11.137. Through the vertices $A, C$ and $D_{1}$ of a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a plane is drawn, forming a dihedral angle of $60^{\circ}$ with the base plane. The sides of the base are 4 and 3 cm. Find the volume of the parallelepiped.
## Solution.

Drop a perpendicular $D K$ from point $D$ to $A C$ (Fig. 11.34). $D K$ is the projection of $D_{1} K$ on the base plane of the parallelepiped. Then $D_{1} K \perp A C, \angle D_{1} K D$ is the linear angle of the dihedral angle, which by condition is $\angle D_{1} K D=60^{\circ}$. From $\triangle A D C\left(\angle A D C=90^{\circ}\right): \quad A C=\sqrt{A D^{2}+D C^{2}}=5 \mathrm{~cm}$, $S_{\triangle A D C}=\frac{1}{2} A C \cdot D K=\frac{1}{2} A D \cdot D C$. Then $\quad D K=\frac{A D \cdot D C}{A C}=2.4 \mathrm{~cm}$. From $\triangle D_{1} D K\left(\angle D_{1} D K=90^{\circ}\right): D D_{1}=D K \operatorname{tg} \angle D_{1} K D=2.4 \sqrt{3}$ cm. The volume of the parallelepiped $V=S_{A B C D} \cdot D D_{1}=A D \cdot D C \cdot D D_{1}=3 \cdot 4 \cdot 2.4 \sqrt{3}=\frac{144 \sqrt{3}}{5}\left(\mathrm{~cm}^{3}\right)$.

Answer: $\frac{144 \sqrt{3}}{5}\left(\mathrm{~cm}^{3}\right)$.