Half of the angles must look like arcsin(3/(2x))

Asking Wolfram for the radius x that solves 

arcsin(3/(2x))+arcsin(11/(2x))+2arcsin(15/(2x))=pi/2

gives 

x=5sqrt(17/2) [This agrees with Charlie.  Also confirmed graphically]

x^2=425/2

The apothem, h, to each side, s, is then of the form 

h^2=x^2-(s/2)^2

s=3, h=29/2

s=11, h=27/2

s=15, h=25/2

Finally sum 1/2*s*h*2 (to go all the way around)

3*29/2+11*27/2+2*15*25/2 = 567

Note to Charlie's solution: The circle will have area pi*425/2 = 667.6

and this octagon is roughly the same as in inscribed regular hexagon which would have an area of about 552.1. The answer should be in this ballpark.

Note on my solution:  I attempted to solve for x analytically using the sum of arcsines formula, but the algebra got to be too much.