25) A cube with side $1 \mathrm{~m}$ and a sphere have the same center, and the surface of the sphere passes through the midpoints of all the sides of the cube. What is the area of the surface of the cube that is outside the sphere?
(A) $\left(6-\frac{3 \pi}{2}\right) \mathrm{m}^{2}$,
(B) $(8-2 \pi) \mathrm{m}^{2}$,
(C) $\left(6-\frac{4 \pi}{3}\right) \mathrm{m}^{2}$
(D) $(12-3 \pi) \mathrm{m}^{2}$,
(E) $\pi \mathrm{m}^{2}$.
25. The answer is (A).

In the drawing, one of the faces of the cube is represented. The hatched part represents the portion of the face inside the sphere.

![](https://cdn.mathpix.com/cropped/2024_04_17_e415f128e14398de33bcg-9.jpg?height=265&width=262&top_left_y=119&top_left_x=954)

The area of the non-hatched part is, by difference,

$$
\left(1-\frac{\pi}{4}\right) \mathrm{m}^{2} .
$$

Overall, the surface area of the part of the cube outside the sphere is given by the area just calculated multiplied by the number of faces of the cube, which is six. The result is then $6\left(1-\frac{\pi}{4}\right) \mathrm{m}^{2}=\left(6-\frac{3}{2} \pi\right) \mathrm{m}^{2}$.

[Problem proposed by K. Kuzmin.]