A circle with a radius of 20 cm. A chord of length 32 cm is drawn. What is the distance from the center of the circle to the chord?

Correct — B. The perpendicular from the centre bisects the chord, giving a right triangle with the radius as hypotenuse.

Half the chord: 32 ÷ 2 = 16 cm
Pythagoras: d² = 20² − 16² = 400 − 256 = 144
Root: d = √144 = 12 cm → option (b)

Why the others are wrong

• (a)
  16 cm — 16 cm is half the chord, one leg of the right triangle. The distance is the remaining leg: √(20² − 16²) = 12.
• (c)
  8 cm — 8 cm is 20 − 12, the gap from the chord out to the arc along the perpendicular. The question asks for the distance from the centre to the chord, which is 12.
• (d)
  4 cm — 4 cm is 20 − 16, lengths subtracted before squaring. Pythagoras subtracts the squares: 400 − 256 = 144, and √144 = 12.

The line from the centre perpendicular to a chord bisects it. Join the centre to one end of the chord and a right triangle appears:

hypotenuse = radius r
one leg = half the chord
remaining leg = distance d from the centre

So r² = d² + (half-chord)², and any two of radius, chord and distance give the third.

Key facts

• The perpendicular from the centre to a chord bisects the chord.
• r² = d² + (half-chord)², where d is the chord's distance from the centre.
• Equal chords are equidistant from the centre.
• The diameter, the longest chord, passes through the centre, so its distance is 0.

Common traps

• Putting the full chord into Pythagoras. 20² − 32² is negative, so the half-chord, 16, must go in.
• Answering 20 − 12 = 8, the gap from the chord to the arc, instead of the distance from the centre.

No directly related past PYQ was found.