11.057. Determine the volume of a rectangular parallelepiped if its diagonal is equal to $d$, and the lengths of the edges are in the ratio $m: n: p$.
Solution.

$V=a b c, D_{1} B=d$. Let $A B=m x, B C=n x, B_{1} B=p x$ (since $m: n: p=A B: B C: B_{1} B$) (Fig. 11.52). In $\triangle D_{1} D B \angle D_{1} D B=90^{\circ}$, then $D_{1} B^{2}=D_{1} D^{2}+D B^{2}$. In $\triangle D A B \angle D A B=90^{\circ}$ and $D B^{2}=A D^{2}+A B^{2}$, so $D_{1} B^{2}=D_{1} D^{2}+A D^{2}+A B^{2}, \quad d^{2}=p^{2} x^{2}+m^{2} x^{2}+n^{2} x^{2}$, hence, $x=$ $=\frac{d}{\sqrt{m^{2}+n^{2}+p^{2}}}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-685.jpg?height=592&width=484&top_left_y=30&top_left_x=142)

Fig. 11.53

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-685.jpg?height=418&width=422&top_left_y=207&top_left_x=745)

Fig. 11.54

From this, $V=A B \cdot B C \cdot B_{1} B=m n p x^{3}=\frac{m n p \cdot d^{3}}{\left(m^{2}+n^{2}+p^{2}\right)^{3 / 2}}$.

Answer: $\frac{m n p \cdot d^{3}}{\left(m^{2}+n^{2}+p^{2}\right)^{3 / 2}}$.