In a circle with centre O, the radius is 5 centimetres and the length of chord AB is 8 centimetres. Find the perpendicular distance of chord AB from the centre O in centimetres.

Introduction / Context:
This circle geometry problem involves finding the perpendicular distance from the centre of a circle to a chord. It uses the relationship between the radius, the chord length and the perpendicular distance. Such questions frequently appear in aptitude and school geometry exams to test your understanding of right triangles inside a circle.

Given Data / Assumptions:

• A circle has centre O.
• The radius of the circle is 5 cm.
• AB is a chord of the circle.
• The length of chord AB is 8 cm.
• We must find the distance from O to chord AB, measured along the perpendicular from O to AB.
• Standard Euclidean geometry is assumed.

Concept / Approach:
The perpendicular drawn from the centre of a circle to a chord bisects the chord. This means the chord is cut into two equal segments. If the full chord length is 8 cm, each half is 4 cm. The radius from the centre to each endpoint of the chord is 5 cm. These three segments form a right triangle with the radius as hypotenuse, half chord as one leg and the perpendicular distance as the other leg. Applying the Pythagoras theorem allows us to find the perpendicular distance from O to AB.

Step-by-Step Solution:
Step 1: Let the perpendicular from O to AB meet AB at point M. Step 2: Since OM is perpendicular to AB, M is the midpoint of AB. Step 3: Half of the chord AB is AM = AB / 2 = 8 / 2 = 4 cm. Step 4: OA is a radius, so OA = 5 cm. Step 5: In right triangle OAM, OA is the hypotenuse, AM is one leg and OM is the other leg. Step 6: Apply Pythagoras theorem: OA^2 = OM^2 + AM^2. Step 7: Substitute values: 5^2 = OM^2 + 4^2, so 25 = OM^2 + 16. Step 8: Rearrange: OM^2 = 25 - 16 = 9. Step 9: Take square root: OM = 3 cm.

Verification / Alternative check:
We can confirm by reversing the calculation: if OM = 3 cm and AM = 4 cm, then OA^2 should equal 3^2 + 4^2 = 9 + 16 = 25, so OA = 5 cm, matching the given radius. Also, doubling AM gives AB = 8 cm, the given chord length. This consistency among radius, chord and distance validates the result.

Why Other Options Are Wrong:
2 cm and 4 cm: These values do not satisfy the relation OA^2 = OM^2 + AM^2 when OA = 5 cm and AM = 4 cm.
5 cm: This would mean the chord collapses to a point (distance equal to radius), which is not possible for an 8 cm chord.
15 cm: This is larger than the radius and cannot be the distance from the centre to a chord lying inside the circle.

Common Pitfalls:
Students sometimes forget that the perpendicular from the centre bisects the chord and incorrectly use the full chord length instead of half. Others misidentify which segment is the hypotenuse in the right triangle, leading to wrong Pythagoras calculations. It is also easy to make sign errors when rearranging equations. Carefully drawing a diagram and labelling all segments helps avoid these issues.

Final Answer:
The perpendicular distance of chord AB from the centre O is 3 cm.