Question 1039827
Cut into two right triangles from point B to midpoint of AC.


Look at either of these right triangles.  BC is opposite a 90 degree angle, a base leg is opposite a 60 degree angle, and altitude of the isosceles is opposite the 30 degree angle.  SPECIAL RIGHT TRIANGLE.


Let 6 be the short leg of the special 30-60-90 right triangle.  
12 is the length of the hypotenuse.
Let y be altitude or the other leg.


{{{y^2+6^2=12^2}}}
{{{y^2=144-36}}}
{{{y^2=108}}}
{{{y^2=2*54=2*2*27=2*2*3*3*3}}}
{{{highlight_green(y=6sqrt(3))}}}


Put attention onto the original isosceles triangle.

You want the area.
{{{(1/2)12*6sqrt(3)}}}, one-half of base times height;
{{{highlight(36*sqrt(3))}}}