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Question

# Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm. Subtending an angle of 90o till the centre.

A
32.14 sq. cm
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B
35.42 sq. cm
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C
38.96 sq. cm
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D
42.43 sq. cm
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Solution

## The correct option is D 32.14 sq. cmLet r be the radius of circle and AB be the chord, which makes 90o angle at centre.AB=5cmIn right △OAB, using pythagoras theoremOA2+OB2=AB2⇒r2+r2=52⇒2r2=25⇒r=5√2Area of circle =πr2=227×5√2×5√2=39.28cm2Area of minor segment =Areaofsector−Areaof△OAB=90o360o×πr2−12×5√2×5√2=14×39.28−254=14.284=3.57cm2Area of major segment =39.28−3.57=35.71cm2Required difference =35.71−3.57=32.14cm2

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