In an isosceles trapezoid with bases of 20 and 12, determine the length of the diagonal and the leg, given that the center of the circumscribed circle lies on the longer base of the trapezoid.
[Find the diagonal and the leg of an isosceles trapezoid with bases of 20 and 12, given that the center of the circumscribed circle lies on the longer base of the trapezoid.]
Solution
Since AD is the diameter of the circle (see the figure), it follows that OD = OC = 10\,\text{cm}.
Draw the perpendicular CL \perp AD. Then OL = 6\,\text{cm}, and from triangle \triangle CLO we get:
CL = \sqrt{OC^2 - OL^2} = \sqrt{100 - 36} = \sqrt{64} = 8\,\text{cm}
Now, from triangles \triangle ALC and \triangle CLD we find:
AC = \sqrt{CL^2 + AL^2} = \sqrt{64 + 256} = \sqrt{320} = 8\sqrt{5}\,\text{cm}
CD = \sqrt{CL^2 + LD^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}\,\text{cm}
Answer:
4\sqrt{5}\,\text{cm};\quad 8\sqrt{5}\,\text{cm}
