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Title: Area related to circles
Description: Provide area related to circles questions and solutions.

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Templates covered in this chapter:
Template 1: Find the circumference of the circle when either radius or area of circle is given
...

Template 5: Find the area of minor and major sectors when angle of sector and radius are given
...

Steps required to follow:
Step 1: Find the radius of the circle from the given information:
If the radius of the circle is not given directly, then calculate the radius of the circle
Case 1: from diameter of the circle, if information of diameter is given
...

Step 2: Find the circumference of circle:
Substitute the value of radius in the formula of circumference of the circle and evaluate it
...
by multiplying the cost of fencing per unit length with circumference of circle
...


Question 1:
The area of a circle is 98
...
Find its circumference
...
56 π‘π‘š2

… (1)

But
Area of circle = πœ‹π‘Ÿ 2
Therefore from eq (1) and (2)
πœ‹π‘Ÿ 2 = 98
...
56
7

π‘Ÿ2 = (

7 Γ— 98
...
92
22

)

)

π‘Ÿ 2 = 31
...
6 cm
Step 2: Find the circumference of circle:
We know that, Circumference of circle = 2 πœ‹ π‘Ÿ
= 2 ( 3
...
6)

… (3)

[ Since π‘Ÿ = 5
...
168
Therefore, circumference of circle is πŸ‘πŸ“
...
Find the circumference of the circle
...
F
...

We know that,
Diameter of circle = 2π‘Ÿ
∴ Equation (1) becomes
C
...
C = 2π‘Ÿ + 45
2π‘Ÿ = 𝐢
...
πΆβˆ’45

……………… (2)

2

Step 2: Find the circumference of circle:
Circumference of circle (C
...
C) = 2πœ‹π‘Ÿ
Put
The value or radius r in equation (3)
C
...
C =

2πœ‹(𝐢
...
𝐢 βˆ’ 45 )

… (3)

……………………
...
𝐢 βˆ’ 45)
7

7 (𝐢
...
𝐢) βˆ’ 990
990 = ( 22 βˆ’ 7 )( 𝐢
...
𝐢
therefore,
C
...
C =

990
15

C
...
C = 66 cm
Therefore, circumference of the circle is equal to πŸ”πŸ” cm
...

Steps required to follow:
Step 1: Find the radius of the circle from the given information:
If the radius of the circle is not given directly, then calculate the radius of the circle
Case 1: from diameter of the circle, if information of diameter is given
...

Step 2: Find the area of circle:
Substitute the value of radius in the formula of area of the circle and evaluate it
...
by multiplying the cost of painting per unit area with the area of the circle
...


Question 1:
The circumference of a circle is 39
...
Find its area
...
6 cm

……………
...
(2)

Now from Equation(1) and equation (2)

2πœ‹π‘Ÿ = 39
...
6
7

39
...
2
44

π‘Ÿ = 12
...
6)2
7

= 498
...
96 π‘π‘š2

Question 2:
Find the area of a circle whose circumference is 66 cm
...
(1)

= 2πœ‹π‘Ÿ
Now from equation
...
(2)

Therefore the area of circle is πŸπŸ‘πŸ–πŸ” π’„π’ŽπŸ
...

If radius of each circle is not given directly, find it when,
Case 1: Diameter of smaller circle or circumference of smaller circle and width of each circle is given
...

Case 2: Diameter of larger circle or circumference of larger circle and width of each circle is given
...

Step 2: Find the width of the circle:
If width of the circle is not given, but only circumference or radii of concentric circles are given
...
If width of ring/concentric circle is asked to find
...


Question 1:
A road which is 7 m wide surrounds a circular park whose circumference is 352 m
...


