Horrible Hexagons 2

The shaded region is composed of two congruent non-overlapping equilateral triangles of side length $2\sec(30^{\circ}) = \dfrac{4}{\sqrt{3}}.$

Their combined areas, and hence the area of the shaded region, is then

$2 \times \dfrac{1}{2} \times \left( \dfrac{4}{\sqrt{3}} \right)^{2} \times \sin(60^{\circ}) = \dfrac{16}{3} \times \dfrac{\sqrt{3}}{2} = \dfrac{8\sqrt{3}}{3}.$

Thus $ab = 8 \times 3 = \boxed{24}.$

The area is a parallelogram with height $2$ and base $2 \times \frac{2}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$ . Therefore, the area is $\boxed{\frac{8\sqrt{3}}{3}}$ .

$\text{All small triangles have an area of}~\dfrac{\sqrt3}{3}\\\text{each of the two shaded big triangles have four such triangles. }\\\therefore~ shaded ~~areas \\= 8*\dfrac{\sqrt3}{3}.~~~~~~~~~~ab=\large. 24$