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Question

What is the area of the minor segment of a circle of radius 14 cm , when the angle of the corresponding sector is 60∘

A
(3083493) cm2
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B
(3203503) cm2
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C
49 cm2
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D
3083 cm2
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Solution

The correct option is A (3083−49√3) cm2Given that , radius of circle (r ) = 14 cm Angle of the corresponding sector. i.e central angle (θ)=60∘ ∵ In Δ AOB OA=OB=Radius of circle ∴Δ AOB is isosceles. ⇒∠OAB=∠OBA=θ Now, in Δ OAB ∠AOB+∠OAB+∠OBA=180∘ [Since , sum of interior angles of any triangle is 180∘] ⇒60∘+θ+θ=180∘ [given,∠AOB=60∘] ⇒2θ=120∘ ⇒θ=60∘ i.e ∠AOB=∠OBA=60∘=∠AOB Since, all angles of ΔAOB are equal to 60∘. So, ΔAOB is an equilateral triangle. Also, area of ΔAOB =√34×(14)2 [∵Area of an equilateral triangle=√34(side)2] =√34×196 =49√3 cm2 Area of sector OBAO =πr2360∘×θ =227×14×14360×60∘ =22×2×146 =22×143 =3083 cm2 ∴ The area of the minor segment = Area of sector OBAO - Area of the equilateral triangle =(3083−49√3) cm2

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