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# Question Bank: Area Related to Circles Notes | EduRev

## Class 10: Question Bank: Area Related to Circles Notes | EduRev

The document Question Bank: Area Related to Circles Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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``` Page 1

AREA RELATED TO CIRCLES

1) Find the circumference of circle of radius 4cm, if (pie = 3.14cm)
Solution:
Given: r = 4cm.
Therefore, Circumference of circle = r ? 2

cm 12 . 25
4 14 . 3 2
?
? ? ?

2) Find the area of circle whose radius is 7cm.
Solution:
Given: r = 7cm
Therefore, Area of circle =
2
r ?
= 7 7
7
22
? ?
=
2
154cm

3)
Find the area of quadrant of circle whose circumference is 33cm.

Solution:
Given: Circumference of the circle = 33cm.
We know that the area of quadrant of circle =

area of circle.
i.e. to find area of quadrant we need area of circle too which is =

.
Now we know the value of   , so to find value of ‘r’ we have,
Circumference of circle = 2
i.e. 33 = 2   ----------------------(given)

33
7
22
2 ? ? ? r

22 2
7 33
?
?
? r
cm r 25 . 5 ?
Area of circle =
2
r ?
= 25 . 5 25 . 5
7
22
? ?
=
2
625 . 86 cm Area of quadrant of circle = 625 . 86
4
1
? =
2
66 . 21 cm
.

Page 2

AREA RELATED TO CIRCLES

1) Find the circumference of circle of radius 4cm, if (pie = 3.14cm)
Solution:
Given: r = 4cm.
Therefore, Circumference of circle = r ? 2

cm 12 . 25
4 14 . 3 2
?
? ? ?

2) Find the area of circle whose radius is 7cm.
Solution:
Given: r = 7cm
Therefore, Area of circle =
2
r ?
= 7 7
7
22
? ?
=
2
154cm

3)
Find the area of quadrant of circle whose circumference is 33cm.

Solution:
Given: Circumference of the circle = 33cm.
We know that the area of quadrant of circle =

area of circle.
i.e. to find area of quadrant we need area of circle too which is =

.
Now we know the value of   , so to find value of ‘r’ we have,
Circumference of circle = 2
i.e. 33 = 2   ----------------------(given)

33
7
22
2 ? ? ? r

22 2
7 33
?
?
? r
cm r 25 . 5 ?
Area of circle =
2
r ?
= 25 . 5 25 . 5
7
22
? ?
=
2
625 . 86 cm Area of quadrant of circle = 625 . 86
4
1
? =
2
66 . 21 cm
.

AREA RELATED TO CIRCLES

4) A bicycle wheel makes 5000 revolutions in moving 11 km.  Find the diameter of the wheel.
Solution:
Given: Number of revolution = 5000
(1km = 1000m and 1m=100cm then 11km=11*1000m also 11*1000*100cm)
Distance = 11 km = cm 100 1000 11 ? ?

To find: ?

Distance moved by the wheel in 1 revolution = (Total distance covered / Total number of revolutions)
=
5000
100 1000 11 ? ?

= 220cm

Therefore, Circumference of the wheel = 220cm
220 2 ? r ?
220
7
22
2 ? ? ? r

22 2
7 220
?
?
? r
cm r 35 ?
Therefore, diameter of the wheel =d= 2r = 2(35) =70cm

5) The diameter of a wheel of a bus is 140cm. How many revolutions per minute must a wheel make in order to move
at a speed of 66km per hour?
Solution:
Distance covered by a wheel in 1 minute = cm cm 110000
60
100 1000 66
? ?
?
?
?
?
? ? ?

Circumference of a wheel = cm cm 440 70
7
22
2 ? ?
?
?
?
?
?
? ?
Number of revolution in 1 min = 250
440
110000
? ?
?
?
?
?
?

6) Find the radius of circle if an arc of angle
?
80 has length of ? 8 cm. Hence, find the area of the sector formulate by
this arc.

