<PRE><B>Question 1210279</B>
&lt;br&gt;
EB is the diagonal of a square; its length is 12.  Since AEB is an isosceles right triangle, the side length of the square is {{{12/sqrt(2)=6*sqrt(2)}}}.&lt;br&gt;
AEB, FEB, and FBC are all congruent.  Using the standard formula one-half base times height for the area of a triangle, the area of each of those triangles is &lt;br&gt;
{{{(1/2)(6sqrt(2))^2=(1/2)(72)=36}}}&lt;br&gt;
Note some students will find it easier to view each of those three triangles as one-quarter of a square with side length 12, making the area of each one&lt;br&gt;
{{{(1/4)(12^2)=144/4=36}}}&lt;br&gt;
Either way, the area of ABCE is 3*36 = 108.&lt;br&gt;
Angle EDF is 30 degrees, so triangle EDF is a 30-60-90 right triangle with a short leg of length {{{6*sqrt(2)}}}.  Using the properties of a 30-60-90 right triangle, the length of DF is&lt;br&gt;
{{{(6sqrt(2))(sqrt(3))=6sqrt(6)}}}&lt;br&gt;
DF is the altitude of triangle CDE.&lt;br&gt;
Then the area of triangle CDE -- again using the standard formula one-half base time height -- is&lt;br&gt;
{{{(6sqrt(2))(6sqrt(6))=36sqrt(12)=72sqrt(3)}}}&lt;br&gt;
And so the area of the composite figure is&lt;br&gt;
ANSWER: {{{108+72sqrt(3)}}}&lt;br&gt;
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