I assume that AB=BC=CD=1, and DE=EF=FA=2AB=BC=CD=1,andDE=EF=FA=2
As shown in the diagram, 3(x+y)=360^@ => x+y=120^@3(x+y)=360∘⇒x+y=120∘ w=(180-x)/2, z=(180-y)/2w=180−x2,z=180−y2 => w+z=(360-(x+y))/2=(360-120)/2=120^@⇒w+z=360−(x+y)2=360−1202=120∘
In DeltaCDE, CE^2=CD^2+DE^2-2(CD)(DE)cos(w+z) CE^2=1^2+2^2-2*1*2*cos120=7
In DeltaOCE, CE^2=OC^2+OE^2-2(OC)(OE)cos(x+y) 7=r^2+r^2-2r^2cos120 => 7=2r^2(1-cos120)=2r^2(1-(-1/2))=3r^2 => r^2=7/3 => r=sqrt(7/3)