Here's the answer, but maybe not in a form you were looking for :

Let the above circle have its center at O, and let its radius = 1/2 m.
Then, AC is the diameter of the above circle = 1 m. Now.... let D be a point on the circle such that OD is perpendicular to AC and locate B at any point on the circle except at D.
Then let triangle ABC be inscribed in a circle such that its base is a diameter of 1.....and since angle ABC inscribes an arc of 180 degrees......it's measure is 1/2 of this arc = 90 degrees and AC forms the hypotenuse of this right triangle = 1 m
So.....triangle ABC is a right triangle.......and it's clear that, if we move point B to D, the area of this right triangle ADC will be greater than any other right triangle ABC inscribed in the circle, because even though their bases will be the same, the height of ADC will always be greater than the height of any other right triangle ABC.
So the base of ADC = 1m ..... and its height will be the radius of the circle = 1/2m....so the maximum area of such a triangle will be (1/2)bh = (1/2)(1m)(1/2 m) = 1/4 m^2......
