Question Number 90440 by ajfour last updated on 23/Apr/20
Commented by ajfour last updated on 23/Apr/20
$${If}\:{BC}\:{is}\:\bot\:{to}\:{CD}\:,\:{find}\:{R}/{r}. \\ $$
Commented by john santu last updated on 23/Apr/20
$$\left({R}+{r}\right)^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} +\mathrm{2}{Rr}+{r}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{Rr}+{r}^{\mathrm{2}} \\ $$$$\mathrm{4}{Rr}\:=\:{R}^{\mathrm{2}} \:\Rightarrow\:\frac{{R}}{{r}}\:=\:\mathrm{4}\::\:\mathrm{1} \\ $$
Commented by ajfour last updated on 23/Apr/20
$$\mathcal{T}{hanks}\:{sir},\:{keen}\:{observation}! \\ $$