Question Number 52533 by Tawa1 last updated on 09/Jan/19
Commented by ajfour last updated on 09/Jan/19
Commented by ajfour last updated on 10/Jan/19
![ρ = r+2R = a(√2) r+R = R(√2) ⇒ r = R((√2)−1) ρ = R((√2)−1+2)= a(√2) shaded area A= 4a^2 −πr^2 −2πR^2 −4R^2 = 2R^2 ((√2)+1)^2 −4R^2 −πR^2 [2+((√2)−1)^2 ] =R^2 {6+4(√2)−4−π(5−2(√2))} = 4{2+4(√2)−π(5−2(√2))} = 8+16(√2)−20π+8(√2)π = 8+(16+8π)(√2)−20π a+b+c+d = 8+16+8−20 = 12 .](Q52551.png)
$$\rho\:=\:{r}+\mathrm{2}{R}\:=\:{a}\sqrt{\mathrm{2}} \\ $$$${r}+{R}\:=\:{R}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:\:{r}\:=\:{R}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\rho\:=\:{R}\left(\sqrt{\mathrm{2}}−\mathrm{1}+\mathrm{2}\right)=\:{a}\sqrt{\mathrm{2}} \\ $$$${shaded}\:{area}\:{A}=\:\mathrm{4}{a}^{\mathrm{2}} −\pi{r}^{\mathrm{2}} −\mathrm{2}\pi{R}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} \\ $$$$\:\:\:\:=\:\mathrm{2}{R}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} −\pi{R}^{\mathrm{2}} \left[\mathrm{2}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$$\:\:={R}^{\mathrm{2}} \left\{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{4}−\pi\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{2}}\right)\right\} \\ $$$$\:\:=\:\mathrm{4}\left\{\mathrm{2}+\mathrm{4}\sqrt{\mathrm{2}}−\pi\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{2}}\right)\right\} \\ $$$$\:\:=\:\mathrm{8}+\mathrm{16}\sqrt{\mathrm{2}}−\mathrm{20}\pi+\mathrm{8}\sqrt{\mathrm{2}}\pi \\ $$$$\:\:=\:\mathrm{8}+\left(\mathrm{16}+\mathrm{8}\pi\right)\sqrt{\mathrm{2}}−\mathrm{20}\pi \\ $$$${a}+{b}+{c}+{d}\:=\:\mathrm{8}+\mathrm{16}+\mathrm{8}−\mathrm{20}\:=\:\mathrm{12}\:. \\ $$
Commented by Tawa1 last updated on 09/Jan/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 09/Jan/19
$$\mathrm{Sir},\:\mathrm{they}\:\mathrm{said}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\:\mathrm{12}.\: \\ $$
Commented by Tawa1 last updated on 10/Jan/19
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{checking}\:\mathrm{sir} \\ $$