Solution:
Step 1: Identify the radius of the circle from the given information:
Let the inner radius of the park be π‘Ÿ1 m and outer radius π‘Ÿ2 m
...
(1)
but circumference of inner circle = 352 ………………(2)
Therefore, from equation (1) and equation (2)
2πœ‹π‘Ÿ1 = 352
22

2 ( ) π‘Ÿ1 = 352
7

π‘Ÿ1 = (
π‘Ÿ1 = (

7 Γ— 352
44

2264
44

)

)

π‘Ÿ1 = 56 cm
But width of road = 7cm
Therefore, radius of outer circle is = π‘Ÿ2 = π‘Ÿ1 + 7
π‘Ÿ2 = 56 + 7
= 63 cm
Step 2: Find the area of each circle in the given concentric circle:
Required area of road = area of outer circle - area of inner circle
= πœ‹π‘Ÿ22 βˆ’ πœ‹π‘Ÿ12
= πœ‹ (π‘Ÿ22 βˆ’ π‘Ÿ12 )
22

= ( ) (632 βˆ’ 562 )
7

22

= ( ) (63 βˆ’ 56 ) Γ— (63 + 56)
7

22

= ( ) (7 Γ— 119)
7

= (22) Γ— (119)
Required area of road = 2618 π‘š2
Therefore, required area of road is πŸπŸ”πŸπŸ– π’ŽπŸ

Question 2:
A race track is in the form of a ring whose inner and outer circumferences are 437 m and 503 m respectively
...


Solution:
Step 1: Identify the radius of the circle from the given information:
Let the inner radius of the park be π‘Ÿ m and outer radius 𝑅 m
...
(1)
but circumference = 437 ………………(2)
Therefore, from equation (1) and equation (2)
We get, 2πœ‹π‘Ÿ1 = 437

2(3
...
28 )
π‘Ÿ1 = 69
...
5 m
Therefore, radius of inner circle is πŸ”πŸ—
...

Similarly, we get circumference = 2πœ‹π‘… ………………
...

Step 2: Find the width of the concentric circle:
Now, Width of the track = 𝑅 βˆ’ π‘Ÿ
Width of the track = 80 βˆ’ 69
...
5
Width of the track = 10
...
52 )
7

22

= ( ) Γ— ( 80 βˆ’ 69
...
5 )
7

22

= ( ) Γ— ( 10
...
5 )
7

= (22) Γ— (224
...
5 π‘š2

Therefore required area of track is πŸ’πŸ—πŸ‘πŸ‘
...
Find the radius of resultant circle which has circumference equal to the sum of the circumferences of the
two circles or area equal to the sum of areas of two circles
...

Step 2: Find the circumference/area of resultant circle:
Case 1: Find the sum of circumference of two given circles to find the circumference of the resultant circle, if
circumference is asked
...

Step 3: Find the radius of the required circle:
Case 1: Equate the circumference of the circle with the circumference of the circle formula to find the radius of
resultant
...


Example 1:
The radii of two circles are 19 cm and 9 cm respectively
...


Solution:
Step 1: Find the circumference\area of two circles using given radius:
Let 𝑅 and π‘Ÿ be the radii of two given circles and radius of resultant circle π‘…π‘Ÿ
Given, 𝑅 = 19 cm, π‘Ÿ = 9 cm and
Now, circumference of circle with radius 𝑅 = 2πœ‹π‘…
Circumference 𝐢1 = 2πœ‹(19)
= 38 πœ‹ cm
and
Circumference of circle with radius π‘Ÿ = 2πœ‹π‘Ÿ
Circumference 𝐢2 = 2πœ‹ (9)
= 18 πœ‹
= 18 πœ‹ cm

Step 2: Find the circumference/area of resultant circle:
Resultant circumference of two circles = 𝐢1 + 𝐢2
= 38 πœ‹ + 18 πœ‹
= 56 πœ‹
Resultant circumference of two circles = 56 πœ‹ cm
Step 3: Find the radius of the required circle:
Circumference of resultant circle = Resultant circumference of two circles =
= 2πœ‹π‘…π‘Ÿ
56 πœ‹ = 2πœ‹π‘…π‘Ÿ
56 = 2π‘…π‘Ÿ
Radius of the circle π‘…π‘Ÿ =

56
2

= 28 cm
...


