In a circle of radius 5 cm if AB and AC are two equal chords of length 6 cm each, then the length of chord BC is

Put the centre O at the origin and A at (0, 5), since A lies on the circle. Both B and C are 6 cm from A and also lie on the circle. Their distance from the centre comes from the perpendicular-from-centre theorem: a chord of length 6 in a circle of radius 5 sits at distance √(25 − 9) = 4 from the centre, so the chords are symmetric about line OA.

Solve directly. For B = (x, y): x² + y² = 25 and x² + (y − 5)² = 36. Subtracting gives 10y = 14, so y = 7/5. Then x² = 25 − 49/25 = 576/25, so x = ±24/5. Hence B and C are (24/5, 7/5) and (−24/5, 7/5), and BC = 48/5 cm = 9.6 cm.

None of A, B or C matches, so the answer is D. The core concept is the perpendicular bisector of a chord passing through the centre, which is also the basis of the chord-length formula used in Mains geometry and mensuration problems.

Trap: Aspirants compute 24/5, which is only the half-chord (the distance from the foot of the perpendicular to B), and pick option A. BC is the full chord.