10,11

In a slanted parallelepiped, the projection of a lateral edge onto the base plane is 5, and the height is 12. A section perpendicular to the lateral edge is a rhombus with an area of 24 and a diagonal of 8. Find the lateral surface area and the volume of the parallelepiped.
Let $P$ be the orthogonal projection of vertex $A$ of the parallelepiped $A B C D A 1 B 1 C 1 D 1$ onto the base plane $A 1 B 1 C 1 D 1$. According to the problem, $A P=12, P A 1=5$. By the Pythagorean theorem in the right triangle $A P A 1$, we find that

$$
A A 1=\sqrt{A P^{2}+P A_{1}^{2}}=\sqrt{144+25}=13
$$

Let $V$ be the volume of the parallelepiped, and $S$ be its lateral surface area. The volume of the prism is the product of the area of the perpendicular section and the lateral edge. Therefore,

$$
V=24 \cdot 13=312
$$

The area of the rhombus is half the product of its diagonals $d 1$ and $d 2$. If $d 1=8$, then

$$
d 2=2 \cdot \frac{21}{d_{1}}=\frac{\frac{18}{8}}{8}=6 \text {. }
$$

Therefore, the side of the rhombus is

$$
\sqrt{\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}}=\sqrt{4^{2}+3^{2}}=5 .
$$

The lateral surface area of the prism is the product of the perimeter of the perpendicular section and the lateral edge.

Therefore,

$$
S=4 \cdot 5 \cdot 13=260
$$

## Answer

$260 ; 312$.

The base of the prism is a trapezoid. Find the volume of the prism if the areas of the parallel lateral faces are $S 1$ and $S 2$, and the distance between them is $h$.

## Solution

First, we will prove that if the area of the lateral face $A A 1 C 1 C$ of the triangular prism $A B C A 1 B 1 C 1$ is $S$, and the distance from the opposite lateral edge $B B 1$ to this plane is $h$, then the volume of the prism is $\frac{1}{2} S h$. For this, we will extend the prism $A B C A 1 B 1 C 1$ (Fig.1) to a parallelepiped $A B K C A 1 B 1 K 1 C 1$, by drawing a plane through the line $B B 1$ parallel to the face $A C C 1 A 1$, and a plane through the line $C C 1$ parallel to the face $A B B 1 A 1$. The height of the parallelepiped, dropped onto the base $A C C 1 A 1$, is equal to $h$. Therefore, its volume is $Sh$. The volume of the prism $A B C A 1 B 1 C 1$ is half the volume of the constructed parallelepiped, i.e., $\frac{1}{2} S h$. This is what we needed to prove. We will divide the given quadrilateral prism into two triangular prisms by a plane passing through the parallel diagonals of the bases (Fig.2). According to the previously proven result, the volumes of these triangular prisms are $\frac{1}{2} 51 h$ and $\frac{1}{2} S 2 h$. Therefore, the volume of the original prism is $\frac{1}{2}(S 1+S 2) h$.

## Answer

$\frac{1}{2}(S 1+S 2) h$.