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Find the area of a sector of a circle with radius 6 cm if angle of the sector is .
Let OACB be a sector of the circle making 60° angle at centre O of the circle.
Area of sector of angle θ =
Area of sector OACB = 
Therefore, the area of the sector of the circle making 60° at the centre of the circle is
Find the area of a quadrant of a circle whose circumference is 22 cm.
Let the radius of the circle be r.
Circumference = 22 cm
2πr = 22
Quadrant of circle will subtend 90° angle at the centre of the circle.
Area of such quadrant of the circle
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate =
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ =
Area of sector of 30° 
Therefore, the area swept by the minute hand in 5 minutes is
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use )
Let AB be the chord of the circle subtending 90° angle at centre O of the circle.
Area of major sector OADB =
Area of minor sector OACB =
Area of ΔOAB =
= 50 cm2
Area of minor segment ACB = Area of minor sector OACB −
Area of ΔOAB = 78.5 − 50 = 28.5 cm2
In a circle of radius 21 cm, an arc subtends an angle of at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord
Radius (r) of circle = 21 cm
Angle subtended by the given arc = 60°
Length of an arc of a sector of angle θ =
Length of arc ACB =
= 22 cm
Area of sector OACB =
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠OAB + ∠AOB + ∠OBA = 180°
2∠OAB + 60° = 180°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB =
Area of segment ACB = Area of sector OACB − Area of ΔOAB
A chord of a circle of radius 15 cm subtends an angle of at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use
and
= 1.73)
Radius (r) of circle = 15 cm
Area of sector OPRQ =
In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2 ∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
Area of ΔOPQ =
Area of segment PRQ = Area of sector OPRQ − Area of ΔOPQ
= 117.75 − 97.3125
= 20.4375 cm2
Area of major segment PSQ = Area of circle − Area of segment PRQ
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use and
= 1.73)
Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,
Area of ΔOST =
Area of sector OSUT =
Area of segment SUT = Area of sector OSUT − Area of ΔOST
= 150.72 − 62.28
= 88.44 cm2
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 (see the figure). Find:
(i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use )
From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius.
Area that can be grazed by horse = Area of sector OACB
Area that can be grazed by the horse when length of rope is 10 m long
Increase in grazing area = (78.5 − 19.625) m2
= 58.875 m2
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find : (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
Total length of wire required will be the length of 5 diameters and the circumference of the brooch.
Radius of circle =
Circumference of brooch = 2πr
= 110 mm
Length of wire required = 110 + 5 × 35
= 110 + 175 = 285 mm
It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.
Therefore, area of each sector =
An umbrella has 8 ribs which are equally spaced (see the figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending 
at the centre of the assumed flat circle.
Area between two consecutive ribs of circle =
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of . Find the total area cleaned at each sweep of the blades.
It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.
Area of such sector =
Area swept by 2 blades =
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use
)
It can be observed from the figure that the lighthouse spreads light across a
sector of 80° in a circle of 16.5 km radius.
Area of sector OACB =
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per . (Use
)
It can be observed that these designs are segments of the circle.
Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute 
at the centre of the circle.
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠AOB = 60°
∠OAB + ∠OBA + ∠AOB = 180°
2∠OAB = 180° − 60° = 120°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB =
= 333.2 cm2
Area of sector OAPB =
Area of segment APB = Area of sector OAPB − Area of ΔOAB
Cost of making 1 cm2 designs = Rs 0.35
Cost of making 464.76 cm2 designs = 
= Rs 162.68
Therefore, the cost of making such designs is Rs 162.68.
Tick the correct answer in the following : Area of a sector of angle (in degrees) of a circle with radius R is
(A) (B)
(C)
(D)
We know that area of sector of angle θ =
Area of sector of angle P =
Hence, (D) is the correct answer.
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