Solution :-
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