Page 3

AREA RELATED TO CIRCLES

1) Find the circumference of circle of radius 4cm, if (pie = 3.14cm)
Solution:
Given: r = 4cm.
Therefore, Circumference of circle = r ? 2

cm 12 . 25
4 14 . 3 2
?
? ? ?

2) Find the area of circle whose radius is 7cm.
Solution:
Given: r = 7cm
Therefore, Area of circle =
2
r ?
= 7 7
7
22
? ?
=
2
154cm

3)
Find the area of quadrant of circle whose circumference is 33cm.

Solution:
Given: Circumference of the circle = 33cm.
We know that the area of quadrant of circle =

area of circle.
i.e. to find area of quadrant we need area of circle too which is =

.
Now we know the value of   , so to find value of ‘r’ we have,
Circumference of circle = 2
i.e. 33 = 2   ----------------------(given)

33
7
22
2 ? ? ? r

22 2
7 33
?
?
? r
cm r 25 . 5 ?
Area of circle =
2
r ?
= 25 . 5 25 . 5
7
22
? ?
=
2
625 . 86 cm Area of quadrant of circle = 625 . 86
4
1
? =
2
66 . 21 cm
.

AREA RELATED TO CIRCLES

4) A bicycle wheel makes 5000 revolutions in moving 11 km.  Find the diameter of the wheel.
Solution:
Given: Number of revolution = 5000
(1km = 1000m and 1m=100cm then 11km=11*1000m also 11*1000*100cm)
Distance = 11 km = cm 100 1000 11 ? ?

To find: ?

Distance moved by the wheel in 1 revolution = (Total distance covered / Total number of revolutions)
=
5000
100 1000 11 ? ?

= 220cm

Therefore, Circumference of the wheel = 220cm
220 2 ? r ?
220
7
22
2 ? ? ? r

22 2
7 220
?
?
? r
cm r 35 ?
Therefore, diameter of the wheel =d= 2r = 2(35) =70cm

5) The diameter of a wheel of a bus is 140cm. How many revolutions per minute must a wheel make in order to move
at a speed of 66km per hour?
Solution:
Distance covered by a wheel in 1 minute = cm cm 110000
60
100 1000 66
? ?
?
?
?
?
? ? ?

Circumference of a wheel = cm cm 440 70
7
22
2 ? ?
?
?
?
?
?
? ?
Number of revolution in 1 min = 250
440
110000
? ?
?
?
?
?
?

6) Find the radius of circle if an arc of angle
?
80 has length of ? 8 cm. Hence, find the area of the sector formulate by
this arc.

AREA RELATED TO CIRCLES

Solution
Length of the arc (l) = ? 8 cm

Angle ? ?
?
80 ? ?
{to find radius ‘r’=? and area of sector formulated by this arc=?}
Let the radius of circle be ‘r’ cm.
Length of the sector (l) =
?
180
? ?r

Therefore,
??
?
180 ?
?
l
r =
?
?
80
180 8
?
?
?
?
= 18cm
Therefore, Radius of circle = 18cm.

Area of sector =
?
360
2
? ?r
=
?
?
360
80 18
2
? ? ?
=
?
?
360
80 18 18 ? ? ? ?
=
2
72 cm ?

7) In the figure find the area of shaded region.
Solution:
(Here to find the area of shaded region we will find the area of
Circle and area of rectangle too and then we will subtract the
Area of rectangle from area of circle because the rectangle
is inside the circle)
Length of rectangle ABCD = 12 cm
Width of rectangle ABCD = 5cm.
By the diagram you can see that diagonal of rectangle ABCD is a
Diameter of the circle and therefore
In right triangle, ABC
2 2
BC AB AC ? ? =
2 2
5 12 ? = 169 =13cm
Therefore, diameter of circle = 13 cm
Therefore, radius of circle = cm 5 . 6
2
13
? …………..{since d = 2r}
Since we know, Area of circle =
2
r ?
Therefore, Area of circle = 5 . 6 5 . 6 14 . 3 ? ? = 132.665 sq.cm
Area of the rectangle ABCD = Length x width
=
2
60 5 12 cm ? ?
Therefore, Area of the shaded portion =Area of the circle – Area of rectangle

Page 4

AREA RELATED TO CIRCLES

1) Find the circumference of circle of radius 4cm, if (pie = 3.14cm)
Solution:
Given: r = 4cm.
Therefore, Circumference of circle = r ? 2

cm 12 . 25
4 14 . 3 2
?
? ? ?

