Question Number 219723 by Spillover last updated on 01/May/25
Answered by mehdee7396 last updated on 03/May/25
$$\:{OB}=\mathrm{4}\Rightarrow{OT}=\mathrm{2}\Rightarrow{r}=\mathrm{3} \\ $$$${BT}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${BT}×{AT}={KT}×{TT}' \\ $$$$\mathrm{12}=\mathrm{6}×{TT}'\Rightarrow{TT}'=\mathrm{2}\Rightarrow{r}'=\mathrm{1} \\ $$$${S}=\mathrm{16}\pi−\mathrm{9}\pi−\mathrm{1}\pi=\mathrm{6}\pi \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 01/May/25
Commented by Spillover last updated on 01/May/25
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 01/May/25
