Question

Please solve this question..

Answer 1

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,

AB2 = AP2 + BP2

⇒ BP2 = 62 - AP2 .............(1)

In right triangle OBP, we have

OB2 = OP2 + BP2

⇒ 52 = (5 - AP)2 + BP2

⇒ BP2 = 25 - (5 - AP)2 ...........(2)

Equating (1) and (2), we get

62 - AP2 = 25 - (5 - AP)2

⇒ 11 - AP2 = -25 - AP2 + 10AP

⇒ 36 = 10AP

⇒ AP = 3.6 cm

putting AP in (1), we get

BP2 = 62 - (3.6)2 = 23.04

⇒ BP = 4.8 cm

⇒ BC = 2BP = 2 × 4.8 = 9.6 cm


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