Inside a square with a side of $4 \mathrm{~cm}$, a circle with a diameter of $1 \mathrm{~cm}$ moves such that it always touches the perimeter of the square. Calculate the area of the part of the square that can never be covered by the moving circle.

(M. Dillingerová)

![](https://cdn.mathpix.com/cropped/2024_04_17_3b8c55268a64867bcc1ag-1.jpg?height=366&width=375&top_left_y=1573&top_left_x=869)
The uncovered part of the square is marked in gray in Fig. 1. It consists of four "corners" and an inner square. If we put the corners $\mathrm{k}$ together, we get Fig. 2. The area of this uncovered region will be $1-\pi \cdot 0.5^{2} \text{ cm}^{2}$.

b) In the middle, an uncovered square with a side length of $4-1-1=2 \text{ cm}$ remains. Its area is $2 \cdot 2=4 \text{ cm}^{2}$.

In total, the uncovered area is $1-\pi \cdot 0.5^{2}+4=\left(5-\frac{1}{4} \pi\right) \text{ cm}^{2}$.