A trapezoid is inscribed in a circle of radius 29, the base of which is 40 and 42, and the center of the circle

A trapezoid is inscribed in a circle of radius 29, the base of which is 40 and 42, and the center of the circle lies outside the trapezoid. find the height of the trapezoid.

From the center of the O circle, draw a perpendicular to the base of the BC of the trapezoid. OK belongs to the diameter of the circle, then OK we divide the chords BC and AD in half. CK = BC / 2 = 40/2 = 20 cm, DH = AD / 2 = 42/2 = 21 cm.

Let’s draw the radii ОD and ОC and from the right-angled triangles DOH and CОК we define the legs ОН and ОК.

OH ^ 2 = OD ^ 2 – DH ^ 2 = 841 – 441 = 400.

OH = 20 cm.

OK ^ 2 = OC ^ 2 – CK ^ 2 = 841 – 400 = 441.

OK = 21cm.

Then KH = OK – OH = 21 – 20 = 1 cm.

Answer: The height of the trapezoid is 1 cm.