2.41. The sides of the bases of a regular truncated quadrilateral pyramid are 2 and 1 cm, and the height is 3 cm. A plane is drawn through the point of intersection of the diagonals of the pyramid, parallel to the bases of the pyramid, dividing the pyramid into two parts. Find the volume of each of them.
2.41. Let $O_{1} O_{2}=x$ (Fig. 2.36); then $O O_{2}=3-x$. Since $\Delta B_{1} O_{2} D_{1} \sim \Delta B O_{2} D$, we have $B_{1} D_{1}: B D = O_{1} O_{2}: O O_{2}$, or $1: 2 = x: (3-x)$, from which $x=1$ (cm). Further, $\Delta B_{1} D_{1} B \sim \Delta L O_{2} B$ and, therefore, $B_{1} D_{1}: L O_{2} = O O_{1}: O O_{2} = 3: 2$, i.e., $L O_{2} = \frac{2}{3} B_{1} D_{1}, L N = \frac{4}{3} B_{1} D_{1}$. Then $S_{K L M N} = \frac{16}{9} S_{A_{1} B_{1} C_{1} D_{1}} = \frac{16}{9} \left(\mathrm{cm}^{2}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-478.jpg?height=463&width=483&top_left_y=426&top_left_x=670)

Fig. 2.36

Finally, we get

$$
V_{1}=\frac{1}{3} \cdot 2\left(4+\frac{16}{9}+\frac{8}{3}\right)=\frac{152}{27}\left(\mathrm{~cm}^{3}\right), V_{2}=\frac{1}{3} \cdot 1\left(\frac{16}{9}+1+\frac{4}{3}\right)=\frac{37}{27}\left(\mathrm{~cm}^{3}\right)
$$

Answer: $\frac{152}{27}$ and $\frac{37}{27} \mathrm{~cm}^{3}$.