Question

Two concentric circles are of radii 5 cm and 4 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Hint:

The radius drawn to the point of contact of a tangent always bisects the chord in this specific geometric configuration.

Answer 1

Step 1: Understanding the Concept:
Concentric circles share the same centre. A chord of the larger circle that touches the smaller circle acts as a tangent to the smaller circle and is bisected by the radius at the point of contact.
Step 2: Key Formula or Approach:
1. Radius of small circle (\(r = 4\) cm) is perpendicular to the chord.
2. Radius of large circle (\(R = 5\) cm) forms the hypotenuse of a right-angled triangle.
3. Pythagoras' Theorem: \((\text{Half-chord})^2 = R^2 - r^2\)
Step 3: Detailed Explanation:
1. Let \(O\) be the centre, \(AB\) be the chord, and \(M\) be the point of contact. 2. In \(\triangle OAM\), \(\angle OMA = 90^\circ\), \(OA = 5\) cm, and \(OM = 4\) cm. 3. Apply Pythagoras' Theorem: \[ AM^2 = OA^2 - OM^2 \] \[ AM^2 = 5^2 - 4^2 = 25 - 16 = 9 \] \[ AM = \sqrt{9} = 3 \text{ cm} \] 4. Total length of chord \(AB = 2 \times AM\): \[ AB = 2 \times 3 = 6 \text{ cm} \]
Step 4: Final Answer:
The length of the chord is 6 cm.