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Question

In the given figure, ∆ABC is a right-angled triangle with ∠B = 90°, AB = 48 cm and BC = 14 cm. With AC as diameter a semicircle is drawn and with BC as radius, A quadrant of a circle is drawn. Find the area of the shaded region.

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Solution

Consider the triangle ABC.

AC=AB2+BC2 =482+142=2304+196=2500=50 cm

Now,

Area of the shaded part =AreaABC+Areasemicircle having diameter AC-Areasector BCD=12×14×48+12×π×25×25-90360×π×14×14=336+π225×25-14×142=336+2214625-98=1164.14 cm2

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