Question Number 150793 by cherokeesay last updated on 15/Aug/21
$$\left({R}−{r}\right)^{\mathrm{2}} \:+\:{R}^{\mathrm{2}} \:=\:\left({R}\:+\:{r}\right)^{\mathrm{2}} \:\Leftrightarrow \\ $$$${R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} −\mathrm{2}{rR}\:+\:{R}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} \:+\mathrm{2}{rR}\:\Rightarrow \\ $$$${R}^{\mathrm{2}} \:=\:\mathrm{4}{rR}\:\Rightarrow\:{r}\:=\:\frac{{R}}{\mathrm{4}}\:=\:\mathrm{1}{cm} \\ $$$$\mathscr{A}_{{S}} \:=\:\pi{r}^{\mathrm{2}} \:=\:\pi×\mathrm{1}{cm}^{\mathrm{2}} \:=\:\pi{cm}^{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Aug/21
$${Nice}! \\ $$$$\:{Pl}\:{write}\:{this}\:{as}\:{an}\:{answer} \\ $$$${of}\:{your}\:{question}.{Don}'{t}\:{start}\:{new} \\ $$$${thread}\:{for}\:{an}\:{answer}.\:{Thanks}. \\ $$
Commented by cherokeesay last updated on 15/Aug/21
$$ \\ $$the answer to this problem has already been processed by a member (see below) what I have established is my own resolution of this same exercise but accompanied by a diagram! that's all ! good to you.