In a circle of radius 13 cm, a chord lies at a perpendicular distance of 12 cm from the centre. Using the Pythagorean theorem, what is the length of the chord in centimetres?

Introduction / Context:
This geometry question deals with chords in a circle and their distance from the centre. It primarily tests whether you can apply the Pythagorean theorem in the right triangle formed by the radius, the perpendicular from the centre to the chord and half the chord length.

Given Data / Assumptions:
- The circle has radius 13 cm.
- A chord is at a perpendicular distance of 12 cm from the centre.
- We must find the length of the chord in centimetres.

Concept / Approach:
When you drop a perpendicular from the centre of a circle to a chord, it bisects the chord. This gives a right triangle where the radius is the hypotenuse, the distance from the centre to the chord is one leg, and half the chord length is the other leg. The Pythagorean theorem then connects these three quantities.

Step-by-Step Solution:
Let the circle have centre O and the chord be AB. Drop a perpendicular from O to chord AB, meeting it at point M. Then OM is perpendicular to AB and OM = 12 cm. The radius OA is 13 cm, and triangle OMA is a right triangle at M. Let AM be half the length of the chord AB. Then AO^2 = OM^2 + AM^2 by the Pythagorean theorem. So 13^2 = 12^2 + AM^2. Compute: 169 = 144 + AM^2, hence AM^2 = 169 - 144 = 25. Therefore AM = 5 cm. Since the perpendicular from the centre bisects the chord, AB = 2 * AM = 2 * 5 = 10 cm.

Verification / Alternative check:
You can also think of a circle with radius 13 forming a 5 12 13 right triangle from the well known Pythagorean triplet. Recognising that 5, 12 and 13 form such a triangle immediately suggests that the half chord is 5 cm, leading again to a chord length of 10 cm.

Why Other Options Are Wrong:
A chord of length 5 cm would correspond to half chord 2.5 cm, which does not satisfy 13^2 = 12^2 + 2.5^2. Similarly, chord lengths 6, 8 or 12 cm produce half chords that do not fit the Pythagorean relation with legs 12 and radius 13. Only 10 cm is consistent with the given distances.

Common Pitfalls:
Students sometimes mistakenly treat the distance from the centre to the chord as the chord length itself or forget that the perpendicular bisects the chord. Others misapply Pythagoras by writing 12^2 = 13^2 + AM^2, which is incorrect. Always identify the hypotenuse correctly as the radius in this setup.

Final Answer:
The length of the chord is 10 cm.