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Question
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 center. 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`
`ND=(CD)/2=5/2`
Let ON be x , so OM will be 6-x in Δ MOB
`OM^2+MB^2=OB^2`
`(6-x)^2+(5/2)^2=OB^2`
`36+x^2-12x+25/4=OB^2`
In Δ NOD
`ON^2+ND^2=OD^2`
`x^2+(11/2)^2=OD^2`
`x^2+121/4=OD^2`
We have OB = OD. (radii of same circle)
So, from equation (1) and (2).
`36+x^2-12x+25/4=x^2+121/4`
`⇒12x=36+25/4-121/34`
`=(144+25-121)/4=48/4=12`
`x=1`
From equation (2)
`(1)^2+(121/4)=OD^2`
`OD^2=1+121/4=121/4`
`OD=(5sqrt5)/2`
So, radius of circle is found to be `(5sqrt5)/2cm`
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