10,11 Find the volume of a rectangular parallelepiped, the areas of the diagonal sections of which are equal to $\sqrt{13}, 2 \sqrt{10}$ and $3 \sqrt{5}$
Let $A B C D A 1 B 1 C 1 D 1$ be a rectangular parallelepiped, and

$$
S_{B B} 1 D 1 D=\sqrt{13}, S_{C D A} 1 B 1=2 \sqrt{10}, S_{A D C} 1 B 1=3 \sqrt{5}
$$

Denote $A B=x, B C=y, A A 1=z$. Then

$$
B D 2=x 2+y 2, A 1 D 2=y 2+z 2, A B 1=x 2+z 2
$$

Since $B B 1 D 1 D, C D A 1 B 1$, and $A D C 1 B 1$ are rectangles with areas $\sqrt{13}, 2 \sqrt{10}$, and $3 \sqrt{5}$, respectively, we have

$$
\begin{aligned}
& \sqrt{13}=S(B B 1 D 1 D)=B D \cdot B B 1=\sqrt{x^{2}+y^{2}} \cdot z \\
& 2 \sqrt{10}=S(C D A 1 B 1)=C D \cdot A 1 D=\sqrt{y^{2}+z^{2}} \cdot x \\
& 3 \sqrt{5}=S(A D C 1 B 1)=A B 1 \cdot A D=\sqrt{x^{2}+z^{2}} \cdot y
\end{aligned}
$$

From this, we obtain the system of equations

$$
\left\{\begin{array}{l}
\left(x^{2}+y^{2}\right) z^{2}=13 \\
\left(y^{2}+z^{2}\right) x^{2}=40 \\
\left(x^{2}+z^{2}\right) y^{2}=45
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x^{2} z^{2}+y^{2} z^{2}=13 \\
x^{2} y^{2}+x^{2} z^{2}=40 \\
x^{2} y^{2}+z^{2} y^{2}=45
\end{array}\right.
$$

Adding the first and second equations term by term and subtracting the third, we get

$$
2 x 2 z 2=8, x 2 z 2=4, x z=2 \text {. }
$$

Similarly, $x y=6$ and $y z=3$. Therefore,

$$
V 2_{A B C D A} 1 B 1 C 1 D 1=x 2 y 2 z 2=x y \cdot y z \cdot x z=6 \cdot 3 \cdot 2=36
$$

$$
V_{A B C D A} 1 B 1 C 1 D 1=6
$$

## Answer

6.00

## Problem