Find the area of an isosceles trapezoid if the bases have lengths 12 and 17 and the base angles measure 60 degrees?
So, if you position the trapezoid with the long base on the bottom, you have to draw an altitude from one of the top vertices down. This creates a 30-60-90 triangle. The side of that triangle that is part of the long base equals half the difference between the bases, or (17-12)/2 = 2.5.
2.5 is the short leg of the triangle, and the altitude drawn is the long leg. The altitude is
h = 2.5√3
So, the area of a trapezoid is (1/2)(b1+b2)h