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.



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