Question Number 196155 by sonukgindia last updated on 19/Aug/23
Commented by mr W last updated on 19/Aug/23
Answered by mr W last updated on 19/Aug/23
$${R}+\sqrt{\mathrm{2}}{R}=\frac{{r}}{\:\sqrt{\mathrm{2}}}+\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){R}−{r}=\sqrt{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){Rr}+{r}^{\mathrm{2}} } \\ $$$${let}\:\lambda=\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){R}}{{r}} \\ $$$$\lambda−\mathrm{1}=\sqrt{\mathrm{2}\lambda+\mathrm{1}} \\ $$$$\Rightarrow\lambda=\mathrm{4}=\frac{\left(\sqrt{\mathrm{2}}+\mathrm{2}\right){R}}{{r}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{4}}{\:\mathrm{2}+\sqrt{\mathrm{2}}}=\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\approx\mathrm{1}.\mathrm{1716} \\ $$
Commented by Bharathi last updated on 19/Aug/23
Commented by mr W last updated on 19/Aug/23
$${many}\:{errors}! \\ $$$$\theta=\frac{\pi}{\mathrm{8}}\:\nRightarrow\:\mathrm{sin}\:\theta=\frac{\pi}{\mathrm{8}}! \\ $$$$\frac{{r}}{\:\sqrt{\mathrm{2}}}\neq{r}\:! \\ $$$${r}−{circle}\:{and}\:{R}−{circle}\:{don}'{t}\:{touch} \\ $$$${in}\:{the}\:{radius}\:{of}\:{quarter}\:{circle}! \\ $$$$… \\ $$