11) A parallelepiped with a square base is inscribed in a sphere. If the side of the base is $\frac{1}{4}$ of the height, what is the ratio between the surface area of the sphere and the total surface area of the parallelepiped?
(A) $\pi$
(B) $\frac{\pi}{4}$
(C) $2 \pi$
(D) $\frac{\pi}{2}$
(E) depends on the radius of the sphere.
11) The answer is (A). Let $l$ be the side length of the base of the parallelepiped, then the height $h$ is $4l$, while the diagonal of the base $d$ is $\sqrt{2} l$. The rectangle obtained by sectioning the parallelepiped with a plane containing two parallel diagonals of the bases is inscribed in the largest circle of the sphere. The diameter of the sphere is the diagonal of this rectangle, which has sides $h$ and $d$. Using the Pythagorean theorem, the radius of the sphere is therefore:

$$
r=\frac{\sqrt{(4 l)^{2}+(\sqrt{2} l)^{2}}}{2}=\frac{\sqrt{18}}{2} l
$$

The surface area of the sphere is thus $4 \pi r^{2}=18 \pi l^{2}$, while the total surface area of the parallelepiped is: $4(4 l \cdot l)+2 l^{2}=18 l^{2}$. The desired ratio is therefore $\pi$.