A chord of length 16 cm is drawn in a circle of radius 10 cm. What is the perpendicular distance (in centimetres) of this chord from the centre of the circle?

Introduction / Context:
This problem examines the basic geometry of a circle involving chords and the perpendicular distance from the centre of the circle to a chord. When a perpendicular is drawn from the centre of a circle to a chord, it always bisects the chord. Using this fact along with the Pythagorean theorem allows us to compute the distance from the centre to the chord when the radius and the chord length are known.

Given Data / Assumptions:
- Radius of the circle r = 10 cm.- Length of the chord = 16 cm.- A perpendicular is drawn from the centre of the circle to the chord.- The perpendicular from the centre to the chord bisects the chord.- The triangle formed is a right angled triangle, so the Pythagorean theorem can be used.

Concept / Approach:
For a circle, the perpendicular from the centre to a chord divides the chord into two equal segments. This creates a right angled triangle whose hypotenuse is the radius of the circle and whose one leg is half the chord length. The other leg is the perpendicular distance from the centre to the chord, which we need to find. By applying the Pythagorean theorem (hypotenuse^2 = base^2 + height^2), we can solve for this distance.

Step-by-Step Solution:
Step 1: Let O be the centre of the circle and AB be the chord of length 16 cm.Step 2: Draw a perpendicular from O to AB, meeting it at point M. Then OM is the distance from the centre to the chord.Step 3: Since OM is perpendicular from the centre to the chord, it bisects AB. Therefore, AM = MB = 16 / 2 = 8 cm.Step 4: Consider right triangle OMA, where OA is the radius of the circle.Step 5: OA = r = 10 cm, AM = 8 cm, and OM is the unknown distance.Step 6: Apply Pythagorean theorem: OA^2 = OM^2 + AM^2.Step 7: Substitute values: 10^2 = OM^2 + 8^2.Step 8: This gives 100 = OM^2 + 64.Step 9: Rearranging, OM^2 = 100 - 64 = 36.Step 10: Therefore, OM = √36 = 6 cm.

Verification / Alternative check:
We can verify by checking that the values satisfy the Pythagorean relation. Using OM = 6 cm and AM = 8 cm, compute OM^2 + AM^2 = 36 + 64 = 100, which equals OA^2 = 10^2. This confirms that the triangle is consistent and that the distance 6 cm is correct. Any other distance value would break this equality and therefore would not fit the geometry of the circle and chord as given.

Why Other Options Are Wrong:
- 8 cm: This would suggest that AM is 8 and OM is also 8, leading to OA^2 = 8^2 + 8^2 = 128, which does not equal 10^2.- 4 cm: Using 4 cm would give OA^2 = 4^2 + 8^2 = 16 + 64 = 80, which is not 100.- 12 cm: This would give OA^2 = 12^2 + 8^2 = 144 + 64 = 208, far larger than 100.- 5 cm: This gives OA^2 = 5^2 + 8^2 = 25 + 64 = 89, which again does not match the radius squared, 100.

Common Pitfalls:
Many students forget that the perpendicular from the centre to a chord bisects the chord, and they mistakenly use the full chord length in the Pythagorean theorem. Others may confuse the positions of the sides in the right triangle and incorrectly assign the radius to a smaller side instead of the hypotenuse. Carefully drawing the diagram and marking all given lengths helps avoid these errors and leads to the correct calculation of the perpendicular distance.

Final Answer:
The perpendicular distance of the chord from the centre of the circle is 6 cm.