in a circle of radius 5cm AB and AC are the two chords. AB=AC=6cm .fin

1.	in a circle of radius 5cm AB and AC are the two chords. AB=AC=6cm .find the length of the chord BC
Answer» 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

in a circle of radius 5cm AB and AC are the two chords. AB=AC=6cm .find the length of the chord BC

Answer»

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