR (both || AC) and PQ = SR =
Q U A D R I L A T E R A L S
1
½
½
1
½
½
P-35
2
QUADRILATERALS
WORKSHEET-17
Section - A
1.
AOB = 90º.
The
1
Section - B
2.
Construction : Join AC.
In DAC, S and R are mid points of DA and DC.
1
By mid-point theorem : SR || AC and SR =
AC
2
1
Similarly, in BAC, PQ || AC and PQ =
AC
2
From (1) and (2), we get SR || PQ and SR = PQ
Hence, PQRS is a parallelogram.
...(1)
1
C
R
D
Q
S
...(2)
A
B
P
1
Section - C
3.
DE =
EF =
DF =
Perimeter of DEF =
=
=
=
1
AC
C
2
1
AB
2
1
BC (Mid point theorem)
2
DE + EF + DF
1
(AC +AB + BC)
2
1
(7·8 + 6 + 7·2)
2
1
(21) = 10·5 cm.
2
1
A
B
1
F
D
C
E
1
Section - D
4.
(i)
In ABC, M is mid-point of AB and MD || BC
1
 D is mid-point of AC.
(ii) MD || BC BCD = MDA = 90°
MD AC
(iii) M is mid-point of AB, i.e.,
1
AB
2
ADM  CDM (SAS)
AM = MB =
CM = MA =
5.
A
1
AB.
2
1
1
M
D
C
B
1
In triangles APD and CQB,
PD = BQ (Given)
AD = BC
P-36
M A T H E M A T I C S --
IX
T E R M -- 2
ADP = QBC (Alt. angles)
APD  CQB
{SAS}
AP = CQ (c.p.c.t.)
AB = CD (Opp. sides of a parallelogram)
BQ = PD (Given)
PDC = BQA

AQB  CPD
{SAS}

AQ = CP
 APCQ is a quadrilateral in which opposite sides are equal.
 APCQ is a parallelogram.


In triangles AQB and CPD,
6.
1
½
1
½
1
In AOD and AOB,
AD = AB
(sides of rhombus)
OD = OB
(given)
AO = AO

AOD = AOB (S.S.S.)
Similarly,
Now,
But

Hence
D
AOD = AOB
Hence,

(common)
C
DOC  AOB
1
O
DOC = BOC
AOD + DOC = AOB + BOC
1
A
1
B
AOD + DOC + AOB + BOC = 360° (complete angles)
AOD + DOC = 180°.
AOC is a straight line.
Q U A D R I L A T E R A L S
Proved. 1
P-37
2
QUADRILATERALS
WORKSHEET-18
Section - A
1.
It is a square.
1
2.
(a) Opposite sides are equal.
½
(b) Opposite angles are equal.
½
Section - B
3.
In  ADC, P and Q are mid points of lines DA and DC respectively.
So,
PQ || AC
DPQ = PAC = 30° (corresponding angles for PQ || AC)
y = 30°.
In  PDQ,

1
y + 120° + DQP = 180°
30° + 120° + DQP = 180°

DQP = 30°

½
x = 180° – DQP =180° – 30° = 150°.
½
Section - C
4.
  l || m

XCA = YAC (Alt. int. angles)
1
1
XCA =
YAC
C
2
2

 1 =  2
But by position they are alternate interior angles

CB || DA
Similarly,
AB || DC
Therefore, ABCD is a parallelogram.
½
p

X
C
l
1
D
B
2
m
A
Y
1
1
½
Section - D
5.
Proof : BAC = BDC = 30° (angles in the same segment)
C
D
In DBC,
Now,



1
BDC + DBC + BCD =
30° + 70° +  BCD =
BCD =
AB =
BAC =
BAC =
ECD =
E
180° (A.S.P.)
70º
180°
30º
80°
A
B
BC
BCA (angles opp. to equal sides)
30°
80° – 30° = 50°.
1
1
Value Based Question
6.
(i) Let the photo-frame be ABC such that BC = a, CA = b and AB = c and the mid-points of
AB, BC and CA are respectively D, E and F.
We have to determine the perimeter of DEF.
P-38
M A T H E M A T I C S --
1
IX
T E R M -- 2
In ABC, DF is the line-segment joining the mid-points of sides AB and AC.
So, DF is parallel to BC and half of it.
and

i.e.,
DF =
BC a

2
2
Similarly,
DE =
AC b

2
2
A
c D
AB c

EF =
2
2
DF + DE + EF =
a b c
 
2 2 2
½
B
abc
2
1
( a  b  c) .
Hence, required perimeter =
2
(ii) Mid point theorem.
(iii) Unity and cooperation or mutual understanding.
F
E
a
b
1
C
=
Q U A D R I L A T E R A L S
½
½
½
P-39
2
QUADRILATERALS
WORKSHEET-19
Section - A
1.
The value of R is 100º.
1
Section - B
2.
Since PQRS is a parallelogram.
So, PS|| QR and PQ || SR, QS is a transversal

4y =

y =

10x =
x =
20° (Alternate interior angle)
5°
60° (Alternate interior angle)
6°.
Section - C
3.
1
A
Given, ED || BC


4.
1
AB = AC
 2 =  3
 3 +  4 = 180°
(Interior angles)
 2 +  4 = 180°
1
B
4
2
EF =
C
1
1
AC
C
2
EF || AC
Similarly,
GH =
...(i)
G
1
DC
AC and GH || AC
2
 EFGH is a parallelogram.
Also,
In  AEH and  BEF,
1
3
{from above two eqns.}
But these are opposite angles of a quadrilateral.
 BCDE is a cyclic quadrilateral.
In  ABC, E and F are mid-points of AB and BC respectively.

1
D
E
C
...(ii)
H
F
A
B
AD = BC AH = BF
E
1
AE = BE, AH = BF and  A =  B = 90°

 AEH   BEF

EH = EF
From (i) and (iii), EFGH is a parallelogram with EF = EH.
...(iii) 1
1
i.e., EFGH is a rhombus.
Section - D
5.
In ADF, E is the mid point of side DF and BE is parallel to AD.
(given)
So by converse of mid point theorem B is the mid point of side AF.
P-40
M A T H E M A T I C S --
IX
T E R M -- 2
So,
AB = BF
But
AB + BF = AF
AF = 2AB.
4
D and E are mid-points of AB and BC respectively.
1
So,
6.
(By figure)

DE || AC
Similarly,
DF || BC and EF || AB
½
 ADEF, BDFE and DFCE are all parallelograms.
1
DE is the diagonal of parallelogram BDFE.
½
Similarly,
~ FED
BDE =
~ FED
DAF =
and
~ FED
EFC =

½
 All the four triangles are congruent.
7.
½
A
Through C draw CE || AD.
 AECD is a parallelogram   A +  2 = 180°
In BCE, BC = CE   1 =  2
Also,
B
2 E
1
 B +  1 = 180°
A = B.
From (i) and (ii), we get
....(i) 1
....(ii) 1
D
C
(As 1  2 )
Again, we get
A +  D =  B +  C = 180°

C = D.
1
In triangles ABC and BAD,
AB = BA
B = A
BC = AD
So,
ABC = BAD (SAS).
Q U A D R I L A T E R A L S
1
P-41
2
QUADRILATERALS
WORKSHEET-20
Section - A
1.
Square.
1
2.
The value of x + y = a + b.
1
Section - B
3.
Let, ABCD is a parallelogram and opposite angles of it are equal.
1

But


 ABCD is a rectangle.
1
B
B + D
2D
D
=
=
=
=
D
180°
180°
90°
(Opp. angles of a cyclic quadrilateral)
Section - C
4.
Let the measures of the angles be 2x, 4x, 5x and 7x.

D
2x + 4x + 5x + 7x = 360°

18x = 360°  x = 20°

 A = 40°,   B = 180°,
As

5x
7x
2x
 C = 100°,  D = 140°
1
C
1
4x
A
B
 A +  D = 180° and  B +  C = 180°
CD || AB  ABCD is a trapezium.
1
Section - D
5.
In ADG, E is the mid-point of AD and EF || DG.
 F is the mid-point of AG (converse of mid point theorem)

AF = FG
In CBF, BF || DG, D is the mid-point of BC.
6.
P-42
1
A
...(i)
F
E
 G is the mid-point of FC

FG = GC
From (i) and (ii),
AF = FG = GC
B
G
D
Proved.
AC = AF + FG + GC = 3AF .
Given : AB and CD are equal chords intersecting at 90°.
To prove : OPQR is a square.
Proof : Since P and R the mid points of AB and CD respectively.
{A line segment from the centre to bisect the chord is to the cuord}

OPB = ORD = 90°

OPQ = ORQ = 90°
Since equal chords on a circle are equidistant from the centre.

OP = OR
M A T H E M A T I C S --
C
1
...(ii) 1
1
1
IX
T E R M -- 2
Thus in OPQ and ORQ, we have
OP

OPQ
and
OQ

OPQ
Thus in quadrilateral OPQR, we have
OP
and
OPQ
Hence OPQR is a square.
Q U A D R I L A T E R A L S
= OR
= ORQ
= OQ {Common}
~ ORQ
=
1
1
= OR, PQ = RQ
= ORQ = 90°
Proved.
1
P-43
2
QUADRILATERALS
WORKSHEET-21
Section - A
1.
The largest angle of the quadrilateral is 140º.
1
2.
The smaller angle is 45º.
1
Section - B
3.
In SOR and QOR,
SO = QO

SOR = QOR = 90°
½
OR = OR
(Common)

SOR  QOR

SR = QR
Similarly,
S
QORQOP  SOP

PQ = QR = RS = SP
 quadrilateral PQRS is a rhombus.
½
R
½
O
P
½
Q
Section - C
4.
Here,
 DAB =  DCB (opp. angles of parallelogram) ...(1) ½
 2 =  1 (given)

But
 PAQ =  QCP
 QCP =  CQB

 PAQ =  CQB
P
D
...(2)
{Subtracting (2) from}
(alternate angles)
1
C
1
2
A
Q
B
But these form corresponding angles.

AP || QC
Also,
½
AQ || PC (as AB || DC of parallelogram ABCD)
AQCP is a parallelogram. (both pairs of opp. sides are parallel).
1
Section - D
5.
 ABC is an isosceles triangle

ABC = BCA
PAC = ABC + BCA = 2 BCA
AD bisects  PAC   
PAC = 2 DAC
From (i) and (ii),
BCA = DAC.
...(ii)
1
These are alternative angles when lines BC and AD are intersected by AC
P-44
M A T H E M A T I C S --
...(i) 1
IX
1
T E R M -- 2

BC || AD.
Also
BA || CD (given)
ABCD is a parallelogram.
1
Value Based Question
6.
(i) R = 80° (Given)
SR || PQ and RQ is a transversal.

R + Q = 180° (co-interior angles)

Q = 180° – 80°
= 100°
S
1
P
g.
80º
R
Q
Similarly,
Q + P = 180°

P = 180° – 100° = 80°
and
S + R = 180°

S = 180° – 80° = 100°
Hence, P = 80°, Q = 100°, R = 80°, S = 100°
1
(ii) Property of sum co-interior angles when a pair of straight lines intersected by another
straight line (Geometry)
1
(iii) Deligence i.e., dedication, determination and hard work.
1
Q U A D R I L A T E R A L S
P-45
2
QUADRILATERALS
WORKSHEET-22
Section - A
1.
The perimeter of DEF is 6.4 cm.
1
2.
The length of CD is 6 cm.
1
Section - B
3.
If a line is drawn parallel to one side of a triangle through the mid-point of the second side,
then it bisects the third side.
 D is the mid-point of AC.
Since,

MD || BC
ADM = ACB
= 90°
MD  AC.

1
(Corresponding angles)
1
Section - C
4.
Since ABCD is a cyclic parallelogram.
A = C
So,
 A +  C = 180°
and

A
B
D
C
 A =  C = 90°
Thus, ABCD is a parallelogram with one angle of 90°
1
1
½
ABCD is a rectangle.
5.
½
(i) In triangles APD and CQB,
AD = BC
(Opp. sides of a parallelogram)
PD = BQ
(Given)
ADP = QBC

 APD   CQB

AP = CQ
(Alt. Angles)
1
(cpct)
(ii) In triangles AQB and CPD,
AB = DC, BQ = DP
and
ABQ = PDC (Alt. angles)

 AQB   CPB AQ = CP (c.p.c.t.)
1
(iii) In quad. APCQ, AP = CQ and AQ = CP
A
 APCQ is a parallelogram.
1
Section - D
6.
P-46
Given :
To prove :
Proof :
D
ECB + EDE = 180°
AB = AC  1 = 2
AD = AE  3 = 4
1
B
M A T H E M A T I C S --
IX
3
4
E
2
C
1
1
T E R M -- 2
7.

A + 1 + 2

21

1
 DE || BC (Corresponding angle)

1 + BDE
1 + CED
 B, C, E, D are concyclic.
Const. : Draw RM || SP meeting PQ in M.
Proof :
 PMRS is a parallelogram

In RMQ,
= 180° = A + 3 + 4
= 23
= 3 = 2 = 4
1
= 180° (BDE = CED, as 3 = 4)
= 180°
S
PQ || SR (given)
PS || MR (construction)
P
Proved. 1
1
R
M
Q
 S = PMR
1
RM = RQ (as RM = SP, opp. sides of a parallelogram
and PS = QR)

RMQ = Q
½
But
RMQ = 180° –  PMR (linear pair)
= 180° –  S.
½

Q = 180° –  S
Hence
Q +  S = 180°
Since, one pair of opposite angles is supplemetary, therefore PQRS is cyclic.
Proved. 1
Q U A D R I L A T E R A L S
P-47
2
QUADRILATERALS
WORKSHEET-23
Section - A
1.
The value of (x + y) is 180º.
2.
Let angles be x, 2x, 3x and 4x
1
So,
x = 36º
1
So, angles are 36º, 72º, 108º, 144º.
Section - B
3.
Given : ABCD is a trapeziom, where AB || CD and AD = BC.
To prove : ABCD is cyclic.
Construction : Draw DL  AB and CM  AB.
Proof : In ALD and BMC,
D
 A
L
C
B
M
AD = BC
(given)
DL = CM
(distance between parallel sides)
ALD = BMC
(90º)
ALD = BMC
(RHS congruence criterion)
DAL = CBM
(CPCT)
1
Since, AB || CD.
DAL + ADC = 180º
(Sum of adjacent interior angles is supplementary)

CBM + ADC = 180º
(from 1)
 ABCD is a cyclic trapezium.
1
Section - C
4.
 PRQ = QPR = 55° (angles opposite to equal sides)
1
 PQR = 180° – 110° = 70°,  PSR = 180° – 70° = 110°
1
  TSR = 180° – 110° = 70°.
1
Section - D
5.
Given a square ABCD in which diagonal AC
To prove : AC = BD and AC  BD.
Proof : In ADB and BCA,
AD =

BAD =

AB =

ADB 

AC =
and BD intersects at O.
C
D
1
BC (sides of a square)
ABC (90° each)
BA (common)
BCA (By SAS)
BD (By cpct)
O
1
B
A
P-48
M A T H E M A T I C S --
IX
T E R M -- 2
In AOB and AOD,
OB
AOB
AOB
AOB + AOD
AOB
AO
Hence AC = BD and AC BD.




