An envelope is pictured below. Solve for the area of the regions
marked XXX.



DE is half of 12 or 6. So we find AD with the Pythagorean theorem.
(Or you may recognize it as a 6-8-10 right triangle, a 3-4-5 right triangle
with all sides doubled):

 

So tha area of right triangle ADE is 
in2.

Triangle BCE is congruent to triangle ADE so it also has area in2. 

Total shaded area (or area marked XXX) = (2)(24) = 48 in2.

Edwin