11.182. The radius of the base of the cone is $R$, and the lateral surface area is equal to the sum of the areas of the base and the axial section. Determine the volume of the cone.
## Solution.

Let $H$ be the height, and $l$ the slant height of the cone. Then $l=\sqrt{H^{2}+R^{2}}$, the area of the axial section $S_{1}=R H$, the area of the base $S_{2}=\pi R^{2}$, and the lateral surface area $S_{3}=\pi R l=\pi R \sqrt{H^{2}+R^{2}}$. According to the condition, $S_{3}=S_{1}+S_{2}$.

$$
\text { Then } \pi R \sqrt{H^{2}+R^{2}}=R H+\pi R^{2} ; \pi \sqrt{H^{2}+R^{2}}=H+\pi R \text {; }
$$

$\pi^{2}\left(H^{2}+R^{2}\right)=H^{2}+2 \pi R H+\pi^{2} R^{2} ; \pi^{2} H^{2}=H^{2}+2 \pi R H ;\left(\pi^{2}-1\right) H=2 \pi R ;$ $H=\frac{2 \pi R}{\pi^{2}-1} \cdot$ The volume of the cone $V=\frac{1}{3} \pi R^{2} H=\frac{1}{3} \pi R^{2} \cdot \frac{2 \pi R}{\pi^{2}-1}=\frac{2 \pi^{2} R^{3}}{3\left(\pi^{2}-1\right)}$.

Answer: $\quad \frac{2 \pi^{2} R^{3}}{3\left(\pi^{2}-1\right)}$.