Example 2:
The radii of two circles are 8 cm and 6 cm respectively
...


Solution:
Step 1: Find the circumference\area of two circles using given radius:
Let 𝑅 and π‘Ÿ be the radii of two given circles and radius of resultant circle π‘…π‘Ÿ
Given, 𝑅 = 8 cm, π‘Ÿ = 6 cm and
Now
Area of circle with radius 𝑅 = πœ‹π‘…2
= πœ‹ (8)2
Thus, area of circle 𝐴1 = 64 πœ‹π‘π‘š2
and
Area of circle with radius 𝑅 = πœ‹π‘…2
= πœ‹ (6)2
Area of circle 𝐴2 = 36 πœ‹π‘π‘š2

Step 2: Find the circumference/area of resultant circle:
Resultant Area of two circles = 𝐴1 + 𝐴2
= 64 πœ‹ + 36 πœ‹
= 100 πœ‹
Resultant Area of two circles is 100 πœ‹ cm2
Step 3: Find the radius of the required circle:
Area of circle = πœ‹π‘…π‘Ÿ2 = Resultant Area of two circles
100 πœ‹ = πœ‹π‘…π‘Ÿ2
100 = π‘…π‘Ÿ2
Radius of the circle π‘…π‘Ÿ = 10
Thus, radius of the circle 𝑹 = 𝟏𝟎 cm
...

Steps required to follow:
Step 1: Find the area of the minor sector:
Substitute the values of angle of sector (πœƒ) and radius (π‘Ÿ) in the area of sector formula

πœƒ
360

Γ— πœ‹π‘Ÿ 2 and simplify it
...

Or
Substitute the values of angle of minor sector(πœƒ) and radius(π‘Ÿ) in area of major sector formula
360βˆ’πœƒ
360

Γ— πœ‹π‘Ÿ 2 to find the area of a major sector
...


Solution:
Step 1: Find the area of the minor sector:
Given, Radius of circle (π‘Ÿ) = 14 cm
Angle of sector (πœƒ) = 60
Now, Area of sector =
=

60
360

Γ— πœ‹(14)2

πœƒ
360

Γ— πœ‹π‘Ÿ 2

=

1
6

Γ— πœ‹(196)
22

= (98) ( ) (3)
7

Therefore, area of minor sector is 𝟏𝟎𝟐
...


Solution:
Step 1: Find the area of major sector:
Given, Radius of circle (π‘Ÿ) = 21 cm
Angle of minor sector (πœƒ) = 120
Now
Area of major sector =
=
=

240
360
2
3

360βˆ’πœƒ
360

Γ— πœ‹π‘Ÿ 2

Γ— πœ‹(21)2

Γ— πœ‹(441)

= 2πœ‹(147)
Therefore, area of major sector is πŸ—πŸπŸ‘
...

Case 1: Substitute the given angle of sector in the area of sector formula and equate it to the given area of sector to
find radius of sector
...


Question 1:
The area of a sector 𝑂 βˆ’ 𝐴𝐡𝐢 is equal to 125 πœ‹ π‘π‘š2 and radius of a sector is 15 cm
...


Solution:

Step 1: Find the angle of sector
...


Question 2:
The area of a sector 𝑂 βˆ’ 𝐴𝐡𝐢 is equal to 250πœ‹ π‘π‘š2 and angle of a sector is 200Β°
...


Solution:
Step 1: Find the radius of sector
...