6.
7.
=

=
=
=

OD, AB = AD, AO = AO (By SSS)
AOD (SSS)
AOD
180°
AOD = 90°
BD, AC BD
1
Proved.
According to question, E and F are the mid-points of sides AB and CD.
1

AE =
AB
2
1
CF =
CD
2
 In the parallelogram opposite sides are equal, so
AB = CD

AE = CF
Again,
AB || CD
So,
AE || FC
Hence, AECF is a parallelogram.
In  ABP,
E is mid-point of AB
EQ || AP
 Q is mid-point of BP
Similarly, P is mid-point of DQ

DP = PQ = QB
 Line segments AF and EC trisect the diagonal BD.
Join AC and BD.
In DAC, S and R are the mid-points of AD and DC.
1

SR =
AC and SR || AC
C
2
Also, in BAC, P and Q are the mid-points of AB and BC.
1

PQ =
AC and PQ || AC
C
2
From (i) and (ii), PQRS is a parallelogram.
From (i), we get
SM || NO
and
SP || BD
S

SN || MO
From (iii) and (iv), we get MSNO is a parallelogram.
A M
Since ABCD is a rhombus, so
DOA = 90°
P

MSN = 90°
Hence, PQRS is a rectangle.
Q U A D R I L A T E R A L S
1
1
1
1
1
...(i) 1
...(ii) 1
...(iii)
D
N
R
...(iv) 1
C
O
Q
B
Proved. 1
P-49
2
QUADRILATERALS
WORKSHEET-24
Section - A
1.
The length of DE is
1
BC.
2
1
Section - B
ADC + BCD = 180°
2.
Divided both side by 2.
or
In ODC,
1
1
ADC + BCD
2
2
1 + 2
1 + 2 + DOC
DOC
½
B
A
= 90°
½
O
= 90°
= 180°
= 90°.
2
D
1
C
1
Section - C
3.
(i) In triangles RSM and PQL,
 RSM = PQL (Alternate interior angles)
S
 M =  L = 90°
R
L
Opposite side of || gm

SR = PQ
By AAS,
(ii)
1
M
P
 RSM   PQL.
Q
1
1
PL = RM (c.p.c.t.)
Section - D
4.
ABCD is a parallelogram.
To show LMNO is a rectangle,
Divide both side by 2
A +  D = 180°
In AOD,
1
1
A +  D = 90°
2
2
OAD + ODA = 90°
A
B
L
O
D
½
M
N
½
C
½
OAD + ADO + DOA = 180°

DOA = 90°
½

LON = 90°
½
Similarly, OLM = LMN = MNO = 90°
½
 A quadrilateral with all angles 90° is a rectangle. Also opposite angles are equal. It is a
parallelogram.
1
Value Based Question
5.
(i) Since, sum of adjacent angles of a parallelogram is 180°.
 We have

P-50
A + B = 180
5x + 7 + 3x – 3 = 180
M A T H E M A T I C S --
IX
T E R M -- 2






1
8x + 4 = 180
8x = 176
176
 22
8
A = (5x + 7)° = (5 × 22 + 7)° = 117°
1
B = (3x – 3)° = (3 × 22 – 3)° = 63°
C = A = 117°
(opposite angles of || gm)
x =
D
C
Save Electricity
A
B
and
D = B = 63°.
(ii) Properties of parallelogram.
(iii) Energy conservation is very necessary for a happy and prosperous future.
Q U A D R I L A T E R A L S
1
½
½

P-51
2
QUADRILATERALS
WORKSHEET-25
Type (A)
1. False.
2. False.
3. True.
4. True.
5. True.
6. False.
7. False.
8. True.
9. False.
Type (B)
1. A quadrilateral ABCD is said to be a parallelogram, if :
(i)
1
Opposite sides are equal or
(ii) Opposite angles are equal or
(iii) Diagonals bisect each other or
(iv) A pair of opposite sides is equal and parallel.
2. No, since sum of all angles in this case will be less than 360º while for a quadrilateral the sum
should be equal to 360º.
1
3. Rectangle, since all the angles are equal i.e., 4 (each angle) = 360º.
1
4. No, diagonals of a rectangle need not be perpendicular but they are equal.
1
5. This statement is false, because diagonals of a rhombus are perpendicular but not equal to each
other.
1
6. ABCD is not a parallelogram, because diagonals of a parallelogram bisect each other. Here
OA  OC.
1
7. It need not be a parallelogram, because we may have A = B = C = 80º and D = 120º.
Here B D.
1
8. Trapezium or cyclic quadrilateral.
1
9. Since the diagonals of a parallelogram ABCD are equal, it is a rectangle. Therefore,
ABC = 90º.
1
10. A triangle is obtained if three out of four points are collinear.
1
11. Mid Point Theorem states that the line segment joining the mid points of two sides of a triangle
is parallel to the third side.
1
12. Quadrilateral is the union of words. Quad meaning four and lateral meaning sides.
1
13. It is a Activity type question. Student must do in the class it self.
Type (C)
1 to 5 are activity type question. Student must do in the class it self.
P-52
M A T H E M A T I C S --
IX
T E R M -- 2
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-26
Section - A
1.
2.
The ratio of areas of GAB and HAB is 2 : 1.
ar (AEC) = 20
1
cm2.
1
Section - B
3.
Given :

4.
BE = 14 cm, AD = 8 cm
Area of (ADB) =
1
× 8 × 14 = 56 cm2
2
1
 ABCD is a parallelogram.

ar (DBC) = ar (ADB) = 56 cm2.
PSR and PSQ are on the same base PS and between same parallels PS and QR.

ar. (PSR) = ar. (PSQ)
Now,
ar. (PSR) – ar. (PSO) = ar. (PSQ) – ar. (PSO)

ar. (ROS) = ar. (POQ).
1
1
½
½
Section - C
5.
In ABC,
Now,
AC =
BC2 – AB2 =
52  32 = 4 cm
1
1
1
× AB × AC =
× 3 × 4 = 6 cm2
1
2
2
2
ar. (DBC) = ar. (ABC) = 6 cm ,
1
(Triangles on same base and between same parallels).
ar. (ABC) =
Section - D
6.
AD is the median of ABC.
A

ar (ABD) = ar (ACD)
F
E
In GBC, GD is the median.
G

ar (GBD) = ar (GCD)
From (i) and (ii), we get
B
D
C
ar (ABD) – ar (GBD) = ar (ACD) – ar (GCD)

ar (AGB) = ar (AGC)
Similarly, we can prove that
ar (AGB) = ar (BGC)
From (iii) and (iv), we get
ar (AGB) = ar (BGC) = ar (AGC)
Now,
ar (ABC) = ar (AGB) + ar (BGC) + ar (AGC)
= ar (AGB) + ar (AGD) + ar (AGD)
= 3 ar (AGB).
...(i) ½
...(ii) ½
...(iii) ½
...(iv) ½
...(v) ½
½
1
ar (ABC).
3
Hence,
ar (AGB) =
Hence,
ar (AGB) = ar (AGC) = ar (BGC) =
AREA OF PARALLELOGRAMS AND TRIANGLES
1
ar (ABC). Proved. 1
3
P-53
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-27
Section - A
1.
2.
1
(ABC).
4
The ratio of ar (ABFE) and ar (EFCD) is (3a + b) : (a + 3b).
The area of ABE =
1
1
Section - B
3.
A
AD is median and AE  BC
E
B
C
D
½
1
Area of ABD =
× BD × AE
2
and
But

½
1
× DC × AE
2
BD = DC, (AD is median)
Area of ABD = area of (ADC).
Area of ADC =
½
½
Section - C
4.
Given : ABQ and parallelogram ABCD are on the same base AB and between the same parallels
DC and AB.
½
To prove :
ar (ABQ) =
1
ar (parallelogram ABCD)
2
Construction : Extend DC to R so that BR || AQ.
Proof : DCBA and QRBA are on the same base and between same parallels.

ar (DCBA) = ar (QRBA)
Q
D
A diagonal divides a parallelogram into two congruent
½
C
R
...(i) ½
triangles with equal area.
1
ar (QCRA)
2

ar (QAB) =
From (i) and (ii), we get
1
ar (QAB) =
ar (DCBA)
2
...(ii) ½
A
B
Proved. 1
D
Section - D
5.
DB is the transversal because DC || AB
P-54
C
4
because CDB = ABD = 90° [form a pair of alternate s]
DC || AB and DC = AB
2.5
A
2.5
M A T H E M A T I C S --
B
IX
2
T E R M -- 2
 ABCD is a parallelogram.
ar (ABCD) = B × H

= 2·5 × 4

= 10 cm2.
2
Value Based Question
6.
A
B
O
D
C
1
E
(i) Let the plot be ABCD.
Construction : Join AC. Draw
Proof :
BE || AC.
1
ar (ADE) = ar (Quad. ABCD)
Health centre can be constructed and the farmer can have the
triangular plot ADE.
1
(ii) Helpful, wise and co-operating.
½
(iii) Yes, constructing a Health Centre is justified and essential also.
½
AREA OF PARALLELOGRAMS AND TRIANGLES
P-55
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-28
Section - A
1.
The area of parallelogram ABCD = 2 ar  (AEB).
2.
The area of QTR is
1
1
AEB.
2
1
Section - B
3.
3 cm
D
C
90º
In the figure,
CDB = ABD = 90°
1
But they are alternate angles
4 cm

AB ||
Also
DC =
A quadrilateral with a pair of equal and
is a parallelogram.

Area =
=
4.
DC
AB = 3 cm.
parallel sides
A
B
3 cm
b × h = (3 × 4) cm2
12 cm2.
½
½
Here, AB || DC

½
½
½
ar (ABD) = ar (ABC)
ar (ABD) – ar (ABO) = ar (ABC) – ar (ABO)
ar (AOD) = ar (BOC)
Section - C
5.
CFXB is a parallelogram,

XF = BC (CFXB is || gm)
Similarly,
EY = BC (CFXB is || gm)

EX = YF
XBE  CFY
A
1
X
E
Y
F
½
1
ar (AXE) = ar (AYF) (BY CPCT)
ar (AEB) = ar (ACF) (BY CPCT).
B
C
½
Section - D
6.
Given,
ar (AOD) = ar (BOC),
Adding ar (ODC) on both sides
ar (AOD) + ar (ODC) = ar (BOC) + ar (ODC)

ar (ADC) = ar (BDC)

A
O
1
1
× DC × AL =
× DC × BM
2
2
AL = BM
AB || DC
D
M A T H E M A T I C S --
1
½
½
Proved.
L
M C
ABCD is a trapezium.
P-56
1
1
B
IX
T E R M -- 2
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-29
Section - A
1.
The area of ABCD is 30 cm2.
1
2.
The area of ABCD is 4a sq. units.
1
Section - B
3.
In rt. ABC, using Pythagoras theorem,
BC =
½
102  62  82 = 8 cm
1
Now,
ar ABC =
× 6 × 8 = 24 cm2.
2
Medians divide a triangle in two parts of equal area.
1

ar BDC =
ar ABC = 12 cm2
2
1
Also,
ar BEC =
ar BDC = 6 cm2.
2
½
½
½
Section - C
4.
ar (QER)
ar (GQR) + ar (EGR)
Also,
ar (PFR)
ar (PFGE) + ar (EGR)
Subtracting eqn. (ii) from eqn. (i),
ar (GQR) – ar (PFGE)
ar (GQR)
=
=
=
=
ar
ar
ar
ar
1
...(i)
1
...(ii)
(QEP)
[QE is median]
(PFGE) + ar (GFQ)
(QFR)
(GQR) + ar (GFQ)
= ar (PFGE) – ar (GQR)
= ar (PFGE).
1
Section - D
5.
Construction : Join CX.
Proof :
AB || DC (Given)
ADX and ACX are on the same base AX and between
the same parallels AB and DC.

ar (ADX) = ar (ACX)
Also,
AC || XY (Given)
ACY and ACX are on the same base AC and between
the same parallels AC and XY.

ar (ACY) = ar (ACX)
From (i) and (ii), we get
ar (ADX) = ar (ACY).
Value Based Question
6.
1
A
X
...(i) 1
B
Y
...(ii) 1
D
C
Proved. 1
(i) Let ABCD be the plot and Naveen decided to donate some portion to construct as home
for orphan girls from one corner say C of plot ABCD. Now, Naveen also purchases equal
amount of land in lieu of land CDO, so that he may have triangular form of plot. BD is
joined. Draw a line through C parallel to DB to meet AB produced to P.
Construction : Join DP to intersect BC at O.
AREA OF PARALLELOGRAMS AND TRIANGLES
1
P-57
A
B
P
O
D
C
Proof : BCD and BPD are on the same base and between same parallels CP || DB.

ar (BCD) = ar (BPD)

ar (COD) + ar (DBO) = ar (BOP) + ar (DBO)

ar (COD) = ar (BOP)

ar (quad. ABCD) = ar (quad. ABOD) + ar (COD)

= ar (quad. ABOD) + ar (BOP),
1
[ ar (COD) = ar (BOP) proved above]
= ar (APD)
Hence, Naveen purchased the portion BOP to meet his requirement.
1
(ii) Area of parallelogram.
½
(iii) We should help the orphans.
½
P-58
M A T H E M A T I C S --
IX
T E R M -- 2
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-30
Section - A
1.
2.
1
1
ABCD is a parallelogram.
The value of x is 12.5 cm2.
Section - B
3.
Since CD is bisected at O.

In ADC, AO is the median.

In CDB, BO is the median.
...(i) ½
ar (AOC) = ar (AOD)
ar (BOC) = ar (BOD)
A
B
O
Adding (i) and (ii), we get
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)

ar (ABC) = ar (ABD).
4.
½
C
CO = OD
D
½
½
Area of parallelogram is divided into four equal parts by the diagonals.

1
ar (ABCD)
4
1
1
ar (BOP) =
ar (AOB) = ar (ABCD)
2
8
1
=
× 80
8
= 10 cm2.
A
ar (AOB) =
...(ii) ½
B
½
P
½
O
D
C
½
Section - C
5.
A median divides a triangle into two triangles of equal area.
1
2
1

2
1

2
1

8
ar (BOE) =
ar (ABE),
BO is median

1
ar (ABD),
AE is median 
2
1
1
×
×
ar (ABC), AD is median ½
2
2
×
ar (ABC).
½
Section - D
6.
Median QT and RT divide PQS and PRS in two triangles of equal areas
and
Adding Eq. (i) and (ii),
Now,
ar(QTS) =
1
ar (PQS)
2
...(i)
ar (RTS) =
1
ar (RPS)
2
...(ii)
ar(QTS + RTS) =
ar(QTR) =
1
[ar(PQS) + ar(PRS)]
2
1
ar(PQR).
2
AREA OF PARALLELOGRAMS AND TRIANGLES
P-59
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-31
Section - A
1.
1:2
1
2.
1:4
1
Section - B
3.
A
In ABC, AD is a median.
In BEC, ED is a median.
ar (ABD) = ar (ACD)
ar (BDE)
Subtracting the equation (ii) from equation
ar (ABD) – ar (BDE)

ar (ABE)
...(i) ½
E
...(ii) ½
= ar (CDE)
(i), we get
B
= ar (ACD) – ar (CDE)
= ar (ACE).
D
C
½
Proved. ½
Section - C
4.
BD || PC

ar (BDP) = ar (BDC)
( Two triangles having the same base and equal
areas lie between the same parallels)
A
Adding ar (ADB) on both sides,
ar (BDP) + ar (ADB) = ar (BDC) + ar (ADB)
ar (ADP) = ar (quad. ABCD).
1
C
D
B
1
P
Proved.
1
Section - D
5.
Given :
AP || BQ || CR
Therefore, BCQ and BQR are on the same base BQ and
A
between the same parallels BQ and CR.
So,
P
...(i) 1
ar (BCQ) = ar (BQR)
Also,
AP || BQ. (Given)
B
Q
Therefore, ABQ and PBQ are in the same base BQ and
between the same parallels BQ and AP.
C

ar (ABQ) = ar (PBQ)
Adding (i) and (ii), we get
ar (BCQ) + ar (ABQ) = ar (BQR) + ar (PBQ)
ar (AQC) = ar (PBR).
R
A
Value Based Question
6.
...(ii) 1
Proved. 1
(i)  D, E and F are mid-points of AB, BC and CA respectively
D
 By mid-point theorem, we have
F
DF || BC and EF || AB

P-60
DF || BE and EF || BD
B
M A T H E M A T I C S --
E
IX
C
1
T E R M -- 2
 BEFD is a parallelogram.
 The diagonal of a parallelogram divide it into two congruent triangles

DEF  BED
Similarly,
DEF  ADF
and
DEF  CEF

1
DEF  BED ADF CEF

ar (DEF) = ar (BED) = ar (ADF) = ar (CEF)

