Question Number 221592 by fantastic last updated on 08/Jun/25
Answered by mr W last updated on 08/Jun/25
Commented by mr W last updated on 08/Jun/25
$$\left(\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{r}\right)\left({R}−{r}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){r}=\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){R} \\ $$$$\frac{{r}}{{R}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${fraction}=\frac{\mathrm{2}×\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}}{\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}}\:=\mathrm{2}\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by fantastic last updated on 08/Jun/25
$${thanks}\:{sir} \\ $$$$ \\ $$
Answered by fantastic last updated on 08/Jun/25
