`A B\ a n d\ C D` are two chords of a circle such that `A B=6\ c m ,\ C D=12\ c m\ a n d\ A B C Ddot` If the distance between `A B\ a n d\ C D` is `3\ c m ,` find the radius of the circle.

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
\and\ 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`
`6x=18=>x=18/6=3cm`
Putting value of `x` in equation in (i),`(OC)^2=(3)^2+36=9+36=45`
`OC=
sqrt45`
`OC=3 sqrt5 cm.`
Thus, radius of circle is `3sqrt 5cm`