The Reuleaux triangle is the figure formed from an equilateral triangle, replacing the sides with circular arcs centered at the vertices of the triangle and with radii equal to the side of the triangle. What is the area of a Reuleaux triangle if the sides of the initial equilateral triangle measure $1 \mathrm{~cm}$?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-035.jpg?height=254&width=257&top_left_y=313&top_left_x=1616)
The Reuleaux triangle is formed by 4 regions, an equilateral triangle, and three caps. Each cap is one-sixth of a circle of radius 1, from which an equilateral triangle of side 1 has been removed. By the Pythagorean Theorem, the height of this equilateral triangle is

$$
\sqrt{1-\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{3}}{2}
$$

therefore, the area of this equilateral triangle is given by

$$
\frac{1 \times \frac{\sqrt{3}}{2}}{2}=\frac{\sqrt{3}}{4}
$$

The area of a sector of a circle that is one-sixth of a circle of radius 1 is given by $\frac{1}{6} \pi$, therefore, the area of each cap is given by the difference

$$
\frac{\pi}{6}-\frac{\sqrt{3}}{4}
$$

Thus, the area of the Reuleaux triangle is

$$
3\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)+\frac{\sqrt{3}}{4}=\frac{\pi}{2}-\frac{\sqrt{3}}{2} \mathrm{~cm}^{2}
$$