Circular arcs in an equilateral triangle

We can tessellate this equilateral triangle to completely fill up the Cartesian plane. We can see that the area of the 3 petals is identical to 6 times the area of a segment of a circle, with radius $r$ , and a central angle $\theta = \frac\pi3$ .

The value of $r$ can be solved using cosine rule , $r^2 + r^2 - 2r^2 \left (1 - \cos\left( \frac {2\pi}3\right)\right) = 200^2 \quad \Rightarrow \quad r^2 = \frac{40000}3 .$ Now, we use the formula of the area of a sector of a circle, $\frac12 r^2 \left(\theta - \sin \theta\right) = \frac{10000}9 \left(2\pi - 3\sqrt3 \right) .$ Multiply this value by 6 and we have our answer, $40000 \left( \frac\pi3 - \frac{\sqrt3}2 \right) \approx \boxed{7246.885}$ .