Question

In two concentric circles centred at O, a chord AB of the larger circle touches the smaller circle at C. If OA = 3.5 cm, OC = 2.1 cm, then AB is equal to

• A. 5.6 cm
• B. 2.8 cm
• C. 3.5 cm
• D. 4.2 cm

Answer 1

The Correct Option is A

Given:
- Two concentric circles centered at \(O\).
- Chord \(AB\) of the larger circle touches the smaller circle at \(C\).
- \(OA = 3.5\, \text{cm}\) (radius of larger circle).
- \(OC = 2.1\, \text{cm}\) (radius of smaller circle).
- Find the length of chord \(AB\).

Solution:
- \(C\) is the midpoint of chord \(AB\).
- \(OC \perp AB\).

Applying Pythagoras Theorem to triangle \(OCA\)
- \(OA = 3.5\, \text{cm}\).
- \(OC = 2.1\, \text{cm}\).
- Find \(AC\).
- \(OA^2 = OC^2 + AC^2\)
- \(AC^2 = OA^2 - OC^2 = (3.5)^2 - (2.1)^2 = 12.25 - 4.41 = 7.84\)
- \(AC = \sqrt{7.84} = 2.8\, \text{cm}\)

Calculating \(AB\)
- \(AB = 2 \times AC = 2 \times 2.8 = 5.6\, \text{cm}\)

Answer:
\[\boxed{5.6\, \text{cm}}\]