2.26. A cone with base radius $R$ is inscribed in an arbitrary pyramid, the perimeter of the base of which is $2 p$. Determine the ratio of the volumes and the ratio of the lateral surface areas of the cone and the pyramid.
2.26. Let the total height of the cone and the pyramid be $H$ (Fig. 2.22). 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 $l-$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-468.jpg?height=400&width=539&top_left_y=965&top_left_x=613)
is the slant height of the cone. Let's find $V_{2}$ and

Fig. 2.22 $S_{2}$. 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 ): Thus,

$$
\frac{V_{1}}{V_{2}}=\frac{1}{3} \pi R^{2} H: \frac{1}{3} p R H=\frac{\pi R}{p}, \frac{S_{1}}{S_{2}}=\pi R l: p l=\frac{\pi R}{p}
$$

Answer: $\frac{\pi R}{p}$ (in both cases).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-469.jpg?height=383&width=369&top_left_y=122&top_left_x=46)

Fig. 2.23