Two chords, AB and CD, are equal in length and are at a distance of 10 cm from the center of a circle. If the radius is 26 cm, what is the length of AB?

Correct — D. The perpendicular from the centre bisects the chord. That makes a right triangle: the 26 cm radius as hypotenuse, the 10 cm distance and half of AB as the legs.

Pythagoras: (AB⁄2)² = 26² − 10²
= 676 − 100 = 576
Square root: AB⁄2 = √576 = 24 cm
Double it: AB = 2 × 24 = 48 cm

AB = 48 cm → option (d)

Why the others are wrong

• (a)
  12 cm — 12 cm would make the half-chord 6 cm, and 6² + 10² = 136, far from 26² = 676. It is a quarter of the true chord.
• (b)
  24 cm — 24 cm is the half-chord, the leg of the right triangle. The perpendicular from the centre cuts AB into two equal parts, so AB = 2 × 24 = 48 cm.
• (c)
  36 cm — 36 cm would make the half-chord 18 cm, and 18² + 10² = 424, not 26² = 676. The legs must fit the 26 cm radius exactly.

The perpendicular from a circle's centre to a chord bisects the chord. Radius, distance and half-chord then form a right triangle: r² = d² + (c⁄2)².

Know any two of radius, distance and chord, and the third follows.

Equal chords sit at equal distances from the centre, and the reverse holds too. That is why CD changes nothing: it is also 48 cm long.

Key facts

• The perpendicular from the centre to a chord bisects the chord.
• Chord length = 2√(r² − d²), where d is the chord's distance from the centre.
• Equal chords of a circle are equidistant from the centre, and chords equidistant from the centre are equal.

Common traps

• Stopping at the half-chord, 24 cm, and marking it as AB.
• Adding the squares instead of subtracting: √(26² + 10²) ≈ 27.9 cm is longer than the radius, which no half-chord can be.

No directly related past PYQ was found.