two chords ab and ac are drawn in circle. ab = ac = 6, then find bc
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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
⇒ 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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