Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Draw OM perpendicular to AB and ON perpendicular to CD
Join OB and OC
BM = 
= 
(Perpendicular from centre bisects the chord)
ND = 
= 
Let,
ON be r so OM will be (6 – x)
In 
,
OM2 + MB2 = OB2
(6 – x)2 + (
)2 = OB2
36 + x2 – 12x + 
= OB2 (i)
In 
ON2 + ND2 = OD2
x2 + (
)2 = OD2
x2 + 
= OD2 (ii)
We have,
OB = OD (Radii of same circle)
So from (i) and (ii), we get
36 + x2 + 2x + 
= x2 + 
12x = 36 + 
- 
= 
= 
= 12
From (ii), we get
(1)2 + (
) = OD2
OD2 = 1 + 
= 
OD = 
So, radius of circle is found to be 
cm
15