Selection tests for the Balkan Mathematical Olympiad 2013

We extend sides $FA$ and $BC$ to intersect at $A^{\prime }$, and sides $BC$ and $DE$ to intersect at $B^{\prime }$, and sides $DE$ and $FA$ to intersect at $C^{\prime }$.



Triangles $A^{\prime }{B}^{\prime }{C}^{\prime }$, $A^{\prime }BA$, $B^{\prime }DC$, and $C^{\prime }FE$ are equilateral with side lengths $11$, $3$, $4$, and $5$ respectively. Therefore, the area of hexagon $ABCDEF$ is

$$\frac{\sqrt{3}}{4}\cdot 11^2-\frac{\sqrt{3}}{4}\cdot 3^2-\frac{\sqrt{3}}{4}\cdot 4^2-\frac{\sqrt{3}}{4}\cdot 5^2=\frac{71\sqrt{3}}{4}.$$