Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution
Draw OM ⊥ AB and ON ⊥ CD. Join OB and OD.
BM = AB/2 = 5/2 (Perpendicular from the centre bisects the chord)
ND = CD/2 = 11/2
Let ON be x. Therefore, OM will be 6− x.
In ΔMOB,
OM2 + MB2 = OB2
(6 - x)2 + (5/2)2 = OB2
36 + x2 - 12x + 25/4 = OB2 ........(1)
In ΔNOD,
ON2 + ND2 = OD2
x2 + (11/2)2 = OD2
x2 + 121/4 = OD2 .........(2)
We have OB = OD (Radii of the same circle)
Therefore, from equation (1) and (2),
`36+x^2-12x+25/4=x^2+121/4`
`12x=36+24/4-121/4`
`=(144+25-121)/4`
`=48/4`
= 12
x = 1
From equation (2),
`(1)^2+(121/4)=OD^2`
`OD^2 = 1+121/4=125/4`
`OD=5/2sqrt5`
Therefore, the radius of the circle is `5/2sqrt5" cm."`
