11.043. The base of a right parallelepiped is a rhombus with an area of $Q$. The areas of the diagonal sections are $S_{1}$ and $S_{2}$. Determine the volume and the lateral surface area of the parallelepiped.
Solution.

We have $V=S_{\text {base }} h$, where $S_{\text {base }}=Q$ (by

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-676.jpg?height=378&width=462&top_left_y=591&top_left_x=785)

Fig. 11.38 condition); thus, we need to find $h$. Since $A B C D$ is a rhombus, then $S_{\text {base }}=0.5 A C \cdot B D$ (Fig. 11.38); considering that $A C \cdot h=S_{1}$, $B D \cdot h=S_{2}$, we find $A \dot{C}=\frac{S_{1}}{h}, B D=\frac{S_{2}}{h}$. From this, we get $2 Q=\frac{S_{1}}{h} \cdot \frac{S_{2}}{h}$, i.e., $h=\sqrt{\frac{S_{1} S_{2}}{2 Q}}$. Then $V=\sqrt{\frac{S_{1} S_{2} Q}{2}}$. From $\triangle C O D: C D^{2}=\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}=$ $=\left(\frac{S_{1}}{2 h}\right)^{2}+\left(\frac{S_{2}}{2 h}\right)^{2}=\frac{S_{1}^{2}+S_{2}^{2}}{4 h^{2}}$. Therefore,

$$
S_{\text {lateral }}=4 C D \cdot h=4 h \cdot \frac{\sqrt{S_{1}^{2}+S_{2}^{2}}}{2 h}=2 \sqrt{S_{1}^{2}+S_{2}^{2}}
$$

Answer: $\sqrt{S_{1} S_{2} Q / 2} ; 2 \sqrt{S_{1}^{2}+S_{2}^{2}}$.