11.006. The diagonal of a rectangular parallelepiped is 13 cm, and the diagonals of its lateral faces are $4 \sqrt{10}$ and $3 \sqrt{17}$ cm. Determine the volume of the parallelepiped.
Solution.

Let the sides of the base be $a$ and $b$, the height of the parallelepiped be $h$, and let $b > a$. The diagonal of the base of the parallelepiped is $d_{0} = \sqrt{a^{2} + b^{2}}$. The diagonal of the lateral face with edges $b$ and $h$ is $d_{1} = \sqrt{b^{2} + h^{2}} = 4 \sqrt{10}$. The diagonal of the lateral face with edges $a$ and $h$ is $d_{2} = \sqrt{a^{2} + h^{2}} = 3 \sqrt{17}$. The diagonal of the parallelepiped is $d = \sqrt{d_{0}^{2} + h^{2}} = 13$. We obtain the system $\left\{\begin{array}{l}b^{2} + h^{2} = (4 \sqrt{10})^{2}, \\ a^{2} + h^{2} = (3 \sqrt{17})^{2}, \text { from which } a = 3, b = 4, h = 12. \text { The volume } V = a b h = 144 \mathrm{~cm}^{3}. \\ a^{2} + b^{2} + h^{2} = 13^{2},\end{array}\right.$

Answer: $144 \mathrm{~cm}^{3}$.