Question Number 177596 by cherokeesay last updated on 07/Oct/22
Answered by mr W last updated on 07/Oct/22
Commented by mr W last updated on 07/Oct/22
$$\sqrt{\mathrm{2}}{r}_{\mathrm{1}} ={R}=\mathrm{4} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} ={R}\left({R}−\mathrm{2}{r}_{\mathrm{2}} \right) \\ $$$$\sqrt{\mathrm{2}}{r}_{\mathrm{2}} =\mathrm{2}−{r}_{\mathrm{2}} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${A}_{{green}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi{r}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}−\pi{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi×\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}−\pi\left(\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{5}\right)\pi\:\approx\:\mathrm{4}.\mathrm{127} \\ $$
Commented by Ar Brandon last updated on 07/Oct/22
$$\mathrm{Sir},\:\mathrm{why}\:\sqrt{\mathrm{2}}{r}_{\mathrm{1}} ={R}\:? \\ $$
Commented by mr W last updated on 07/Oct/22
Commented by Tawa11 last updated on 07/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Ar Brandon last updated on 07/Oct/22
Thanks Sir