Two chords AB and CD of lengths 5cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6cm, find the radius of the circle.

                                                          
Let the radius of the circle be r cm. Let OM = x cm.
Then ON = (6 - x) cm.
Given,  OM ⊥ CD
∴ M is the mid-point of CD.      (∵ The perpendicular from the centre of a circle to a chord bisects the chord)
∴ MD = MC = CD =(11)cm= cm
Given,  ON ⊥ AB
∴ N is the mid-point of AB   (∵ The perpendicular from the centre of a circle to a chord bisects the chord) 
∴ NB = AN = AB = (5) = cm
In right triangle ONB,
OB² = ON² + NB²                   (By Pythagoras Theorem) 
⇒ r² = (6 – x)² +   …..(1)
In right triangle OMD, ( By Pythagoras Theorem)
⇒ r² = x² +       …..(2)
From (1) and (2), we get
(6 – x)² +    = x² +    
⇒ 36 – 12x + x² + = x² +
⇒ 12x = 36 + -
⇒ 12x = 12
⇒ x = = 1
Putting x = 1 in (2), we get
r² = (1)² +    
= 1 + =
⇒ r =
Hence the radius of the circle is  cm.