11.149. The base of a right parallelepiped is a parallelogram, one of the angles of which is $30^{\circ}$. The area of the base is 4 dm $^{2}$. The areas of the lateral faces of the parallelepiped are 6 and 12 dm ${ }^{2}$. Find the volume of the parallelepiped.
Solution.

The volume of the parallelepiped $V=S_{\text {base }} h$, where $h$ is the height of the parallelepiped. Since the parallelepiped is a right one, the heights of the lateral faces are also equal to $h$ (Fig. 11.44). According to the problem, $\angle B A D=30^{\circ}, A B \cdot h=6$ dm $^{2}, A D \cdot h=12$ dm $^{2}$, i.e., $A D=2 A B$. Let $B K \perp A D$; then $B K=\frac{1}{2} A B$.

Since $A D \cdot B K=4$ dm ${ }^{2}$, we have $2 A B \cdot \frac{1}{2} A B=4, A B=2$ (dm); therefore, $h=3$ dm. Finally, $V=4 \cdot 3=12$ dm $^{3}$.

Answer: 12 dm $^{3}$.