Question:

Step 1: Find the inradius of the equilateral triangle. For an equilateral triangle with side $a$: The area is given by $\text{Area} = \frac{\sqrt{3}}{4}a^2$. The semi-perimeter is $s = \frac{3a}{2}$. The inradius $r$ is calculated using the formula $r = \frac{\text{Area}}{s}$: $$r = \frac{\frac{\sqrt{3}}{4}a^2}{\frac{3a}{2}} = \frac{\sqrt{3}a^2}{4} \cdot \frac{2}{3a} = \frac{\sqrt{3}a}{6}$$ Step 2: Relate the inscribed square to the circle's diameter. The circle has radius $r = \frac{\sqrt{3}a}{6}$. The diameter of the circle is $d = 2r = 2 \cdot \frac{\sqrt{3}a}{6} = \frac{\sqrt{3}a}{3}$. For a square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. If $x$ represents the side of the square, its diagonal is $x\sqrt{2}$. Step 3: Calculate the side of the inscribed square. Equating the diagonal of the square to the diameter of the circle: $$x\sqrt{2} = \frac{\sqrt{3}a}{3}$$ Solving for $x$: $$x = \frac{\sqrt{3}a}{3\sqrt{2}} = \frac{\sqrt{3}a}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}a}{6}$$ Step 4: Find the area of the square. The area of the square is $x^2$: $$\text{Area} = x^2 = \left(\frac{\sqrt{6}a}{6}\right)^2 = \frac{6a^2}{36} = \frac{a^2}{6}$$