A circle is inscribed in an equilateral triangle and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is

It is clear from the question that the centers of all three figures coincide. 

Let x be the length side of the equilateral triangle, hence the area (A) is ($$\sqrt{\ 3}$$/4)$$x^2$$

From this, the radius of the inscribed circle is= ($$\sqrt{\ 3}$$/2)x * (1/3) = x/2$$\sqrt{\ 3}$$

So, the diameter is x/$$\sqrt{\ 3}$$

Now, the diameter of the circle coincides with the diagonal of the square.

Hence, assuming the sides of square to be y, we have y$$\sqrt{\ 2}$$= x/$$\sqrt{\ 3}$$

y= x/$$\sqrt{\ 6}$$

From this, the area of square (B) is $$y^2$$, which is $$x^2$$/6

Now, the ratio of A and B is = $$3\sqrt{\ 3}$$/2