We know that each angle in the hexagon is \(120 ^ \circ\)
By symmetry, we know that \(\overline {AC} = \overline{CE} = \overline { AE}\), meaning \(\triangle ACE\)(red in diagram) is equilateral.
Now, let the center of the triangle be G, and draw lines (blue) connecting it to Points A, E, and C.
This divides it into 3 isosceles triangles, each with 1 \({120 ^ \circ}\) and 2 \({30 ^ \circ}\) angles.
Note that these triangles are congruent to \(\triangle AFE, \triangle CDE, \) and \(\triangle ABC\).
This means that \(\triangle ACE\) contains 3 out of the 6 triangles, meaning its area is \(0.5 \times 36 = \color{brown}\boxed{18}\)