11.026. The base of a right parallelepiped is a rhombus. A plane, passing through one side of the lower base and the opposite side of the upper base, forms an angle of $45^{\circ}$ with the base plane. The area of the resulting section is equal to $Q$. Determine the lateral surface area of the parallelepiped.
## Solution.

Draw $B K \perp D C$ (Fig. 11.21), then $B_{1} K \perp D C$ as well. Hence, $\angle B K B^{\prime}=45^{\circ}$ and $B K=B B^{\prime}=h, B^{\prime} K=\sqrt{2} h$. Let the sides of the rhombus be $a$. Then the lateral surface area

$S_{6}=4 a \cdot B B_{1}=4 a h=\frac{4}{\sqrt{2}} a \cdot \sqrt{2} h=$

$=2 \sqrt{2} \cdot C D \cdot B^{\prime} K=2 \sqrt{2} Q$, since the area of the section $A_{1} B_{1} C D$ is given to be $Q$.

Answer: $2 Q \sqrt{2}$.