Question Number 171677 by cherokeesay last updated on 19/Jun/22
Answered by mr W last updated on 19/Jun/22
$${R}=\mathrm{2}{r} \\ $$$${rectangle}\:{a}×{b}\:{with}\:{b}=\mathrm{2}{r} \\ $$$${a}=\sqrt{\left({r}+{R}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}}{r} \\ $$$${shaded}\:{area}={ab}−\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}={ab}−\frac{\mathrm{3}\pi{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\%=\mathrm{1}−\frac{\mathrm{3}\pi{r}^{\mathrm{2}} }{\mathrm{2}{ab}}=\mathrm{1}−\frac{\mathrm{3}\pi{r}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}{r}×\mathrm{2}\sqrt{\mathrm{2}}{r}} \\ $$$$\:\:\:\:=\mathrm{1}−\frac{\mathrm{3}\pi}{\mathrm{8}\sqrt{\mathrm{2}}}=\mathrm{17\%} \\ $$
Commented by Tawa11 last updated on 19/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cherokeesay last updated on 19/Jun/22
$${thank}\:{you}\:{sir}\:! \\ $$