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