Since \triangle ABD is a right triangle,

BD = \sqrt{3^2+4^2} =5

\because 13^2 = 12^2+5^2 =169

The measure of three sides of \triangle BCD has the following relationship

BC^2 = BD^2+CD^2

Therefore, \triangle BCD is also a right triangle.

So the area of quadrilateral ABCD is,

A_{ABCD} =A_{\triangle ABD }+A_{\triangle BCD }

=\dfrac{1}{2}\cdotp 3\cdotp 4+\dfrac{1}{2}\cdotp 5\cdotp 12

=36