Area of $\triangle BQE$

In the above parallelogram, we observe that $\triangle DQC$ $\sim$ $\triangle BQE$ by AA similarity condition.

Thus, $\frac{Ar \triangle BQE}{Ar \triangle DQC}$ = $(\frac{BE}{DC})^{2}$

As $3BE = 2DC$ and area of $\triangle DQC$ = 36 $units^{2}$ :

$\frac{Ar \triangle BQE}{36}$ = $(\frac{2}{3})^{2}$

$\implies$ Area of $\triangle BQE$ = 36 x $\frac{4}{9}$ = 16