Question-177013

Question Number 177013 by mr W last updated on 29/Sep/22

Commented by mr W last updated on 29/Sep/22

$${find}\:{the}\:{area}\:{of}\:{the}\:{sector}. \\ $$

Answered by Rasheed.Sindhi last updated on 29/Sep/22

The diameter=(√(4^2 +(3+5)^2 )) =4(√5) Rsdius=2(√5) Area of circle=π(2(√5) )^2 =20π Area of sector=((20π)/4)=5π

$$\mathcal{T}{he}\:{diameter}=\sqrt{\mathrm{4}^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} }\:=\mathrm{4}\sqrt{\mathrm{5}} \\ $$$${Rsdius}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${Area}\:{of}\:{circle}=\pi\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} =\mathrm{20}\pi \\ $$$${Area}\:{of}\:{sector}=\frac{\mathrm{20}\pi}{\mathrm{4}}=\mathrm{5}\pi \\ $$

Commented by mr W last updated on 29/Sep/22

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Commented by Rasheed.Sindhi last updated on 29/Sep/22

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Commented by mr W last updated on 29/Sep/22