## Task 14/90

Given an equilateral arc triangle, whose arcs touch each other at the corners. What is its area $A$ if the arc radius $r=1$ LE?
If one draws a perpendicular from each corner of the arc triangle to the arc tangent, a regular triangle with side length $a=2 r=2$ units is formed. The area sought is then the difference between the area $A_{D}=r^{2} \sqrt{3}=\sqrt{3} \text{ units}^2$ of the triangle and the sum of the areas $A_{S}=\frac{1}{6} r^{2} \pi=\frac{\pi}{6} \text{ units}^2$ of three circular sectors with radius $r=1 \text{ unit}$ and central angle $\alpha=60^{\circ}$:

$$
A=A_{D}-3 A_{S}=\left(\sqrt{3}-3 \cdot \frac{\pi}{6}\right)=\left(\sqrt{3}-\frac{\pi}{2}\right) \quad \text{sq units} \approx 0.161 \quad \text{sq units}
$$