The ratio of the areas of a square and a regular hexagon, both inscribed in a circle is?

let the radius of the circle = r

Then the diameter of the circle = the diagonal of the square = 2r

Let a side of the square = x

Then $$2x^2 = (2r)^2 = 4r^2$$

The area of the square(A) $$A= x^2 = (2r)^2$$

The hexagon can be partitioned into 6 equilateral triangles whose sides = r

If the sides of an equilateral triangle = r then its area(B) $$B= r^2 \sqrt \frac{3}{4}$$

So the area of the hexagon $$(B)=6\times r^2 \sqrt \frac{3}{4}=3\times r^2 \sqrt {3}$$

Thus the ratio of the area of the square to that of the hexagon =

$$A:B=\frac{(2r)^2}{3\times r^2 \sqrt {3}}$$

$$A:B=\frac{(4)}{3 \sqrt {3}}$$