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Question Number 175717 by ajfour last updated on 05/Sep/22

Answered by mr W last updated on 05/Sep/22

$${(R-r)}^2=r^2+{(\frac{R}{2}+r)}^2$$$$r^2+3Rr-\frac{3R^2}{4}=0$$$$\Rightarrow \frac{r}{R}=\sqrt{3}-\frac{3}{2}\approx 0.232$$

Commented by mr W last updated on 05/Sep/22

Commented by ajfour last updated on 05/Sep/22

$$Excellentnoticesir!$$

Commented by ajfour last updated on 05/Sep/22

$$\frac{R}{r}=\frac{2}{2\sqrt{3}-3}=\frac{2(2\sqrt{3}+3)}{3}=\frac{4}{\sqrt{3}}+2$$$$=\frac{4\times 1.732}{3}+2=\frac{6.928}{3}+2\approx 4.31$$

Commented by Tawa11 last updated on 05/Sep/22

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

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