In an isosceles trapezoid, the bases are 5 cm and the angle at the base is 17 cm equal to 45 degrees. Find the area.

The area of the trapezoid is equal to the product of half the sum of the bases and the height. S = (BC + AD) / 2 * BM

We need to find the height of BM. Consider the triangle ABM. It is rectangular and isosceles, because angle M = 90 degrees, angle A = angle B = 45. So AM = BM.

Because the trapezoid is isosceles, then AM = ND, and BM = (AD – BC) / 2.

BM = (17 – 5) / 2 = 12/2 = 6 (cm)

S = (5 + 17) / 2 * 6 = 22/2 * 6 = 11 * 6 = 66 (cm ^ 2)

Answer. 66 cm ^ 2.