Question

In a circle of radius 10 cm, length of two parallel chords are 12 cm and 16 cm respectively. Find the distance between AB and CD of chords are 

(a) same side of center 

(b) on opposite sides of center.

Answer 1

Let O is center and r is radius of circle r = 10 cm chord AB = 12 cm and chord CD = 16 cm. 

Draw OP ⊥ AB which cuts chord CD at Q

Since AB || CD

Thus, OQ || CD

AP = BP

= \(\frac { 1 }{ 2 }\)AB

= \(\frac { 1 }{ 2 }\) × 12 = 6 cm

and CQ = QD = \(\frac { 1 }{ 2 }\) × CD = \(\frac { 1 }{ 2 }\) × 16 = 8 cm.

In right angled triangle OPA By Pythagoras theorem

OA² = AP² + OP²

(10)² = (6)² + (OP)²

100 = 36 + (OP)²

OP² = 100 – 36 = 64

OP = \(\sqrt { 64 }\) = 8 cm

Similarly on right angled triangle OCQ 

By Pythagoras theorem

OC ² = CQ² + OQ²

(10)² = (8)² + OQ²

100 = 64 + OQ²

OQ² = 100 – 64 = 36

OQ = \(\sqrt { 36 }\) = 6 cm

(a) Hence, distance between AB and CD

PQ = OP – OQ = 8 – 6 cm = 2 cm

(b) Hence, distance  between two chords AB and CD

PQ = OP + OQ = 8 + 6 cm PQ = 14 cm

Thus, Distance between two chords is 14 cm