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.
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
CN = DN = 1/2CD
= 1/2 x 11
In right triangles OAM and OCN, we have,
OA2 = 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
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