Hence,
ar (DEF) + ar (BED) + ar (ADF) + ar (CEF) = ar (ABC)
ar (DEF) = ar (BED) = ar (ADF) = ar (CEF)
=
1
ar (ABC)
4
1
Each child will get equal share of property.
(ii) Area of parallelogram and triangle and mid-point theorem.
½
(iii) Every child, boy or gild have equal right, so avoid discrimination in boy and girl.
½
AREA OF PARALLELOGRAMS AND TRIANGLES
P-61
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-32
Section - A
1.
The area of BFEC =
3
ar (ABC).
4
1
Section - B
2.
F
Area of parallelogram = AD × CF = CD × AE
E
C
D
Putting the values in equation = AD × 10
= 16 × 8, ( CD = AB)

3.
16  8
10
AD = 12·8 cm.
A
AD =

In figure, PS is median of PQR,

Now, QT is median of PQS,
ar (PQS) =
1
ar (PQR)
2

From (i) and (ii), we get
ar (QTS) =
1
ar (PQS)
2
ar (QTS) =
1
ar (PQR).
4
B
1
1
½
P
...(i) ½
T
Q
...(ii) ½
R
S
½
Section - C
4.
Proof :

1
ar (ABED)
2
and parallelogram on same base and between same parallels
1
ar (BEC) =
ar (ABED)
2
[a triangle and a parallelogram on equal base and between same parallels]
ar (ABF) = ar (BEC).
1
ar (ABF) =
Section - D
5.
Through O, draw
Also
PABS is a parallelogram.
A
P
AB || PS.
PA || BS
S

P-62
ar (QOR) =
½
O
B
1
So,
ar (POS) =
ar (PABS)
2
and a parallelogram on same base and between same parallels]
Similarly,
Q
R

1
1
ar (QABR)
2
1
1
ar (PABS) + ar (QABR)
2
1

ar (PQRS).
2
ar (POS) + ar (QOR) =
M A T H E M A T I C S --
IX
½
T E R M -- 2
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-33
Section - A
1.
1
ar (APQ) = ar (RBQ)
Section - B
2.
P
D
C
Q
A
B
APB and parallelogram ABCD are on same base AB and between same parallel lines AB and
DC.
1
So,
ar (APB) =
ar (ABCD)
...(i) 1
2
1
Similarly,
ar (BCQ) =
ar (ABCD)
...(ii) ½
2
From (i) and (ii),
ar (APB) = ar (BCQ).
½
Section - C
3.
Diagonal of a parallelogram divides it into two congruent triangles.
So,
ABC  ACD
...(i) ½
Diagonals AC and BD bisect each other at O.
Now, in ABC, ABO & DBC have equal base and triangles having same vertex.
D
C
So,
ar (ABO) = ar (BCO)
...(ii) ½
Similarly,
ar (ADO) = ar (CDO)
...(iii)

ar (ADO) = ar (ABO)
...(iv)
O

ar (BCO) = ar (CDO)
...(v) 1
From (ii), (iii), (iv) and (v),
A
B
ar (ABO) = ar (BCO) = ar (CDO) = ar (DAO)
Hence proved. 1
Section - D
4.
Given : ABCD and PBQR are parallelograms, CP || AQ
To prove :
ar (||gm ABCD) = ar (||gm PBQR)
½
Construction : Join AC and PQ.
Proof :
D
C
½
ar (CAQ) = ar (PAQ)
       same base AQ and same parallels AQ || CP
A
Subtracting ar (BAQ) from both sides,
D
P
B
C
A
P
B
Q
Q
R
1
R
AREA OF PARALLELOGRAMS AND TRIANGLES
P-63
ar (CAQ) – ar (BAQ) = ar (PAQ) – ar (BAQ)

ar (ABC) = ar (BQP)

2 ar (ABC) = 2 ar (BQP)
½

ar (ABCD) = ar (PBQR)
(diagonal of parallelogram divides it into two triangles of equal areas)
½
1
Value Based Question
5.
(i)
Ar. of triangular plot =
1
×b×h
2
1
× 120 × 90
2
= 5400 m2
(ii) In ABC draw median AD on base BC and divide it into
two equal areas ABD and ACD. Take any point E on AD and join BE
and CE. The brothers get areas ar (ABE) and ar (ACE) ar (BCE) is
donated to school.
=
(iii) Any positive value is accceptable. Both brothers know
importance of education, love their community.
P-64
M A T H E M A T I C S --
1
A
E
B
IX
D
2
C
1

T E R M -- 2
3
AREA OF PARALLELOGRAMS AND TRIANGLES
WORKSHEET-34
1.
(i) Quadrilateral ABCD and triangle PDC lie on the same base CD and between the same
parallels AB and CD.
1
(ii) Quadrilateral PQRS and quadrilateral ABRS lie on the same base RS and but they do not
lie between the same parallels.
1
(iii) There is no figure between the same base and between same parallels as there is only one
triangle.
1
(iv) There is no figure between the same base and between same parallels since the square and
the triangle CDQ lie on the same base DQ but not between the same parallels.
1
(v) Quadrilaterals AEGC and AEHD lie on the same base AE but not between the same parallels.
1
hb
2
h
(a  b)
2
2.
(i)
3.
(a) Corresponding parts.
1
(b) equal.
1
4.
(iii) lb
(iii) ah
(iv)
2
(a) The distance between opposite sides of a parallelogram. Formally, the shortest line segment
between opposite sides and altitude refers to the length of this segment.
1
(b) According to congruent area axiom, if ABC  PQR, then Area (ABC)  Area (PQR).
(c) If R1, R2 are two polygonal regions such that R1  R2, then area (R1)  area (R2).
1
(d) A triangular region is the union of a triangle and it’s interior.
1
(e) Two figures are said to be on the same base and between the same parallels if they have
a common base (side) and the vertices (or the vertex) opposite to the common base of each
figure lie on a line parallel to the base.
1
(f) Activity type question. Student must do in the class itself.
5.
6.
(a) True.
(b) False.
(d) False.
(e) True.
(a) 88
(b) 168
(d) 120
(e) 70.
AREA OF PARALLELOGRAMS AND TRIANGLES
(c) False.
1
(c) 24
1
P-65
4
CIRCLES
WORKSHEET-35
Section - A
BCD = 60º
1.
2.
1
1
The values of x and y are 36º and 60º.
Section - B
3.
In a cyclic quadrilateral,


2x + 4° + 4x – 64° = 180°
6x – 60° = 180°
6x = 180° + 60° = 240°


1
240
 40
6
x = 40°.
x =
1
Section - C
ACP = ABP
4.
Similarly,
But,
From (i), (ii) and (iii), we get
...(i)
(Angles in the same segment of a circle are equal) 1
QCD = QBD
...(ii) ½
ABP = QBD
(Vertically opposite angles) ...(iii) 1
ACP = QCD.
Proved. ½
Section - D
5.
Given : AB and CD are chords of a circle with centre O such that AB = CD.
To prove :
Proof : In AOB and COD,
6.
Hence,
(i)
(ii)
(iii)
(iv)

P-66
AOB = COD.
AO
BO
AB
AOB
AOB
QRP
QPR
QPS
QRS
PRS
=
=
=

=
=
=
=
=
=
=
PSR =
PTQ =
PQT =
B
D
1
(Radii of same circle)
CO
O
DO
A
CD
(given)
C
COD (S.A.S.)
2
COD (cpct)
1
90°
(angle in the semi-circle)
25°
(angle sum property)
QPR + RPS = 50°
180° – 50° = 130° (PQRS is a cyclic quad.) 1
130° – QRP
130° – 90° = 40°.
1
180° – 65° = 115°.
1
90°
90° – 60° = 30°.
1
M A T H E M A T I C S --
IX
T E R M -- 2
4
CIRCLES
WORKSHEET-36
Section - A
1.
Clearly,
OPA = 90º.
1
Section - B
2.
Given,

In DAB,


ACB = 70°
ADB =
DAB + ADB + DBA =
60° + 70° + DBA =
DBA =
70° (angles in the same segment of a circle) ½
180°
1
180°
50°.
½
Section - C
3.
Given,
In DBC,


BAC = BDC = 45°, (angles in the same segment) 1
DBC + BCD + CDB = 180°, (A.S.P.)
55° + BCD + 45° = 180°
BCD = 80°.
1
1
Section - D
4.
Construction : Join OC, OD and BC.
CD = Radius of the circle
CD = OD = OC

1
 ODC is an equilateral triangle

COD = 60°
1
Now,
CBD =
COD
½
2

CBD = 30°
½
Also,
ACB = 90°
(Angle in a semicircle is 90°) ½

BCE = 180° – ACB = 180° – 90°
= 90°
½
E
D
C
A
B
O
Now, in triangle BCE,
CBE + BCE + CEB
30° + 90° + CEB

CEB

AEB
C I R C L E S
=
=
=
=
180°
180°
60°
60°
½
Proved. ½
P-67
Value Based Question
5.
(i) Let the butter-chords of the biscuit be AB and CD; and centre of the biscuit be O.
Join each of A, B, C, D to O
A
C
O
B
D
In OAB and OCD,
AB = CD
(Given)
OA = OC
(Each equal to radius)
OB = OD
(Each equal to readius)

OAB  OCD [SSS]

AOB = COD (CPCT)
Therefore, the butter-chords subtend equal angles at the centre of the biscuit.
Proved. 1
(ii) We are given the length of either chord is greater than the radius and less than diameter of the
circle.
½
Let length of either chord = l, radius = r and angles subtended by either butter-chord = 
Two cases arise :
Case I. If l = r,
½
In this case, the chord and corresponding radius form an equilateral triangle with side r.

 = 60°.
Case II. If l = 2r,
½
In this case, the butter chord passes through the centre.

 = 180°
Consequently, we arrive at the following inequality :
60° <  < 180°, (As r < l < 2r)
Thus, the required range is from 60° to 180° excluding both.
½
(iii) (a) Congruence of triangles
(b) cpct (Corresponding parts of congruent traingles) are equal
(c) Equilateral triangle and its angles.
1
(iv) Industrialist, thoughtfulness, selfconfidence, rationality.
1
P-68
M A T H E M A T I C S --
IX
T E R M -- 2
4
CIRCLES
WORKSHEET-37
Section - A
1.
The value of DBC is 40º.
1
2.
Only one circle can pass.
1
Section - B
3.
In EDC,
EDC + 20º = 130º

EDC = 110º or BDC = 110º
1

BAC = BDC = 110º.
1
Section - C
4.
In OMB,
1
OM = 4 cm, MB = 3 cm
OB2
=
OM2 +
MB2 =
9 + 16 = 25

In OND,
25 = 5 cm
OB =
2
ON = OD2 – DN2
DN = 3cm, OD = OB = 5 cm
ON2 = 52 – 32 = 16 ON = 4 cm
 The other chord is at a distance of 4 cm from the centre.
1
O
C
D
N
A
B
M
1
Section - D
5.
6.
A
Construction : Join AB,
½
ABD = 90° (Angle in a semi-circle)
1

ABC = 90° (Angle in a semi-circle)
1
C
ABD + ABC = 180°
½
D
B
 DBC is a line.
 D, B and C are collinear.
Proved. 1
Consider a circle with centre O and radius r. Let M be a point in a circle such that, a chord
AB passes through it and (i) M is mid-point of AB.
Let CD be another chord passes through M.
To Prove : CD > AB
Proof : Join OM and draw ON  CD
In right triangle ONM, OM is the hypotnuse
 OM > ON
 Chord CD is nearer to O in comparision to AB
O
D
N
B
A
C
 CD > AB {Q of any two chords of a circle, the one which is nearer to the centre is louger.]
1
C I R C L E S
P-69
4
CIRCLES
WORKSHEET-38
Section - A
AOB = 15º
1.
1
Section - B
2.
Draw OPperpendicular to xy from the centre to a chord bisecting it.
OP to chord BC

BP = PC
Similarly,
AP = PD
x
From (i) and (ii), we get
A B
AP – BP = PD – PC
or
AB = CD
...(i)
O
C
P
D
y
1
...(ii)
½
½
Section - C
3.
In ABC,

A + B + ACB = 180º
1
A = 180º – (69º + 31º)
1
= 80º

BDC = A = 80º.
1
Section - D
4.
Since, perpendicular from the centre of the circle to a chord bisects the chord.
P and Q are the mid-points of AB and CD
So,
AP =
1
AB = 3 cm
2
and,
CQ =
1
CD = 4 cm
2
In right triangles
1
OAP and OCQ,
OA2 = OP2 + AP2 and OC2 = OQ2 + CQ2
Putting value,
52 = OP2 + 32 and 52 = OQ2 + 42

OP2 = 52 – 32 and OQ2 = 52 – 42

OP2 = 16 and OQ2 = 9

OP = 4 and OQ = 3

1+1
1
PQ = OP + OQ = 4 + 3 = 7 cm.
Value Based Question
5.
P-70
(i) Let us assume that A, B and C are the position of Priyanka, Sania and David respectively
on the boundary of circular park with centre O.
Draw AD  BC.
1
Since the centre of the circle coincides with the centroid of the equilateral ABC.
2

Radius of circumscribed circle =
AD
3
M A T H E M A T I C S --
IX
T E R M -- 2
2
AD
3
3

AD = 20 ×
2

AD = 30 m
Now, AD  BC and let AB = BC = CA = x
1
x

BD = CD =
BC =
2
2

20 =
1
A (Priyanka)
20
B
(Sania)
m O
D
C
(David)
In rt. BDA, D = 90°

By Pythagoras Theorem, we have
AB2 = BD2 + AD2
2
x
2
x 2 =    (30)
2
or


x2 
x2
= 900
4
3 2
x = 900
4

x 2 = 900 

x 2 = 1200
x =
4
3
1200  20 3 m
Hence, distance between each of them is 20 3 m .
(ii) Properties of circle, equilateral triangle and Pythagoras theorem.
(iii) Live and let live.
C I R C L E S
½
½
½
½
P-71
4
CIRCLES
WORKSHEET-39
Section - A
1.
The length of chord AB is 4 cm.
1
2.
The ABC is 50º.
1
Section - B
3.
A
Construction : Join AB.
Now,
ABD
Also,
ABC
Adding both the angle
ABD + ABC
= 90° (angle in semi-circle)
= 90° (angle in semi-circle)
we get,
= 90° + 90°
= 180°
Hence, DBC is a line or B lies on the line segment DC.
1
D
B
C
½
Proved. ½
Section - C
4.
E
C
D
A
Given,
B
AE = BE
A = B (Angle opposite to equal sides of a  are equal)
EDC = B; ECD = A (Exterior angle of cyclic quad.)
1½

EDC = A
{But by position they arc corresponding angles}

Proved. 1½
AB || DC
Section - D
5.
Proof :
P
C
D
Q
E
A
O
B
AEB = 90° = AED (Semi-circle)
P-72
M A T H E M A T I C S --
1
IX
T E R M -- 2
EAC + ACD + CDE + AED = 360° (Sum of angles in quad.)
EAC + 90° + 90° + 90° = 360°
EAC = 360° – 270°
= 90°
So,
each angle = 90°
 EACD is a rectangle
AC = ED.
Given : O is the centre of the circle.
To prove :
BOC = 2BAC
Construction : Join O to A.
In AOB,
B

OA = OB (Radii of same circle)

1 = 2
Simlarly, in AOC,
3 = 4
Now, by exterior angle property,
5 = 1 + 2
6 = 3 + 4

5 + 6 = 1 + 2 + 3 + 4

5 + 6 = 2 2 + 2 3
B
= 2 (2 + 3)

BOC = 2 BAC.
Now,


6.
C I R C L E S
1
1
Proved.
A
½
O
C
1
A
1
2 3
O
5
1½
1
4
6
C
Proved. 1
P-73
4
CIRCLES
WORKSHEET-40
Section - A
1.
DBA is 80º.
1
Section - B
2.
We know that angle subtended by an arc of a circle at the centre is double the angle subtended
by it at any point on the remaining part of circle.
1
AOC
2
AOC = 2 ABC = 2 × 45° = 90°
OA  OC.
ABC =


1
Proved. 1
Section - C
3.
AB is the chord of a circle with centre O and
AD =
In s ODA and ODB,
OA =
OD =
ODA =

ODA 

AD =
OD  AB. We have to prove that
DB
OB
OD
ODB
ODB
DB
1
(radii)
(common)
(each in a rt. angle)
1
(R.H.S.)
(CPCT)
Proved. 1
Section - D
4.
B
Draw : OM  AB and ON  CD.
C
A
M
E
N
1
O
D
In OME and ONE,
 By A.A.S. congruence rule,

Hence
OEM
OME
OE
OME
OM
AB
= OEN
(Given)
= ONE = 90°
= OE
(common)
1½
 ONE
= ON
(CPCT)
½
= CD (Chords equidistant from the centre are equal).
1
Value Based Question
5.
(i) Given :
So,

P-74
d = 50 m
50
 25 m
2
perimeter = 2r
= 2 × 3·14 × 25
= 157 m.
radius r =
M A T H E M A T I C S --
1
IX
T E R M -- 2
(ii) In the figure, if we draw a perpendicular line from centre of circle to the line segment , it bisects
the line segment.
1
AD = BD = 7 cm
In right triangle AOD,
AO2 = AD2 + DO2
(25)2 = (7)2 + (DO)2

(DO)2 = 625 – 49
= 576

(DO)2 = (24)2

DO = 24 m
1
(iii) Neeraj shows the quality of intelligence and smartness.
½
(iv) Yes, sport is very important part of our life as it involves physical activities, that gives you a
relaxing break from your daily routine and makes you more focused and active. Also, it keeps
our body fit.
½
C I R C L E S
P-75
4
CIRCLES
WORKSHEET-41
Section - A
1.
2.
1
CD = 2 cm.
1
The value of x is 85º.
Section - B
3.
It is given that OP  AB and OQ  CD
1
But,
OP = OQ
(given)

AB = CD.
(equal chords are equidistant from the centre) 1
Section - C
4.
Given : Radius of circle (r) = 5 cm
And
AB = 8 cm
OM  AB
AM = MB = 4 cm ( drawn from the centre of the circle bisect the chord)
In rt. OMA, by Pythagoras theorem,
OA2 = OM2 + AM2
Putting values,
5 2 = OM2 + (4)2

OM2 = 9

OM = 3 cm.
½
1
½
A
½
O
5 cm
B
4 cm M
½
Section - D
5.