2) Find the area of circle whose radius is 7cm.
Solution:
Given: r = 7cm
Therefore, Area of circle =
2
r ?
= 7 7
7
22
? ?
=
2
154cm

3)
Find the area of quadrant of circle whose circumference is 33cm.

Solution:
Given: Circumference of the circle = 33cm.
We know that the area of quadrant of circle =

area of circle.
i.e. to find area of quadrant we need area of circle too which is =

.
Now we know the value of   , so to find value of ‘r’ we have,
Circumference of circle = 2
i.e. 33 = 2   ----------------------(given)

33
7
22
2 ? ? ? r

22 2
7 33
?
?
? r
cm r 25 . 5 ?
Area of circle =
2
r ?
= 25 . 5 25 . 5
7
22
? ?
=
2
625 . 86 cm Area of quadrant of circle = 625 . 86
4
1
? =
2
66 . 21 cm
.

AREA RELATED TO CIRCLES

4) A bicycle wheel makes 5000 revolutions in moving 11 km.  Find the diameter of the wheel.
Solution:
Given: Number of revolution = 5000
(1km = 1000m and 1m=100cm then 11km=11*1000m also 11*1000*100cm)
Distance = 11 km = cm 100 1000 11 ? ?

To find: ?

Distance moved by the wheel in 1 revolution = (Total distance covered / Total number of revolutions)
=
5000
100 1000 11 ? ?

= 220cm

Therefore, Circumference of the wheel = 220cm
220 2 ? r ?
220
7
22
2 ? ? ? r

22 2
7 220
?
?
? r
cm r 35 ?
Therefore, diameter of the wheel =d= 2r = 2(35) =70cm

5) The diameter of a wheel of a bus is 140cm. How many revolutions per minute must a wheel make in order to move
at a speed of 66km per hour?
Solution:
Distance covered by a wheel in 1 minute = cm cm 110000
60
100 1000 66
? ?
?
?
?
?
? ? ?

Circumference of a wheel = cm cm 440 70
7
22
2 ? ?
?
?
?
?
?
? ?
Number of revolution in 1 min = 250
440
110000
? ?
?
?
?
?
?

6) Find the radius of circle if an arc of angle
?
80 has length of ? 8 cm. Hence, find the area of the sector formulate by
this arc.

AREA RELATED TO CIRCLES

Solution
Length of the arc (l) = ? 8 cm

Angle ? ?
?
80 ? ?
{to find radius ‘r’=? and area of sector formulated by this arc=?}
Let the radius of circle be ‘r’ cm.
Length of the sector (l) =
?
180
? ?r

Therefore,
??
?
180 ?
?
l
r =
?
?
80
180 8
?
?
?
?
= 18cm
Therefore, Radius of circle = 18cm.

Area of sector =
?
360
2
? ?r
=
?
?
360
80 18
2
? ? ?
=
?
?
360
80 18 18 ? ? ? ?
=
2
72 cm ?

7) In the figure find the area of shaded region.
Solution:
(Here to find the area of shaded region we will find the area of
Circle and area of rectangle too and then we will subtract the
Area of rectangle from area of circle because the rectangle
is inside the circle)
Length of rectangle ABCD = 12 cm
Width of rectangle ABCD = 5cm.
By the diagram you can see that diagonal of rectangle ABCD is a
Diameter of the circle and therefore
In right triangle, ABC
2 2
BC AB AC ? ? =
2 2
5 12 ? = 169 =13cm
Therefore, diameter of circle = 13 cm
Therefore, radius of circle = cm 5 . 6
2
13
? …………..{since d = 2r}
Since we know, Area of circle =
2
r ?
Therefore, Area of circle = 5 . 6 5 . 6 14 . 3 ? ? = 132.665 sq.cm
Area of the rectangle ABCD = Length x width
=
2
60 5 12 cm ? ?
Therefore, Area of the shaded portion =Area of the circle – Area of rectangle

AREA RELATED TO CIRCLES

=
60 665 . 132 ?

