In a circle of radius 5 cm,AB and AC are two chords such that AB = AC = 6 cm.Find the length of chord BC.

Given AB and AC are two equal words of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

⇒ OA is the bisector of ∠BAC.

Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.

∴ P divides BC in the ratio = 6 : 6 = 1 : 1.

⇒ P is mid-point of BC.

⇒ OP ⊥ BC.

In ΔABP, by pythagoras theorem,

In right triangle OBP, we have

Equating (1) and (2), we get

Putting AP in (1), we get

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