Question Number 166281 by Tawa11 last updated on 17/Feb/22
Answered by mr W last updated on 18/Feb/22
Commented by mr W last updated on 18/Feb/22
$${area}\:{of}\:{segment}\:{A}_{\mathrm{1}} \\ $$$${A}_{\mathrm{1}} =\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$${a}=\mathrm{2}{r}\:\mathrm{sin}\:\frac{\pi}{\mathrm{12}}=\frac{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right){r}}{\mathrm{2}} \\ $$$${shaded}\:{area}={a}^{\mathrm{2}} +\mathrm{4}{A}_{\mathrm{1}} \\ $$$$=\frac{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{4}×\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left(\frac{\pi}{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{3}}\right){r}^{\mathrm{2}} \\ $$$$=\left(\frac{\pi}{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{3}}\right)×\mathrm{20}^{\mathrm{2}} \\ $$$$\approx\mathrm{126} \\ $$
Commented by Tawa11 last updated on 18/Feb/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$