`A B\ a n d\ C D` are two parallel chords of a circle with centre `O` such that `A B=6\ c m\ a n d\ C D=12\ c mdot` The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is `6\ c m`

Explanation

Let `O` is center and `r` is radius of circle, whose chords `AB=6cm,CD=12cm.`
Draw `OM_|_AB` which cuts `CD \at\ N,AB∣∣CD`
`ON_|_CD`

`OM_|_AB,AM=MB`(Perpendicular drawn from center of circle bisects the chord.)
`AM=MB= 1/2AB =1/2xx6=3cm.`
`ON_|_CD
CN=ND=1/2xxCD=1/2xx12=6cm.`
Distance between chords `AB \and\ CDMN=3cm.`
Let `ON=x cm.`
In right angled `/_\OCN,` by Pythagoras theorem
`OC^2=ON^2+CN^2`
`(OC)^2=(x)^2+(6)^2`
`(OC)^2=x^2+36` …..(i)
Now in right angled `/_\OAM` By Pythagoras

`OA^2=AM^2+OM^2`
`OA
^2=AM^2+(ON+MN)^2`
`OA^2=AM^2+(x+3)^2`
`OA^2=(3)^2+(x+3)^2`
`OA^2=9+x^2+9+6x` …..(ii)
`OA=OC`
`OA^2=OC
^2`
`x^2+36=9+x^2+9+6x` [from equation (i) and (ii)]
`36=6x+18`
`6x=36–18=18`
`x= 18\6=3cm.`
Putting value of x in equation (i)`OC= sqrt45`
`OC=3sqrt5cm.`
Thus, radius of circle is `3
sqrt5cm`