11.214 The faces of the parallelepiped are rhombuses, the diagonals of which are 3 and 4 cm. The parallelepiped has trihedral angles composed of three acute angles of the rhombuses. Find the volume of the parallelepiped.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-268.jpg?height=301&width=365&top_left_y=209&top_left_x=113)

Let $A$ be the vertex of the given parallelepiped $T=A B C D A_{1} B_{1} C_{1} D_{1}$, to which the acute angles of its faces adjoin.

Since a rhombus is uniquely determined by its diagonals, all faces of $T$ are equal to each other, and its edges are equal to

$$
\sqrt{\left(\frac{3}{2}\right)^{2}+2^{2}}=\frac{5}{2}
$$

The pyramid $A B C D$ is regular because $\triangle B C D$ is equilateral with side 3 and $A C=A B=A D=\frac{5}{2}$. Let $A P$ be the height of this pyramid.

Then $A P=\sqrt{A C^{2}-C P^{2}}=\frac{\sqrt{13}}{2}\left(C P=\frac{B C}{\sqrt{3}}=\sqrt{3}\right), V_{A B C D}=\frac{1}{3} A P \cdot S_{B C D}=\frac{9 \sqrt{39}}{24}$.

On the other hand, if $h$ is the height of the pyramid $A B C D$ drawn from vertex $C$ to the base $A B D$, then $V_{A B C D}=\frac{1}{3} h \cdot S_{A B D}=\frac{1}{6} h \cdot S_{A B A, D}=\frac{1}{6} V_{T} \Rightarrow$

$$
\Rightarrow V_{T}=6 \cdot V_{A B C D}=\frac{9 \sqrt{39}}{4}
$$

Answer: $\frac{9 \sqrt{39}}{4}$.