Explanation

This uses the property that the radius to a point of tangency is perpendicular to the tangent. The chord MN (24 cm) touches the inner circle, so the segment from the centre to the point of tangency is perpendicular to MN and bisects it into two 12 cm halves. Joining the centre to M forms a right triangle with hypotenuse equal to the outer radius (26/2 = 13 cm) and one leg 12 cm. The inner radius is the other leg: √(13² − 12²) = √(169 − 144) = √25 = 5 cm. Hence option (a). The distractors 6, 8, 10 cm result from using the diameter instead of the radius or from forgetting to halve the chord.