#### Answer

$33 \sqrt3$

#### Work Step by Step

We first find the height of the trapezoid. It is a leg of the 30-60-90 triangle containing the side of length 6, so we obtain:
$h= 6 \times \frac{\sqrt3}{2} = 3\sqrt3$
We also find the lengths of AE and FD, which are equal:
$AE=FD = 6 \times \frac{1}{2} = 3$
Thus:
$A = 1/2(b_1+b_2)(h) = 1/2(8+14)(3\sqrt3)=33\sqrt3$