Question Number 200457 by cortano12 last updated on 19/Nov/23
Commented by cortano12 last updated on 19/Nov/23
$$\:\:\: \\ $$
Answered by mr W last updated on 19/Nov/23
$${R}={bigger}\:{radius} \\ $$$${r}={smalle}\:{radius} \\ $$$${we}\:{have}\:{R}=\sqrt{\mathrm{2}}{r} \\ $$$$\frac{{x}}{\mathrm{2}{r}}=\frac{\mathrm{1}+{x}}{\mathrm{2}{R}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\sqrt{\mathrm{2}}+\mathrm{1}\:\checkmark \\ $$
Commented by cortano12 last updated on 19/Nov/23
$$\mathrm{very}\:\mathrm{simple}\: \\ $$