11.115. The base of the prism is a trapezoid. Express the volume of the prism in terms of the areas $S_{1}$ and $S_{2}$ of the parallel lateral faces and the distance $h$ between them.
## Solution.

Trapezoid $M N K P (N K \| M P)$ is a perpendicular cross-section of prism $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.13). Then the height of the trapezoid is $h$. Let $S_{1}$ be the area of face $A A_{1} D_{1} D$, $S_{2}$ be the area of face $B B_{1} C_{1} C$, $l$ be the length of the lateral edge of the prism, $M P=a$, and $N K=b$. Then $S_{1}=a l, S_{2}=b l$, from which $a=\frac{S_{1}}{l}, b=\frac{S_{2}}{l}$.

The area of trapezoid $M N K P S=\frac{a+b}{2} h=\frac{\left(S_{1}+S_{2}\right) h}{2 l}$.

The volume of the prism $V=S l=\frac{\left(S_{1}+S_{2}\right) h}{2}$.

Answer: $\frac{\left(S_{1}+S_{2}\right) h}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0830.jpg?height=484&width=489&top_left_y=77&top_left_x=448)

Fig. 11.14