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4. In a circle of radius 5 cm, AB and CD are two parallel chords of lengths8 cm and 6 cm respectively. Calculate the distance between the chordsif they are(i) on the same side of the centre,(ii) on the opposite sides of the centre. |
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Answer» GivenAB 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, AB^2= AP^2+ BP^2 ⇒ BP^2= 6^2- AP^2.............(1) In right triangle OBP, we haveOB^2= OP^2+ BP^2 ⇒ 5^2= (5 - AP)^2+ BP^2 ⇒ BP^2= 25 - (5 - AP)^2...........(2) Equating (1) and (2), we get 6^2- AP2= 2^5 - (5 - AP)^2 ⇒ 11 - AP^2= -25 - AP^2+ 10AP ⇒ 36 = 10APAP = 3.6 cm putting AP in (1), we get BP^2= 6^2- (3.6)^2= 23.04 ⇒ BP = 4.8 cm ⇒ BC = 2BP = 2× 4.8 = 9.6 cm ⇒ AP = 3.6 cm putting AP in (1), we get BP^2= 6^2- (3.6)^2= 23.04 ⇒ BP = 4.8 cm |
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