Areas related to circles - Area of sector and segment of a circle (Class 10 maths)

Areas Related To Circles
Problems based on
Area of sector and segment of a circle
Chapter : Areas Related To Circles Website: www.letstute.com

Q) A horse is tied to a peg at one corner of a square shaped
grass field of side 15 m by means of a 5 m long rope. Find:
a) The area of that part of the field in which the horse can graze
b) The increase in the grazing area if the rope were 10 m long
instead of 5 m
Use [ π = 3.14]
Given: Side of a square = 15m
Length of the rope = 5 m
To Find:
a) Area of that part of the field in which
the horse can graze.
b) Increase in the grazing area if 10 m
long rope is used
Problems based on
C
15m
15m
5 m
5m
A B
D
Q 15m
15m
P
10 m
Q

Problems based on
Solution: a) Let ABCD represent the square shaped grass field
of side 15 m.
Let D be the corner to which the horse is tethered and let DP
(= 5m) be the rope by which it is tied.
Area of the part of the field over
which the horse can graze = Area
of quadrant DPQ
Thus, radius r (= 5m) and
sector angle θ (= 90⁰)
C
15m
15m
5 m
5m
A B
D
Q 15m
15m
P
10 m
Q

Problems based on
= θ x πr2
360
=
= 19.625 m2
Hence, the area of the part of the
field in which the horse can graze
is 19.625 m2




 5514.3
360
90
m2




 5514.3
4
1
= m2
4
5.78= m2
Area of quadrant DPQ
C
15m
15m
5 m
5m
A B
D
Q 15m
15m
P
10 m
Q

Problems based on
b) If r = 10 m, then the grazing area
= m2
= 78.5 m2
= m2




 101014.3
360
90
= θ x πr2
360




 101014.3
4
1
= m2
4
314
C
15m
15m
5 m
5m
A B
D
Q 15m
15m
P
10 m
Q

Problems based on
Increase in grazing area = (78.5 - 19.625) m2
=
Hence, the increase in the grazing area is 58.875 m2
58.875m2
C
15m
15m
5 m
5m
A B
D
Q 15m
15m
P
10 m
Q

Problems based on
Q) An umbrella has 8 ribs which are equally spaced. Assuming
umbrella to be a flat circle of radius 45 cm, find the area
between the two consecutive ribs of the umbrella.
Given: Number of ribs = 8
Radius = 45 cm
To find: Area between two consecutive ribs = ?

Problems based on
Solution: Angle made by the two consecutive ribs of the umbrella
at the centre
= Angle of the full circle
Number of ribs
= 360⁰
8
= 45⁰

Problems based on
Area between two consecutive ribs
= Area of a sector of the circle of radius r (=45cm) and sector
angle θ (=450)
= θ x πr2
360
= cm2
= cm2




 4545
7
22
360
45
= cm2




 4545
7
11
4
1
28
22275

Problems based on
Hence, the area between two consecutive ribs of the umbrella
is 795.53 cm2
= 795.53 cm2

Problems based on
Q) The diagram represents the area swept by the wiper of a car
with the dimensions given in the figure, calculate the shaded
area swept by the wiper.
O
300
7 cm
14 cm
B
D
C
A
Given: OD = 7 cm
DC = 14 cm
∠ 𝐂𝐎𝐁 = ∠ 𝐃𝐎𝐀 = 𝛉 = 𝟑𝟎 𝟎
To find: Area swept by the wiper??

Problems based on
Solution: Let the radii of the sectors COB and DOA be ‘R’ and ‘r’
respectively
Then, R = (7+14) cm = 21 cm and r = 7 cm -------- (1)
∠ 𝐂𝐎𝐁 = ∠ 𝐃𝐎𝐀 = 𝛉 = 𝟑𝟎 𝟎
−−−−−−−− (2)
300
7 cm
14 cm
O
B
D
C
A

Problems based on
300
O
B
D
C
A
Area of the shaded = [Area of sector COB – Area of sector DOA]
region
= θ x πR2 - θ x πr2
360 360
= θ x π (R2 - r2)
360
= 30 x 22 (212 – 72) cm2
360 7
Using (1) and (2), we get,
= 1 x 22 (441 - 49) cm2
12 7
= 22 x 392 cm2
84
7 cm
14 cm

Problems based on
300 7 cm
14 cm
O
B
D
C
A
= 11 x 392 cm2
42
Hence, the area swept by the wiper
is 102.67 cm2
102.67cm2
= 11 x 56 cm2
6
= 11 x 28 cm2
3
= 308 cm2
3
=

Problems based on
Q) In the given figure, O is the center of the concentric circles.
Radius of the inner circle is half the radius of the outer circle.
Given∠AOC = 1350 and OA = 14 cm, Calculate the area of the
shaded region. (Leave your answer in terms of π)
Given:∠AOC = 1350
OA = 14 cm
To find: Area of the shaded region = ?
S R
O1350
A D
C B
P Q
14 cm

Problems based on
Solution: Let the radii of the outer circle and inner concentric
circles be ‘R’ and ‘r’ respectively.
Then, R = 14 cm and r = 7 cm -------- (1)
∠ 𝐀𝐎𝐃 = 𝟏𝟖𝟎 𝟎 − 𝟏𝟑𝟓 𝟎
= 450 ------- (Linear pair)
𝐀𝐥𝐬𝐨 ∠ BOC =∠ AOD = 450 (=θ)
----- V. opp.∠’s ---------- (2) S R
O1350
A D
C B
P Q
14 cm

Problems based on
Area of the shaded region
= 2[Area of the sector AOD – Area of sector POQ]
= 2
𝛉
𝟑𝟔𝟎
× 𝛑𝐑 𝟐
−
𝛉
𝟑𝟔𝟎
× 𝛑𝐫 𝟐
= 2 ×
𝟒𝟓
𝟑𝟔𝟎
𝛑 𝟏𝟒 𝟐 − 𝟕 𝟐
𝐜𝐦 𝟐
By using (1) and (2), we get,
= 𝟐 ×
𝟓
𝟒𝟎
𝛑 𝟏𝟗𝟔 − 𝟒𝟗 𝐜𝐦 𝟐
=
𝟐
𝟖
𝛑 × 𝟏𝟒𝟕 cm2
= 𝟐 ×
𝛉
𝟑𝟔𝟎
× 𝛑(𝐑 𝟐
− 𝐫 𝟐
)
S R
O1350
A D
C B
P Q
14 cm

Problems based on
Result: Area =
Hence, the area of the shaded region
is
𝟏𝟒𝟕
𝟒
𝛑 cm2
𝟏𝟒𝟕
𝟒
𝛑cm2
=
𝛑
𝟒
× 𝟏𝟒𝟕 𝐜𝐦 𝟐
=
𝟏𝟒𝟕
𝟒
𝛑cm2
S R
O1350
A D
C B
P Q
14 cm

Now we know…
19
Problems based on
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Problems based on
Areas of combination of plane figures
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Areas related to circles - Area of sector and segment of a circle (Class 10 maths)

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