The diagonal of a rectangular parallelepiped is 13, and the diagonals of the side faces are $4 \sqrt{10}$ and $3 \sqrt{17}$.

Find its volume.
Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a rectangular parallelepiped, in which $B D_{1}=13, A D_{1}=3 \sqrt{17}, C D_{1}=4 \sqrt{10}$.

Apply the Pythagorean theorem to the right triangles $A B D_{1}, B C D_{1}$, and $C D D_{1}$.

## Solution

Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a rectangular parallelepiped, in which $B D_{1}=13, A D_{1}=3 \sqrt{17}, C D_{1}=4 \sqrt{10}$.

The line $A B$ is perpendicular to the intersecting lines $A D$ and $A A_{1}$ in the plane $A A_{1} D_{1} D$, so the line $A B$ is perpendicular to the plane $A A_{1} D_{1} D$. Therefore, $A B \perp A D_{1}$. Similarly, we can prove that $B C \perp C D_{1}$. By the Pythagorean theorem from the right triangles $A B D_{1}, B C D_{1}$, and $C D D_{1}$, we find that

$$
\begin{gathered}
A B=\sqrt{B D_{1}^{2}-A D_{1}^{2}}=\sqrt{169-153}=\sqrt{16}=4 \\
B C=\sqrt{B D_{1}^{2}-C D_{1}^{2}}=\sqrt{169-160}=\sqrt{9}=3 \\
D D_{1}=\sqrt{C D_{1}^{2}-C D^{2}}=\sqrt{C D_{1}^{2}-A B^{2}}=\sqrt{160-16}=\sqrt{144}=12
\end{gathered}
$$

Therefore,

$$
V\left(A B C D A_{1} B_{1} C_{1} D_{1}\right)=A B \cdot B C \cdot D D_{1}=4 \cdot 3 \cdot 12=144
$$

## Answer

## Problem