6.
BOD = 180° – 75° = 105°
1
CED = 90° (angle in semi-circle)
CDE = 90° – OCE = 90° – 40° = 50°
OBD = OBE = 180° – (105° + 50°)
= 25°.
1
1
1
According to the question,
 OAB is an equilateral triangle,
and,

P-76
OA = AB = OB
AOB = 60°
ACB = 30°
1
=
AOB
2
ADB = 180° – 30° = 150°.
1
C
1
O
60
A
M A T H E M A T I C S --
D
IX
B
1
1
T E R M -- 2
4
CIRCLES
WORKSHEET-42
Section - A
1.
1
The value of x is 20º.
Section - B
2.
We know that the angle subtended by an arc of a circle at the centre is double the angle
subtended by it at any point on the remaining part of circle. Hence,
1
× 140° = 70°
½
2
ABC = 180° – 70° = 110° (cyclic quadrilaterals)
½
OCB = = 360° – (140° + 50° + 110°) = 360° – 300° = 60°.
1
APC =
Now,

Section - C
3.
Proof : In ABC and ADC,
and,
A
2
B
1
D
45º
4
3
C
1
3
AC
ABC
B
B + D
B
=
=
=

=
=
=
2
4
1
AC
ADC
(A.S.A.)
1
D
(CPCT)
180°
(opp. s of cyclic quad.)
D = 90°   ABC = 90°.
Proved. 1
Section - D
4.
Construction :
Join OE. Draw OL  AB and OM  CD.
Given,
AB = CD

OL = OM
OLE  OME (R.H.S.)

LE = ME (c.p.c.f.)
Since,
AB = CD
1
1
AB =
CD
2
2

BL = DM
Subtracting (ii) from (i)
LE – BL = ME – DM

BE = DE.
AB = CD and BE = DE

AB + BE = CD + DE

AE = CE
Hence, BE = DE and AE = CE.

A
L
B
E
O
C
M
1½
...(i)
D
...(ii) 1
1½
Value Based Question
5.
(i) Given : A circle C(O, r) and chord AB = chord AC. AD is bisector of CAB.
To prove : Centre O lies on the bisector of BAC.
C I R C L E S
P-77
Construction : Join BC, meeting bisector AD of BAC at M.s
B
A
O
M
D
C
Proof : In triangles BAM and CAM,
AB = AC
[Given]
BAM = CAM
[Given]
and
AM = AM
[Common]

BAM  CAM
[SAS]

BM = CM and BMA = CMA
1½
As
BMA + CMA = 180°
[Linear pair]

BMA = CMA = 90°
 AM is the perpendicular bisector of the chord BC.
AM passes through the centre O.
[ Perpendicular bisector of chord of a circle passes through the centre of the circle]
1½
Hence, the centre of the park likes on the angle bisector of BAC.
Proved.
(ii) Congruence of triangles by SAS axiom (Geometry).
½
(iii) Cleanliness and respect for labour.
½
P-78
M A T H E M A T I C S --
IX
T E R M -- 2
4
CIRCLES
WORKSHEET-43
Section - A
1.
The measure of ACD is equal to 20º.
1
2.
BOC is 160º.
1
Section - B
3.
We know that the angle subtended by an arc of a circle at the centre is double the angle
subtended by it any point on the remaining part of the circle.
½
1
BOC
2
1
=
× 80° = 40°
2
BAC =
So,

½
ABCD is a cyclic quadrilateral,
So,
BAC + BDC = 180°

BDC = 180° – 40° = 140°.
1
Section - C
4.
Let O and O be the centres of the circle of radii 10 cm and 8 cm respectively. Let PQ be their
common chord. We have
OP = 10 cm, OP = 8 cm, PQ = 12 cm
1
PQ = 6 cm
2
OP2 = OL2 + PL2

1
PL =
In right OLP, we have

OL =

=
OP 2  OL2
(10) 2  (6) 2  64 = 8 cm
1
P
O'
O
L
Q
In right OLP, we have




Similarly
So,
C I R C L E S
(OP)2 = (PL)2 + (OL)2
(8)2 = (6)2 + (O’L)2
(O’L)2 = 64 – 36
O’L =
32 = 4 2 cm
OL = 8 cm
OL + O’L = 4( 2 ) + 8 = 4


2  2 cm.
1
P-79
Section - D
5.

DN =
and


1
BM = x,
82
+
62
=
62
+
1
x2
x = 8 cm
AB = 16 cm.
C
12
8
O
A
N
1
D
6
Mx
B
RST forms an equilateral triangle.
S
In an equilateral triangle the median and perpendicular bisector of a side are same line.
Hence O is circumcentre and centroid.
Centroid divides the median in ratio 2 : 1
Hence, SM =




P-80
1
CD = 6 cm
2
OD2 = OB2
ON2 + ND2 = OM2 + MB2 = (Radius)2


6.
½
Perpendicular from the centre bisects the chord.
3
a, where a is side of the triangle
2
2
SO =
× SM
3
2
3
40 =
×
a
3
2
40  3
a =
3
= 40 3 cm
40
R
M
T
1
1
1
Required distance = 40 3 m.
M A T H E M A T I C S --
1
O
IX
T E R M -- 2
4
CIRCLES
WORKSHEET-44
Section - A
1.
The BCD is equal to 100º.
1
Section - B
AOB = 80°
½
ADB = 40°
(AOB = 2 ADB)
1
ACB = ADB = 40°. (angles in the same segment) ½
2.


Section - C
3.
Since, sum of the opposite pairs of angles in a cyclic quadrilateral is 180°.
Hence,
B + D = 180°

B = 180° – 70° = 110°
A
Again, AB | | CD and AD is its transversal, so
A + 70° = 180°

A = 180° – 70° = 110°
and
A + C = 180°
70°

110° + C = 180°
D

C = 180° – 110 = 70°.
1
B
1
C
1
Section - D
4.
A
C
1
2
D
O
B
To show : CD is bisector of AB.
Proof : In ACD and BCD,
CD is common


In ACO and BCO,
AC = BC          radii of same circle
AD = BD
ACD  BCD
1 = 2
(by S.S.S.)
(by CPCT)
AC = BC
1 = 2
OC is common


But,

C I R C L E S
ACO
AO
AOC
AOC + BOC
CD

=
=
=

BCO
OB
BOC,
180°
BA.
1
½
1
(by S.A.S.)
(CPCT)
½
Proved. 1
P-81
4
CIRCLES
WORKSHEET-45
Section - A
1.
The value of x is 115º.
1
2.
The value of CBD = 20º.
1
Section - B
3.
Join OB.
In OAB,
In OCB,
Now,

OAB
OCB
ABC
AOC
=
=
=
=
OBA = 30° (Isosceles Property)
OBC = 40°
OBA + OBC = 30° + 40° = 70°
2 ABC = 140°
½
½
1
Section - C
4.
In OBD and ODC
OB = OC
[radius of the circle]
OD is common
ODB = ODC
 By right angle hypotenuse side congruency, OBD and ODC are congruent
BOD = COD
A

BOD =
O

B
D
[CPCT]
1
BOC
2
1
Since the angle subtended by the chord at the centre is twice
C
the angle subtended by the chord at the circumference.
BOD = BAC
Proved.
Section - D
5.
Given : O is the centre of the circle.
To prove :
BOC
Construction : Join O to A.
In AOB,
OA

1
Similarly, in AOC
3
Now, by exterior angle property,
5
6

5 + 6

5 + 6

A
= 2BAC.
= OB (Radii of same circle)
= 2
= 4
1 + 2
3 + 4
1 + 2 + 3 + 4
2 2 + 2 3
2 (2 + 3)
2 BAC
C
½
1
A
1
M A T H E M A T I C S --
1½
1
2 3
O
5
4
6
Proved. 1
C
B
P-82
O
B
=
=
=
=
=
BOC =
2
IX
T E R M -- 2
6.
A
20
B
O
20
D
20
C
Here A, B, C are the three points where three girls are sitting.
ABC is an equilateral triangle.
In an equilateral triangle, the circum-centre is the point of intersection of medians.
O divides AD in the ratio 2 : 1
Hence, if
AO = 20 m
Then,
OD = 10 m
Also median is same as an altitude for an equilateral triangle.
In ODC,
OC2 = OD2 + DC2

20 2 = 102 + DC2

DC2 = 400 – 100 = 300

DC = 10 3 m

BC = 2DC = 20 3 m
Length of the string of each phone = 20 3 m.
C I R C L E S
1
½
1
½
1
P-83
4
CIRCLES
WORKSHEET-46
Section - A
1.
The value of PQR is 125º.
1
Section - B
2.
In ACB,

Also,
ACO =
1
1
× AOB =
× 90° = 45°
2
2
CAB =
=
CAO =
=
180° – (30° + 45°)
105°
105° – OAB
105° – 45° = 60°.
½
½
1
Section - C
3.
OL  AB and OM  CD are drawn and OP is joined



Also

D
A
M
P
C
O
L
B
OPL
PM
AL
AL – PL
AP
AB – AP
BP

=
=
=
=
=
=
OPM
PL
CM
CM – PM
CP
CD – CP
DP.
½
(R.H.S.)
(CPCT)
½
½
½
Proved. 1
Section - D
4.
Arc DC subtends DBC and DAC in the circle.
1

DBC = DAC = 55°
Similarly, BC subtends BAC and BDC on the circle.

BAC = BDC = 45°

BAD = 45° + 55° = 100°
ABCD is a cyclic quadrilateral.

BAD + BCD = 180°

BCD = 180° – 100°
= 80°.
1
½
1
½
D
A
45º
55º
B
P-84
C
M A T H E M A T I C S --
IX
T E R M -- 2
4
CIRCLES
WORKSHEET-47
Type (A)
1.
(i)
BEC = 45.7º, BDC = 45.7º
Relationship = Equal because BEC = BDC being angles in same segment of a circle. 1
(ii) BEC = 55.52º, BDC = 55.52º
Relationship = Equal because BEC = BDC being angles in same segment of a circle. 1
(iii) BEC = 24.98º, BDC = 24.98º
Relationship = Equal because BEC = BDC being angles in same segment.
1
Type (B)
2.
(a) True, (b) False, (c) False, (d) False, (e) True.
1
3.
(a) 90, (b) 12 (c) exterior, (d) 90, (e) ≅ .
1
4.
(a) The quadrilateral ABCD is called a cyclic quadrilateral if all its four vertices lie on
a circle.
1
5.
(b) We can draw infinite number of pair of equal chords for a give circle.
1
(c) Two or more circles having common centre are called as concentric circles.
1
(d) Circles having equal radii are called congruent circles.
1
(e) One and only one circle can pass.
1
(i) arcAB, (ii) chord AB, (iii) centre, (iv) Interior of circle, (v) minor arc.
2
C I R C L E S
P-85
5
GEOMETRIC CONSTRUCTIONS
WORKSHEET-48
Section - A
1.
The bisector of an angle divide in two equal parts.
1
2.
The measure of each angle that is constructed is 15º.
1
Section - B
3.
If A = 30o, then 2A = 60o, 3A = 90o and 4A = 120o
½+½+½+½
H
m DBC = 120º
m EBC = 30º
m ABC = 60º
B
C
Section - C
4.
Let a be the length of each side. Since perimeter, 3a = 12 cm so each side of triangle is a = 4 cm.
Steps of construction :
1. Draw AB = 4 cm.
2. At A and B draw angles of 60°.
3. Mark the point of intersection of two
4. Join AC and BC.
ABC is the required triangle.
Justification of Construction : As CAB
So,
CAB + CBA + ACB
So,
ACB
Also,
AC
AB
So,
AB
C
angles as C.
1
=
=
=
=
=
=
60o
CBA =
180o
60°
60o
A
BC
(as CAB = CBA)
AC
(as CAB = ACB)
BC = AC = 4 cm.
60°
B
2
5.
M
L
D
45°
45°
P
6.
P-86
E
3
45°
45°
F
Q
Steps of construction :
(i) Draw line segment AB = 12 cm. Draw a ray AX making an angle of 90 with AB.
(ii) Cut a line segment AD of 18 cm. (As sum of other two sides is 18 cm) from ray AX.
(iii) Join DB and make an angle DBY equal to ADB.
1
(iv) Let BY intersects AX at C. Join AC and BC.ABC is the required triangle.
1
M A T H E M A T I C S --
IX
T E R M -- 2
X
D
Y
1
C
A
Section - D
7.
B
Steps of Construction :
(i) Draw the line segment BC = 8 cm and at point B make an angle = 45 XBC = 45°.
(ii) Cut the line segment BD = 3·5 cm (equal to AB – AC) on ray BX.
1
(iii) Join DC and draw the perpendicular bisector PQ of DC.
(iv) The perpendicular bisector intersects BX at point A. Join AC. ABC is the required triangle.
1
X
A
P
D
45°
B
C
Q
2
G E O M E T R I C
C O N S T R U C T I O N S
P-87
5
GEOMETRIC CONSTRUCTIONS
WORKSHEET-49
Section - A
1.
1
The angle of 40º is not possible.
Section - B
2.
Steps of construction :
(i) Draw a line segment of length 4·7 cm.
½
(ii) With centre A and radius more than half of AB, draw arcs, on both sides of AB.
½
(iii) With B as centre and the same radius as before, draw arcs, cutting the previously drawn
½
arcs at P and Q respectively.
(iv) Join PQ intersecting AB at O. Then PQ is the bisector of the line segment AB which divides
½
line AB at point O.
P
A
O
4.7 m
B
Q
3.
Steps of construction :
R
½
(i) Draw any line segment PQ = 5·5 cm.
(ii) With P as centre and radius 5·5 cm draw an arc.
(iii) With Q as centre and radius 5·5 cm draw an arc
to cut the previous arc at R.
(iv) Join PR and QR, then PQR is the required triangle.
5.5 cm
5.5 cm
½
½
5.5 cm
P
Q
½
Section - C
4.
D
x
y'
A
1
x'
B
P-88
5 cm
C
M A T H E M A T I C S --
IX
T E R M -- 2
5.
Steps for construction :
(a) Draw BC = 5 cm.
(b) Draw CBX = 60o and cut off BD = 7.7 cm.
1
(c) Join CD and draw its perpendicular bisector meeting BD at A.
(d) Join AC, then ABC is the required triangle.
1
Given : In right ABC, base BC = 4·5 cm and perimeter = 11·7 cm
i.e., AB + AC + BC = 11·7 cm or AB + AC = 7·2 cm, ABC = 90o.
½
o
Required : To construct the ABC with ABC = 90 , base BC = 4·5 cm and sum of other two
sides as 7·2 cm.
Y
D
A
90°
B
2
C
4.5 cm
Steps of construction :
1. Draw BC = 4.5 cm.
2. Draw ray BY such that CBY = 90o.
3. From ray BY cut off BD = 7·2 cm.
4. Join DC.
5. Draw the perpendicular bisector of DC intersecting BD at A.
6. Join AC, then ABC is the required triangle.
½
6.
X
M
A
30º
B
6 cm
C
2
Steps of construction :
(a) Draw a line segment BC = 6 cm.
(b) Draw ray BX such that CBX = 30°.
(c) From ray BX cut off BM = 10 cm.
(d) Join MC.
(e) Draw the perpendicular bisector of MC, intersecting BM at A.
(f) Join AC, then ABC is the required triangle.
G E O M E T R I C
C O N S T R U C T I O N S
1
1
P-89
Section - D
7.
Following steps will be followed to construct an angle of 90.
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which
intersect PQ at R.
½
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc at S.
½
(iii) Taking S as centre and with the same radius as before, drawn an arc intersecting the arc
at T (see figure).
½
(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.
(v) Join PU, which is the required ray making 90 with given ray PQ.
½
U
S
T
90º
P
R
Q
1
Justification of construction :
We can justify the construction, if we can prove UPQ = 90.
For this let us join PS and PT.
We have SPQ = TPS = 60. In (iii) and (iv) steps of the construction, we have drawn PU
as the bisector of TPS.

