Question-177596

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

(√2)r_1 =R=4 ⇒r_1 =2(√2) (r_1 +r_2 )^2 −(r_1 −r_2 )^2 =(R−r_2 )^2 −r_2 ^2 4r_1 r_2 =R(R−2r_2 ) (√2)r_2 =2−r_2 ⇒r_2 =(2/( (√2)+1))=2((√2)−1) A_(green) =((πR^2 )/4)−((πr_1 ^2 )/4)−πr_2 ^2 =((π×4^2 )/4)−((π(2(√2))^2 )/4)−π(2((√2)−1))^2 =2(4(√2)−5)π ≈ 4.127

$$\sqrt{\mathrm{2}}{r}_{\mathrm{1}} ={R}=\mathrm{4} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{4}{r}_{\mathrm{1}} {r}_{\mathrm{2}} ={R}\left({R}−\mathrm{2}{r}_{\mathrm{2}} \right) \\ $$$$\sqrt{\mathrm{2}}{r}_{\mathrm{2}} =\mathrm{2}−{r}_{\mathrm{2}} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${A}_{{green}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi{r}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}−\pi{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi×\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}−\pi\left(\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{5}\right)\pi\:\approx\:\mathrm{4}.\mathrm{127} \\ $$

Commented by Ar Brandon last updated on 07/Oct/22