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Question Number 218494 by Spillover last updated on 10/Apr/25

Answered by vnm last updated on 10/Apr/25

$$\frac{R}{\sqrt{R^2+{\left(2R\right)}^2}}=\frac{r}{\sqrt{R^2+{\left(2R\right)}^2}-(R+r)}$$$$\frac{1}{\sqrt{5}}=\frac{r}{R\sqrt{5}-R-r}=\frac{r/R}{\sqrt{5}-1-r/R}$$$$\sqrt{5}=\frac{\sqrt{5}-1}{r/R}-1$$$$\frac{r}{R}=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$$$$A=\frac{1}{2}{\left(2R\right)}^2\tan (\frac{\pi }{2}-{2\tan }^{-1}\frac{1}{2})=$$$$\frac{2R^2}{\tan \left({2\tan }^{-1}\frac{1}{2}\right)}=\frac{3R^2}{2}$$$$$$

Commented by Spillover last updated on 11/Apr/25

$$correct$$

Answered by mr W last updated on 10/Apr/25

Commented by mr W last updated on 10/Apr/25

$$\frac{R-r}{\sqrt{{(R+r)}^2-{(R-r)}^2}}=\frac{R}{2R}$$$$\frac{{(R-r)}^2}{4Rr}=\frac{1}{4}$$$$r^2-3Rr+R^2=0$$$$\Rightarrow \frac{r}{R}=\frac{3-\sqrt{3^2-4}}{2}=\frac{3-\sqrt{5}}{2}✓$$

Answered by Spillover last updated on 11/Apr/25

Answered by Spillover last updated on 11/Apr/25

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