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# Question 10 Find the area of the minor segment of a circle of radius 14 cm , when the angle of the corresponding sector is 60∘

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Solution

## Given that , radius of circle (r ) = 14 cm Angle of the corresponding sector. i.e central angle (θ)=60∘ Since, in ΔAOB, OA=OB=Radius of circle. i.e., ΔAOB is isosceles. ⇒∠OAB=∠OBA=θ Now, in ΔOAB∠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∘ i.e ΔAOB is an equilateral triangle. Also, OA=√34(side)2 =√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=3083cm2 ∴ The area of the minor segment = Area of sector OBAO - Area of the equilateral triangle (3083−49√3)cm2

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