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Question Number 161733 by Tawa11 last updated on 21/Dec/21

Answered by FongXD last updated on 21/Dec/21

$$\mathrm{let}\mathrm{O}\mathrm{be}\mathrm{the}\mathrm{point}\mathrm{where}\mathrm{the}2\mathrm{intersect}$$$$\mathrm{let}\mathrm{T}\in \left(\mathrm{QR}\right),\mathrm{where}\left(\mathrm{PT}\right)\mathrm{\perp }\left(\mathrm{QR}\right)$$$$\bullet \mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}\mathrm{PQT}$$$$\mathrm{we}\mathrm{have}:{\mathrm{PQ}}^2={\mathrm{PT}}^2+{\mathrm{TQ}}^2$$$$\Rightarrow \mathrm{PT}=\sqrt{{(3+1)}^2-{(3-1)}^2}=2\sqrt{3}$$$$\bullet \cos \left(\mathrm{\angle }\mathrm{OQR}\right)=\frac{\mathrm{TQ}}{\mathrm{PQ}}=\frac{3-1}{3+1}=\frac{1}{2},\Rightarrow \mathrm{\angle }\mathrm{OQR}=\frac{\pi }{3}$$$$\bullet \mathrm{\angle }\mathrm{QPT}=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6},\Rightarrow \mathrm{\angle }\mathrm{OPS}=\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}$$$$\bullet \mathrm{Area}\mathrm{of}\mathrm{PQRS}=\frac{(\mathrm{PS}+\mathrm{QR})\mathrm{PT}}{2}=\frac{(1+3)\times 2\sqrt{3}}{2}=4\sqrt{3}\left(1\right)$$$$\bullet \mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{OQR}=\frac{\mathrm{\angle }\mathrm{OQR}}{2}\times {\mathrm{R}}^2=\frac{\pi }{6}\times 3^2=\frac{3\pi }{2}\left(2\right)$$$$\bullet \mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{OPS}=\frac{\mathrm{\angle }\mathrm{OPS}}{2}\times {\mathrm{r}}^2=\frac{\pi }{3}\left(3\right)$$$$\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}=\left(1\right)-\left(2\right)-\left(3\right)$$$$=4\sqrt{3}-\frac{3\pi }{2}-\frac{\pi }{3}=4\sqrt{3}-\frac{11\pi }{6}$$

Commented by Tawa11 last updated on 21/Dec/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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