=
2
67 . 72 cm

8) A paper is in the form of rectangle ABCD in which AB =18cm and BC =14cm A semi-circular portion with BC
as diameter is cut off. Find the area of remaining paper.
Solution:
Required Area = (area of rectangle ABCD) – (area of semi-circle with r = 7cm)
=
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
2
1
14 18 r ?

=
?
?
?
?
?
?
? ? ? ? 7 7
7
22
2
1
252
= ? ? 77 252 ?
=
2
175cm

9) ABCP is a quadrant of a circle of radius 14cm. With AC as diameter, semicircle is drawn. Find the area of
Solution:
In right triangle, ABC
2 2
BC AB AC ? ? =
2 2
14 14 ? = 2 14
AC is the diameter
So, radius = cm 2 7

Area of shaded region =Area of region(ABCQA) – Area of quadrant (ABCPA)
Area of region ABCQA = (Area of triangle ABC) + (Area of semi-circle ACQA)
= [

base height]  +[

= 2 7 2 7
7
22
2
1
14 14
2
1
? ? ? ? ? ?
= (98 + 154)
=
2
252cm

Page 5

AREA RELATED TO CIRCLES

1) Find the circumference of circle of radius 4cm, if (pie = 3.14cm)
Solution:
Given: r = 4cm.
Therefore, Circumference of circle = r ? 2

cm 12 . 25
4 14 . 3 2
?
? ? ?

2) Find the area of circle whose radius is 7cm.
Solution:
Given: r = 7cm
Therefore, Area of circle =
2
r ?
= 7 7
7
22
? ?
=
2
154cm

3)
Find the area of quadrant of circle whose circumference is 33cm.

Solution:
Given: Circumference of the circle = 33cm.
We know that the area of quadrant of circle =

area of circle.
i.e. to find area of quadrant we need area of circle too which is =

.
Now we know the value of   , so to find value of ‘r’ we have,
Circumference of circle = 2
i.e. 33 = 2   ----------------------(given)

33
7
22
2 ? ? ? r

22 2
7 33
?
?
? r
cm r 25 . 5 ?
Area of circle =
2
r ?
= 25 . 5 25 . 5
7
22
? ?
=
2
625 . 86 cm Area of quadrant of circle = 625 . 86
4
1
? =
2
66 . 21 cm
.

AREA RELATED TO CIRCLES

4) A bicycle wheel makes 5000 revolutions in moving 11 km.  Find the diameter of the wheel.
Solution:
Given: Number of revolution = 5000
(1km = 1000m and 1m=100cm then 11km=11*1000m also 11*1000*100cm)
Distance = 11 km = cm 100 1000 11 ? ?

To find: ?

Distance moved by the wheel in 1 revolution = (Total distance covered / Total number of revolutions)
=
5000
100 1000 11 ? ?

= 220cm

Therefore, Circumference of the wheel = 220cm
220 2 ? r ?
220
7
22
2 ? ? ? r

22 2
7 220
?
?
? r
cm r 35 ?
Therefore, diameter of the wheel =d= 2r = 2(35) =70cm

5) The diameter of a wheel of a bus is 140cm. How many revolutions per minute must a wheel make in order to move
at a speed of 66km per hour?
Solution:
Distance covered by a wheel in 1 minute = cm cm 110000
60
100 1000 66
? ?
?
?
?
?
? ? ?

Circumference of a wheel = cm cm 440 70
7
22
2 ? ?
?
?
?
?
?
? ?
Number of revolution in 1 min = 250
440
110000
? ?
?
?
?
?
?

