2.45. 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.
2.45. Since $C D=3 \text{ cm}, A D=$ $=4 \text{ cm}$ (Fig. 2.39), from $\triangle A D C$ we find $\quad A C=\sqrt{3^{2}+4^{2}}=5 \quad$ (cm). Draw $D K \perp A C$, then $D_{1} K \perp A C$ and $\angle D K D_{1}=60^{\circ}$ (by condition). But $\triangle A D C \sim \triangle C D K$, hence $\frac{A C}{C D}=\frac{A D}{D K}$,

i.e., $D K=\frac{3 \cdot 4}{5}=\frac{12}{5}$ (cm). Therefore, $D D_{1}=D K \sqrt{3}=\frac{12 \sqrt{3}}{5}$ (cm) and

$$
V=C D \cdot A D \cdot D D_{1}=3 \cdot 4 \cdot \frac{12 \sqrt{3}}{5}=\frac{144 \sqrt{3}}{5}\left(\text{ cm}^{3}\right)
$$

Answer: $28.8 \sqrt{3} \text{ cm}^{3}$.