Answer
36 + 36$\sqrt 3$ $units^{2}$
Work Step by Step
We know in kite diagonals are perpendicular to each other
$\angle$ BOA = $\angle$ BOC = 90$^{\circ}$
As we know one diagonal is perpendicular bisector of other
OD=OB = 6
Lets take $\triangle$ BOA
It is a 45$^{\circ}$ - 45 $^{\circ}$ - 90$^{\circ}$ triangle
therefore OA = 6 units
$\triangle$ BOC
It is a 30$^{\circ}$ - 60 $^{\circ}$ - 90$^{\circ}$ triangle
therefore OC = 6 $\sqrt 3$units
The area of Kite ABCD = $\frac{1}{2}$* d1*d2
= $\frac{1}{2}$* BD* AC
=$\frac{1}{2}$* 12* (OA+OC)
=6*(6+6$\sqrt 3$)
= 36 + 36$\sqrt 3$ $units^{2}$
