11.148. The areas of the bases of a truncated pyramid are $S_{1}$ and $S_{2}\left(S_{1}<S_{2}\right)$, and its volume is $V$. Determine the volume of the complete pyramid.
## Solution.

Let $H$ be the height of the complete pyramid, $h$ be the height of the truncated pyramid, and $x=H-h$.

We have $\frac{S_{1}}{S_{2}}=\frac{x^{2}}{H^{2}}$ or $\frac{\sqrt{S_{1}}}{\sqrt{S_{2}}}=\frac{x}{x+h}$, from which $x \sqrt{S_{1}}+h \sqrt{S_{1}}=x \sqrt{S_{2}}$, $x=\frac{h \sqrt{S_{1}}}{\sqrt{S_{2}}-\sqrt{S_{1}}} \cdot$ Since $H=x+h=\frac{h \sqrt{S_{2}}}{\sqrt{S_{2}}-\sqrt{S_{1}}}$, the volume of the complete pyramid $V_{n p y a m}=\frac{1}{3} S_{2} H=\frac{1}{3} \frac{h S_{2} \sqrt{S_{1}}}{\sqrt{S_{2}}-\sqrt{S_{1}}}$.

By the given condition $V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)$, and thus, $h=\frac{3 V}{S_{1}+S_{2}+\sqrt{S_{1} S_{2}}}$.

We obtain, $V_{n p 2 m}=\frac{1}{3} \frac{S_{2} \sqrt{S_{2}}}{\sqrt{S_{2}}-\sqrt{S_{1}}} \cdot \frac{3 V}{S_{1}+S_{2}+\sqrt{S_{1} S_{2}}}=\frac{V S_{2} \sqrt{S_{2}}}{S_{2} \sqrt{S_{2}}-S_{1} \sqrt{S_{1}}}$.

Answer: $\frac{V S_{2} \sqrt{S_{2}}}{S_{2} \sqrt{S_{2}}-S_{1} \sqrt{S_{1}}}$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0857.jpg?height=363&width=613&top_left_y=79&top_left_x=81)

Fig. 11.45