Let the radius of the circle be R. Area = π R²
The maximum square ABCD that can be inscribed in the circle is when the diameter of the circle is the diagonal (AC or BD) of the square.
Hence, AC = BD = √2 AB = 2 R => AB = √2 R.
Area of square = 2 R²
Regarding the triangle inscribed inside the square, there are possibilities. Let us examine the triangles right angle ΔDCB, isosceles Δ DCE and the isosceles Δ AFG.
Area of ΔDCB = 1/2 DC*BC = 1/2 AB² = 2 R²/2 = R²
Area of ΔDCE = 1/2 DC * AD = 1/2 AB² = R²
Area of Δ AFG:
We see that it is half of the product of base FG and altitude from A on to FG. Let us move the point F from B towards C and at the same time move G from D towards C. If the altitude increases by an amount "h", then the base decreases by "2h" due to the linearity of the geometry.
Hence, the area of Δ AFG, will be less than the area of ABD.
Maximum area of a triangle inscribed in the square = R²
Hence the ratios = π R² : 2 R² : R² = π : 2 : 1.