Two chords AB,

Let r be the radius of the given circle and its center be O. Draw OM ⊥ AB and ON⊥ CD.

Since, OM perpendicular AB, ON perpendicular CD.

and AB||CD

Therefore, points M, O and N are collinear.

So, MN = 6cm

Let, OM = x cm.

Then,ON = (6 - x)cm.

Join OA and OC.

Then OA = OC = r

As the perpendicular from the centre to a chord of the circle bisects the chord.

∴   AM = BM = 1/2 AB

= 1/2 x 5

= 2.5cm

CN = DN = 1/2CD

= 1/2 x 11

= 5.5cm
In right triangles OAM and OCN, we have,

 OA^{2 =} OM²+ AM²and OC²= ON² + CN²







From (i) and (ii), we have




  

⇒   4x² + 25 = 144 + 4x²- 48x + 121
⇒    48x = 240
⇒  x  = 240/48

⇒ x = 5
Putting the value of x in euation (i), we get
     r² = 5² + (5/2)² 
⇒ r² = 25 + 25/4
 ⇒  r²= 125/4
⇒    r = 5√5/2 cm