The area of the shaded segment is:
A=(1/2) r^2 [theta-sin(theta)]
(this is a standard result). Now we want this to be 1/3 of the area
of the circle, or:
so we have:
(1/2) r^2 [theta-sin(theta)] = (1/3) pi*r^2,
or after a bit of simplification:
theta - sin(theta) - (2/3)*pi = 0.
Now this is a mixed algebraic-transcendental equation so really
has to be solved numerically. Doing so gives:
theta ~= 0.264932.
Finally we can see from the diagram that:
cos(theta/2) = h/r,
h = r*cos(theta/2) ~= 0.264932 r.
Finally we may observe that your B = 2h, and A = (2r-B)/2, and as
your r= 30 mm, we have:
h ~= 7.94796 mm,
B ~= 15.8959 mm
A ~= 22.052 mm.