Question

10. AB and CD are two parallel chords of lengths 5 cm 11 cm respectively of a circle. If
distance between the chords be 3 cm, then find the radius of the circle.
​

Answer 1

Let there is a circle having center O and let radius is r .

Draw OP perpendicular to AB and OQ perpendicular to CD.

Now since  OP perpendicular to AB and OQ perpendicular to CD and AB || CD

So P, O,Q are collinear.

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 to 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 equation 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

So radius of the circle is 5.6 cm