11.199 The edge of a slanted parallelepiped is equal to $l$. To it, two adjacent faces are attached, with areas equal to $m^{2}$ and $n^{2}$, and their planes form an angle of $30^{\circ}$. Calculate the volume of the parallelepiped.
## Solution.

$B_{I}$
Let $T=A B C D A_{1} B_{1} C_{1} D_{1}$ be the given pa-

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-259.jpg?height=274&width=438&top_left_y=1082&top_left_x=176)
rallelepiped with base $A B C D$,

$$
S_{A A_{1} D_{1} D}=m^{2}, S_{C C_{1} D_{1} D}=n^{2}, D D_{1}=l, \text { angle }
$$

between the faces $A A_{1} D_{1} D$ and $C C_{1} D_{1} D$ is $30^{\circ}$.
Draw through an arbitrary point $K$ of the edge $D D_{1}$ a plane perpendicular to $D D_{1}$. Then the section $K L M N$ of the body $T$ by this plane is a parallelogram (Theorem 13), and $\angle L K N=30^{\circ}$ (by condition).
$V_{T}=S_{K L M N} \cdot D D_{1}=L K \cdot N K \cdot \sin 30^{\circ} \cdot D D_{1}=\frac{1}{2} \cdot \frac{\left(L K \cdot D D_{1}\right)\left(N K \cdot D D_{1}\right)}{D D_{1}}=\frac{m^{2} n^{2}}{2 l}$. Answer: $\frac{m^{2} n^{2}}{2 l}$.