Question 1172767
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Triangles ABD and ACD are 30-60-90 right triangles, with long leg AD having length 12.  The side length of triangle ABC is then {{{12*(2/sqrt(3)) = 24/sqrt(3) = 8*sqrt(3)}}}.<br>
AD is half that side length, or {{{4*sqrt(3)}}}; ED is half again as much, or {{{2*sqrt(3)}}}.<br>
So the base of triangle AEC has length {{{6*sqrt(3)}}}.<br>
Then the area is one-half base time height:<br>
{{{(1/2)(6*sqrt(3))(12) = 36*sqrt(3)}}}<br>