2.14. 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 lateral surface area of the parallelepiped.
2.14. We have $V=S_{\text {osn }} h$, where $S_{\text {osn }}=Q$ (by condition); thus, we need to find $h$. Since $A B C D$ is a rhombus, $S_{\text {osn }}=\frac{1}{2} A C \cdot B D$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-462.jpg?height=409&width=650&top_left_y=884&top_left_x=256)

Fig. 2.13

(Fig. 2.13); considering that $A C \cdot h=S_{1}, B D \cdot h=S_{2}$, we find $A C=\frac{S_{1}}{h}, B D=\frac{S_{2}}{h}$. 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 $V=\sqrt{\frac{S_{1} S_{2} Q}{2}}$. Further, from $\triangle C O D$ it follows that

$$
C D^{2}=\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}=\left(\frac{S_{1}}{2 h}\right)^{2}+\left(\frac{S_{2}}{2 h}\right)^{2}=\frac{S_{1}^{2}+S_{2}^{2}}{4 h^{2}}
$$

## Finally, we get

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

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

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-463.jpg?height=419&width=477&top_left_y=473&top_left_x=46)

Fig. 2.14