11.114. In a truncated triangular pyramid, the height is 10 m, the sides of one base are 27, 29, and 25 m, and the perimeter of the other base is 72 m (Fig. 11.12). Determine the volume of the truncated pyramid.
Solution.

Let $S_{1}$ and $S_{2}$ be the areas of the bases of the given truncated pyramid, and $P_{1}$ and $P_{2}$ be the perimeters of the bases, with $P_{2}=72 \text{ m}$. Then $P_{1}=27+29+52=108 \text{ (m)}$, from which, using Heron's formula, $S_{1}=270 \text{ m}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0829.jpg?height=537&width=569&top_left_y=79&top_left_x=93)

Fig. 11.12

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0829.jpg?height=457&width=492&top_left_y=159&top_left_x=739)

Fig. 11.13

Since the bases of the truncated pyramid are similar, $\frac{S_{2}}{S_{1}}=\frac{P_{2}^{2}}{P_{1}^{2}}$; $S_{2}=S_{1} \cdot\left(\frac{P_{2}}{P_{1}}\right)^{2}=120 \text{ m}^{2}$

The volume of the truncated pyramid $V=\frac{1}{3} H\left(S_{1}+\sqrt{S_{1} S_{2}}+S_{2}\right)=1900 \text{ m}^{3}$.

Answer: $1900 \text{ m}^{3}$.