Question Number 192765 by Mingma last updated on 26/May/23
Answered by ajfour last updated on 26/May/23
Commented by ajfour last updated on 26/May/23
$$\mathrm{tan}\:\theta=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${r}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${r}\mathrm{sin}\:\theta={h}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:{r}=\frac{\mathrm{1}/\mathrm{9}}{\mathrm{1}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}}=\frac{\mathrm{13}−\mathrm{3}\sqrt{\mathrm{13}}}{\mathrm{36}} \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\sqrt{\mathrm{13}}}{\mathrm{9}\left(\sqrt{\mathrm{13}}+\mathrm{3}\right)}=\frac{\mathrm{25}−\mathrm{3}\sqrt{\mathrm{13}}}{\mathrm{36}} \\ $$$${G}=\frac{\mathrm{1}}{\mathrm{81}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{r}+{r}\mathrm{cos}\:\theta\right)\left({h}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−\left(\pi−\theta\right)\frac{{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by Mingma last updated on 28/May/23
Nice work!