# Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle.

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Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle.

answered Sep 28, 2018 by (-2,499 points)

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
OA2 = OM2 + AM2 and OC2 = ON2 + CN2
r2 = x2 + (5/2)2  ...(i)
r2 = (6-x)2 + (11/2)2 ....(ii)
From (i) and (ii),we have
x2 + (5/2)2 = (6-x)2 + (11/2)2
x2 + 25/4 = 36 + x2 - 12x + 121/4
⇒   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