2. Given that the base of a right parallelepiped is a rhombus, and the areas of the two diagonal faces are $Q_{1}$ and $Q_{2}$, find the lateral surface area of this right parallelepiped.
2. Let the side length of the rhombus base of the right parallelepiped $A_{1} C$ be $a$, and the side length of the lateral edge be $l$, then we have
$$
S_{\text {side }}=4 a l \text {. }
$$

According to the problem, we can obtain the system of equations
$$
\left\{\begin{array}{l}
Q_{1}=l \cdot A C, \\
Q_{2}=l \cdot B D, \\
4 a^{2}=A C^{2}+B D^{2} .
\end{array}\right.
$$
$(1)^{2}+(2)^{2}$, we get
$$
Q_{1}^{2}+Q_{2}^{2}=l^{2}\left(A C^{2}+B D^{2}\right) \text {. }
$$

From (3) and (4), eliminating $A C$ and $B D$, we get
$$
Q_{1}^{2}+Q_{2}^{2}=4 l^{2} a^{2} \text {, }
$$

thus $2 a l=\sqrt{Q_{1}^{2}+Q_{2}^{2}}$,
hence $S_{\text {side }}=4 a l=2 \sqrt{Q_{1}^{2}+Q_{2}^{2}}$.