Question 253197
First, draw an isosceles trapezoid with the question information included.
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Second, draw an altitude from the top vertex on the left to the bottom, long base, and then do the same on the right top vertex. We have created 2 right (30-60-90) triangles. The height becomes 5sqrt(3) and the long base will become 20; remember that the side opposite the 30 degree angle is 1/2 the hypotenuse of 10 = 5.
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Third find the area as
{{{A =(1/2)*(10+20)*5*sqrt(3)}}}
{{{A = 75sqrt(3)}}}