In the drawing below, $\triangle A B C$ is an equilateral triangle and $C D$ is both a height of the triangle and a diameter of the circle. If $A B=10 \mathrm{~cm}$, determine the shaded area.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-02.jpg?height=426&width=491&top_left_y=495&top_left_x=680)
Solution

Since $CD$ is the diameter, its midpoint $H$ is the center of the circle. Let $I$ and $J$ be the other intersections of the circumference with the sides $AC$ and $BC$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-02.jpg?height=468&width=537&top_left_y=1254&top_left_x=657)

Since $\angle ICH = \angle HCJ = 30^{\circ}$ and $IH = CH = HJ$, it follows that triangles $\triangle CHI$ and $\triangle CHJ$ are isosceles with the vertex angle equal to $120^{\circ}$. If $l$ is the radius of the circle, since the height of the triangle and the diameter of the circle coincide, $2l = \frac{10 \sqrt{3}}{2} \text{ cm}$ and consequently $l = \frac{5 \sqrt{3}}{2} \text{ cm}$. Each of the shaded regions corresponds to the area of a circular sector of $120^{\circ} = 2\pi / 3$ minus an isosceles triangle, that is,

$$
\begin{aligned}
\frac{(2\pi / 3) l^{2}}{2} - \frac{l^{2} \sin 120^{\circ}}{2} & = \frac{\pi l^{2}}{3} - \frac{\sqrt{3} l^{2}}{4} \\
& = \frac{(4\pi - 3\sqrt{3}) l^{2}}{12} \\
& = \frac{(4\pi - 3\sqrt{3})}{12} \cdot \left(\frac{5 \sqrt{3}}{2}\right)^{2} \\
& = \frac{100\pi - 75\sqrt{3}}{8} \text{ cm}^{2}
\end{aligned}
$$

Since we have two equal regions, the area sought is twice the value found, that is, $\frac{100\pi - 75\sqrt{3}}{4} \text{ cm}^{2}$.