11.165. The diagonals of the faces of a rectangular parallelepiped are equal to $a, b$ and $c$. Determine its total surface area.
## Solution.

In the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.59) $A C=a, A D_{1}=b, C D_{1}=c$. Introduce the unknowns $A D=x, C D=y$, $D D_{1}=z$. Consider the system of equations: $\left\{\begin{array}{l}a^{2}=x^{2}+y^{2} ; \\ b^{2}=x^{2}+z^{2} ; \\ c^{2}=y^{2}+z^{2} ;\end{array}\right.$ By adding the equations of the system, we get: $a^{2}+b^{2}+c^{2}=2\left(x^{2}+y^{2}+z^{2}\right)$, from which $a^{2}+b^{2}+c^{2}=2\left(a^{2}+z^{2}\right) ; a^{2}+b^{2}+c^{2}=2\left(b^{2}+z^{2}\right) ; a^{2}+b^{2}+c^{2}=2\left(c^{2}+x^{2}\right)$.

Therefore, $x=\sqrt{\frac{a^{2}+b^{2}-c^{2}}{2}} ; y=\sqrt{\frac{a^{2}+c^{2}-b^{2}}{2}} ; z=\sqrt{\frac{c^{2}+b^{2}-a^{2}}{2}}$.

The total surface area of the parallelepiped:

$$
\begin{gathered}
S=2(x y+x z+y z)=\sqrt{a^{4}-\left(b^{2}-c^{2}\right)^{2}}+\sqrt{b^{4}-\left(a^{2}-c^{2}\right)^{2}}+\sqrt{c^{4}-\left(a^{2}-b^{2}\right)^{2}} \\
\text { Answer: } \sqrt{a^{4}-\left(b^{2}-c^{2}\right)^{2}}+\sqrt{b^{4}-\left(a^{2}-c^{2}\right)^{2}}+\sqrt{c^{4}-\left(a^{2}-b^{2}\right)^{2}}
\end{gathered}
$$