A chord of length 32 cm is drawn at a distance of 12 cm from the centre of a circle. Find the radius of the circle.

Correct — D. The perpendicular dropped from the centre to a chord bisects that chord, so it forms a right triangle whose legs are half the chord and the given distance, with the radius as hypotenuse.

Half-chord = 32⁄2 = 16 cm
Distance from centre = 12 cm
r² = 16² + 12² = 256 + 144 = 400
r = √400 = 20 cm → option (d).

It is the 3-4-5 triple scaled by 4: 12, 16, 20.

Why the others are wrong

• (a)
  18 cm — A radius of 18 cm with the centre 12 cm away cuts a chord of 2√(18² − 12²) = 2√180 ≈ 26.8 cm. The paper's chord is 32 cm, so 18 cm is too small.
• (b)
  22 cm — 22 cm overshoots the other way: it would cut a chord of 2√(22² − 12²) = 2√340 ≈ 36.9 cm at that distance, longer than the 32 cm given.
• (c)
  28 cm — 28 = 16 + 12 — the two legs added instead of squared. Pythagoras needs 16² + 12² = 400, and √400 is 20, not 28.

One theorem carries the whole question: the perpendicular from the centre of a circle to a chord bisects the chord.

That gives a right triangle with legs d (centre-to-chord distance) and c⁄2 (half the chord), and hypotenuse r. So r² = d² + (c⁄2)².

The same relation answers the question from any two of the three quantities — find the radius, the chord, or the distance, depending on what is withheld.

Key facts

• The perpendicular from the centre to a chord bisects the chord.
• r² = d² + (c⁄2)², where d is the distance from centre to chord and c is the chord length.
• Here 16² + 12² = 400, so the radius is 20 cm.
• 12, 16, 20 is the 3-4-5 right-triangle triple multiplied by 4.

Common traps

• Using the full 32 cm as a leg instead of halving it first
• Adding 16 and 12 rather than applying Pythagoras
• Treating the 12 cm as a radius or a diameter rather than a perpendicular distance

No directly related past PYQ was found.