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Question Number 166281 by Tawa11 last updated on 17/Feb/22

Answered by mr W last updated on 18/Feb/22

Commented by mr W last updated on 18/Feb/22

$$areaofsegment{A}_1$$$$A_1=\frac{r^2}{2}(\frac{\pi }{6}-\sin \frac{\pi }{6})$$$$a=2r\sin \frac{\pi }{12}=\frac{(\sqrt{6}-\sqrt{2})r}{2}$$$$shadedarea=a^2+4A_1$$$$=\frac{{(\sqrt{6}-\sqrt{2})}^2{r}^2}{4}+4\times \frac{r^2}{2}(\frac{\pi }{6}-\frac{1}{2})$$$$=(\frac{\pi }{3}+1-\sqrt{3})r^2$$$$=(\frac{\pi }{3}+1-\sqrt{3})\times {20}^2$$$$\approx 126$$

Commented by Tawa11 last updated on 18/Feb/22

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

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