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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^2 =a^2 +b^2 2R=((BC)/(sin 45°))=(√(2(a^2 +b^2 ))) ⇒a^2 +b^2 =2R^2 blue area=((π(a^2 +b^2 ))/4)=((πR^2 )/2) =((area of red circle)/2)=((4π)/2)=2π

$${BC}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{2}{R}=\frac{{BC}}{\mathrm{sin}\:\mathrm{45}°}=\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} \\ $$$${blue}\:{area}=\frac{\pi\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\mathrm{4}}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{{area}\:{of}\:{red}\:{circle}}{\mathrm{2}}=\frac{\mathrm{4}\pi}{\mathrm{2}}=\mathrm{2}\pi \\ $$

Commented by Tawa11 last updated on 27/Jun/24

Thanks sir. I appreciate sir.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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