Determine the area of the gray moon from the image, given that the diameter $AB$ of the larger semicircle is $2 \mathrm{~cm}$, the diameter $CD$ of the smaller semicircle is $1 \mathrm{~cm}$, and $AB \| CD$.

(P. Tlustý)

![](https://cdn.mathpix.com/cropped/2024_04_17_371419af3500cb7ea530g-2.jpg?height=363&width=580&top_left_y=498&top_left_x=772)
Let $S$ be the midpoint of segment $A B$. Triangle $S C D$ is equilateral (with side length $1 \, \text{cm}$). The area of the gray crescent can be calculated by subtracting the area of the segment of the larger circle from the area of the smaller semicircle (with diameter $1 \, \text{cm}$).

Area of the smaller semicircle:

$$
S_{M}=\frac{1}{2} \pi \cdot\left(\frac{1}{2}\right)^{2}=\frac{\pi}{8} \, \text{cm}^{2}
$$

Area of the segment: We calculate it as the difference between the area of the corresponding circular sector (one-sixth of the area of the larger circle) and the area of the equilateral triangle $S C D$.

Area of the sector:

$$
S_{V}=\frac{1}{6} \pi \cdot 1^{2}=\frac{\pi}{6} \, \text{cm}^{2}
$$

Area of the triangle: the side measures $1 \, \text{cm}$, the height $\frac{1}{2} \sqrt{3} \, \text{cm}$,

$$
S_{\triangle}=\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} \, \text{cm}^{2}
$$

Area of the gray crescent:

$$
S=S_{M}-\left(S_{V}-S_{\triangle}\right)=\frac{\pi}{8}-\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\pi}{24}\right) \, \text{cm}^{2}
$$