12.263. The height of a regular truncated triangular pyramid is $H$ and is the mean proportional between the sides of the bases. The lateral edge makes an angle $\alpha$ with the base. Find the volume of the pyramid.
Solution.

Let $a$ and $\beta$ be the sides of the bases of the given truncated pyramid, with $a > \beta$.

Then $H = \left(\frac{a \sqrt{3}}{3} - \frac{b \sqrt{3}}{3}\right) \tan \alpha = \frac{(a - b) \sqrt{3}}{3} \tan \alpha, a - b = H \sqrt{3} \cot \alpha$, the areas of the bases $S_{1} = \frac{a^{2} \sqrt{3}}{4}, S_{2} = \frac{b^{2} \sqrt{3}}{4}, H^{2} = a b$.

The volume of the truncated pyramid

\[
\begin{aligned}
& V = \frac{H}{3}\left(S_{1} + S_{2} + \sqrt{S_{1} S_{2}}\right) = \frac{H}{3}\left(\frac{a^{2} \sqrt{3}}{4} + \frac{b^{2} \sqrt{3}}{4} + \frac{a b \sqrt{3}}{4}\right) = \\
& = \frac{H \sqrt{3}}{12}\left(a^{2} + b^{2} + a b\right) = \frac{H \sqrt{3}}{12}\left((a - b)^{2} + 3 a b\right) = \\
& = \frac{H \sqrt{3}}{12}\left(3 H^{2} \cot^{2} \alpha + 3 H^{2}\right) = \frac{H^{3} \sqrt{3}}{4}\left(\cot^{2} \alpha + 1\right) = \frac{H^{3} \sqrt{3}}{4 \sin^{2} \alpha} \\
& \text{Answer: } \frac{H^{3} \sqrt{3}}{4 \sin^{2} \alpha}
\end{aligned}
\]