[ $\quad$ Volume of a prism $]$  $[\quad$ Regular tetrahedron $\quad]$

Find the volume of a parallelepiped, all faces of which are equal rhombuses with a side length of $a$ and an acute angle of $60^{\circ}$.
Let the edge $A A_{1}$ of the given parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ form an angle of $60^{\circ}$ with the edges $A B$ and $A D$. Then the triangular pyramid $A_{1} A B D$ is a regular tetrahedron with an edge equal to $a$.

## Solution

Let the edge $A A_{1}$ of the given parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ form an angle of $60^{\circ}$ with the edges $A B$ and $A D$. Since $A A_{1} B, A A_{1} D$, and $A B D$ are equilateral triangles, the triangular pyramid $A_{1} A B D$ is a regular tetrahedron with an edge equal to $a$. We find its height $A O$ from the right triangle $A O A_{1}:$

$$
A_{1} O=\sqrt{A A^{2}-A O^{2}}=\sqrt{a^{2}-(a \sqrt{3} / 3)^{2}}=a \sqrt{2 / 3}
$$

Since $A_{1} O$ is the height of the given parallelepiped,

$$
V\left(A B C D A_{1} B_{1} C_{1} D_{1}\right)=S(A B C D) \cdot A_{1} O=\left(a^{2} \sqrt{3} / 2\right) \cdot a \sqrt{2 / 3}=a^{2} \sqrt{2} / 2
$$

## Answer

$$
a^{2} \sqrt{2} / 2
$$