## Tessellated Arcs

The tessellated figure below consists of 3 equal arcs in an equilateral triangle. Each side measures 2 inches. What is the area of the region labeled $X$?
Source: Julia Robinson Mathematics Festival

SOLUTION

Draw altitude $\overline{AH}$ to side $\overline{BC}$. Because $\triangle {AHC}$ is a $30^\circ-60^\circ-90^\circ$ triangle with a hypotenuse of length 2 inches, $AH=1\sqrt 3$.

Area of $\triangle {ABC}=\frac{1}{2}\left (BC\right )\left (AH\right )$
$=\frac{1}{2}\left (2\right )\left (\sqrt 3\right )$
$=\sqrt 3$

Each arc lies on a circle of radius 1 inch centered at one of the three vertices of $\triangle {ABC}$. Each arc forms a sector with a central angle measuring $60^\circ$. The three sectors if placed contiguously next to each other would therefore form a semicircle.

Area of the three sectors = $\frac{180^\circ}{360^\circ}\pi \left (1^2\right )=\frac{\pi}{2}$

Area of region $X$= area of $\triangle{ABC}-$ area of the three sectors
$=\sqrt 3-\frac{\pi}{2}$

Answer: $\sqrt 3-\frac{\pi}{2}$ square inches.