Now,
P-90
1
1
TPS =
× 60° = 30°
2
2
UPQ = SPQ + UPS
= 60 + 30 = 90
UPS =
M A T H E M A T I C S --
IX
1
T E R M -- 2
5
GEOMETRIC CONSTRUCTIONS
WORKSHEET-50
Section - A
1.
The following informations are :
(i) AB + BC
(ii) BC + CA
Section - B
2.
X
Steps of construction :
(a) Draw a line BC = 4 cm.
(b) At B, draw a ray BY such that YBC = 30o.
(c) At C, draw a ray CX such that BCX = 50o and
this ray intersects BY at A.
(d) Join A to B and C.
ABC is the required triangle.
1
Y
A
30°
B
3.
C
1
(iii) CA + AB.
1
50°
C
4 cm
R
2
Q
S
P
A
B
Section - C
4.
Steps of construction :
(i) Draw BC = 11 cm.
(ii) At B draw an angle of
1
1
× 30° = 15° and at C draw an angle of
× 90° = 45°.
2
2
(iii) Let the arms meet at X.
(iv) Draw right bisector of BX and CX which intersect BC at point Y and Z respectively.
(v) Join XY and XZ. Thus XYZ is required triangle.
1
X
30°
15°
B
Y
45°
Z
C
2
5.
Steps of construction :
1. Draw a line segment BC = 4 cm.
2. Bisect BC at D.
G E O M E T R I C
C O N S T R U C T I O N S
1
P-91
3.
4.
6.
From B and D draw arcs at a distance 5 cm each on the
same side of BC, cutting each other at A.
Join AB and AC.
Then, ABC is the required triangle.
A
1
5 cm
Let each of the base angles = xo.
the vertical angle = 2xo.
xo + xo + 2xo = 80o
4xo = 180o
xo = 45o
o
Each of the base angles is 45 and the vertical
angle is 90o.
Steps of construction :
(a) Draw BC = 7·5 cm.
(b) At B construct CBA = 45o and at C construct
BCA = 45o so that BA and CA intersect at A.
ABC is the required triangle.
1




B
4 cm
C
1
A
1
1
B
Section - D
7.
1
5 cm
45°
45°
7.5 cm
C
Steps of construction :
(i)
(ii)
(iii)
(iv)
(v)
Draw a line segment AB = 11 cm. (As XY + YZ + ZX = 11 cm)
Construct an angle PAB of 30 at point A and an angle QBA = 90 at point B.
1
Bisect PAB and QBA. These bisectors intersect each other at point X.
Draw perpendicular bisectors ST of AX and UV of BX.
Perpendicular bisector ST intersects AB at Y and UV intersects AB at Z. Join XY, XZ.XYZ
is the required triangle.
1
Q
S
X
P
U
30°
A
Y
Z
T
B
V
2
P-92
M A T H E M A T I C S --
IX
T E R M -- 2
5
GEOMETRIC CONSTRUCTIONS
WORKSHEET-51
Section - A
1.
1
30º.
Section - B
2.
C
P
N
A
M
B
1
Steps of construction :
(1) BAC = 110o is drawn with the help of a protractor.
(2) Taking A as centre, draw an arc of any radius to cut AB and AC at M and N respectively..
(3) Taking M and N as centre, arcs of equal radii are drawn to cut at P.
(4) AP is joined which is the required angle bisector of BAC.
1
Section - C
3.
Steps of construction :
(i) Draw a line segment BC = 7 cm. At point B draw an XBC = 75°.
(ii) Cut a line segment BD = 12 cm (that is equal to AB + AC) from the ray BX.
(iii) Join DC and make an angle DCY equal to BDC.
(iv) Line CY intersects BX at A. ABC is the required triangle.
1
1
X
D
Y
A
1
75°
B
G E O M E T R I C
C
C O N S T R U C T I O N S
P-93
4.
Steps of construction :
(i) Draw line segment QR = 6 cm. At point Q draw an angle = 60, i.e., XQR = 60°.
(ii) Cut a line segment QS = 2 cm from the line segment QT extended on opposite side of
line segment XQ. (As PR > PQ and PR – PQ = 2 cm). Join SR.
(iii) Draw perpendicular bisector AB of line segment SR, which intersects QX at point P. Join
PQ and PR. PQR is the required triangle.
1
X
P
Q
60°
A
R
S
B
T
5.
2
A =
3
× 180° = 45°
12
B =
4
× 180° = 60°
12
5
C =
× 180° = 75°
12
A
S
P
60º
B
R
C
75º
Q
Steps of construction :
(a) Draw a line PQ = 12·5 cm.
(b) At P construct SPQ = 60o and at Q construct RQP = 75o.
(c) Draw the bisectors of SPQ and RQP, intersecting at A.
(d) Draw the right bisector of AP and AQ intersecting PQ at B and C respectively.
(e) Join A to B and A to C.
ABC is the required triangle.
1
1
1
Section - D
6.
P-94
Given : Base BC = 7·5 cm, the difference of the other two sides AB – AC or AC – AB = 2·5
cm and one base angle 45o.
Let
AB > AC
AB – AC = 2·5 cm.
Steps of construction :
(i) Draw a ray BX and cut off a line segment BC = 7·5 cm from it.
(ii) Construct YBC = 45o
(iii) Cut off a line segment BD = 2·5 cm from BY.
(iv) Join CD.
(v) Draw a perpendicular bisector RS of CD intersecting BY at a point A.
M A T H E M A T I C S --
IX
T E R M -- 2
1
(vi) Join AC, then ABC is the required triangle.
Y
R
A
D
2.5 cm
45°
7.5 cm
B
C
X
S
2
Justification of construction :
RS is the perpendicular bisector of DC.
So,
AD = AC
and,
BD = AB – AD = AB – AC.
G E O M E T R I C
C O N S T R U C T I O N S
1

P-95
5
GEOMETRIC CONSTRUCTIONS
WORKSHEET-52
Type (A)
1.
(A).
1
2.
(A).
1
3.
(D).
1
Type (B)
4.
5.
P-96
(a) 40º.
1
(b) 6.8 cm.
1
(c) <
1
(d) Perimeter
1
(e) 22½º
1
(a) True.
1
(b) True.
1
(c) True.
1
(d) True.
1
(e) True.
1
M A T H E M A T I C S --
IX
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-53
Section - A
1.
The ratio of their heights is 2 : 1.
2.
The volume is
1
a 3
.
12
1
Section - B
T.S.A. of cube = 6 (side)2
3.
½
= 6 × 10.5 × 10.5
½
= 661.5
mm2
½
= 6.615
cm2.
½
Section - C
4.
Let the radius of the cone
Let the height of the cone
4x = r
3x = h
Volume of the cone = 2156 cm3

1
r2h = 2156
3

1 22

× 4x × 4x × 3x = 2156
3 7

22
× 16x3 = 2156
7

x3 =

x =

1
3
7 7 7
7
= 
222
2
7
= 3·5 cm
2
radius of the cone, r = 4x = r × 3.5 = 14 cm
height of the cone h = 3x = 3 × 3.5 = 10.5 cm
slant height of the cone, l =
h2  r 2 =
10  5 2  142
l = 17·5 cm
C.S.A. of the cone = rl
=
22
× 14 × 17·5 cm2
7
= 44 × 17.5 cm2
C.S.A. of the cone = 770 cm2.
Surface
Areas
and
Volumes
1
1
P-97
Section - D
Area (l × b) = 20 × 16 = 320 m2 of floor.
5.



Area of floor + Area of roof = S.A. of 4 walls
lb + lb = 2(l + b) × h
320 + 320 = 2(20 + 16)h
½
½
640
= 8·88 m.
72
Volume of the hall = (20 × 16 × 8·88) m3
= 2841·6 m3.

½
h =

6.
½
Given,
1
1
Total height of tent = 20 m
Radius = 8 m
Height of cylindrical portion of tent h = 20 – 6 = 14 m
Height of conical portion = 6 cm
Slant height of cone, l =

=
Total surface area of tent =
=
=
=
=
=
Total cost of canvas =
=
P-98
1
62  82
100  10 m
C.S.A. of cone + C.S.A. of cylinder
rl + 2rh
22
22
× 8 × 10 + 2 ×
× 8 × 14
7
7
1760
+ 704
7
251·42 + 704
955·42 m2
` 955·42 × 50
` 47,771.
M A T H E M A T I C S --
IX
1
2
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-54
Section - A
1.
1
The volume of cuboid is abc units.
Section - B
T.S.A. of cube = 864 m2 = 6a2
2.
6a2 = 864  a = 12 m

Volume of cube = a3 = (12)3 = 1728 m3.
1
1
Section - C
3.
Height of cone = 24 m
Circumference = 44 m
2r = 44
44  7
r =
=7m
2  22
l =
h2  r 2 = 576  49


Now,
1
=
1
22
× 7 × 25
7
= 550 m2.
1
625 = 25 m
Curved surface area = rl
=
Section - D
4.
For cylinder : H = 5 cm and R = 6 cm.
1
For sphere : r = 2 cm
Volume of water in cylinder =
R2H
cu. units
22
×6×6×5
7
110  36
=
7
= 565·71 cm3
Let rise in water level in cylinder be 'h' units

Volume of sphere = volume of water displaced in cylinder
4 3
r = R 2 h
3
3R
h =
4r 3
3 6  6
=
4 2 2 2
= 3·37 cm.
6
Volume of wall = 12 ×
× 4·5 m3
10
=
5.
Surface
Areas
and
Volumes
1
1
1
P-99
Volume of wall for bricks =
Volume of 1 brick =

No. of bricks =
9
6
× 12 ×
× 4·5 m3
10
10
1
18  12  10
m3
100 100 100
Volume of wall for bricks
Volume of 1 brick
1
9  12  6  4·5
10
10
= 18
12

 10
100 100 100
=
9
6
45
100
100
100
× 12 ×
×
×
×
×
10
10
10
18
12
10
1
= 13500
Cost of bricks at Rs. 225 per 100 bricks =
13500  225
100
1
= ` 30375.
P-100
M A T H E M A T I C S --
IX
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-55
Section - A
T.S.A. of cube = 6a2, here, a =
1.
a
T.S.A. = 6( a ) ( a ) = 6a.
1
Section - B
2.
l = Radius of quadrant of cone
l = 28 cm
Given,
1
×  × (28)2 = rl
4
28  28
= r × 28  r = 7 cm
4
C.S.A. of cone = rl
22
=
× 7 × 28
7
= 22 × 28 = 616 cm2.
or,
1
1
Section - C
3.
Mass = Density × Volume
4  22
Volume of sphere =
× 4·9 × 4·9 × 4·9
3 7
4 22 49  49  49  75
Mass of the shot putt = 3  7 
g
104
22  7  49  49
=
g
100
= 3697·54 g.

4.
Let,
Given :

radius of base = r
CSA, 2rh = 94·2
2 × 3·14 × r × 5 = 94·2
r =
=
volume =
=
=
Now,
94  2
10  3  14
3
r 2h
3·14 × 3 × 3 × 5
141·3 cm3.
½
½
1
1
1
1
1
Section - D
5.
Inner radius (r)
R
h
C.S.A. (Inner)
(i)
Surface
Areas
and
Volumes
= 2 cm
= 2·2 cm
= 77 cm
= 2rh
1
P-101
= 2×
(ii)
= 968 cm2
C.S.A. (Outer) = 2Rh
= 2×
(iii)
22
× 2 × 77
7
1
22
× 2·2 × 77
7
= 1064.8 cm2
Area of top =  (R + r) (R – r)
=
1
22
× 4·2 × 0·2
7
= 2·64 cm2 = Area of the bottom
T.S.A.= Inner (C.S.A.) + Outer (C.S.A.) + Area of top + Area of bottom
= 968 + 1064·8 + 2 × 2·64
= 2038·08 cm2.
P-102
M A T H E M A T I C S --
IX
1
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-56
Section - A
1.
It is same.
1
2.
The T.S.A. is r(4r + l).
1
Section - B
3.
Let the side of new cube be x cm
x3
8x 3
Volume of 8 new cubes
16 × 16 × 16
16  16  16
3
x =
= 83
8
x = 8
Volume of new cube
Volume of 8 new cubes
Volume of solid
8x 3

 Side of new cube is 8 cm.
=
=
=
=
½
½
½
½
Section - C
4.
2(lb + bh + hl) = 1078

lb + bh + hl
Let dimensions be x, 2x, 3x.

2x 2 + 6x2 + 3x2

11x 2

x2

x
 l = 7 m, b = 14 m, h = 21 m.

Volume of the box
½
½
½
½
= 539
=
=
=
=
539
539
49
7
½
= lbh = 7 × 14 × 21 = 2058 m3.
½
Section - D
5.
6.
 
2 36 3

cm3
3
2
Vol. of cylindrical bottle = (3)2 × 6 cm3
Suppose required bottles are x.
2 (18)3

x = 3
= 24.
(3)3  6
If r be the radius of cone, then slant height
l =
h2  r 2
1

c = rl; V = r2h
3
L.H.S. = 3Vh2 – c2h2 + 9V2
1
= 3 r 2 h h3 – (rl)2h2 + 9 1 r 2 h
3
3
= 2r2h4 – 2r2l2h2 + 2r4h2
= 2r2h4 – 2r2 (h2 + r2) h2 + 2r4h2
= 0.
Vol. of hemispherical bowl =

Surface
Areas
and
Volumes


1
1
2
1
1

2
1
1
P-103
6
SURFACE AREAS AND VOLUMES
WORKSHEET-57
Section - A
1.
The ratio is 21 : 11.
1
2.
The diameter is 42 cm.
1
Section - B
3.
Let the radius of hemisphere be r.

