[ Special cases of parallelepipeds (other).] Law of Cosines

In a right parallelepiped, the sides of the base are equal to $a$ and $b$, and the acute angle between them is $60^{\circ}$. The larger diagonal of the base is equal to the smaller diagonal of the parallelepiped. Find the volume of the parallelepiped.
Let $A B C D A 1 B 1 C 1 D 1$ be a right parallelepiped with base $A B C D$ and lateral edges $A A 1, B B 1, C C 1$, and $D D 1$, perpendicular to the plane of the base $A B C D$, where $A B=a, A D=b, \angle B A D=60^{\circ}$. Then $A C$ is the larger diagonal of the parallelogram $A B C D$, and $B D 1$ is the smaller diagonal of the parallelepiped $A B C D A 1 B 1 C 1 D 1$, since the orthogonal projection of $B D$ of the inclined line $B D 1$ on the base plane $A B C D$ is less than the orthogonal projection of $A C$ of the inclined line $C A 1$ on the base plane $A B C D$. According to the problem, $B D 1=A C$. Using the cosine theorem in triangles $A B C$ and $A B D$, we find that

$$
\begin{aligned}
& A C 2=A B 2+B C 2-2 A B \cdot B C \cos \angle A B C=a 2+b 2-2 a b \cos 120^{\circ}= \\
& =a 2+b 2+a b, \\
& B D 2=A B 2+A D 2-2 A B \cdot A D \cos \angle B A D=a 2+b 2-2 a b \cos 60^{\circ}= \\
& =a 2+b 2-a b \text {. }
\end{aligned}
$$

From the right triangle $B B 1 D$, we find the height $B B 1$ of the parallelepiped:

$$
\begin{aligned}
& B B 1=\sqrt{B D^{2}-B D^{2}}=\sqrt{B D^{2}-B D^{2}}= \\
& =\sqrt{a^{2}+b^{2}+a b-\left(a^{2}+b^{2}-a b\right)}=\sqrt{2 a b}
\end{aligned}
$$

Therefore,

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot B B 1=a b \sin 60^{\circ} \cdot \sqrt{2 a b}=\frac{1}{2} a b \sqrt{6 a b}
$$

## Answer

$\frac{1}{2} a b \sqrt{6 a b}$.

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