Simple enough.. again.

Extend AB to B' such that BB'=5 and join C to B'. It's evident that CB'=16. Moreover, ACB' is a rt triangle since AB'²=AC²+CB'² and thus area of tr. ACB'=12*16/2=96. We claim that area of tr. ACB' = Area of trapezium ABCD (Why?) and thus the required area =96 s.u.

We set up LoC equations in terms of 5,15,12,16,y,x, where x is the length of the left leg and y is the angle formed between the left leg and the segment of length 5. We eliminate the xy term to get x = sqrt97. Using basic trigonometry with Pythag Theorem gives the height as 48/5. The area is then 96.

Let's denote a=15,b=5,d1=12,d2=16 You should solve following system of equation: $a-b=x-y$ $d1^2-(b+x)^2=d2^2-(b+y)^2$