11.175. A right parallelepiped is described around a sphere, with the diagonals of the base being $a$ and $b$. Determine the total surface area of the parallelepiped.
Solution.

Let the radius of the sphere be $R$. In the section of the sphere by a plane passing through its center and parallel to the base of the parallelepiped, we obtain a parallelogram circumscribed about a circle of radius $R$. Since the sums of the opposite sides of such a circumscribed parallelogram are equal, it is a rhombus. Let the side of the rhombus be $m$; then the total surface area $S=2 S_{\text {base }}+S_{\text {lateral }}=2 m \cdot 2 R+4 m \cdot 2 R=6 m \cdot 2 R=6 S_{\text {base }}$. But $S_{\text {base }}=\frac{1}{2} a b$ and finally we get $S=3 a b$.

Answer: $3 a b$.