If a chord of a circle of radius 11 cm is a tangent to another circle of radius 7 cm, both the circles being concentric, then the length of the chord is:

Given:

Radius of the outer circle (R) = 11 cm

Radius of the inner circle (r) = 7 cm

The circles are concentric.

A chord of the outer circle is a tangent to the inner circle.

Formula Used:

The radius to the point of tangency is perpendicular to the tangent.

Pythagorean theorem: In a right-angled triangle, (hypotenuse)² = (base)² + (perpendicular)²

The perpendicular from the center of a circle to a chord bisects the chord.

Calculation:

Let the outer circle have center O.

Let the chord of the outer circle be AB, which is tangent to the inner circle at point M.

Since OM is the radius to the point of tangency, OP is perpendicular to the chord AB. Thus, ∠OMA = 90°.

Consider the right-angled triangle OMA.

OA is the radius of the outer circle, so OA = R = 11 cm (hypotenuse).

OM is the radius of the inner circle, so OM = r = 7 cm (perpendicular).

AM is half the length of the chord AB (since the perpendicular from the center bisects the chord).

Using the Pythagorean theorem in triangle OMA:

OA² = OM² + AM²

11² = 7² + AM²

121 = 49 + AM²

AM² = 121 - 49

AM² = 72

AM = \(√{72} = √{36 \times 2} = 6√{2}\) cm

The length of the chord AB = 2 x AM

Length of the chord = 2 x 6√2 = 12√2 cm

The length of the chord is 12√2 cm.