\left.\begin{array}{l}{[\text { Parallelepipeds (other). }]} \\ {[\text { Volume of a parallelepiped }]}\end{array}\right]

The edges of the parallelepiped are equal to $a, b$, and $c$. The edges equal to $a$ and $b$ are mutually perpendicular, and the edge equal to $c$ forms an angle of $60^{\circ}$ with each of them. Find the volume of the parallelepiped.
Let $A B C D A 1 B 1 C 1 D 1$ be the given parallelepiped, in which

$$
\begin{gathered}
A B=a, A D=b, A A 1=c \\
\angle B A D=90^{\circ}, \angle A 1 A B=\angle A 1 A D=60^{\circ}
\end{gathered}
$$

From the base $H$ of the height $A 1 H$ of the parallelepiped, we drop perpendiculars $H P$ and $H Q$ to the lines $A B$ and $A D$ respectively. By the theorem of three perpendiculars, $A 1 P \perp A B$ and $A 1 Q \perp A D$. From the equality of right triangles $A 1 A P$ and $A 1 A Q$, it follows that $A 1 P=A 1 Q$. Therefore, $H P=H Q$. Hence, $A H$ is the bisector of angle $B A D$. Let $\angle A 1 A H=\alpha$. Then

$$
A H=A A 1 \cos \alpha=c \cos \alpha, A 1 H=A A 1 \sin \alpha=c \sin \alpha
$$

$$
\begin{aligned}
& A P=A A 1 \cos \angle A 1 A P=A A 1 \cos \angle 60^{\circ}=\frac{c}{2} \\
& A P=A H \cos \angle P A H=c \cos \alpha \cos 45^{\circ}=\frac{c \cos \alpha}{\sqrt{2}} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_41a6f93ed2bb2ffd440eg-36.jpg?height=88&width=1278&top_left_y=2723&top_left_x=1)

$$
A 1 H=c \sin \alpha=\frac{c}{2}
$$

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot A 1 H=\frac{a b c \sqrt{2}}{2} .
$$

## Answer