Two circles of radii 13 cm and 12 cm are concentric. Calculate the length of the chord of the outer circle which touches the inner.

Given:

Radius of the outer circle (R) = 13 cm

Radius of the inner circle (r) = 12 cm

The circles are concentric.

A chord of the outer circle touches 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 concentric circles have center O.

Let the chord of the outer circle be AB, which touches the inner circle at point M.

Since OM is the radius to the point of tangency, OM 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 = 13 cm (hypotenuse).

OM is the radius of the inner circle, so OM = r = 12 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

13² = 12² + AM²

169 = 144 + AM²

AM² = 169 - 144

AM² = 25

AM = \(\sqrt{25}\) = ± 5 cm

The length of the chord AB = 2 × AM

Length of the chord = 2 × 5 = 10 cm

The length of the chord is 10 cm.