The radius of a circle is 12.5 cm and the length of one its chords is 11 cm. What is the distance of the chord from the centre? (correct to one decimal place)

Correct — C. The perpendicular from the centre to a chord bisects it, so the 11 cm chord splits into two halves of 5.5 cm.

That perpendicular, the half-chord and the radius form a right triangle.

d² = r² − (half-chord)²
= 12.5² − 5.5²
= 156.25 − 30.25 = 126

d = √126 ≈ 11.2249
To one decimal place the distance is 11.2 cm → option (c).

Why the others are wrong

• (a)
  10 cm — 10 cm would need a half-chord of √(156.25 − 100) = 7.5 cm, that is a chord of 15 cm. The chord given is 11 cm.
• (b)
  12 cm — 12 cm belongs to a chord of 7 cm: √(156.25 − 144) = 3.5, doubled. A shorter chord sits further from the centre, not this one.
• (d)
  13.2 cm — 13.2 cm is larger than the radius of 12.5 cm. No chord can lie further from the centre than the radius, so this option is impossible before any arithmetic.

A chord, the radius to one of its ends and the perpendicular from the centre make a right triangle. That single picture links the three quantities.

The relation is r² = d² + (c⁄2)², where c is the chord length and d the distance from the centre.

Give any two and the third follows. Do not skip the halving step: the perpendicular meets the chord at its midpoint, so 5.5 goes into the triangle, never 11.

Key facts

• A perpendicular dropped from the centre of a circle to a chord bisects that chord.
• For radius r, chord c and centre-distance d: r² = d² + (c⁄2)².
• The distance from the centre to a chord is always less than the radius, and is 0 only for a diameter.
• √126 ≈ 11.2249, which rounds to 11.2 at one decimal place.

Common traps

• Putting the full chord 11 cm into the Pythagorean step instead of the half-chord 5.5 cm
• Choosing a distance greater than the radius without noticing
• Rounding √126 up to 11.3 by mis-squaring 11.2

No directly related past PYQ was found.