3r 2 = 5940

r2 =

r =
5940  7
= 630
3  22
1
1
630 = 3 70 cm
Section - C
4.
Base radius (r) =
14
= 7 cm
2
½
l = 25 m
C.S.A. = rl =
22
× 7 × 25 = 550 m2
7
½+1
Cost of white washing @ Rs. 210 per 100 m2
= `
5.
550  210
= ` 1155.
100
1
Diameter of the base = 20 cm

height
Margin at the top and bottom
Total height
Curved surface area
½
½
1
½
=
=
=
=
30 cm
2·5 × 2 = 5 cm
30 + 5 = 35 cm
2rh
22
=
× 20 × 35
7
= 2200 cm2.
½
Section - D
6.
7.
1
Inner dimensions of the box are 21 cm, 14 cm, 11 cm

Inner volume or capacity of the box = 21 × 14 × 11 cm3
= 3234 cm3
Outer volume of the box = 25 × 18 ×15 cm3
= 6750 cm3.

Volume of the wood used = 6750 – 3234
= 3516 cm3.
Let the radii of two spheres be r1, r2

r1 + r2 = 21
P-104
M A T H E M A T I C S --
1
1
1
1
IX
T E R M -- 2
or,
r1 = 21 – r2
4
Vol. of I sphere =
r 3 cm3
3 1
4
Vol. of II sphere =
r 3 cm3
3 2
4 π (21 − r )3
2
Vol. of I sphere
64
3
=
Vol. of II sphere =
27
4 πr 3
3 2
(21  r2 )3
64
=
r23
27

21  r2
r2
63 – 3r2
63
r2
r1





Surface
Areas
and
Volumes
=
=
=
=
=
4
3
4r2
7r2
9 cm.
12 cm.
½
½
½
½
½
½
P-105
6
SURFACE AREAS AND VOLUMES
WORKSHEET-58
Section - A
T.S.A. of hemisphere = 27 cm2
1.
So,
r =
9
cm.

1
Section - B
2.
Circumference of the base = 22 m

2r = 22
7
r =
m
2
Total surface area = 2rh + r2

Now,
= 22 × 6 +
= 170·5 m2.
½
½
22
49
×
7
4
1
Section - C
3.
h = 1 m = 100 cm.
Volume = 15·4 litres = 15·4 ×103 cm3
Now,



r 2h = 15·4 × 103 =
22
× r2 × 100
7
154  100  7
= 72
22  100
r = 7 cm
Total surface area = metal sheet required
= 2r (r + h)
½
r2 =
1
h
r
22
× 7(7 + 100)
7
= 44 × 107
= 4708 cm2
= 0·4708 m2.
½
= 2×
1
Section - D
4.
Volume of wall = l × b × h
Length
Thickness
Height
Volume of wall
Volume of bricks
No. of bricks × Volume of one brick
=
=
=
=
=
=
No. of bricks =
P-106
½
25 m = 2500 cm
0·3 m = 30 cm
6 m = 600 cm
2500 × 30 × 600 cm3
50 × 15 × 10 cm3
Volume of the wall
½
½
½
2500  30  600
= 6000
50  15  10
M A T H E M A T I C S --
1
IX
T E R M -- 2
Mortar occupies = 1/10th of volume
But.,
No. of bricks used =
6000 –
6000
= 5400.
10
1
Value Based Question
5.
Volume of glass, A = r 2h
= 3·14 × 2·5 × 2·5 × 10 = 196·25 cm3
(i)
Volume of hemisphere in glass B =
½
2
r3
3
2
× 3·14 × 2·5 × 2·5 × 2·5 = 32·71 cm3
3
Volume of glass B = Volume of glass A – Volume of hemisphere
= 196·25 – 32·71 = 163·54 cm3
=

1 2
1
r h =
× 3·14 × 2·5 × 2·5 × 1·5
3
3
= 9·81 cm3
Volume of glass C = 196·25 – 9·81
= 186·44 cm3.
(ii) The glass of type B has minimum capacity of 163·54 cm3.
(iii) Volume of solid figures.
(iv) Honesty.
Now,
Surface
Areas
1
volume of cone of glass C =
and
Volumes
½
1
1
P-107
6
SURFACE AREAS AND VOLUMES
WORKSHEET-59
Section - A
1.
The increased surface is 125%.
1
2.
The radius of the sphere is 3 cm.
1
Section - B
3.
Height of the cylinder formed, h = 22 cm
Circumference of the base, 2r = 22 cm

r =
11  7 7
11
 cm
=

22
2
1
22  7  7
× 22
7 2 2
= 847cm3.
Volume of the cylinder = r2h =
1
Section - C
4.
Circumference of base, 2r = 132 cm

132
× 7 = 21 cm
2  22
Capacity of cylindrical vessel = r 2h
=

1
r =
Volume in litres =
½
22
× 212 × 25 = 34650 cm3
7
1
34650
= 34·65 litre.
1000
½
Section - D
5.
Let,

side of new cube = x
(12)3 = 8x 3
(12)3
or,
= x3
8
12 3 = x 3
2
or
6 = x

side of new cube = 6 cm.
Ratio of surface areas of 8 new cubes to original cube
 
8  6 x2
8  6(6)2
=
2
6(12)
6  12  12
8666
2
=
=
6  12  12
1

Ratio = 2 : 1.
Given : lb = 15 sq. cm, bh = 20 sq. cm., lh = 12 sq cm.

(lbh)2 = 15 × 20 × 12 = (5 × 4 × 3)2

lbh = 5 × 4 × 3 cm3
Volume = 60 cm3.
=
6.
P-108
M A T H E M A T I C S --
IX
2
1
1
1
1
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-60
Section - A
1.
The ratio of their volumes is 20 : 27.
1
2.
The C.S.A. of hemisphere is 308 cm2.
1
Section - B
Volume of one math box = 4 × 2.5 × 1.5 cm3
3.
= 15 cm3
½
½
Volume of 12 such boxes = 15 × 12 = 180
cm3.
1
Section - C
4.
C.S.A. = 94.2 = 2rh
2


22
× r × 5 = 94.2
7
r =
1
94.2  7
= 3 cm.
2  22  5
1
Volume = r 2h
=
5.
Volume of the shot-put =
=
Mass of shot-put =
22
× 3 × 3 × 5 = 141.3 cm3.
7
4 22  35 
  
3 7  10 
1
3
11  49
cm3
3
11
78
 49 
= 1401.10 g.
3
10
½
1
1½
Section - D
6.
Let, the dimensions of the box are 2x m, 3x m, 4x m
Total surface area = 2 (2x × 3x + 3x × 4x + 4x × 2x) = 52x2
C1 = Cost of covering @ ` 8·00 per m2 = 52x2 × 8
C2 = Cost of covering @ ` 9·50 per m2 = 52x2 × (9·50)
Now,
C2 – C1= 52x2 (9·50 – 8) = 1248 (given)
1
1
1248
 16
52  1·5

x = 4
Dimensions of the box are 8 m, 12 m, 16 m.
1
Let r and R be the inner and outside radius of the cylindrical metallic pipe respectively.
Height of the metallic pipe, h = 14 cm
Curved surface area of the outer cylinder – curved surface area of the inner cylinder = 44 cm2
(given)
x2 =

7.
1
Surface
Areas
and
Volumes
P-109







2Rh – 2rh = 44 cm2
2 (R – r) × 14 cm = 44 cm2
44 × 7
1
(R – r) =
cm =
cm
44 × 14
2
Volume of the metal in the pipe = 99 cm3 (given)
R2h – r2h = 99 cm3
2
 (R – r2) × 14 cm = 99 cm3
22
× (R + r) (R – r) × 14 cm = 99 cm3
7
(R + r) =
Adding 1 and 2 we get
R–r+R+r =

1
99
9
cm =
cm
22
2
2
1 9
+
2 2
½
10
= 5 cm
2
5
R =
cm
2
2 R =

From 2 we have
P-110
1
22
1
× (R – r) ×
cm × 14 cm = 99 cm3
7
2


...(i)
5
9
cm + r =
cm
2
2
r = 2 cm
M A T H E M A T I C S --
½
IX
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-61
Section - A
1.
The volume is 144 cm3.
1
Section - B
2.
Given,
Circumference 2r = 22
and
h = 3m

1
C.S.A. = 2r × h
= 22 × 3
= 66 m2.
1
Section - C
3.
Let radius = r and height = h,
r
5
12
=
h=
r
h
12
5
given that
1 2
r h
3
1
12
× 3·14 × r2 ×
r
3
5
1
314
12
×
× r3 ×
3
100
5
r3
r
h
Given,





Now,
= 314 cm3
= 314
= 314
= 25 × 5 = (5)3
= 5 cm
= 12 cm
Slant height, l =

1
Curved Surface Area =
=
=
=
r 2  h2 = 52  12 2 = 13 cm.
rl sq. units
3·14 × 5 × 13
15·70 × 13
204·10 cm2.
1
1
Section - D
4.
Thickness of sheet = 1 cm = 0.01 m
Inner radius, r = 1 m
Outer radius, R = 1 + 0·01 m
= 1·01 m
2 3 2 3
R  r
3
3
2 22
  [(1·01)3  (1)3 ]
=
3 7
Now,
Volume of iron used =
=
1 cm
Surface
Areas
and
Volumes
2 22
  (1·030301  1)
3 7
1
1
1
P-111
2 22
  0·030301
3 7
= 0·06348 m3.
=
1
Value Based Question
5.
(i)


Also,

Radius of a cylinderical candle (r) = 2 cm
Height of a cylinderical candle (h) = 7 cm
Volume of a cylinderical candle = r 2h
22
=
× 2 × 2 × 7 = 88 cm3
7
Volume of 12 cylinderical candles = 12 × 88 = 1056 cm3
radius of a spherical fire cracker (r) = 1·5 cm =
Volume of a spherical fire cracker =
1
3
cm
2
4 3
r
3
=
4 22 3 3 3

  
3 7 2 2 2
=
99
cm3
7
1
99
× 14 = 198 cm3.
7
(ii) Lipsa has better project work because candles do not pollute the environment.
(iii) Avoid pollution and save energy.
Volume of 14 spherical fire crackers =
P-112
M A T H E M A T I C S --
IX
1
½
½
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-62
Section - A
1.
The volume is 125 cm3.
1
2.
The ratio of surface area of football and cricket ball is 25 : 1.
1
Section - B
3.
Volume of sphere =
4 3
r
3
½
4
22
×
× 4·9 × 4·9 × 4·9
3
7
= 493 cm3
=
½
Mass of 1 cm3 of metal is 7·8 gm
Mass of the shot-putt = 7·8 × 493 gm
= 3845·44 gm
= 3·85 kg (approx.)
½
½
Section - C
4.
Let, r1 and r2 are the radii of two cones.
l1
1
r1 4
 and l =
2
2
r2 1
1
CSA1
r1l1  r1   l1  4 1 2
     
=
CSA2
r2 l2  r2   l2  1 2 1
1
r1 : r2 = 4 : 1 
Now,

CSA1 = 2CSA2
Section - D
5.
Internal radius, r = 12 cm
External radius, R = 12·5 cm
S.A. = 2r2 + 2R2 + (R2 – r2)
= 2(r2 + R2) + (R – r) (R + r)
= 2 (144 + 156·25) +  (12·5 + 12) ( 12·5 – 12)
=
Cost of painting 1925.79
6.
Surface
cm2
I cone
r = radius
Areas
and
Volumes
1
22
(600·50 + 12·25)
7
= 1925·79 cm2
@ ` 0·05/cm2
= 1925·79 × 0·05
= ` 96·29.
(½)
1
II cone
r = radius
1
1
½
P-113
l=
h=
C.S.A. =
Given, rl =
rl =
slant height
height
rl
2rl
2r2l
r
2  2l
4
=
=
r'
l
1
r : r = 4 : 1.
P-114
(½)
l = slant height
h = height
C.S.A. = rl
½
½
½
½
½
M A T H E M A T I C S --
IX
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-63
Section - A
1.
The volume of cube is 81 3a3 cubic units.
1
2.
cm3.
1
The volume is 84
Section - B
Volume of cuboid = l × b × h = 880 cm3
3.
Area = l × b = 88 cm2
and,
So,
1
88 × h = 880

h = 10 cm.
1
Section - C
4.
Given,
Cost of painting at ` 10/m2 =
` 13200
13200
= 1320 m2
10
Perimeter of base = 110 m.
Area of 4 walls = Perimeter × height
1320 = 110h

Area painted =
Given,

1320
110
= 12 m.

1
1
h =
Total Cost
= Vol. of sphere
Cost/m 3
5.
=
4 3
r
3
1
1
1
33957
4 22 3
 r
=
7
3 7

101871
= r3  r3 = 1157.625
88
r = 10.5 cm.


1
Section - D
6.
Let, Outer radius of hemispherical cell R = 10/2 cm = 5 cm
and,
Inner radius r = 6/2 cm = 3 cm
Now,
Radius of cylinder =
14
= 7 cm
2
1
Let H be the height of the cylinder.

Vol. of hemispherical cell = Vol. of cylinder
Surface
Areas
and
Volumes
P-115
2
(R3 – r3) = (7)2 H
3
2

(53 – 33) = 49H
3
2
1
 98  = 4 cm

H =
3
49 3
1
= 1 cm.
3
Inside surface area of the hemispherical dome = 2r2
Cost of white washing = ` 498·96

7.
Inside surface area of the hemispherical dome = 2r2 =

P-116
1
1
1
1
1
1
498·96
m2
2
1
r = 6·3 m
2 3
Volume of air inside dome =
r = 523·91 m3.
3
M A T H E M A T I C S --
1
IX
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-64
Section - A
1.
The diameter of largest sphere that is curved out of a cube is 7 cm.
1
Section - B
2.
½
2 (lb + bh + hl) = 1372
Now,
l = 4x, b = 2x, h = x

2(8x2 + 2x2 + 4x2) = 1372

28x 2 = 1372

x = 7

1
Length = 4 × 7 = 28 cm.
½
Section - C
3.
Let,
the radius of the cone, 4x = r
Let,
Now,
the height of the cone, 3x = h
Volume of the cone = 2156 cm3

1
r2h = 2156
3

1 22

× 4x × 4x × 3x = 2156
3 7

22
× 16x3 = 2156
7

x3 =

x =

(given area = 154 cm2)
1
3
7 7 7
7
= 
222
2
7
= 3·5 cm
2
radius of the cone, r = 4x = r × 3.5 = 14 cm
height of the cone, h = 3x = 3 × 3.5 = 10.5 cm
slant height of the cone, l =
h2  r 2 =
10  5 2  142
l = 17·5 cm
C.S.A. of the cone = rl
=
22
× 14 × 17·5 cm2
7
= 44 × 17.5 cm2
= 770 cm2.
Surface
Areas
and
Volumes
1
1
P-117
Section - D
Vol. of water = r 2h
4.
=  × 9 × 9 × 40
Vol. of sphere =
=
Now,
No. of spheres =
=
1
4 3
r
3
4
 × 3 × 3 × 3 = 4 × 9
3
1
  9  9  40
4  9
1
90.
1
Value Based Question
5.
1
(i) We have, radius of cylinderical tank (r) = 1 m
and
height of cylinderical tank (h) = 3.5 m
Now,
Volume of cylinderical tank = r 2h
=
22
× 1 × 1 × 3.5 = 11 m3
7
1
Let the rain fall be h m, then
Volume of water on the roof = Volume of cylinderical tank

22 × 20 × h = 11
11
1
100


h =
m=
= 2·5 cm.
22  20 40
40
(ii) Save water to save earth.
P-118
M A T H E M A T I C S --
IX
1
1
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-65
Section - A
1.
The volume of a cube is 64 cm3.
1
2.
The largest rod in metres is 27 m.
1
Section - B
3.

V =
1
r2h
3
½
V =
3.5 3.5
1 22

×
×
× 12
2
2
3 7
½
= 38.5 m3
½
= 38500 litre.
½
Section - C
4.
Here r = 5 m, h = 12 m.
Now
Slant height l =
Curved surface area of tent =
=
=

Area of cloth required =
Given,
Width of cloth =

Length of cloth =
=
52  122 = 13 m
rl
22
× 5 × 13 m2
7
1430 2
m
7
1430 2
m
7
4
11
1 m=
m
7
7
1430
11
÷
7
7
130 m.
1
1
1
Section - D
5.

