Question Number 197582 by cherokeesay last updated on 22/Sep/23
Answered by som(math1967) last updated on 23/Sep/23
Commented by som(math1967) last updated on 23/Sep/23
$${let}\:{rad}\:{of}\:{big}\:{semicircle}={R} \\ $$$$\:{BO}={R}−\mathrm{1} \\ $$$${r}={R}−\mathrm{1}+{R}−\mathrm{1}=\mathrm{2}{R}−\mathrm{2} \\ $$$$\Rightarrow{R}=\frac{{r}+\mathrm{2}}{\mathrm{2}} \\ $$$$\:{OA}^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({R}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} ={r}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{R}+\mathrm{1} \\ $$$$\Rightarrow{r}^{\mathrm{2}} −{r}−\mathrm{2}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{cm}\:\:\left({golden}\:{ratio}\right) \\ $$$$\:{R}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\:+\mathrm{1}=\frac{\sqrt{\mathrm{5}}+\mathrm{5}}{\mathrm{4}}{cm} \\ $$$${Green}\:{area} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{1}^{\mathrm{2}} ×\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}×\pi×\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{\pi}{\mathrm{4}}\left\{\frac{\left(\mathrm{5}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{8}}−\mathrm{2}−\frac{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\right\}{cm}^{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{4}}×\frac{\mathrm{2}+\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{8}}{cm}^{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{16}}×\left(\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{1}\right){cm}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 23/Sep/23
$${so}\:{nice}\:! \\ $$$${thank}\:{you}\:! \\ $$