Question Number 218886 by Spillover last updated on 17/Apr/25
Answered by mr W last updated on 17/Apr/25
Commented by Spillover last updated on 17/Apr/25
$${perfect} \\ $$
Commented by mr W last updated on 17/Apr/25
$$\mathrm{2}{R}−\mathrm{2}{r}=\mathrm{18}\:\Rightarrow{R}−{r}=\mathrm{9} \\ $$$${R}−\sqrt{{r}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }=\mathrm{10} \\ $$$${R}−\mathrm{10}=\sqrt{\mathrm{2}{Rr}−{R}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\mathrm{50}\:\Rightarrow{r}=\mathrm{41} \\ $$$${A}_{{shade}} =\pi\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)=\mathrm{819}\pi\:\checkmark \\ $$
Answered by Spillover last updated on 17/Apr/25
Answered by Spillover last updated on 17/Apr/25
Answered by Spillover last updated on 17/Apr/25
