In the circle in which the trapezoid is inscribed with bases of 30 and 40 cm, a diameter is drawn

In the circle in which the trapezoid is inscribed with bases of 30 and 40 cm, a diameter is drawn that is perpendicular to the bases of the trapezoid. The part of this diameter located outside the trapezium between the large base and the circumference is 4/5 of its length. Find the height of the trapezoid.

Let the trapezoid ABCD, point O – the center of the circle, K – the middle of the lower base, L – the middle of the upper base.
The part of the segment from the circle to the diameter is equal to (4/5) D = (8/5) r = (1 + 3/5) r, r is the radius of the circle.
That is, the center of the circle is outside the trapezoid, and the bottom base is 3/5 from the center of the circle.
Consider a triangle AKO – rectangular, AK = AD / 2 = 20 cm, AO = r, OK = (3/5) r.
By the Pythagorean theorem:
(AO) ^ 2 = (AK) ^ 2 + (OK) ^ 2;
(r) ^ 2 = (20) ^ 2 + ((3/5) r) ^ 2;
((16/25) r ^ 2 = (20) ^ 2;
(4/5) r = 20;
r = 25 cm.
Consider now the triangle OBL – rectangular, BO – the radius of the circle, BL = BC / 2 = 15 cm, LO = h + KO = h + (3 * 25/5) = h + 15, h – the desired height.
Then, by the Pythagorean theorem:
(BO) ^ 2 = (BL) ^ 2 + (LO) ^ 2;
(r) ^ 2 = (15) ^ 2 + (h + 15) ^ 2;
(25) ^ 2 = (15) ^ 2 + (h + 15) ^ 2;
(h + 15) ^ 2 = 400;
h + 15 = 20;
h = 5 cm.
The height of the trapezoid is 5 cm.