2.40. In a truncated triangular pyramid, a plane parallel to the opposite lateral edge is drawn through the side of the upper base. In what ratio is the volume of the truncated pyramid divided if the corresponding sides of the bases are in the ratio $1: 2$?
2.40. Since the sides of the bases are in the ratio $1: 2$, the areas of the bases are in the ratio $1: 4$ (Fig. 2.35). Then the volume of the truncated pyramid is

Fig. 2.35

$$
V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)=\frac{1}{3} h\left(4 S_{2}+S_{2}+2 S_{2}\right)=\frac{7}{3} S_{2} h
$$

where $S_{2}$ is the area of the upper base, and $h$ is the height. The volume of the prism $A D E A_{1} B_{1} C_{1}$ is $V_{1}=S_{2} h$, and thus the volume of the remaining part of the pyramid is $V_{2}=V-V_{1}=\frac{7}{3} S_{2} h-S_{2} h=\frac{4}{3} S_{2} h$. Therefore, $V_{1}: V_{2}=3: 4$.

Answer: $3: 4$.