A trapezoid is inscribed in a circle of radius 41, the bases of which are equal to 18 and 80, and the center of the circle

| July 1, 2021 | education

A trapezoid is inscribed in a circle of radius 41, the bases of which are equal to 18 and 80, and the center of the circle lies inside the trapezoid. Find the height of this trapezoid.

Let’s draw the height KH through the center O of the circle and the radii OA, OB, OC, OD.

Triangles BOC and AOD are isosceles, then their heights OK and OH are also the medians of the triangles.

Then BK = CK = BC / 2 = 18/2 = 9 cm, AH = DH = AD / = 80/2 = 40 cm.

From a right-angled triangle AOD, according to the Pythagorean theorem, OH ^ 2 = AO ^ 2 – AH ^ 2 = 41 ^ 2 – 40 ^ 2 = 1681 – 1600 = 81.

OH = 9 cm.

From the right-angled triangle BOC, according to the Pythagorean theorem, OK ^ 2 = BO ^ 2 – BK ^ 2 = 41 ^ 2 – 9 ^ 2 = 1681 – 81 = 1600.

OK = 40 cm.

Height KH = OH + OK = 9 + 40 = 49 cm.

Answer: The height of the trapezoid is 49 cm.



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