11.081. A cone with base radius $R$ is inscribed in an arbitrary pyramid, where the perimeter of the base of the pyramid is $2 p$. Determine the ratio of the volumes and the ratio of the lateral surface areas of the cone and the pyramid.
## Solution.

Let the total height of the cone and the pyramid be $H$ (Fig. 11.73). Denote the volumes of the cone and the pyramid by $V_{1}$ and $V_{2}$, and their lateral surfaces by $S_{1}$ and $S_{2}$; then $V_{1}=\frac{1}{3} \pi R^{2} H, S_{1}=\pi R l$, where $l$ is the slant height of the cone. Let's find $V_{2}$ and $S_{2}$. Since the perimeter of the base of the pyramid is $2 p$, and the base of the cone is a circle inscribed in the base of the pyramid, the area of the base of the pyramid is $p R$, from which $V_{2}=\frac{1}{3} p R H, S_{2}=p l$ (the height of any face is $l$). Therefore, $\frac{V_{1}}{V_{2}}=\frac{1}{3} \pi R^{2} H: \frac{1}{3} p R H=\frac{\pi R}{p}, \frac{S_{1}}{S_{2}}=\pi R l: p l=\frac{\pi R}{p}$.

$$
\text { Answer: } \frac{\pi R}{p}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-700.jpg?height=424&width=466&top_left_y=90&top_left_x=143)

Fig. 11.73

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-700.jpg?height=348&width=370&top_left_y=178&top_left_x=799)

Fig. 11.74