6) Find the radius of circle if an arc of angle
?
80 has length of ? 8 cm. Hence, find the area of the sector formulate by
this arc.

AREA RELATED TO CIRCLES

Solution
Length of the arc (l) = ? 8 cm

Angle ? ?
?
80 ? ?
{to find radius ‘r’=? and area of sector formulated by this arc=?}
Let the radius of circle be ‘r’ cm.
Length of the sector (l) =
?
180
? ?r

Therefore,
??
?
180 ?
?
l
r =
?
?
80
180 8
?
?
?
?
= 18cm
Therefore, Radius of circle = 18cm.

Area of sector =
?
360
2
? ?r
=
?
?
360
80 18
2
? ? ?
=
?
?
360
80 18 18 ? ? ? ?
=
2
72 cm ?

7) In the figure find the area of shaded region.
Solution:
(Here to find the area of shaded region we will find the area of
Circle and area of rectangle too and then we will subtract the
Area of rectangle from area of circle because the rectangle
is inside the circle)
Length of rectangle ABCD = 12 cm
Width of rectangle ABCD = 5cm.
By the diagram you can see that diagonal of rectangle ABCD is a
Diameter of the circle and therefore
In right triangle, ABC
2 2
BC AB AC ? ? =
2 2
5 12 ? = 169 =13cm
Therefore, diameter of circle = 13 cm
Therefore, radius of circle = cm 5 . 6
2
13
? …………..{since d = 2r}
Since we know, Area of circle =
2
r ?
Therefore, Area of circle = 5 . 6 5 . 6 14 . 3 ? ? = 132.665 sq.cm
Area of the rectangle ABCD = Length x width
=
2
60 5 12 cm ? ?
Therefore, Area of the shaded portion =Area of the circle – Area of rectangle

AREA RELATED TO CIRCLES

=
60 665 . 132 ?

=
2
67 . 72 cm

8) A paper is in the form of rectangle ABCD in which AB =18cm and BC =14cm A semi-circular portion with BC
as diameter is cut off. Find the area of remaining paper.
Solution:
Required Area = (area of rectangle ABCD) – (area of semi-circle with r = 7cm)
=
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
2
1
14 18 r ?

=
?
?
?
?
?
?
? ? ? ? 7 7
7
22
2
1
252
= ? ? 77 252 ?
=
2
175cm

9) ABCP is a quadrant of a circle of radius 14cm. With AC as diameter, semicircle is drawn. Find the area of
Solution:
In right triangle, ABC
2 2
BC AB AC ? ? =
2 2
14 14 ? = 2 14
AC is the diameter
So, radius = cm 2 7

Area of shaded region =Area of region(ABCQA) – Area of quadrant (ABCPA)
Area of region ABCQA = (Area of triangle ABC) + (Area of semi-circle ACQA)
= [

base height]  +[

= 2 7 2 7
7
22
2
1
14 14
2
1
? ? ? ? ? ?
= (98 + 154)
=
2
252cm

AREA RELATED TO CIRCLES

Area of quadrant ABCPA = 14 14
7
22
2
1
? ? ?
=
2
154cm

Therefore, Area of shaded region = 154 252 ?
=
2
98cm

10)  A road which is 7m wide surrounds a circular park whose circumference is 352m. Find the area of the road.
Solution:
Let the radius of the park be ‘r’ m
Then, its circumference = r ? 2
So, 352 2 ? r ?
352
7
22
2 ? ? ? r

22 2
7 352
?
?
? r = 56 m
Thus, inner radius is 56 m
Therefore outer radius = (56 + 7) = 63 m

Area of the road = ? ?
2 2
) 56 ( ) 63 ( ? ?

=
? ? ) 56 63 )( 56 63 (
7
22
? ?

= 119 7
7
22
? ?
=
2
2618m

11) A circular pond of diameter 40 metres is surrounded by a grass path 5 metre wide. Find the area of the path.
Solution:
The radius of the inner circle, which is pond, is 20 m

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