Question Number 224282 by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
$${ABCD}\:{is}\:{a}\:{square}. \\ $$$${What}\:{is}\:{the}\:{total}\:{area}\:{of}\: \\ $$$${two}\:{semicircles} \\ $$
Answered by fantastic last updated on 31/Aug/25
Commented by fantastic last updated on 31/Aug/25
$$\mathscr{A}=\frac{\mathrm{1}}{\mathrm{2}}\pi\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \right) \\ $$$${r}_{\mathrm{1}} \sqrt{\mathrm{2}}=\sqrt{\mathrm{16}−\mathrm{2}{r}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\Rightarrow{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{8} \\ $$$$\mathscr{A}=\frac{\mathrm{1}}{\mathrm{2}}\pi×\mathrm{8}=\mathrm{4}\pi\checkmark \\ $$
Answered by mr W last updated on 31/Aug/25
Commented by mr W last updated on 31/Aug/25
$${A}_{{semicircles}} =\frac{\pi\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\pi\left(\sqrt{\mathrm{2}}{c}\right)^{\mathrm{2}} }{\mathrm{8}}=\frac{\pi{c}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{4}\pi \\ $$$${c}=\mathrm{4} \\ $$