Q4 of 100 Page 16

Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.

Construction: Draw OP perpendicular to CD

Chord AB = 5 cm


Chord CD = 11 cm


Distance PQ = 3 cm


Let,


OP = x cm


OC = OA = r cm


We know that,


The perpendicular distance from centre to chord bisects the chord


Therefore,


CP = PD = cm


And,


AQ = BQ = cm


In by using Pythagoras theorem


OC2 = OP2 + CP2


r2 = x2 + ()2 (i)


In by using Pythagoras theorem


OA2 = OQ2 + AQ2


r2 = (x + 3)2 + ()2 (ii)


Compare (i) and (ii), we get


(x + 3)2 + ()2 = x2 + ()2


x2 + 9 + 6x + = x2 +


x2 + 6x – x2 = - – 9


6x = 15


x =


=



More from this chapter

All 100 →
2

Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.

 3

Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.

 5

Give a method to find the centre of a given circle.

 6

Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.