An equilateral triangle is inscribed in a circle. The radius of the circle is (20sqrt(3))/3 inches. What is the side...

1. TRANSLATE the problem information

• Given information:
  • Equilateral triangle is inscribed in a circle
  • Circle radius: \(\mathrm{R = \frac{20\sqrt{3}}{3}}\) inches
  • Need to find: side length of the triangle

2. INFER the mathematical relationship needed

• For an equilateral triangle inscribed in a circle, we need the circumradius formula
• The key insight: circumradius relates directly to side length through a specific formula
• Strategy: Use the formula to solve for side length from the given radius

3. SIMPLIFY to solve for side length

• Start with circumradius formula: \(\mathrm{R = \frac{a\sqrt{3}}{3}}\)
• Solve for \(\mathrm{a}\) by algebraic manipulation:
  • Multiply both sides by 3: \(\mathrm{3R = a\sqrt{3}}\)
  • Divide both sides by \(\mathrm{\sqrt{3}}\): \(\mathrm{a = \frac{3R}{\sqrt{3}} = R\sqrt{3}}\)

4. SIMPLIFY the final calculation

• Substitute the given radius: \(\mathrm{a = \frac{20\sqrt{3}}{3} \cdot \sqrt{3}}\)
• Multiply the radicals:
  \(\mathrm{a = \frac{20\sqrt{3} \cdot \sqrt{3}}{3}}\)
  \(\mathrm{a = \frac{20 \cdot 3}{3}}\)
  \(\mathrm{a = 20}\)

Answer: A. 20

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize they need the specific circumradius formula for equilateral triangles and instead try to use general circle relationships or basic trigonometry inappropriately.

Without the correct formula, they might attempt to use \(\mathrm{C = 2\pi r}\) or try to set up right triangles without the proper geometric insight. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the circumradius formula but make errors when manipulating radicals, particularly not recognizing that \(\mathrm{\sqrt{3} \cdot \sqrt{3} = 3}\).

They might incorrectly calculate \(\mathrm{\frac{20\sqrt{3}}{3} \cdot \sqrt{3}}\) as \(\mathrm{\frac{20\sqrt{6}}{3}}\) or make other radical arithmetic mistakes. This may lead them to select Choice D (\(\mathrm{\frac{40\sqrt{3}}{3}}\)) if they double the answer incorrectly.

The Bottom Line:

This problem requires students to connect geometry (inscribed triangles) with algebra (radical manipulation). The key challenge is recognizing the specific circumradius relationship and then executing the algebraic steps accurately with radicals.