Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. Distance between AB and CD is 6 cm. 

To Prove: Radius of the circle, OP = ? 

Construction: Join OP and OQ, OB and OD. 

Proof: Chord AB || Chord CD. 

AB = 5 cm, and CD =11 cm. OP⊥AB 

∴ BP = AP = \(\frac{5}{2}\) = 2.5 cm. 

OQ⊥CD 

∴ CQ = QD = \(\frac{11}{2}\) = 5.5 cm. 

PQ = 6 cm. (Data) 

Let OQ = 2 cm then, OP = (6 – x) cm. 

In ∆BPO, ∠P = 90° 

As per Pythagoras theorem, 

OB2 = BP² + PO² = (2.5)² + (6 – x)² 

= 6.25 + 36 – 12x + x² 

OB² = x²– 12x + 42.25 …………….. (i) 

In ∆OQD, ∠Q= 90° 

∴ OD² = OQ² + QD² = (x)² + (5.5)² 

OD² = x² + 30.25 ……………….. (ii) 

OB = OD (∵ radii of same circle) 

From (i) and (ii). 

x² – 12x + 42.25 

= x² + 30.25 -12x 

= 30.25 – 42.25 -12x 

= -12 

12x = 12 

∴ x = \(\frac{12}{12}\)

∴ x = 1 cm. 

From (ii), OD² = x² + 30.25 

= (1)2 + 30.25 

= 1 + 30.25 

∴ OD² = 31.25 

OD = √(31.25)

∴ OD = 5.59 cm. 

∴ Radius of circle OP = OD = 5.59 cm.