Regular Partitioning

We draw equilateral triangles as shown above. We can split up $WCXYFZ$ into two trapezoids. $BD=\sqrt{3}$ (since it is twice the altitude of an equilateral triangle with side length $1$ ).

Zoom into the top half of $ABCDEF$ :

The smaller equilateral triangle is similar to the larger ones. We are given that $AB$ , $WZ$ , $XY$ , and $ED$ are parallel and equidistant, so they divide $ABCDEF$ (and therefore $BD$ ) into thirds. So, the altitude of the larger equilateral triangles is $\dfrac{\sqrt{3}}{3}$ , and the altitude of the smaller equilateral triangle is $\dfrac{\sqrt{3}}{6}$ . From that, we know that the smaller equilateral triangle is similar to the larger ones with a similarity ratio of $1:2$ .

The base of the smaller equilateral triangle is $\dfrac{1}{3}$ (because of 30-60-90 triangle ratios). Since the similarity ratio between the smaller and larger equilateral triangles is $1:2$ , the larger ones each have bases of $\dfrac{2}{3}$ . Therefore, $ZW=\dfrac{5}{3}$ . Since we know the height ( $\dfrac{\sqrt{3}}{6}$ ) and two bases of trapezoid $ZWCF$ (since $FC=2$ ) we can compute its area to be $\dfrac{11\sqrt{3}}{36}$ .

Since $WCXYFZ$ is made up of two of these trapezoids, its area is $2\cdot\dfrac{11\sqrt{3}}{36}=\boxed{\dfrac{11\sqrt{3}}{18}}$ .