Given, Angle of sector = 200Β°
Area of sector = 250πœ‹ π‘π‘š2
Now
Area of sector =
250πœ‹ =
250 =

𝟐𝟎𝟎
360
πŸ“
9

πœƒ
360

Γ— πœ‹π‘Ÿ 2

Γ— πœ‹π‘Ÿ 2

Γ— π‘Ÿ2

250 Γ— 9 = 5π‘Ÿ 2
2250 = 5π‘Ÿ 2
π‘Ÿ 2 = 450
π‘Ÿ = 21
...
𝟐𝟏 cm
...

Steps required to follow:
Step 1: Find the length of sector or radius of circle:
Case 1: Substitute the values of angle of sector and radius in the length of arc formula

πœƒ
360

Γ— 2πœ‹r and simplify it to find

the length of the Arc
...


Question 1:
The angle described by sector 0 βˆ’ 𝐴𝐡𝐢 is 45 degree and radius of sector is 14 cm , then find the value of length of arc

Solution:
Step 1: Find the length of sector or radius of circle:
Given,
Angle of sector πœƒ = 45
And radius π‘Ÿ = 14 cm
We have, Length of arc =
=
=

45
360
1
8

πœƒ
360

Γ— 2πœ‹r

Γ— 2πœ‹14

Γ— 28 Γ—

22
7

= 11 cm
Therefore, the length of arc is 𝟏𝟏 cm
...


Solution:
Step 1: Find the radius of circle:
Given, angle of sector (πœƒ) = 135
And the length of arc = 28 cm
We have, Length of arc =

πœƒ
360

Γ— 2πœ‹r

28 =

135
360

Γ— 2πœ‹r

3

28 = Γ— 2πœ‹r
8

Radius( π‘Ÿ ) = 11
...
πŸ–πŸ• cm
...

Steps required to follow:
Step 1: Find the area of sector:
Substitute the radius and angle of sector in the area of corresponding sector using formula

πœƒ
360

Γ— πœ‹π‘Ÿ 2
...

Step 3: Find the area of the segment:
Subtract area of corresponding triangle from area of sector to find the area of segment
...

Find (i) the length of the arc, (ii) the area of the sector, (iii) the area of the minor segment
...

Given, radius of circle (π‘Ÿ) = 21 cm
and angle of sector (πœƒ) = 60Β°
We know that, (i) Length of the arc 𝐴𝐢𝐡 = 2πœ‹π‘Ÿ(

πœƒ
360

)

= (2 Γ—

22
7

Γ— 21 Γ—

60
360

) cm

= 22cm
...

Step 2: Find the area of the corresponding triangle 𝑢𝑨𝑩:
We have, Area of triangle 𝑂𝐴𝐡 =
Area of triangle 𝑂𝐴𝐡 =
=

1
2

1
2

1
2

π‘Ÿ 2 sin πœƒ

(21)2 sin 60

(441) sin 60

= 190
...
πŸ•πŸ‘ π’„π’ŽπŸ
Step 3: Find the area of the segment
...
73
= 40
...


Question 2:
In a circle of radius 14 cm, an arc subtends an angle of 120 at the center
...


Solution:
Step 1: Find the area of sector
Let 𝐴𝐢𝐡 be the given arc subtending an angle of 120° at the center
...
34 π‘π‘š2
...

Step 2: Find the area of the corresponding triangle 𝑢𝑨𝑩:
1

Area of triangle 𝑂𝐴𝐡 = π‘Ÿ 2 sin πœƒ
2
1

Area of triangle 𝑂𝐴𝐡 = (14)2 sin 120
2

1

= (196) sin 120
2

= 49(3
...
86 π‘π‘š2
Therefore , the area of triangle = 153
...

Area of segment = area of sector 𝐴𝑂𝐡𝐴 βˆ’ area of triangle 𝑂𝐴𝐡
= 205
...
86
= 51
...
πŸ’πŸ– π’„π’ŽπŸ
Template 9: Find the area of the quadrant of a circle whose circumference or radius are given
...

Step 2: Find the area of quadrant:
Substitute the radii in the area of the circle formula and find the area of the circle
...


