11.167. The base of a right parallelepiped is a parallelogram with an angle of $120^{\circ}$ and sides of 3 and 4 cm. The smaller diagonal of the parallelepiped is equal to the larger diagonal of the base. Find the volume of the parallelepiped.
Solution.

In the right parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.61) $A B=3$ cm, $A D=4$ cm, $\angle A B C=120^{\circ}$. Then $A C$ is the larger diagonal of the base, $B_{1} D$ is the smaller diagonal of the parallelepiped, and $B_{1} D=A C$. The area of the base $A B C D: S=A B \cdot B C \cdot \sin \angle A B C=6 \sqrt{3} \text{ cm}^{2}$. From $\triangle A B C$:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0871.jpg?height=432&width=632&top_left_y=113&top_left_x=96)

Fig. 11.61

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0871.jpg?height=474&width=602&top_left_y=74&top_left_x=674)

Fig. 11.62

$A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos \angle A B C=9+16-24 \cos 120^{\circ}=37$ (cm). From $\triangle B A D: B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos \angle B A D=9+16-24 \cos 60^{\circ}=13$ (cm).

From $\triangle B_{1} B D\left(\angle B_{1} B D=90^{\circ}\right): \quad B B_{1}=\sqrt{B_{1} D^{2}-B D^{2}}=\sqrt{24}=2 \sqrt{6}(\text{ cm})$. The volume of the parallelepiped $V=S \cdot B B_{1}=6 \sqrt{3} \cdot 2 \sqrt{6}=36 \sqrt{2}\left(\text{ cm}^{3}\right)$.

Answer: $\quad 36 \sqrt{2}\left(\text{ cm}^{3}\right)$.