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 configuration always creates a right triangle with the larger radius as the hypotenuse (\(R\)), the smaller radius as one side (\(r\)), and half the chord length as the other side (\(L/2\)).
Formula: \(L = 2\sqrt{R^2 - r^2}\).

Answer 1

Step 1: Understanding the Concept:
A chord of the larger circle that touches the smaller circle is a tangent to the smaller circle. The radius of the smaller circle to the point of contact is perpendicular to this chord.
Step 2: Key Formula or Approach:
Pythagorean theorem in the right triangle formed by the radius of the small circle, the radius of the large circle, and half of the chord.
Step 3: Detailed Explanation:
Let \(O\) be the common center of the two circles.
Let \(AB\) be the chord of the larger circle which touches the smaller circle at point \(P\).
Since \(P\) is the point of contact, \(OP \perp AB\) and \(OP\) is the radius of the smaller circle (\(OP = 4\) cm).
Join \(OA\). \(OA\) is the radius of the larger circle (\(OA = 5\) cm).
In right-angled triangle \(OPA\), by Pythagoras theorem:
\[ OA^2 = OP^2 + AP^2 \]
\[ 5^2 = 4^2 + AP^2 \]
\[ 25 = 16 + AP^2 \]
\[ AP^2 = 25 - 16 = 9 \implies AP = 3 \text{ cm} \]
Since the perpendicular from the center to a chord bisects the chord:
\[ AB = 2 \times AP = 2 \times 3 = 6 \text{ cm} \]
Step 4: Final Answer:
The length of the chord is 6 cm.