[ Rectangular parallelepipeds ] [ Angles between lines and planes $]$

The sides of the base of a rectangular parallelepiped are equal to $a$ and $b$. The diagonal of the parallelepiped is inclined to the base plane at an angle of $60^{\circ}$. Find the lateral surface area of the parallelepiped.
Let $A B C D A 1 B 1 C 1 D 1$ be the given rectangular parallelepiped with bases $A B C D, A 1 B 1 C 1 D 1$ and lateral edges $A A 1, B B 1, C C 1$ and $D D 1$, where $B C=a, C D=b$. Since $B B 1$ is perpendicular to the plane of the base $A 1 B 1 C 1 D 1, B 1 D 1$ is the orthogonal projection of the diagonal $B D 1$ onto this plane. Therefore, $B D 1 B 1$ is the angle between the line $B D 1$ and the plane of the base $A 1 B 1 C 1 D 1$. According to the problem, $\angle B D 1 B 1=60^{\circ}$. From the right triangles $B 1 C 1 D 1$ and $B B 1 D 1$ we find that

$$
\begin{gathered}
B 1 D 1=\sqrt{B C^{2}+C D^{2}}=\sqrt{a^{2}+b^{2}} \\
B B 1=B 1 D 1 \operatorname{tg} \angle B D 1 B 1=B 1 D 1 \operatorname{tg} 60^{\circ}=\sqrt{3\left(a^{2}+b^{2}\right)}
\end{gathered}
$$

If $S$ is the lateral surface area of the parallelepiped, then

$$
S=2(A B+B C) \cdot B B 1=2(a+b) \sqrt{3\left(a^{2}+b^{2}\right)}
$$

## Answer

$2(a+b) \sqrt{3\left(a^{2}+b^{2}\right)}$