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Question Number 218735 by Spillover last updated on 14/Apr/25

Answered by som(math1967) last updated on 15/Apr/25

$$letradoflargecircle=R$$$$rad.ofsmallcircle=r$$$$AD=2\sqrt{2}$$$$\frac{1}{2}\times R\times (4+2\sqrt{2})=\frac{1}{2}\times 2\times 2$$$$\Rightarrow R=2-\sqrt{2}$$$$againr\sqrt{2}=(2-\sqrt{2}-r)$$$$\Rightarrow r(\sqrt{2}+1)=(2-\sqrt{2})$$$$r=\frac{2-\sqrt{2}}{\sqrt{2}+1}=\sqrt{2}{(\sqrt{2}-1)}^2=\sqrt{2}(3-2\sqrt{2})$$$$Areaofsmallcircle$$$$=\pi \times 2{(3-2\sqrt{2})}^2{cm}^2$$

Commented by Spillover last updated on 16/Apr/25

$$greatwork.thanks$$

Answered by A5T last updated on 15/Apr/25

$$\sqrt{2}+\mathrm{R}=2\Rightarrow \mathrm{R}=2-\sqrt{2}$$$$\mathrm{r}\sqrt{2}=\mathrm{R}-\mathrm{r}\Rightarrow \mathrm{r}=\frac{\mathrm{R}}{1+\sqrt{2}}=\frac{2-\sqrt{2}}{1+\sqrt{2}}$$$$\Rightarrow \mathrm{r}=3\sqrt{2}-4$$$$\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{small}\mathrm{circle}=2\pi (17-12\sqrt{2}){\mathrm{cm}}^2$$

Commented by Spillover last updated on 16/Apr/25

$$greatwork.thanks$$

Answered by Spillover last updated on 16/Apr/25

Answered by Spillover last updated on 16/Apr/25

Answered by Spillover last updated on 16/Apr/25

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