Exposition

Let $EG=KM=r$ & $HF=NI= x$ ;

Since, $AD=EF=6$ ( given ), therefore $OE=OF= 3$ ,

now, $OG = OE - EG = (3 - r)$ & $OH = OF - HF = (3 - x)$ ,

by Pythagoras theorem ;

$(GM)^2 = (OM)^2 - (OG)^2$

or, $(GM)^2 = (3 + r)^2 - (3 - r)^2$

or, $(GM) = \sqrt{12r}$ ----------------- eq.(1) similarly,

$(HN)^2 = (ON)^2 - (OH)^2$

or, $(HN)= \sqrt{12x}$ ------------------eq.(2)

now, considering triangle $MNJ$ ;

$MJ = GH = GO + OH = (6 - r - x)$ ;

& $JN = HN - GM = (\sqrt{12x} - \sqrt{12r})$ & $MN= (r + x)$ ; by Pythagoras theorem;

$(MN)^2 = (MJ)^2 + (JN)^2$ ;

or, $(r + x)^2 = (6 - r -x)^2 + (\sqrt{12x} - \sqrt{12r})^2$

or, $\sqrt{xr} = 3/2$

or, $xr = 9/4$ ----------- eq.(3)

now, since $DC= (2\sqrt{6} + 5)$ given..

$DC = (DF + FI + IC) = (3 + \sqrt{12x)} + x) = (2\sqrt{6} + 5)$

or, $\sqrt{12x} + x = 2\sqrt{6} + 5 - 3$

or, $\sqrt{12x} + x = 2\sqrt{6} + 2$

by comparing we get;

$x = 2$

now putting the value of x in eq.(3) we get

$2r = 9/4 =\boxed{ 2.25}$ (ans.)

Mod: ${\LaTeX}$ 'd