In a square grid, where the squares have a side length of $10 \, \text{cm}$, a circle with center $S$ at a marked grid point and radius $20 \, \text{cm}$ is drawn.

![](https://cdn.mathpix.com/cropped/2024_04_17_cd0342d3a611cc15345bg-3.jpg?height=557&width=494&top_left_y=447&top_left_x=812)

Points $A, B, C$, and $D$ are the intersections of the circle with the grid lines.

Determine the area of the shaded region $A B C D$.
Segment $AB$ has a length equal to the radius of the given circle. We can therefore consider it as a side of a regular hexagon inscribed in this circle. This hexagon is divided into six equilateral triangles, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_cd0342d3a611cc15345bg-3.jpg?height=497&width=491&top_left_y=1373&top_left_x=814)

First, we calculate the area $S_{1}$ of the equilateral triangle with side $r=20 \mathrm{~cm}$. The height of this triangle can be expressed using the Pythagorean theorem $r^{2}=\left(\frac{r}{2}\right)^{2}+v^{2}$. We get $v=\frac{\sqrt{3}}{2} r$, thus

$$
S_{1}=\frac{\sqrt{3}}{4} r^{2}
$$

The shaded area consists of the pentagon $ABCS D$, whose area is $3 S_{1}$, and the circular sector $DSC$. The circular sector forms one-sixth of the circle, so its area is

$$
S_{2}=\frac{1}{6} \pi r^{2}
$$

The total area of the shaded region is

$$
S=3 S_{1}+S_{2}=\frac{3 \sqrt{3}}{4} r^{2}+\frac{1}{6} \pi r^{2}=\frac{9 \sqrt{3}+2 \pi}{12} r^{2} .
$$

After substitution, we get approximately $S \doteq 729 \mathrm{~cm}^{2}$.