Question 1:
Find the area on the quadrant for a circle whose circumference is 198 cm
...

Circumference of circle = 2πœ‹π‘Ÿ
= 198
Therefore, radius π‘Ÿ = (

198 Γ— 7
2 Γ— 22

)

π‘Ÿ = 15
...
5 cm
Step 2: Find the area of quadrant:
Area of circle = πœ‹π‘Ÿ 2
= (3
...
5)2
= 754
...
38 π‘π‘š2
Now
Required area of quadrant =
=

π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘–π‘Ÿπ‘π‘™π‘’
4

754
...
6 π‘π‘š2
Hence, area of quadrant is πŸπŸ–πŸ–
...
then find the area of the quadrant for the given
circle
...
14)(14)2
= 616
Thus, Area of circle = 616 π‘π‘š2

Now , required area of quadrant =
=

π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘–π‘Ÿπ‘π‘™π‘’
4

616
4

= 154 π‘π‘š2
Hence, area of quadrant is πŸπŸ“πŸ’ π’„π’ŽπŸ
...

Step 2: Find the area of combined plane figure:
According to the given problem, find the area of the shaded region in the combined figure by
adding or subtracting corresponding plain figures
...
If asked to find the cost of painting etc
...

Find the area of the shaded region
...

Let these regions meet at a common point 𝑂
...
14 Γ— 3
...
25)
= 100 βˆ’ 78
...
5 π‘π‘š2
Similarly, (area of II) + (area of IV) = 21
...

Step 2: Find the area of combined plane figure:
Area of shaded region = π‘Žπ‘Ÿ(π‘ π‘ž 𝐴𝐡𝐢𝐷) – π‘Žπ‘Ÿ(𝐼 + 𝐼𝐼 + 𝐼𝐼𝐼 + 𝐼𝑉)
= 100 βˆ’ (21
...
5 )
= 100 βˆ’ 43
Thus, area of shaded region is πŸ“πŸ• π’„π’Ž^𝟐
...
If the center of each circular flower bed is the point of intersection 𝑂 of the diagonals of the square lawn, find (i)
the sum of the areas of the lawn and the flower beds, (ii) the sum of the areas of two flower beds
...

So, 𝐴𝐢 = 𝐡𝐷
Diagonal of square (𝐴𝐢) = √2 Γ— 𝐴𝐡
= √2 Γ— 56 = 56 √2
So, 𝑂𝐴 = 𝑂𝐡 =

1
2

1

𝐴𝐢 = 56 √2 = 28√2π‘š
2

Step 2: Find the area of combined plane figure:
Let 𝑂𝐴 = 𝑂𝐡 = π‘Ÿ π‘š [radius of the sector]
Area of sector 𝑂𝐴𝐡 = [
1

1

22

4

4

7

= ( ) πœ‹π‘Ÿ 2 = Γ—
1

22

4

7

=[ Γ—

90Β°
360Β°

] πœ‹π‘Ÿ 2

28 2

Γ— ( ) π‘š2
√2

Γ— 28 Γ— 28 Γ— 2] π‘š2

= 1232 π‘š2
Area of flower bed 𝐴𝐡 = area of sector 𝑂𝐴𝐡 βˆ’ area of Δ𝑂𝐴𝐡
1

1

2

2

β‡’ 1232 βˆ’ Γ— 𝑂𝐡 Γ— 𝑂𝐴 β‡’ 1232 βˆ’ Γ— 28√2 Γ— 28√2
β‡’ 1232 βˆ’ 784 = 448 π‘š2
Similarly, area of square 𝐴𝐡𝐢𝐷 + area of flower bed 𝐴𝐡 + area of flower bed 𝐢𝐷
= (56 Γ— 56) + 448 + 448 = 4032 π‘π‘š2

Template 11: Find the area of the remaining portion after removing certain shapes from a given shape
...

