Distance of a chord from the centre · 2 marks

The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.

Answer: Radius = 13 cm and half-chord = 12 cm, so the distance = √(13² − 12²) = √25 = 5 cm.

Step-by-step solution

Given: Diameter = 26 cm, so radius OB = 13 cm; Chord AB = 24 cm
To find: Distance OM of the chord from the centre

Idea: Halve both: the diameter gives the radius, and the perpendicular from the centre bisects the chord. Then use the Baudhāyana–Pythagoras theorem.

Radius = 26 ÷ 2 = 13 cm. Let OM ⟂ AB; then M is the midpoint of AB, so MB = 24 ÷ 2 = 12 cm.½ mark
In right ΔOMB: OM² = OB² − MB² = 169 − 144 = 25.1 mark
OM = 5 cm.½ mark

The chord is 5 cm from the centre.

Check: 5² + 12² = 13² ✓.

Answer to write in the exam

Radius OB = 26 ÷ 2 = 13 cm

OM ⟂ AB ⇒ MB = ½AB = 12 cm (perpendicular from the centre bisects the chord)

OM² = OB² − MB² = 169 − 144 = 25 (Baudhāyana–Pythagoras theorem)

∴ Distance OM = 5 cm

Common mistakes that cost marks

Using 26 cm as the radius: √(26² − 12²). Halve the diameter first.
Using the whole chord (24) in the right triangle instead of 12.
Writing the distance as 13 − 12 = 1 cm (subtracting lengths instead of squares).

How this can come in the exam

MCQ (1 mark)

A chord of length 30 cm is drawn in a circle of diameter 34 cm. Its distance from the centre is

4 cm
8 cm
15 cm
17 cm

Show answer

(B) 8 cm
Radius 17 cm, half-chord 15 cm: √(289 − 225) = √64 = 8 cm.

Try one yourself

A circle has diameter 50 cm. How far from the centre is a chord of length 48 cm?