The center of a circle of radius R circumscribed about a trapezoid lies on one of the bases.

The center of a circle of radius R circumscribed about a trapezoid lies on one of the bases. Find the perimeter of a trapezoid if one of its corners is 60.

Connect the vertices B and C of the trapezoid with the center of the circle.

The trapezoid is divided into three triangles.

Consider a triangle ABO, in which the angle A = 600 by condition, the side OA = OB as the radii of the circle, then the triangle ABC is equilateral, since it is isosceles with an angle of 600.

OA = AB = BO = R.

Consider a triangle BOC, in which ОВ = OC = R, the angle ОВС = BOA = 60, as the angles lying crosswise at the intersection of parallel lines AD and BC of the secant ОВ, then the triangle BOC is equilateral, ОВ = OC = BC = R.

Consider a triangle DОС, in which ОD = OC = R, the angle COD = BCO = 60, as the angles lying crosswise at the intersection of parallel lines AD and BC of the secant OC, then the triangle DOC is equilateral, OD = OC = CD = R.

Then the perimeter of the trapezoid is: P = AB + BC + CD + AO + DO = 5 * R.

Answer: The perimeter of the trapezoid is 5 * R.