In the trapezoid, the bases AD and BC are 36 and 12, respectively, and the sum of the angles at the base is 90 degrees. Find the radius of the circle passing through points A and B and touching the straight line CD if AB = 10.
Extend the lateral sides of the trapezoid to their intersection at point K.
Since, by condition, the sum of the acute angles of the trapezoid at the base AD is 90, the AKD angle is also 90.
Rectangular triangles AKD and BKC are similar in acute angle.
Let the length of the segment BK = X cm, then AK = AK + BK = 10 + X cm.
In similar triangles: AD / BC = AK / BK.
36/12 = (10 + X) / X.
36 * X = 120 + 12 * X.
24 * X = 120.
X = BK = 120/24 = 5 cm.
The OAB triangle is isosceles, since OA and OB are the radii of the circle, then OM is its height and median, which means AM = BM = AM / 2 = 10/2 = 5 cm.
Then MK = BM + BK = 5 + 5 = 10 cm.
Quadrangle ONKM is a rectangle, then OH = MK = R = 10 cm.
Answer: The radius of the circle is 10 cm.
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