Question Number 204310 by BaliramKumar last updated on 12/Feb/24
Answered by mr W last updated on 12/Feb/24
Commented by mr W last updated on 12/Feb/24
$${R}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${OT}=\mathrm{5} \\ $$$${OA}=\sqrt{\mathrm{5}^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{91}}}{\mathrm{2}} \\ $$$$\frac{{r}}{{R}}=\frac{{OA}}{{OT}} \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{91}}}{\mathrm{2}×\mathrm{5}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{91}}}{\mathrm{20}} \\ $$$${h}={R}−\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{91}}}{\mathrm{20}}\right)^{\mathrm{2}} }=\frac{\mathrm{21}}{\mathrm{20}} \\ $$$${A}_{{cap}} =\mathrm{2}\pi{Rh}=\mathrm{2}\pi×\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{21}}{\mathrm{20}}=\frac{\mathrm{63}\pi}{\mathrm{20}}\:\checkmark \\ $$
Commented by mr W last updated on 12/Feb/24
Commented by BaliramKumar last updated on 12/Feb/24
$$\mathrm{thanks}\: \\ $$🙏🙏🙏🙏🙏