Question Number 111135 by ajfour last updated on 02/Sep/20
Answered by mr W last updated on 02/Sep/20
$$\mathrm{2}{r}+{R}−\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{2}{R} \\ $$$$\mathrm{2}{r}−{R}=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\mathrm{4}{r}^{\mathrm{2}} −\mathrm{4}{Rr}+{R}^{\mathrm{2}} ={R}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\mathrm{5}{r}=\mathrm{4}{R} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 02/Sep/20
$${R}^{\mathrm{2}} ={r}^{\mathrm{2}} +\left(\mathrm{2}{r}−{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4}{rR}=\mathrm{5}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\frac{{R}}{{r}}=\frac{\mathrm{5}}{\mathrm{4}}\:. \\ $$