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