The lateral side of an isosceles trapezoid circumscribed about a circle is equal to a, and one of the angles is 60 °.

The lateral side of an isosceles trapezoid circumscribed about a circle is equal to a, and one of the angles is 60 °. Find the area of the trapezoid.

Let the trapezoid ABCD be given, where AD and BC are the bases, AB = CD = a, angle A = 60. Let’s draw the Height of ВK, then from the right-angled triangle AВK we find this height:
ВK = AB * sinA = a * √3 / 2. The area of the trapezoid is equal to the half-sum of the bases multiplied by the height, i.e.:
S = (AD + BC) / 2 * ВK.
By the property of the described quadrangle, the sums of the opposite sides are equal, which means:
BP + BC = AB + СD = 2 * a
S = 2 * a / 2 * a * √3 / 2 = (a ^ 2) * (√3 / 2).