A circle is inscribed in a square. An equilateral triangle of side $4\sqrt{3}$ cm is inscribed in that circle. The length of the diagonal of the square is

### Concept & Geometry Relationships The problem links three geometric figures: an equilateral triangle inscribed in a circle, and that same circle inscribed in a square. 1. The radius $R$ of a circle circumscribing an equilateral triangle of side $a$ is given by: $$R = \frac{a}{\sqrt{3}}$$ 2. For a circle inscribed in a square, the diameter of the circle is equal to the side of the square ($s = 2R$). 3. The diagonal of a square with side $s$ is $s\sqrt{2}$. ### Step-by-Step Solution 1. **Find the radius of the circle:** Given the side of the inscribed equilateral triangle $a = 4\sqrt{3}$ cm. Radius $R = \frac{4\sqrt{3}}{\sqrt{3}} = 4$ cm. 2. **Find the side of the square:** The diameter of the circle is the side of the square. Side $s = 2 \times R = 2 \times 4 = 8$ cm. 3. **Find the diagonal of the square:** Diagonal $= s\sqrt{2} = 8\sqrt{2}$ cm. ### Exam Strategy & Shortcut Work from the "inside out". Mentally divide the triangle side $4\sqrt{3}$ by $\sqrt{3}$ to get the radius $4$. Double it for the square's side $8$, and tack on a $\sqrt{2}$ for the diagonal. You can solve this entirely in your head in under 10 seconds. ### Common Pitfall A common mistake is confusing the formula for the *inradius* of a triangle ($a/2\sqrt{3}$) with the *circumradius* ($a/\sqrt{3}$). Since the triangle is inside the circle, the circle is its circumcircle. ### Final Answer Therefore, the correct answer is **$8\sqrt{2}$ cm**.