Question Number 194073 by Rupesh123 last updated on 27/Jun/23
Answered by BaliramKumar last updated on 27/Jun/23
$$\mathrm{10}\pi \\ $$
Answered by mr W last updated on 27/Jun/23
$${method}\:{I}: \\ $$$$\sqrt{{R}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\mathrm{6} \\ $$$$\sqrt{{R}^{\mathrm{2}} −\mathrm{4}}=\mathrm{4} \\ $$$$\Rightarrow{R}^{\mathrm{2}} =\mathrm{20} \\ $$$${semicircle}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{10}\pi \\ $$