OA: C

\(O_s\) be center of small circle, \(O_b\) be center of big circle.
\(O_sD\) = Radius of small circle=\(r\) , \(O_bB\)=Radius of big circle(R)=\(9\)cm

\(\angle DAO_s= \angle BAO_b =\frac{60^{\circ}}{2}=30^{\circ}\)

that means \(\triangle DAO_s\) and \(\triangle BAO_b\) are \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle.
their sides will be ratio of \(1:\sqrt{3}:2\)

In \(\triangle BAO_b\)
\(AO_b=2O_bB=2*9=18\)
\(AO_b =O_bO_s +O_sA\)
\(O_sA =AO_b-O_bO_s=18-(9+r) =9-r\) [Using \(O_bO_s =R+r =9+r\)]

In \(\triangle DAO_s\)
\(O_sA=2O_sD\)
\(9-r = 2r\)
\(r=\frac{9}{3}=3\)
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________