Let $\mathcal K$ be the kite formed by joining two right triangles with legs $2$ and $2\sqrt3$ along a common hypotenuse. Six copies of $\mathcal K$ are used to form the polygon shown below, in which the two sides of length $2\sqrt3$ of every kite meet at the point $O$. What is the area of triangle $\Delta ABC$?

1. The kite has hypotenuse $\sqrt{2^2+(2\sqrt3)^2}=4$. At the vertex where the two sides of length $2\sqrt3$ meet, each right triangle contributes an angle of $30^\circ$ (the leg $2$ is opposite it), so the kite has angle $60^\circ$ there.
2. Six such angles fill the full angle around $O$, so the hypotenuses of the kites lie along six rays from $O$ that are $60^\circ$ apart, and the far end of each hypotenuse is at distance $4$ from $O$. These six points form a regular hexagon of side $4$ (the polygon), and $A$, $B$, $C$ are three of its vertices with $\angle AOB=60^\circ$ and $A$, $O$, $C$ collinear.
3. Since $OA=OB=4$ and $\angle AOB=60^\circ$, triangle $AOB$ is equilateral and $AB=4$. Also $AC=8$ is a diameter of the circle through the vertices, so $\angle ABC=90^\circ$ and $BC=\sqrt{8^2-4^2}=4\sqrt3$.
4. $[ABC]=\dfrac12\cdot AB\cdot BC=\dfrac12\cdot4\cdot4\sqrt3=\boxed{8\sqrt3}$.