Question

ABCP is a quadrant of a circle of radius 14 cm.With AC as diameter semicircle is drawn find the area of shaded portion​

Answer 1

$AC = \sqrt{AB+BC} = \sqrt{14 + 14} = 14 \sqrt{2} \: cm \\ radius \: of \: semicircle \: ACQ = \frac{14 \sqrt{2} }{2} = 7 \sqrt{2} \: cm\\ \\ Area \: of \: shaded \: region = \\ ar(ABC)+ar(ACQ)-ar(APCB) \\ = \frac{1}{2} bh + \frac{\pi {r}^{2} }{2} - \frac{90}{360} \times \pi {r}^{2} \\ = \frac{1}{2} \times 14 \times 14 + \frac{22 \times 7 \sqrt{2} \times 7 \sqrt{2} }{7 \times 2} - \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \\ = 98 + 154 - 154 \\ = 98 \: {cm}^{2} \\ \\ \mathsf{ So, \: the \: area \: of \:t he \: shaded \: region \: = \: 98 \: {cm}^{2} }$

Answer 2

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