11.071. Find the ratio of the volume of a regular hexagonal pyramid to the volume of a regular triangular pyramid, given that the sides of the bases of these pyramids are equal, and their apothems are twice the length of the sides of the base.
Solution.

The volume of the hexagonal pyramid: $V_{1}=\frac{1}{3} S_{1} \cdot H_{1}$. The area of the base of the hexagonal pyramid $S_{1}=\frac{3}{2} a^{2} \sqrt{3}$, and the height is $H_{1}=\sqrt{l^{2}-r_{1}^{2}}$, where the slant height $l=2 a$, and the radius of the inscribed circle $r_{1}=\frac{3 a}{2 \sqrt{3}}$. Then $H=\frac{a}{2} \sqrt{13}$. Substituting, we get $V_{1}=\frac{1}{3} \frac{3}{2} a^{2} \sqrt{3} \cdot \frac{\sqrt{13}}{2}=\frac{1}{4} a^{3} \sqrt{39}$.

The volume of the triangular pyramid $V_{2}=\frac{1}{3} S_{2} H_{2}$. The area of the base of this pyramid $S_{2}=\frac{a^{2} \sqrt{3}}{4}$, and its height $H_{2}=\sqrt{l^{2}-r_{2}^{2}}$, where the radius of the inscribed circle $r_{2}=\frac{a \sqrt{3}}{6}$. Substituting, we get $H_{2}=\frac{a}{6} \sqrt{141}$. Then $V_{2}=\frac{1}{3} \frac{a^{3} \sqrt{3}}{4} \cdot \frac{\sqrt{141}}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-694.jpg?height=386&width=504&top_left_y=223&top_left_x=138)

Fig. 11.65

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-694.jpg?height=518&width=406&top_left_y=93&top_left_x=749)

Fig. 11.66

We get the ratio:

$$
\begin{aligned}
& \frac{V_{1}}{V_{2}}=\frac{a^{3} \sqrt{39} \cdot 4 \cdot 6}{4 a^{3} \sqrt{3} \sqrt{141}}=\frac{3 \cdot 6^{2} \sqrt{13}}{\sqrt{141}}=\frac{18 \sqrt{1833}}{141}=\frac{6 \sqrt{1833}}{47} \\
& \text { Answer: } \quad \frac{6 \sqrt{1833}}{47}
\end{aligned}
$$