The Reuleaux triangle is a disk formed from an equilateral triangle by adding circular arcs centered at the vertices of the triangle and with radii equal to the side of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-09.jpg?height=233&width=242&top_left_y=973&top_left_x=1470)

What is the area of a Reuleaux triangle if the equilateral triangle has a side length of $1 \mathrm{~cm}$?
The Reuleaux triangle is formed by 4 regions: an equilateral triangle and three segments. Each segment 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 the equilateral triangle is:

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

thus the area of the triangle is:

$$
\frac{1 \times \frac{\sqrt{3}}{2}}{2}=\frac{\sqrt{3}}{4} \mathrm{~cm}^{2}
$$

The area of a circular sector is one-sixth of the area of the circle, that is, equal to $\frac{\pi}{6}$. Therefore, the area of the segment is the difference:

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

Thus, the area of the Reuleaux triangle is

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