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Question Number 156940 by amin96 last updated on 17/Oct/21

Answered by mr W last updated on 17/Oct/21

Commented by mr W last updated on 17/Oct/21

$$r=radiusofsemicircle$$$$A_{green}=\frac{\pi {\left(2r\right)}^2}{4}-\frac{{\left(2r\right)}^2}{2}-[\frac{\pi r^2}{4}-\frac{r^2}{2}]$$$$A_{green}=\frac{3(\pi -2)r^2}{4}$$$$$$$$\theta =60°$$$$OA=\frac{r}{\tan \frac{\theta }{2}}$$$$OB=2r$$$$AB=b=10$$$${OB}^2={\left(2r\right)}^2={\left(\frac{r}{\tan \frac{\theta }{2}}\right)}^2+b^2-2b\times \frac{r}{\tan \frac{\theta }{2}}\cos \theta$$$$(4-\frac{1}{{\tan }^{2}30})r^2+20\times \frac{r}{\tan 30}\cos 60-100=0$$$$r^2+10\sqrt{3}r-100=0$$$$r=5(\sqrt{7}-\sqrt{3})$$$$A_{green}=\frac{75(\pi -2)(5-\sqrt{21})}{2}\approx 17.8698$$

Commented by Tawa11 last updated on 17/Oct/21

$$\mathrm{Great}\mathrm{sir}$$

Commented by amin96 last updated on 17/Oct/21

$$coolsir.bravo$$

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