OpenSAT

Let's break down the solution: 1. **Equilateral Triangle Properties:** An equilateral triangle has three equal sides and three equal angles of 60 degrees. The height of an equilateral triangle divides it into two 30-60-90 right triangles. 2. **30-60-90 Triangle Properties:** The sides of a 30-60-90 triangle are in the ratio of 1:√3:2. Since the side length of the equilateral triangle is 6, the height of the triangle is 3√3. The radius of the inscribed circle is equal to 1/3 of the height, so the radius of the inscribed circle is √3. 3. **Area of a Circle:** The area of a circle is given by the formula $A = \pi r^2$, where r is the radius. Substituting √3 for r, we get $A = \pi (\sqrt{3})^2 = 3\pi$. 4. **Final Answer:** Since the circle is inscribed in the equilateral triangle, the diameter of the circle is equal to the height of the triangle, which is 3√3. Therefore, the radius of the circle is (3√3)/2, and the area of the circle is $\pi (\frac{3\sqrt{3}}{2})^2 = \frac{27\pi}{4}$, which is equivalent to .