# Chords AB and CD intersect at point E so that AE = 3 CM BE = 36 CE: DE = 3: 4.

**Chords AB and CD intersect at point E so that AE = 3 CM BE = 36 CE: DE = 3: 4. Find CD and the smallest value for the radius of that circle.**

By the property of intersecting chords, the product of the segments formed at the intersection of one chord is equal to the intersection of the segments of the other chord.

CE * DE = AE * BE.

Let the length of the segment CE = 3 * X cm, then DE = 4 * X cm.

3 * X * 4 * X = 3 * 36.

12 * X ^ 2 = 108.

X ^ 2 = 108/12 = 9.

X = 3.

CE = 3 * 3 = 9 cm, DE = 4 * 3 = 12 cm.

Chord length AB = 3 + 36 = 39 cm.

Chord length СD = 9 + 12 = 21 cm.

Then the smallest diameter of the circle is equal to half the length of the larger chord.

Rmin = AB / 2 = 39/2 = 19.5 cm.

Answer: The length of the CD is 21 cm, the smallest radius is 19.5 cm.