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Question Number 19415 by Tinkutara last updated on 10/Aug/17

$$PS\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{segment}\mathrm{of}\mathrm{length}4\mathrm{and}O$$$$\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}PS.\mathrm{A}\mathrm{semicircular}$$$$\mathrm{arc}\mathrm{is}\mathrm{drawn}\mathrm{with}PS\mathrm{as}\mathrm{diameter}.\mathrm{Let}$$$$X\mathrm{be}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{this}\mathrm{arc}.Q\mathrm{and}R$$$$\mathrm{are}\mathrm{points}\mathrm{on}\mathrm{the}\mathrm{arc}PXS\mathrm{such}\mathrm{that}QR$$$$\mathrm{is}\mathrm{parallel}\mathrm{to}PS\mathrm{and}\mathrm{the}\mathrm{semicircular}$$$$\mathrm{arc}\mathrm{drawn}\mathrm{with}QR\mathrm{as}\mathrm{diameter}\mathrm{is}$$$$\mathrm{tangent}\mathrm{to}PS.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}$$$$\mathrm{region}QXROQ\mathrm{bounded}\mathrm{by}\mathrm{the}\mathrm{two}$$$$\mathrm{semicircular}\mathrm{arcs}?$$

Answered by ajfour last updated on 11/Aug/17

Commented by ajfour last updated on 11/Aug/17

$${\mathrm{A}}_1=\mathrm{Area}\mathrm{of}△\mathrm{QOR}=\frac{1}{2}\times 2\sqrt{2}\times \sqrt{2}=2$$$${\mathrm{A}}_2=\mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{OQXR}=\frac{1}{4}\times \pi \times 2^2=\pi$$$${\mathrm{A}}_3=\mathrm{Area}\mathrm{of}\mathrm{semicircle}\mathrm{OQR}=\frac{1}{2}\times \pi \times {\left(\sqrt{2}\right)}^2=\pi$$$$\mathrm{Required}\mathrm{area}={\mathrm{A}}_2-{\mathrm{A}}_1+{\mathrm{A}}_3$$$$=\mathit{\pi }-2+\mathit{\pi }=2\mathit{\pi }-2.$$

Commented by Tinkutara last updated on 11/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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