A chord with a length of 48 cm is perpendicular to the diameter and divides it into segments

A chord with a length of 48 cm is perpendicular to the diameter and divides it into segments at a ratio of 9: 16. Find the radius of the circle.

Since the chord is perpendicular to the diameter of the circle, the chord CD, at the intersection point K, is divided in half. Then CK = DK = CD / 2 = 48/2 = 24 cm.

Let the length of the segment AK = 9 * X cm, then, by condition, the length of the segment BK = 16 * X cm.

By the property of intersecting chords, AK * BK = DK * CK.

24 * 24 = 9 * X * 16 * X.

576 = 144 * X ^ 2.

X^2 = 576/144 = 4.

X = 2.

Then AK = 2 * 9 = 18 cm, BK = 2 * 16 = 32 cm.

AB = D = AK + BK = 18 + 32 = 50 cm.

Then AO = R = AB / 2 = 50/2 = 25 cm.

Answer: The radius of the circle is 25 cm.