Find the area of an isosceles trapezoid if the measure of one angle is 135o and the lengths of the bases are 10 and 18.
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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
=(1/2)(12+17)(2.5√3)
=36.25*√3
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