11.046. The sides of the base of a rectangular parallelepiped are equal to $a$ and $b$. The diagonal of the parallelepiped is inclined to the lateral face containing the side of the base equal to $b$ at an angle of $30^{\circ}$. Find the volume of the parallelepiped.
Solution.

$$
V=S \cdot H, \quad S=a \cdot b . \quad \text { In } \quad \Delta D_{1} A B \quad \angle D_{1} A B=90^{\circ}, \quad \angle A D_{1} B=30^{\circ}
$$

(Fig. 11.41). Hence $A D_{1}=a \cdot \operatorname{ctg} 30^{\circ}=a \sqrt{3}, H=\sqrt{A D_{1}^{2}-A_{1} D_{1}^{2}}=\sqrt{3 a^{2}-b^{2}}$;

$$
V=S \cdot H=a b \sqrt{3 a^{2}-b^{2}}
$$

Answer: $a b \sqrt{3 a^{2}-b^{2}}$.