In the figure, PQSO is a trapezium in which PQ is parallel to OS,∠POS=120∘ and ∠PQS=90∘. Points P,Q and R lie on a circle with center O and radius 10. Then the area of the shaded region, is
A
100π3−25√32
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B
100π3−50√32
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C
100π3−75√32
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D
100π3−100√32
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Solution
The correct option is C100π3−75√32 Area(□PQSO)=Area(ΔPQO)+Area(ΔQOS)=√34×r2+12×√3r2×r2=√34×(10)2+12×√3×102×102=75√32 Required Area=12×r2×2π3−Area(□PQSO)=100π3−75√32