I THINK THIS IS THE SPECIFIC DIAGRAM NEEDED

Let r be the radius of given circle and its centre be O. Draw OM

perpendicular AB and ON perpendicular 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 = 1/2AB = 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 =} OM

^{2 }+ AM

^{2 }and OC

^{2 }= ON

^{2} + CN

^{2}
r

^{2 }= x

^{2} + (5/2)

^{2} ...(i)

r

^{2} = (6-x)

^{2 }+ (11/2)

^{2} ....(ii)

From (i) and (ii),we have

x

^{2 }+ (5/2)

^{2 }= (6-x)

^{2} + (11/2)

^{2}
x

^{2} + 25/4 = 36 + x

^{2} - 12x + 121/4

⇒ 4x

^{2} + 25 = 144 + 4x

^{2 }- 48x + 121

⇒ 48x = 240

⇒ x = 240/48 ⇒ x = 5

Putting the value of x in euation (i), we get

r

^{2} = 5

^{2} + (5/2)

^{2} ⇒ r

^{2} = 25 + 25/4

⇒ r

^{2 }= 125/4 ⇒ r = 5√5/2 cm