Question Number 221626 by fantastic last updated on 08/Jun/25
Answered by mr W last updated on 08/Jun/25
$${side}\:{length}\:{of}\:{square}\:=\mathrm{1} \\ $$$${shaded}\:{area}\:=\frac{\pi×\mathrm{1}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\pi}{\mathrm{4}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${fraction}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 08/Jun/25
Commented by fantastic last updated on 09/Jun/25
$${what}??\:{I}\:{did}\:{the}\:{same}\:{way} \\ $$
Answered by fantastic last updated on 09/Jun/25
Answered by mr W last updated on 09/Jun/25
![shaded area =(1/2)∫_(π/4) ^(π/2) [1^2 −((√2) cos θ)^2 ]dθ =(1/2)∫_(π/2) ^(π/4) cos 2θ dθ =(1/4)[sin 2θ]_(π/2) ^(π/4) =(1/4) fraction =(1/4)](https://www.tinkutara.com/question/Q221659.png)
$${shaded}\:{area}\: \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right]{d}\theta \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{cos}\:\mathrm{2}\theta\:{d}\theta \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{sin}\:\mathrm{2}\theta\right]_{\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{4}}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${fraction}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 09/Jun/25
