Question Number 218318 by Tawa11 last updated on 06/Apr/25
Commented by Tawa11 last updated on 06/Apr/25
$$\mathrm{Area}\:\mathrm{of}\:\mathrm{circle}. \\ $$
Commented by AlagaIbile last updated on 06/Apr/25
$$\:{A}\:=\:\mathrm{200}\pi\left(\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}}\right) \\ $$
Commented by Tawa11 last updated on 06/Apr/25
$$\mathrm{please}\:\mathrm{workings}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 06/Apr/25
Commented by mr W last updated on 06/Apr/25
$${R}=\mathrm{10}/\mathrm{2}=\mathrm{5} \\ $$$$\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{2}\:\mathrm{tan}\:\mathrm{15}°}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{15}°}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\mathrm{15}°+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{15}°−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{15}°=−\sqrt{\mathrm{3}}+\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\frac{{r}}{\mathrm{tan}\:\mathrm{15}°}={R}+\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){r}−{R}=\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {r}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){Rr}+{R}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{Rr} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {r}=\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){R} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){R}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}\right){R}=\mathrm{10}\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{5}\right) \\ $$$$\pi{r}^{\mathrm{2}} =\mathrm{200}\pi\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right) \\ $$
Commented by Tawa11 last updated on 07/Apr/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$