Cost of white washing = ` 498·96
498  96 2
m
2
2r 2 = 249·48 m2

C.S.A. of hemisphere =


2×


22
× r2 = 249·48
7
r2 = 39·69
r = 6·3 cm
Volume =
Surface
Areas
and
Volumes
1
1
2
r3
3
P-119
2 22

× 6·3 × 6·3 × 6·3
3 7
= 523·908 cm3.
Volume of metal contained in this pipe = [R2 – r2] h
1
=
6.
Here,
r = 5 cm, h = 24 cm, R = 5 + 0·5 = 5·5 cm
1
1
5·5
22
22
[(5·5)2 – 52] × 24 =
× 5·25 × 24 cm3
7
7
Mass = Volume × Density
Now,
V =
=
P-120
5
22
× 5·25 × 24 × 7 = 2772 g = 2·772 kg.
7
M A T H E M A T I C S --
1
1
1
IX
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-66
Section - A
1.
 r
The total surface area is r  l   .
 4
1
Section - B
2.
Diagonal of a cube =
3x 2
x2  x2  x2 = 6 3
1
= 6 3
3x = 6 3
x =
6 3
= 6 cm.
3
1
Section - C
Volume of the tank = 5 × 104 litres = 5 × 104 × 103 cm3
3.
Given,
Now,


l
l×b×h
250 × b × 1000
b
= 2·5 m = 250 cm, b = 10 m = 1000 cm
= 5 × 104 × 103
= 5 × 104 × 103
= 200 cm = 2 m.
1
1
1
Section - D
4.
Total surface area = 2R2 + 2r2 + (R2 – r2)
= 3R2 + r2
=
=
=
Cost of painting the vessel all over =
=
Given,
Radius of each pillar =

5.
=
Volume of each pillar =
=
=
Now,
Volume of 14 pillars =
=
So, 14 pillars would need 17·6
Surface
Areas
and
Volumes
m3
=
of concrete
 (3R2 + r2) =
1
22
[3 (8)2 + 62]
7
22
× [228] cm2
7
716·57 cm2
` (716·57 × 2)
` 1433·14.
20 cm
20
m
100
2
r h
22  20  20
× 10 m3
7 100 100
8·8 3
m
7
volume of one pillar × 14
8·8
× 14 m3
7
17·6 m3
mixture.
1
1
1
½
½
½
½
½
½
½
½
P-121
6
SURFACE AREAS AND VOLUMES
WORKSHEET-67
Section - A
1.
The radius of the sphere is 3.5 cm.
1
2.
The diameter of the base is 12 cm.
1
Section - B
3.
n =
Vcuboid 5  2  1

= 10 cubes.
Vcube
1 1 1
2
Section - C
9.375  100
× 100
22.5  10  7.5
= 5555 Bricks.
4.
5.
2
Number of Bricks required =
Given,
and
So,
Internal radius = 5 cm
13
cm
2
4
Volume =
 (R3 – r3)
3
 13 3

4
 53 
=
× 3·14 
3
 2

4
2197  1000
=
× 3·14
3
8
1197
1
=
× 3·14 ×
3
2
Volume of metal = 626·43 cu. cm.
½
External radius =


1
 
½
½

1
½
Section - D
6.
Here,
Slant height l =
=
h2  r 2
1
(2·1)2  202
=
401  41 cm
= 20·11 cm
Therefore, the curved surface area of corn cob
= rl
½
22
× 2·1 × 20·11 cm2
7
= 132·726 cm2
= 132·73 cm2 (approx.)
2
Number of corn cobs on 1 cm of the surface of the corn cob
= 4
Therefore, number of grains on the entire curved surface of the corn cob
= 132.73 × 4
= 530.92 = 531.
=
P-122
M A T H E M A T I C S --
IX
½
½
½
1
T E R M -- 2
6
SURFACE AREAS AND VOLUMES
WORKSHEET-68
1.
This Worksheet Cousists of activities which are to be done in class itself.
Surface
Areas
and
Volumes
P-123
7
STATISTICS
WORKSHEET-69
Section - A
1.
1
The mean of first five prime numbers is 5.6.
Section - B
2.
Given,
Mean of 40 observations =
Sum of 40 observations =
New sum =
=
1
6440
= 161.
40
New mean =
3.
½
160
160 × 40 = 6400
6400 + 165 – 125
6400 + 40 = 6440
½
Arranging the data in ascending order,
Here,
15, 28, 31, 32, 43, 44, 51, 56, 72.
n = 9, which is odd
½
½
th
 9  1  
So, Median = 
  term = 5th term = 43
 2  
½
If 32 is replaced by 23, then the new order is
15, 23, 28, 31, 43, 44, 51, 56, 72.

New median = 5th term = 43.
½
Section - C
4.
16
16
15
3
13
11
Number of Students
10
9
7
5
4
0
10
20
30
40
50
60
70
Marks
P-124
M A T H E M A T I C S --
IX
T E R M -- 2
Section - D
5.
By calculation we get the histogram as,
y
44
44
40
36
32
30
4
Frequency
28
24
20
16
16
12
8
4
6
4
0
1
2
4
6
8
10 12
14
16
18 20
x
Classes
S T A T I S T I C S
P-125
7
STATISTICS
WORKSHEET-70
Section - A
1.
The mean of perimeters of two squares having sides x and y units is 2(x + y) units.
1
Section - B
2.
Mean =

3.
Sum of observations
No. of observations
½
Total all of observations = 145 × 5 = 725
½
Correct total of all observations = 725 – 45 + 25 = 705
½
705
= 141.
5
Sum of 100 observations = 60 × 100 = 6000
Correct mean =
½
½
After replacements sum of new obs. = 6,000 – 50 + 110 = 6060

New mean =
½
6060
= 60.6.
100
1
Section - C
4.
Mean of data = (41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52
+ 98 + 40 + 52 + 52 + 60)/15
=
832
= 55.5.
15
1½
Arranging the data in ascending order
39, 40, 40, 41, 46, 48, 52, 52, 52, 52, 54, 60, 62, 96, 98
1½
In the data 52 occurs most frequently (four times).
Section - D
5.
Mean =

P-126
Sum of observations
No. of observations
1
Sum of observations = 150 × 30 = 4500
1
Correct sum of observations = 4500 – 135 + 165 = 4530
1
Correct mean =
4530
= 151.
30
M A T H E M A T I C S --
1
IX
T E R M -- 2
7
STATISTICS
WORKSHEET-71
Section - A
1.
1
The value of the lower limit is 2m-l.
Section - B
2.
Total of 40 observations = 16.5 × 40 = 660
1
correct total = 660 – 16.4 + 20.4 = 664
664
= 16.6.
40
Class size = 42 – 37 = 5
Now,
1
Correct mean =
3.
Lower limit of last class mark = 57 –
5
= 54.5
2
1
Upper limit of last class mark = 57 +
5
= 59.5.
2
1
Section - C
4.
Median = Mean of 5th and 6th observations

x  1  2 x  13
2
48 = 3x – 12

3x = 60

x = 20.
1
24 =
1
1
Section - D
5.
C.I.
25
30
35
40
45
50
S T A T I S T I C S
-
29
34
39
44
49
54
f
5
15
23
20
10
7
Continuous C.I.
24·5
29·5
34·5
39·5
44·5
49·5
-
29·5
34·5
39·5
44·5
49·5
54·5
Class marks
27
32
37
42
47
52
2
P-127
y
25
Scale :
x-axis
1 big unit=5
y-axis
1 big unit = 5
Frequency
20
15
10
2
5
19.5
24.5
29.5
34.5
39.5
44.5
49.5
54.5 59.5
x
C.I.
P-128
M A T H E M A T I C S --
IX
T E R M -- 2
7
STATISTICS
WORKSHEET-72
Section - A
1.
1
The value of x3 is 9.
Section - B
2.
Given data in ascening order is 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.
n = 10 (even)
Median = Mean of

63 =

63 =
 
n
2
th
+
 n2  1  obs.
th
Mean of 5th and 6th obs.
x x2
2
63 = x + 1
x = 62.


1
1
Section - C
3.
(i) No. of drivers :
(ii) No. of drivers :
1½
1½
15 + 10 = 25.
13 + 17 = 30.
Section - D
4.
Marks
Number of Students
0 – 20
4
20 – 40
10
40 – 60
15
60 – 80
18
80 – 100
20
4
Frequency polygon graph is ABCDEFG.
20
18
Number of students
16
14
12
10
8
6
4
2
–10
S T A T I S T I C S
0
20
40
60
Marks
80
100
x
P-129
7
STATISTICS
WORKSHEET-73
Section - A
1.
1
The class representing is 25–35.
Section - B
2.
Class interval
Tally
84-88
88-92
92-96
96-100
||
|||| ||||
|||| ||||
Marks
Frequency
||||
|||
|
2
4
13
11
½×4 =2
3.
The no. are 7, 8, 9, 9, x.
When, x = 9, mode = 9
When, x = 8, mode = 8
1
Difference between the mode = 9 – 8 = 1.
1
Section - C
4.
(i) Frequency Table :
No. of hours
Tally marks
No. of students
1
2
3
4
5
6
8
9
10
||
|||
|||
||||
|||
|||
|||| |
||
||||
2
3
3
4
3
3
6
2
4
Total
2
30
(ii) Mode = 8, as 8 occurs maximum number of times i.e., 6 times.
1
Section - D
5.
P-130
Marks
Frequency
Class marks
37 – 41
41 – 45
45 – 49
0
4
10
39
43
47
M A T H E M A T I C S --
IX
T E R M -- 2
49 – 53
53 – 57
57 – 61
61 – 65
65 – 69
69 – 73
15
18
20
12
13
0
51
55
59
53
67
71
2
y
20
18
Scale :
x-axis - 1 cms = 4 units
y-axis - 1 cms = 2 units
16
Frequency
14
12
10
2
8
6
4
2
0
37 41
45 49
53 57 61 65
69 73
x
Marks
S T A T I S T I C S
P-131
7
STATISTICS
WORKSHEET-74
Section - A
1.
1
Data.
Section - B
2.
Arrange the data in increasing order
2, 12, 12, 15, 17, 18, 26, 32, 32, 39, 42
Here n is odd

Median =
FG n  1IJ th observation
H 2K
FG 11  1IJ = 6 observation
H 2 K
1
th
=

th
1
Median = 18.
Section - C
3.
First ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
2  3  5  7  11  13  17  19  23  29
Mean =
10
129
=
= 12·9
10
10
 ( xi  x )
Now,
i 1
10
i 1

P-132
½
1
= (2 – 12·9) + (3 – 12·9) + (5 – 12·9) + (7 – 12·9)
+ (11 – 12·9) + (13 – 12·9) + (17 – 12·9) + (19 – 12·9)
+ (23 – 12·9) + (29 – 12·9)
= – 10·9 – 9·9 – 7·9 – 5·9 – 1·9 + 0·1 + 4·1
+ 6·1 +10·1 +16.
= – 36·5 + 36·5 = 0.
 ( xi  x )
4.
½
=
1
0.
x
f
fx
5
6
30
15
4
60
25
9
225
35
6
210
45
5
225
Total
f = 30
fx = 750
Mean ( x ) =
2
fx 750

= 25.
f
30
M A T H E M A T I C S --
1
IX
T E R M -- 2
Section - D
5.
Intervals
No. of lamps
300 – 400
400 – 500
500 – 600
600 – 700
700 – 800
800 – 900
900 – 1000
14
56
60
86
74
62
48
2
Frequency polygon is ABCDEFGHI.
y
No. of Lamps
90
E
F
75
60
C
D
G
H
45
2
30
B
15
0
I
A
x
Time in Hours
S T A T I S T I C S
P-133
7
STATISTICS
WORKSHEET-75
Section - A
1.
1
The mode is 9.
Section - B
2.
Here, we have Prime numbers 7, 11, 13, 17, 19
So,
7  11  13  17  19 67

= 13.4
5
5
Total marks of boys = 60 × 75 = 4500
½
Total marks of girls = 40 × 65 = 2600
½
Mean =
3.
½
Sum for class = 4500 + 2600 = 7100
Mean marks of the class =
2
7100
= 71.
100
½
Section - C
4.
y
200
180
165
152
No. of consumers
150
152
136
3
126
100
50
x
0
A
B
C
D
E
F
Product
Section - D
5.
ABCDEFGHI is the frequency polygon.
y
y-axis = One square = 2 persons
x-axis = One square = 4 years.
12
No. of Persons
10
D
8
2
F
C
6
4
4
E
G
B
H
A
4
I
8
12
16
20
24
28
32
x
Age (in years)
P-134
M A T H E M A T I C S --
IX
T E R M -- 2
7
STATISTICS
WORKSHEET-76
Section - A
1.
1
The class mark i.e., 140.
Section - B
2.
Arranging the data in increasing order,
40, 50, 65, 70, 75, 75, 95, 100
Here, n = 8 (even)
Median =
 n2 
3.
=

obs. + n  1
2
2

th
obs.
1
70  75
4 th obs. + 5 th obs.
=
2
2
=

th
145
= 72·5.
2
1
Ascending order of terms, 1, 4, 9, 16, 25, 36, 49, 64
½
No. of terms = 8 (even)
Median = Mean of 4th and 5th terms
=
y
4.
½
16  25 41

= 20.5.
2
2
1
Section - C
15
13
10
Frequency
10
9
3
6
5
5
2
0
10
20
30
40
50
60
70
x
Class - Interval
S T A T I S T I C S
P-135
Section - D
5.
y-axis : 1 big sq. = 2 runs
x-axis : 1 big sq. = 3 balls
y
10
9
Team A
Team B
8
Runs
7
6
4
5
4
3
2
1
0
3
9
15 21
27
33
39 42
x
Age (in years)
P-136
M A T H E M A T I C S --
IX
T E R M -- 2
7
STATISTICS
WORKSHEET-77
Section - A
1.
1
The class size is 5.
Section - B
2.
Arranging the data in the following form, we get
2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10
In the data 9 occurs most frequently (four times)
1
 Mode of data is 9.
1
3.
Sum of marks obstained by 30 girls = 30 × 73 = 2190
½
Sum of marks obatained by 20 boys = 20 × 71 = 1420
½
Sum of marks of 50 students of class =
2190 + 1420 = 3610
½
Mean for class =
3610
= 72·2 marks.
50
½
Section - C
4.
Mean =

23 =



Note :
253
6m
m
Take mean
=
=
=
=
2  5  4  10  6  13  10  9  (m  5)  6
5  10  13  9  6
10  40  78  90  6m  30
11
238 + 6 m
253 – 238
3.5
23 instead of 6
1
1
1
Section - D
5.
Mean =
Sum of observations =
Sum of observations
No. of observations
1
150 × 30 = 4500
1
Correct sum of observations = 4500 – 135 + 165 = 4530

S T A T I S T I C S
Correct mean =
4530
= 151.
30
1
1
P-137
7
STATISTICS
WORKSHEET-78
Section - A
1.
1
The mode of the following score is 14.
Section - B
Median of the observations = mean of 3rd and 4th observations
2.
x  ( x  2)
2
2x + 2
2x
x
3x + 1





=
½
9
= 9 ×2
= 16
= 8
= 3 × 8 + 1 = 25.
1
½
Section - C
3.
3
0
Miscellaneous
1
Entertainment
2
Rent
3
Education
4
Fuel
5
Medicine
7
6
Grocery
Expenditure
(in thousand rupees)
y
x
Heads
4.
y
14
12
10
3
Number of
Students
8
6
4
2
0
P-138
20
40
60
Marks
80
100
x
M A T H E M A T I C S --
IX
T E R M -- 2
Value Based Question
5.
(i) Required number of students = 8 + 32 = 40
½
(ii) Here, we notice that classes are continuous but class-size is not the same for all the classes. We
notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of
rectangle (adjusted frequency) for each class.
Length of rectangle (adjusted frequency) =
Frequency of class
× Minimum class-size
Width of class
Marks
(C.I.)
Number of students
(f)
Width of class
(Class-size)
0-10
8
10
10-30
32
20
30-45
18
15
45-50
10
5
1
Length of rectangle
8
×5=4
10
32
×5=8
20
18
×5=6
15
10
× 5 = 10
5
Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies
as heights.
1
Histogram representing marks obtained by students in unit test of Mathematics.
Number of students
y
10
9
8
7
6
5
4
3
2
1
O
(iii) Hardwork and Dilligence.
S T A T I S T I C S
10
8
6
4
1
10 20 30 40
Marks obtained
50
x
½
P-139
7
STATISTICS
WORKSHEET-79
Section - A
1.
The median of the given data is 149.
1
2.
The median of first 8 prime numbers is 19.
1
Section - B
3.
½
½
½
Arranging the data in ascending order
3, 5, 7, 7, 8, 9, 12, 14, 15, 17, 24, 27, 27, 27, 30
Total number of observations = 15
th
 15  1 
So, Median 
 i.e., 8th observation = 14.
 2 
In the data 27 occurs most frequently (three times)

Mode = 27.
4.
xx2x4x6x8
5
5x  20
13 =
=x+4
5
x = 9
1
Mean = 13 =

5.
21  16  24  x  29  15
6
105  x
23 =
6
105 + x = 138  x = 33.
½
½
A.M. =

6.
½
1

Arranging in ascending order,
144, 145, 147, 148, 149, 150, 152, 154, 155, 160
These are 10 observations.
Now,
Median = Mean of 5th and 6th observations
149  150
2
= 149.5 cm.
1
½
½
=
1
Section - C
7.
1
Writing the given data in ascending order :
2, 3, 3, 4, 4, 5, 6, 6, 7, 7, 7, 7, 7, 8, 9
Here n = 15
Mean =
=
x
n
85
= 5·6.
15
1
Mode = 7.
1
Median = 8th term = 6.
P-140
M A T H E M A T I C S --
IX
T E R M -- 2
8.
x
f
fx
10
30
50
70
90
Total
17
5p +3
32
7p – 11
19
60 + 12p
170
150p + 90
1600
490p – 770
1710
2800 + 640p
1
 fx
Mean =  f
2800  640 p
50 =
60  12 p
3000 + 600 p = 2800 + 640 p
3000 – 2800 = 640 p – 600 p
200 = 40p
p = 5.




