I've been tutoring a high school kid in geometry recently, and discovered 
that these days, schools are really stressing special right triangles, using
ratio and proportion between the "standard" special right triangles, and the
right triangles they are given.  So I will use that approach entirely.  



The figure is composed of 
three 45-45-90 congruent right triangles, BAE, BFE, and BFC.
and
two 30-60-90 congruent right triangles, DFE and DFC

Each 45-45-90 right triangle has hypotenuse 12.

We set up a ratio between the "standard" 45-45-90 right triangle and
the given 45-45-90 right triangle, letting the leg be x:








So the area of each 45-45-90 right triangle is 



Each 30-60-90 right triangle has shorter leg 

We set up a ratio between the standard 30-60-90 right triangle and
your 30-60-90 right triangles, letting the longer leg be x:









Since the figure is composed of three 45-45-90 congruent right triangles, and
two 30-60-90 congruent right triangles,

The area of the composite figure is 

or 

Edwin