10,11 Find the volume of a parallelepiped, two faces of which are rhombuses with a side of 1 and an acute angle of $60^{\circ}$, and the other faces are squares.

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Let the faces $A B C D$ and $A 1 B 1 C 1 D 1$ of the parallelepiped $A B C D A 1 B 1 C 1 D 1$ be rhombuses with side length 1 and acute angle $60^{\circ}$, and the other faces be squares. Then the sides of these squares are also equal to 1. Since $A A 1 \perp A B$ and $A A 1 \perp A D$, the lateral edge $A A 1$ is perpendicular to the base plane $A B C D$ of the parallelepiped. Therefore, the parallelepiped is a right parallelepiped. Hence, its volume $V$ is equal to the product of the base area and the lateral edge, i.e.,

$$
V=S_{A B C D} \cdot A A 1=1 \cdot 1 \cdot \sin 60^{\circ} \cdot 1=\frac{\sqrt{3}}{2}
$$

Answer

## $\frac{\sqrt{3}}{2}$