Step 2: Find the area of remaining plane figure:
According to the given problem, find the area of the given figure and subtract the area of removed
portions from it to find the area of the remaining figure
...


If asked to find the cost of painting etc
...
Find the area of the shaded region
...

As 𝐴𝐡𝐢 is a quadrant of the circle, ∠𝐡𝐴𝐢 will be measured 90°
...
So substitute radius of sector
and perimeter of sector and simplify it to find length of arc
...


Question 1:
The perimeter of a sector of a circle of radius 14 cm is 68 cm
...


Solution:
Step 1: Find the length of arc:
Let 𝑂 be the center of a circle of radius 14 cm
...

Then, 𝑂𝐴 + 𝑂𝐡 + π‘Žπ‘Ÿπ‘(𝐴𝐢𝐡) = 68 cm
14 π‘π‘š + 14 π‘π‘š + π‘Žπ‘Ÿπ‘(𝐴𝐢𝐡) = 68
π‘Žπ‘Ÿπ‘(𝐴𝐢𝐡) = 68 βˆ’ 14 βˆ’ 14
∴ π‘Žπ‘Ÿπ‘(𝐴𝐢𝐡) = 40 cm
Step 2: Find the area of sector:
1

∴ π‘Žπ‘Ÿ(π‘ π‘’π‘π‘‘π‘œπ‘Ÿ 𝑂𝐴𝐢𝐡𝑂) = ( Γ— π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  Γ— π‘Žπ‘Ÿπ‘ π‘™π‘’π‘›π‘”π‘‘β„Ž )
2

= ( Γ— 14 Γ— 40 )
= 280 π‘π‘š2

∴ π‘Žπ‘Ÿ(π‘ π‘’π‘π‘‘π‘œπ‘Ÿ 𝑂𝐴𝐢𝐡𝑂) = 280 π‘π‘š2

Example 2:
In a circle of radius 21 cm, an arc subtends an angle of 60 at the center
...

Given radius of circle (π‘Ÿ) = 21 cm
and angle of minor sector (πœƒ) = 60Β°
(i) Length of the arc 𝐴𝐢𝐡 = 2πœ‹r (
= (2 Γ—

22
7

Γ— 21 Γ—

πœƒ
360

)

60
360

) cm

= 22cm
...

∴ Area of the sector 𝐴𝐢𝐡𝑂𝐴 = 231 π‘π‘š2
...
e
...
Find the angle made by a minute hand

in the given time
...


Question 1:
The length of the minute hand of a clock is 14cm
...


Solution:
Step 1: Find the angle made by minute hand for the given time:
Given, length of the minute hand = 14 cm
...
πŸπŸ–πŸ” π’„π’ŽπŸ

Question 2:
The length of the minute hand of a clock is 10cm
...
25am
...
25am= 30π‘œ
Step 2: Substitute angle and radius to find required area:

Area of face of clock described = Area of sector = πœ‹π‘Ÿ 2 (
=
=

30
360
1
12

πœƒ
360

)

Γ— πœ‹ Γ— 102

Γ— 100πœ‹

Thus, required area of sector is πŸπŸ”
...


β€’

Then find the distance covered by the wheel in given time
...
As the
circumference of a circular shaped wheel is nothing but the distance travelled by car in one
complete revolution
...


Question 1:
A car has wheels which are 80 cm in diameter
...


Question 2:
The wheel of a motor cycle is of radius 35 cm
...
35 cm
The speed it must keep = 𝑠 = 66 km/Hence, the answer is =

66 Γ—100
60

m/min = 1100 m/min

Step 2: Find the circumference of the wheel:
The circumference = 𝐢 of the wheel = 2πœ‹π‘Ÿ = 2 Γ—

22
7

Γ— 035 π‘š = 2
...

Now the distance = 𝑑 covered by a wheel in one revolution = the circumference of the wheel
...


=

1100
2
Title: Area related to circles
Description: Provide area related to circles questions and solutions.