Question Number 145701 by alcohol last updated on 07/Jul/21
Commented by MJS_new last updated on 07/Jul/21
$$\mathrm{1}. \\ $$$$\mathrm{square}−\mathrm{2}×\left(\mathrm{square}−\mathrm{quarter}\:\mathrm{circle}\right)= \\ $$$$=−\mathrm{square}+\mathrm{2}×\mathrm{quarter}\:\mathrm{circle}= \\ $$$$=\mathrm{half}\:\mathrm{circle}−\mathrm{square}=\mathrm{4}^{\mathrm{2}} \pi/\mathrm{2}−\mathrm{4}^{\mathrm{2}} = \\ $$$$=\mathrm{8}\pi−\mathrm{16} \\ $$$$\mathrm{2}. \\ $$$$\mathrm{trapezoid}−\mathrm{rectangular}−\mathrm{half}\:\mathrm{circle}= \\ $$$$=\frac{\mathrm{6}+\mathrm{8}}{\mathrm{2}}×\mathrm{4}−\mathrm{4}×\mathrm{2}−\mathrm{2}^{\mathrm{2}} \pi/\mathrm{2}=\mathrm{20}−\mathrm{2}\pi \\ $$
Answered by iloveisrael last updated on 07/Jul/21
$$\left(\mathrm{1}\right)\:{shaded}\:{area}\:=\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)^{\mathrm{2}} \right) \\ $$$$\:=\:\mathrm{2}\left(\mathrm{4}\pi−\mathrm{8}\right)=\mathrm{8}\pi−\mathrm{16} \\ $$
Answered by iloveisrael last updated on 07/Jul/21
$$\left(\mathrm{2}\right)\:{shaded}\:{area}=\left(\frac{\mathrm{4}+\mathrm{6}}{\mathrm{2}}\right).\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\:\mathrm{20}−\mathrm{2}\pi\: \\ $$
Commented by alcohol last updated on 07/Jul/21
$${please}\:{explain}\:{this}\: \\ $$$${thanks} \\ $$