2.39. In a truncated triangular pyramid, the height is $10 \mathrm{M}$, the sides of one base are 27, 29, and 52 m, and the perimeter of the other base is 72 m. Determine the volume of the truncated pyramid.
2.39. The desired volume $V=\frac{1}{3} H\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)$, where $S_{1}$ is found using Heron's formula. We have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-477.jpg?height=452&width=415&top_left_y=532&top_left_x=46)

Fig. 2.34. Thus, $2 p_{1}=27+29+52=108$ (m), from which $S_{1}=\sqrt{54 \cdot 27 \cdot 25 \cdot 2}=270\left(\mathrm{M}^{2}\right)$. Since $\triangle A B C \sim \Delta A_{1} B_{1} C_{1}$ (Fig. 2.34), then

$$
\frac{S_{1}}{S_{2}}=\frac{\left(2 p_{1}\right)^{2}}{\left(2 p_{2}\right)^{2}}=\frac{108^{2}}{72^{2}}=\frac{9}{4}
$$

i.e.

$$
S_{2}=\frac{4 S_{1}}{9}=\frac{4 \cdot 270}{9}=120\left(\mathrm{M}^{2}\right)
$$

Finally, we get

$$
V=\frac{1}{3} \cdot 10(270+120+\sqrt{270 \cdot 120})=1900\left(\mathrm{M}^{3}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-477.jpg?height=402&width=396&top_left_y=1199&top_left_x=56)

Answer: $1900 \mathrm{M}^{3}$.