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Question Number 67615 by TawaTawa last updated on 29/Aug/19

Commented by TawaTawa last updated on 29/Aug/19

$$\mathrm{i}\mathrm{did}.\mathrm{A}\mathrm{of}\mathrm{Big}\mathrm{semi}\mathrm{circle}=\frac{\pi {\mathrm{r}}^2}{2}=\frac{\pi .2^2}{2}=2\pi$$$$\mathrm{A}\mathrm{of}\mathrm{second}\mathrm{big}\mathrm{semicircle}=\frac{\pi {\mathrm{r}}^2}{2}=\frac{1}{2}\pi$$$$\mathrm{But}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{what}\mathrm{to}\mathrm{do}\mathrm{again}$$

Answered by mr W last updated on 29/Aug/19

Commented by mr W last updated on 29/Aug/19

$$radiusofbigsemicircle=R=a$$$$radiusofsmallcircles=r=\frac{a}{2}$$$$\alpha =\frac{\pi }{3}$$$$\beta =\frac{2\pi }{3}$$$$AD=2R=2a$$$$BC=r=\frac{a}{2}$$$$A_{ABCD}=\frac{\frac{a}{2}+2a}{2}\times \frac{a}{2}\times \frac{\sqrt{3}}{2}=\frac{5\sqrt{3}{a}^2}{16}$$$$A_{\stackrel{⌢}{AB}}=A_{\stackrel{⌢}{CD}}=\frac{r^2}{2}(\beta -\sin \beta )=\frac{a^2}{8}(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})$$$$A_{\stackrel{⌢}{BC}}=\frac{r^2}{2}(\alpha -\sin \alpha )=\frac{a^2}{8}(\frac{\pi }{3}-\frac{\sqrt{3}}{2})$$$$A_{shade}=\frac{\pi R^2}{2}-A_{ABCD}-2\times A_{\stackrel{⌢}{AB}}-A_{\stackrel{⌢}{BC}}$$$$=\frac{\pi a^2}{2}-\frac{5\sqrt{3}{a}^2}{16}-2\times \frac{a^2}{8}(\frac{2\pi }{3}-\frac{\sqrt{3}}{2})-\frac{a^2}{8}(\frac{\pi }{3}-\frac{\sqrt{3}}{2})$$$$=\frac{a^2}{8}(\frac{7\pi }{3}-\sqrt{3})$$$$=\frac{a^2(7\pi -3\sqrt{3})}{24}$$

Commented by TawaTawa last updated on 29/Aug/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by mr W last updated on 29/Aug/19

Commented by mr W last updated on 29/Aug/19

$$A_1={\int }_0^{\frac{\pi }{3}}\frac{1}{2}(a^2-a^2{\cos }^2\theta )d\theta$$$$=\frac{a^2}{2}{\int }_0^{\frac{\pi }{3}}{\sin }^2\theta d\theta$$$$=\frac{a^2}{4}{\int }_0^{\frac{\pi }{3}}(1-\cos 2\theta )d\theta$$$$=\frac{a^2}{4}{[\theta -\frac{\sin 2\theta }{2}]}_0^{\frac{\pi }{3}}$$$$=\frac{a^2}{4}(\frac{\pi }{3}-\frac{\sqrt{3}}{4})$$$$A_2=\frac{1}{2}\times \frac{\pi }{6}\times (a^2-\frac{a^2}{4})=\frac{a^2\pi }{16}$$$$$$$$A_{shaded}=2(A_1+A_2)$$$$=\frac{a^2}{2}(\frac{\pi }{3}-\frac{\sqrt{3}}{4})+\frac{\pi a^2}{8}$$$$=\frac{a^2(7\pi -3\sqrt{3})}{24}$$

Commented by TawaTawa last updated on 29/Aug/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

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