In a circle of radius 5√13 cm, a chord is at a distance of 10 cm from the centre of the circle. Find the length (in cm) of the chord.

Correct — D. Drop a perpendicular from the centre to the chord. It bisects the chord, so you now have a right triangle.

legs: the distance from the centre = 10, and half the chord = h
hypotenuse: the radius = 5√13

Apply Pythagoras:
(5√13)² = 10² + h²
25 × 13 = 100 + h²
325 = 100 + h²

h² = 225
h = 15

h is only half the chord, so the chord = 2 × 15 = 30 cm → option (d).

Why the others are wrong

• (a)
  36 — 36 is essentially the diameter: 2 × 5√13 = 10√13 = 36.06. That is the chord drawn through the centre, i.e. one at distance 0, not at 10 cm.
• (b)
  15 — 15 is the half-chord — the h that Pythagoras returns. The question asks for the whole chord, so it must be doubled.
• (c)
  28 — 28 needs a half-chord of 14, which would mean 196 = r² − 100 and r² = 296. Here r² = (5√13)² = 325, so the difference is 225 and the half-chord is 15.

One theorem does all the work: the perpendicular from the centre to a chord bisects that chord. It turns every chord question into a right triangle.

The three parts of that triangle are the radius r, the centre-to-chord distance d, and half the chord c⁄2, related by r² = d² + (c⁄2)². Know any two and the third follows.

It also explains the behaviour students half-remember: as d grows the chord shrinks, and at d = 0 the chord becomes the diameter, the longest a circle has.

Key facts

• The perpendicular from the centre of a circle to a chord bisects that chord.
• For radius r, centre-to-chord distance d and chord c, r² = d² + (c⁄2)².
• (5√13)² = 25 × 13 = 325, and 325 − 100 = 225, whose square root is 15.

Common traps

• Reporting 15, the half-chord, as the final answer.
• Squaring 5√13 as 5 × 13 = 65, which makes h² negative and should stop you at once.
• Treating the 10 cm as a radius or a diameter rather than as the perpendicular distance.

No directly related past PYQ was found.