Connect BD and divide the quadrilateral into two right triangles. In triangle BCD, we get (1) BC 2+CD 2 = BD 2 from Pythagorean theorem. In the triangle fault, there is also AB 2+AD 2 = BD 2, because AB=AD, so

(2)bd^2=2ab^2； Then there is (3) BC 2+CD 2 = 2ab 2;

And because (BC+CD) 2 = 8 2 = 64 = BC 2+2bc * CD+CD 2, combined with (3), AB 2+BC * CD = 32.

Quadrilateral area s =1/2ab2+1/2bc * cd =1/2 * 32.

= 16