2
Section - D
9.
Consider the classes 10 – 17 and 18 – 25.
The lower limit of 18 – 25 = 18
The upper limit of 10 – 17 = 17
The difference = 18 – 17 = 1

Half the difference =
1
 0·5
2
y
Age group
Frequency
1100
(in years)
–
–
–
–
–
–
17·5
25·5
33·5
41·5
49·5
57·5
Total
300
980
740
580
260
140
3000
1000
980
900
800
740
4
700
Frequency
9·5
17·5
25·5
33·5
41·5
49·5
600
580
500
400
300
300
260
200
140
100
0
9.5 17.5
25.5 33.5 41.5 49.5
57.5
x
Age Group
S T A T I S T I C S
P-141
7
STATISTICS
WORKSHEET-80
Section - A
1.
The value of x is 20.
1
Section - B
2.
3.
4.
The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
2  3  5  7  11  13  17  19  23  29
Mean =
10
129
=
= 12·9.
10
Arranging the data in ascending order
0·02, 0·03, 0·03, 0·04, 0·05, 0.05, 0·05, 0·07, 0·08, 0·08, 1·00, 1·03, 1·04
(i) Mode = 0·05
(ii) Range = 1·04 – 0·02 = 1·02.
Here,
n = 10, Median = average of 5th and 6th observations
( x  2)  ( x  4)
=
2
2x  6

24 =
2
x = 21.
½
½
1
1.
1
1
½
½
Section - C
5.
(i) Frequency Distribution Table :
Blood group
Tally Marks
No. of students (frequancy)
A
B
O
AB
|||| ||||
|||| |
|||| |||| ||
|||
9
6
12
3
Total
2
30
(ii) Blood group 'O' is most common as it has highest frequency i.e., 12.
(iii) Blood group AB is rarest.
½
½
Section - D
6.
x
55
50
49
81
48
57
65
f
8
3
10
7
3
7
2
fx
440
150
490
567
144
399
130
Total
f = 40
fx = 2320
2
 fx
2320
Mean ( x) =  f =
= 58.
40
P-142
M A T H E M A T I C S --
2
IX
T E R M -- 2
7
STATISTICS
WORKSHEET-81
Section - A
1.
1
The mean of first 10 natural numbers is 5.5.
Section - B
2.
Mean =
fi xi 4  5  6  10  9  10  10  7  15  8

fi
5  10  10  7  8
=
20  60  90  70  120
40
=
360
40
= 9.
1
1
Section - C
3.
3
Frequency distribution table is as follows :
C.I.
Tally marks
0 – 5
5 – 10
10 – 15
15 – 20
20 – 25
||||
||||
||||
||||
||||
Frequency
4
11
8
9
8
|||| |
|||
||||
|||
Section - D
4.
Frequency polygon is ABCDEFGHI.
y
No. of Lamps
90
E
F
75
60
C
D
G
H
45
4
30
B
15
I
A
x
Time in Hours
S T A T I S T I C S
P-143
Value Based Question
(i) The bar graph of the data below :
31.8
16
12
8
(2)
12.4
(3)
4.1
(4)
(5)
Causes
22
Other causes
20
4.3
Respratory conditions
(1)
24
25.4
Cardiovascular conditions
0
28
Injuries
4
Reproductive health conditions
Female Fatality Rate (%)
y
32
Neuropsychiatric conditions
5.
(6)
x
In the graph drawn causes of illnes and death among women between the ages 15-44 (in years)
worldwide is denoted on X-axis and female fatality rate (%) is denoted on the Y-axis.
2
(ii) The major cause of women’s ill health and death worldwide is reproductive health
condition.
1
(iii) Two other factors which play a major role in the cause in (ii) above are neuropsychiatric
conditions and other causes.
1
P-144
M A T H E M A T I C S --
IX
T E R M -- 2
7
STATISTICS
WORKSHEET-82
Section - A
1.
1
The range is 26.
Section - B
2.
Expenditure on pulses and ghee = 10% + 20% = 30%
Expenditure on wheat = 35%

½
Excess ependiture on wheat = 35% – 30% = 5%
3.

1
½
Mean =
Sum of observations
No. of observations
½
Mean =
9 + 18 + 27 + 36 + 45 + 54 + 63
7
1
= 36.
½
Section - C
4.
Mean =
10 + 20 + 30 + 40 + 50 150

 30
5
5
1½
Arranging in ascending order, we get 12, 16, 23, 24, 29, 30, as
a = 30, a –1 = 29

5.
Median of data =
½
23  24
= 23.5
2
1
Total score of 9 innings = 58 × 9 = 522
1
For mean score of 61, the total needed = 61 × 10 = 610
1
Score to be added in the 10th inning = 610 – 522 = 88.
1
Section - D
6.
Number of persons
y
20
18
15
12
10
12
8
4
4
5
0
S T A T I S T I C S
15
6
12 18 24 30
Age (in years)
36
x
P-145
7
STATISTICS
WORKSHEET-83
Section - A
1.
1
The median is 149.
Section - B
2.
First 5 prime numbers are 2, 3, 5, 7, 11
2  3  5  7  11 28

5
5
= 5.6.
1
Mean =
1
Section - C
3.
x
f
fx
4
6
8
10
12
4
8
14
11
3
16
48
112
110
36
Total
f = 40
fx = 322
Mean =
=
1½
 fx
f
½
322
= 8.05.
40
1
Section - D
4.
Frequency polygon is with class marks :
Class Marks
Frequency
155
165
175
185
195
205
5
15
20
25
15
5
1
y
30
25
Frequency
20
15
10
3
5
0
145 155 165 175 185 195 205 215
Class Marks
x
Frequency polygon showing various frequency for the class intervals given.
P-146
M A T H E M A T I C S --
IX
T E R M -- 2
Value Based Question
(i) The required graph is given below :
In the graph, different sections of the society is taken on X-axis and number of girls per
thousand boys is taken on the Y-axis.
Scale : 1 cm = 10 girls.
2
(ii) From the graph, the number of girls to the nearest per thousand boys are maximum in scheduled tribes whereas they are minimum in urban.
1
(iii) Pre-natal sex determination should strictly banned in urban.
1
Y
970
900
930
910
Urban
910
920
Rural
930
920
920
940
ST
940
950
Non-backward districts
950
Backward districs
960
SC
No. of girls per thousand boys
970
Non SC/ST
5.
X
Sections of Indian Society

S T A T I S T I C S
P-147
7
STATISTICS
WORKSHEET-84
The Worksheet consists of activities which are to be done in class itself.
P-148
M A T H E M A T I C S --
IX
T E R M -- 2
8
PROBABILITY
WORKSHEET-85
Section - A
1.
The probability is
27
.
50
1
2.
The probability is
2
.
5
1
Section - B
3.
(i)
(ii)
P(Likes the detergent) =
375 3
 k.
500 4
P(does not likes the detergent) = 1 – k = 1 
3 1
 .
4 4
1
1
Section - C
4.
(i)
P(more than 40 seed) =
(ii)
P(Less than 41 seed) =
(iii)
2
.
5
3
.
5
P(49 seeds in a bag) = 0.
1
1
1
Section - D
5.
(i)
Total number of families = 1500
Number of families with 2 girls = 475
475
19
=
.
1500
60
Number of families with no girl = 211
211
P (family with no girl) =
·
1500
P (family with 2 girls) =
(ii)
P R O B A B I L I T Y
1
1½
1½
P-149
8
PROBABILITY
WORKSHEET-86
Section - A
1.
The probability of getting an even number in a single throw of dice is
3 1
 .
6 2
1
Section - B
2.
Probability of getting less than 2 heads
1
Number of times 0 or 1 head appeared = 70
Hence,
3.
P(E) =
70
= 0.7.
100
1
(i) Total number of cases = 100
1
Number of cases favourable to an odd number = 20 + 20 + 20 = 60
60
3
= .
100
5
(ii) Number of cases favourable to the event ‘a prime number’ = 15 + 20 + 20 = 55
P (odd number) =
P (prime number) =
55
11
=
·
100
20
1
Section - C
4.
(i)
(ii)
5.
P(2 heads) =
72
200
=
9
.
25
1½
72  23 95 19


.
200
200 40
1½
P(at least 2 heads) =
This question is the part of statistics chapter. So Please ignore the question.
P-150
M A T H E M A T I C S --
IX
T E R M -- 2
8
PROBABILITY
WORKSHEET-87
Section - A
3
.
13
1.
The probability is
1
2.
The value of P(B) is 0.68.
1
Section - B
3.
(i)
(ii)
4.
P (the forecast was correct) =
P (the forecast was incorrect) = 1 
(i)
(ii)
175 7

.
300 12
P(2 heads) =
P(at least 2 heads) =
7
5

.
12 12
1
1
72
9

.
200 25
1
72  23 95 19


.
200
200 40
1
Section - C
5.
(i)
(ii)
(iii)
P(more than 40 seeds in a bag) =
4
.
5
P(49 seeds in a bag) = 0.
P(Less than 41 seeds in a bag) =
1
.
5
1
1
1
Section - D
6.
(i)
(ii)
(iii)
P(5) =
P(more than 10) =
P(between 5 and 10) =
P R O B A B I L I T Y
45
9

.
400 80
1
18  5 23

.
400
400
1
62  65  60  43
400
=
230
400
=
23
.
40
2
P-151
8
PROBABILITY
WORKSHEET-88
Section - A
1.
The probability is
7
.
11
1
Section - B
2.
Total number of students = 80
(i) Number of students getting less than 40 marks = 8 + 16 = 24
P(less than 40 marks) =
24
3
=
.
80
10
1
(ii) Number of students getting 60 or more than 60 marks = 10 + 6 = 16
P (60 or more than 60 marks) =
16
1
= ·
80
5
1
Section - C
3.
(i)
(ii)
(iii)
P(Earning ` 100 and more) =
2
.
25
1
P(at least ` 60 but < 80) =
2
.
25
1
5 1
 .
25 5
1
P(less than ` 40) =
Section - D
4.
(i)
240
3
=
.
4000
50
1
(ii)
760
38
=
.
4000
200
1
20  10  0  0  0
3
=
.
4000
400
2
(iii)
P-152
M A T H E M A T I C S --
IX
T E R M -- 2
8
PROBABILITY
WORKSHEET-89
Section - A
1.
The probability is
40  8 32 4

 .
40
40 5
1
2.
The probability is
7
7

7  3 10
1
Section - B
3.
(i)
(ii)
4.
P (for correct forecast) =
175 7

.
250 10
P(for incorrect forecast) = 1 
1
7
3

.
10 10
1
Total bags = 11
More than 5 kg of flour = 6
Prob. of more than 5 kg of flour =
1
6
.
11
1
Section - C
5.
½
No. of students obtained marks 60 or above= 15 + 8 = 23
23
90
Students who obtained marks less than 40 = 7 + 10 + 10 = 27

1
P(marks 60 or above) =
P(marks class than 40) =
1
27
.
90
½
Section - D
6.
Total number of students = 38
(i) Number of students whose weight is at most 60 kg
= 9 + 5 + 14 + 3 + 1 + 2 = 34
 Probability that weight of a student is at most 60 kg
34 17

½
38 19
(ii) No. of students whose weight is at least 36 kg = 5 + 14 + 3 + 1 + 2 + 2 + 1 + 1 = 29
½
=
29
.
38
(iii) No. of students whose weight is not more than 50 kg
= 9 + 5 + 14 + 3 = 31

Probability that weight is at least 36 kg =
 Probability that the weight of a student is not more than 50 kg =
P R O B A B I L I T Y
½
½
1
31
·
38
½
P-153
8
PROBABILITY
WORKSHEET-90
Section - A
1.
The probability is
25  16 9

.
25
25
1
Section - B
2.
(i)
Students getting 60 marks = 4
(ii)
4
2

.
30 15
Students getting less than 60 marks = 5 + 7 = 12
1
Probability of getting 60 marks =
Probability of getting less than 60 marks =
12 2
 .
30 5
1
Section - C
3.
(i)
P(3 heads) =
23
.
200
1
(ii)
P(no head) =
22
11

.
200 100
1
23  84 107

.
200
200
1
(iii)
P(at least 2 heads) =
Section - D
4.
400
4
=
.
700
7
(i)
1
(ii)
48  41  18  8  3
118 59

=
.
700
700 350
2
(iii)
400  180  48  41
669
=
.
700
700
1
Value Based Question
5.
(i) Marks 98·11% and 98·89% are associated with the months February and May respectively.
So the number of favourable outcomes = 2.
Number of all possible outcomes = Total number of marksheets = 6
2 1
 .
6 3
(ii) There is no marksheet associated with the month of July.


2
Required probability =
1
Required probability = 0.
(iii) The probability of an impossible event is zero.
½
(iv) Brilliant student.
½
P-154
M A T H E M A T I C S --
IX
T E R M -- 2
8
PROBABILITY
WORKSHEET-91
Section - A
1.
The probability is 0.49.
1
2.
The balls played is 50.
1
Section - B
3.
Frequency of outcomes of number more than 4
= 60 + 30 = 90
1
90
3

300 10
1
P(getting a number more than 4) =
4.
No. of women = 70 – (15 + 20 + 30) = 5
P(women) =
5
1

.
70 14
1
1
Section - C
5.
(i)
Prob. of less than 41 =
(ii)
Prob. of more than 50 =
(iii)
Prob. of marks between 11 and 80 =
=
8  12 20 2

 .
90
90 9
1
20  13  17  5 11

.
90
18
1
15  20  13  17
90
65
13
=
.
90
18
1
Section - D
6.
(i)
P(exactly 5 occupants) =
(ii)
P(more than 2 occupants) =
(iii)
P(less than 5 occupants) =
P R O B A B I L I T Y
5
1

.
100 20
2
23  17  5 9

.
100
20
1
95 19

.
100 20
1
P-155
8
PROBABILITY
WORKSHEET-92
The Worksheet consists of activities which are to be done in class itself.
P-156
M A T H E M A T I C S --
IX
T E R M -- 2