Question 1210156
<pre>
An envelope is pictured below. Solve for the area of the regions
marked XXX.

{{{drawing(400,2000/7,-1,13,-1,9,

locate(2,3,XXX), locate(9,3,XXX),

locate(0,8.7,A),
locate(12,8.7,B),
locate(12,0,C),
locate(6,0,E),
locate(0,0,D),


line(0,0,12,0),
line(12,0,12,8),
line(12,8,0,8),
line(0,8,0,0),
line(0,8,6,0),
line(6,0,12,8),
locate(5,8.7,matrix(1,2,12,in)),
locate(7.5,4.5,matrix(1,2,10,in)),
locate(3,4.5,matrix(1,2,10,in))
   )}}}

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):

{{{AD = sqrt(AE^2-DE^2) =sqrt(10^2-6^2) = sqrt(100-36) = sqrt(64) = 8}}} 

So tha area of right triangle ADE is {{{expr(1/2)*base*height}}}{{{""=""}}}
{{{expr(1/2)*DE*AD}}}{{{""=""}}}{{{expr(1/2)*6*8)}}}{{{""=""}}}{{{24}}}in<sup>2</sup>.

Triangle BCE is congruent to triangle ADE so it also has area {{{24}}}in<sup>2</sup>. 

Total shaded area (or area marked XXX) = (2)(24) = 48 in<sup>2</sup>.

Edwin</pre>