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.

Answer 1

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`