Question Number 28508 by ajfour last updated on 26/Jan/18
Commented by ajfour last updated on 26/Jan/18
$${Q}.\mathrm{28479}\:\left({solution}\right) \\ $$
Answered by ajfour last updated on 26/Jan/18
![Required area= 8×Area CHM^(⌢) =8(Area OHM^(⌢) −△OHC) =8[(π/(24))−(1/2)(((√3)/2)−(1/2))((1/2))] =(π/3)−(√3)+1 . =1−(√3)+(π/3) .](https://www.tinkutara.com/question/Q28510.png)
$${Required}\:{area}=\:\mathrm{8}×{Area}\:{C}\overset{\frown} {{HM}} \\ $$$$\:\:\:\:=\mathrm{8}\left({Area}\:{O}\overset{\frown} {{HM}}−\bigtriangleup{OHC}\right) \\ $$$$\:\:\:=\mathrm{8}\left[\frac{\pi}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:=\frac{\pi}{\mathrm{3}}−\sqrt{\mathrm{3}}+\mathrm{1}\:. \\ $$$$\:\:=\mathrm{1}−\sqrt{\mathrm{3}}+\frac{\pi}{\mathrm{3}}\:. \\ $$