Is there a better and faster way to solve this? It seems pretty hard to me. Here is I how I solve it:


Call XA and YA the diagonals of the parallelogram A, and XB and YB the diagonals of the parallelogram B. We have

(1) XA+YA=XB+YB.
(2) XA=YA+20.
(3) XB=YB+8.

Since the parallelograms are equilateral, the diagonals intersect with a 90º angle. Therefore, the area of the parallelograms are given by:
A: $XA*YA/2=YA*(YA+20)/2=(YA^2+20YA)/2$
B: $XB*YB/2=YB(YB+8)/2=(YB^2+8YB)/2$
B-A : $(YB^2+8YB-(YA^2+20YA))/2$

Substitute (2) and (3) in (1). This gives you: 2YA+20=2YB+8.
Simplify: YA+10=YB+4. Square both sides:
$YA^2+20YA+100=YB^2+8YB+16$
This leads to $YB^2+8YB-(YA^2+20YA)=84$. Dividing by two we have the difference between the two areas:

Answer is D