11.153. In a truncated triangular pyramid, a plane is drawn through the side of the upper base parallel to the opposite lateral edge. In what ratio will the volume of the truncated pyramid be divided if the corresponding sides are in the ratio $1: 2$?
Solution.

The sides of the bases are in the ratio $1: 2$, so the areas of the bases are in the ratio $1: 4$ (Fig. 11.49). Then the volume of the truncated pyramid $V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)=\frac{1}{3} h\left(4 S_{2}+S_{2}+2 S_{2}\right)=\frac{7}{3} S_{2} h$, where $S_{2}$ is the area of the upper base, $h$ is the height. The volume of the prism $A D E A_{1} B_{1} C_{1}$ is $V_{1}=S_{2} h$ and 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: $3: 4$.