Question Number 219087 by fantastic last updated on 19/Apr/25
Answered by mahdipoor last updated on 19/Apr/25
$${DE}=\mathrm{2}{r}−{AD}−{EB}=\mathrm{2}{r}−\left({d}\right)−\left({r}−{d}\right)={r} \\ $$$$\frac{\pi}{\mathrm{2}}\left(\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left({r}\right)^{\mathrm{2}} \right)=\mathrm{1}.\mathrm{5}\pi{r}^{\mathrm{2}} \\ $$
Commented by fantastic last updated on 20/Apr/25
$${thank}\:\:{you} \\ $$