Question Number 181658 by a.lgnaoui last updated on 28/Nov/22
$${Area}\:{of}\:{Circle}\:? \\ $$
Commented by a.lgnaoui last updated on 28/Nov/22
Commented by BaliramKumar last updated on 28/Nov/22
Commented by BaliramKumar last updated on 28/Nov/22
$${BF}×{FC}\:=\:{EF}×{FG} \\ $$$$\mathrm{4}×\mathrm{4}\:=\:\mathrm{8}×\left({d}−\mathrm{8}\right) \\ $$$$\mathrm{2}\:=\:{d}−\mathrm{8} \\ $$$${d}\:=\:\mathrm{10} \\ $$$$\mathrm{2}{r}\:=\:\mathrm{10} \\ $$$${r}\:=\:\mathrm{5} \\ $$$${Area}\:=\:\pi{r}^{\mathrm{2}} \:=\:\mathrm{25}\pi \\ $$$${Second}\:{method} \\ $$$${OB}^{\mathrm{2}} \:=\:{OF}^{\mathrm{2}} \:+\:{BF}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \:=\:\left(\mathrm{8}−{r}\right)^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \:=\:\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{8}×{r}+{r}^{\mathrm{2}} \:+\:\mathrm{16} \\ $$$$\mathrm{16}{r}\:=\:\mathrm{80} \\ $$$${r}\:=\:\mathrm{5} \\ $$$${Area}\:=\:\pi{r}^{\mathrm{2}} \:=\:\pi\centerdot\mathrm{5}^{\mathrm{2}} \:=\:\mathrm{25}\pi \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 28/Nov/22
$${thank}\:{you} \\ $$