Relating Triangle Side to Circle Radius

The problem involves an equilateral triangle with side length $x$, a circle inscribed within it, and a square inscribed within that circle. First, we find the radius ($r$) of the circle inscribed in the equilateral triangle. The formula for the inradius of an equilateral triangle with side $x$ is:

$r = \frac{x}{2\sqrt{3}}$

Connecting Circle Diameter to Square Diagonal

The square is inscribed inside the circle. This means the diagonal ($d$) of the square is equal to the diameter ($D$) of the circle. The diameter is twice the radius:

$D = 2r$

Therefore, the diagonal of the square is:

$d = 2r = 2 \times \frac{x}{2\sqrt{3}} = \frac{x}{\sqrt{3}}$

Calculating the Square's Side Length

Let the side length of the square be $s$. The relationship between a square's diagonal ($d$) and its side length ($s$) is $d = s\sqrt{2}$. We can now find $s$:

$s\sqrt{2} = d = \frac{x}{\sqrt{3}}$

Solving for $s$:

$s = \frac{x}{\sqrt{3} \times \sqrt{2}} = \frac{x}{\sqrt{6}}$

Determining the Area of the Square

The area ($A$) of a square is given by $A = s^2$. Substituting the value of $s$ we found:

$A = \left( \frac{x}{\sqrt{6}} \right)^2 = \frac{x^2}{(\sqrt{6})^2} = \frac{x^2}{6}$

Thus, the area of the square is $\frac{1}{6}x^{2}$.