The main problem is: how can I find AB? If I find it, I will know the height.
Joining $AC$, crossing $BD$ at $G$ (see @Benson Lin), then since $BG$ and $AF$ are median lines in triangle $ABC$ crossing at $E$, and $CK$ is drawn through $E$, therefore $CK$ is also a median line and $AK = KB = 5$. And since the angle of the parallelogram at $A = 30$ degrees, the perpendicular from $B$ to $AD = \frac12AB = 5$. Thus the area of $ABCD = 12*5 = 60$.