11.008. In a right parallelepiped, the sides of the base are equal to $a$ and $b$, and the acute angle between them is $60^{\circ}$. The larger diagonal of the base is equal to the smaller diagonal of the parallelepiped. Find the volume of the parallelepiped.
## Solution.

The larger diagonal of the base lies opposite the larger angle, i.e., $180^{\circ}-60^{\circ}=120^{\circ}$. From this, by the cosine theorem, $d_{1}=\sqrt{a^{2}+b^{2}-2 a b \cos 120^{\circ}}=\sqrt{a^{2}+b^{2}+a b}$. The smaller diagonal of the base of the parallelepiped $d_{2}=\sqrt{a^{2}+b^{2}-2 a b \cos 60^{\circ}}=\sqrt{a^{2}+b^{2}-a b}$. According to the condition, the smaller diagonal of the parallelepiped is equal to $d_{1}$. On the other hand, $d^{2}=h^{2}+d_{2}^{2}$, where $h$ is the height of the parallelepiped. This means that $h=\sqrt{d_{1}^{2}-d_{2}^{2}}=\sqrt{2 a b}$. The area of the base

$$
S_{\mathrm{base}}=a b \sin 60^{\circ}=\frac{\sqrt{3} a b}{2} \cdot \text { Volume } V=S_{\mathrm{base}} \cdot h=\frac{\sqrt{6} a b \sqrt{a b}}{2} .
$$

Answer: $\frac{\sqrt{6} a b \sqrt{a b}}{2}$.