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
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).
- Need to find length of chord \(AB\).
Step 1: Understand the problem setup
- Since \(C\) lies on the smaller circle, and \(AB\) is a chord of the larger circle touching the smaller circle at \(C\), \(C\) is the midpoint of chord \(AB\).
- \(OC\) is perpendicular from center \(O\) to chord \(AB\), so \(OC \perp AB\).
Step 2: Use right triangle \(OCA\)
- \(OA = 3.5\, \text{cm}\) is the radius of larger circle.
- \(OC = 2.1\, \text{cm}\) is the perpendicular distance from center \(O\) to chord \(AB\).
- We need to find half the chord length \(AC\).
- By Pythagoras theorem:
\[
OA^2 = OC^2 + AC^2
\]
\[
AC^2 = OA^2 - OC^2 = (3.5)^2 - (2.1)^2 = 12.25 - 4.41 = 7.84
\]
\[
AC = \sqrt{7.84} = 2.8\, \text{cm}
\]
Step 3: Calculate full chord length \(AB\)
\[
AB = 2 \times AC = 2 \times 2.8 = 5.6\, \text{cm}
\]
