Question Number 208915 by Tawa11 last updated on 27/Jun/24
Answered by mr W last updated on 27/Jun/24
Commented by mr W last updated on 27/Jun/24
$${BC}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{R}=\frac{{BC}}{\mathrm{sin}\:\mathrm{45}°}=\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} \\ $$$${blue}\:{area}=\frac{\pi\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\mathrm{4}}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{{area}\:{of}\:{red}\:{circle}}{\mathrm{2}}=\frac{\mathrm{4}\pi}{\mathrm{2}}=\mathrm{2}\pi \\ $$
Commented by Tawa11 last updated on 27/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$