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

Chords AB and CD intersect at one point E so that AE = 3cm, BE = 36cm, CE: DE = 3; 4. Find the value of the chord CD and the smallest value of the radius of this circle.

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

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

CE * DE = AE * CE.

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

12 * X2 = 108.

X2 = 108/12 = 9.

X = 3 cm.

DE = 3 * 3 = 9 cm.

CE = 4 * 3 = 12 cm.

Chord СD = 9 + 12 = 21 cm.

The circle will have the smallest radius if its diameter matches the larger chord.

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

Answer: The chord of the СD is 21 cm, Rmin = 19.5 cm.

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