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Question Number 218318 by Tawa11 last updated on 06/Apr/25

Commented by Tawa11 last updated on 06/Apr/25

$$\mathrm{Area}\mathrm{of}\mathrm{circle}.$$

Commented by AlagaIbile last updated on 06/Apr/25

$$A=200\pi (26-15\sqrt{3})$$

Commented by Tawa11 last updated on 06/Apr/25

$$\mathrm{please}\mathrm{workings}\mathrm{sir}$$

Answered by mr W last updated on 06/Apr/25

Commented by mr W last updated on 06/Apr/25

$$R=10/2=5$$$$\tan 30°=\frac{2\tan 15°}{1-{\tan }^{2}15°}=\frac{1}{\sqrt{3}}$$$${\tan }^{2}15°+2\sqrt{3}\tan 15°-1=0$$$$\Rightarrow \tan 15°=-\sqrt{3}+2=\frac{1}{2+\sqrt{3}}$$$$\frac{r}{\tan 15°}=R+\sqrt{{(R-r)}^2-r^2}$$$$(2+\sqrt{3})r-R=\sqrt{R^2-2Rr}$$$${(2+\sqrt{3})}^2{r}^2-2(2+\sqrt{3})Rr+R^2=R^2-2Rr$$$${(2+\sqrt{3})}^{2}r=2(1+\sqrt{3})R$$$$\Rightarrow r=\frac{2(1+\sqrt{3})R}{{(2+\sqrt{3})}^2}=2(3\sqrt{3}-5)R=10(3\sqrt{3}-5)$$$$\pi r^2=200\pi (26-15\sqrt{3})$$

Commented by Tawa11 last updated on 07/Apr/25

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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