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 $Q$. Find the lateral surface area of the parallelepiped.
Let $A B C D A 1 B 1 C 1 D 1$ be a right parallelepiped with base $A B C D$ and lateral edges $A A 1, B B 1, C C 1$ and $D D 1$, where $A B C D$ is a rhombus. The section $A D C 1 B 1$ forms an angle of $45^{\circ}$ with the plane of the base $A B C D$, and $S(A D C 1 B 1) = Q$. Drop a perpendicular $A M$ from vertex $A$ to the line $B 1 C 1$. Since the parallelepiped is right, $A A 1$ is perpendicular to the plane of the base $A 1 B 1 C 1 D 1$, so $A 1 M$ is the orthogonal projection of the inclined line $A M$ onto the plane of the base $A 1 B 1 C 1 D 1$. By the theorem of three perpendiculars, $A 1 M \perp B 1 C 1$, so $A M A 1$ is the linear angle of the dihedral angle formed by the planes of the section and the base $A 1 B 1 C 1 D 1$. According to the problem, $\angle A M A 1 = 45^{\circ}$. Let $A D = a, A A 1 = h$. From the right triangle $A M A 1$, we find that $A M = \frac{A A_{1}}{\sin 45^{\circ}} = h \sqrt{2}$. Therefore,

$$
Q = S_{A D C 1 B 1} = B 1 C 1 \cdot A M = a h \sqrt{2}
$$

from which $a h = \frac{Q}{\sqrt{2}}$. Let $S$ be the area of the lateral surface of the parallelepiped $A B C D A 1 B 1 C 1 D 1$. Then

$$
S = 4 S_{A D D 1 A 1} = 4 A D \cdot A A 1 = 4 a h = \frac{4 Q}{\sqrt{2}} = 2 Q \sqrt{2}
$$

## Answer

$2 Q \sqrt{2}$.