Question Number 32659 by Joel578 last updated on 30/Mar/18
Commented by Joel578 last updated on 30/Mar/18
$$\mathrm{Red}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cm} \\ $$$$\mathrm{Blue}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{4}\:\mathrm{cm} \\ $$$${C}\:\mathrm{is}\:\mathrm{center}\:\mathrm{of}\:\mathrm{blue}\:\mathrm{circle} \\ $$$${AB}\:\mathrm{is}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{red}\:\mathrm{circle} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part} \\ $$
Answered by MJS last updated on 30/Mar/18
$${AC}={BC}=\mathrm{4}\:\wedge\:{AB}=\mathrm{4}\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:{AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} \:\Rightarrow \\ $$$$\Rightarrow\:{black}\:{area}=\frac{\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \pi}{\mathrm{2}}−\left(\frac{\mathrm{4}^{\mathrm{2}} \pi}{\mathrm{4}}−\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}\right)= \\ $$$$=\mathrm{4}\pi−\left(\mathrm{4}\pi−\mathrm{8}\right)=\mathrm{8} \\ $$
Commented by Joel578 last updated on 01/Apr/18
$${thank}\:{you}\:{very}\:{mucb} \\ $$