Question

in a circle of radius 5 ,AB and CD are two parallel chord's of lengths 8cm and 6cm respectively.calculate the distance between the chords if they lie on the opposite sides of the centre

Answer 1

Given distance between AB and CD is 6.

So PQ = 6

Again let OP = x, then OQ = (6-x)

Join OA and OC.

Then OA = OC = r.

Since we know that perpendicular from the center to a chord fo the circle bisects the chord

So AB =PB = 5/2 = 2.5

and CQ = QD = 11/3 = 5.5

From ΔOAP and ΔOCQ

     OA2 = OP2 + AP2

=> r2 = x2 + (2.5)2 .........1

and OC2 = OQ2 + CQ2

=> r2 = (6-x)2 + (5.5)2 ......2

from eqaution 1 and 2, we get

      x2 + (2.5)2 = (6-x)2 + (5.5)2

=> x2 + 6.25 = 36 + x2 - 12x + 30.25

=> 6.25 = -12x + 66.25

=> 12x = 66.25 - 6.25

=> 12x = 60

=> x = 60/12

=> x = 5

Put x = 5 in equation 1, we get

      r2 = 52 + (2.5)2

=> r2 = 25 + 6.25

=> r2 = 31.25

=> r = √31.25

=> r = 5.6 (approximatily)