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Question

A chord of a circle subtends an angle of θ at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that 8 sin θ2 cos θ2+π=πθ45


Solution

The area of the minor segment is given by πr2×θ360r2sin θ2 cos θ2=πr28

πθ360sin θ2 cos θ2=π8

πθ458sin θ2 cos θ2=π

8 sin θ2 cos θ2+π=πθ45

Hence, proved.


Mathematics
RD Sharma
Standard X

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