In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5…

Concept: The Intersecting Chords Theorem states that when two chords of a circle intersect at an interior point, the product of the lengths of the two segments of one chord equals the product of the two segments of the other chord — if chords AB and CD meet at E inside the circle, then AE × EB = CE × ED.

1. CD is a diameter of the circle, so CD = 2 × radius = 2 × 11 = 22 cm.

2. E lies on CD, so ED = CD − CE = 22 − 7 = 15 cm.

3. By the Intersecting Chords Theorem, AE × EB = CE × ED = 7 × 15 = 105.

4. AB is a chord of length 20.5 cm, and E lies on AB, so AE + EB = 20.5.

5. AE and EB are the two roots of t2 − 20.5t + 105 = 0.

6. Discriminant = 20.52 − 4 × 105 = 420.25 − 420 = 0.25, so the square root of 0.25 is 0.5.

7. t = (20.5 ± 0.5) / 2, giving t = 10.5 or t = 10 — so {AE, EB} = {10, 10.5} cm.

8. The difference of the two lengths is 10.5 − 10 = 0.5 cm.

Cross-check: 10 × 10.5 = 105, which equals CE × ED = 7 × 15, and 10 + 10.5 = 20.5, matching AB — confirming the values satisfy both conditions of the theorem.