A triangular garden is laid out as an equilateral triangle, with a circular fence that passes exactly through all three...

1. TRANSLATE the problem information

• Given information:
  • Equilateral triangular garden with area \(784\sqrt{3}\) square feet
  • Circular fence passes through all three vertices (circumscribed circle)
  • Radius can be written as \(w\sqrt{3}\) feet
• What we need: Find the value of w

2. INFER the solution strategy

• Key insight: We need the circumradius of the triangle, but we're only given the area
• Strategy: First find the side length from the area, then use the side length to find the circumradius
• This is a two-step process connecting area → side length → circumradius

3. SIMPLIFY to find the side length

• Use the area formula for an equilateral triangle: \(\mathrm{A} = \frac{\mathrm{s}^2\sqrt{3}}{4}\)
• Set up the equation:
  \(\frac{\mathrm{s}^2\sqrt{3}}{4} = 784\sqrt{3}\)
• Multiply both sides by 4:
  \(\mathrm{s}^2\sqrt{3} = 3136\sqrt{3}\)
• Divide both sides by \(\sqrt{3}\):
  \(\mathrm{s}^2 = 3136\)
• Take the square root:
  \(\mathrm{s} = 56\) feet

4. SIMPLIFY to find the circumradius

• For an equilateral triangle, the circumradius is: \(\mathrm{R} = \frac{\mathrm{s}\sqrt{3}}{3}\)
• Substitute \(\mathrm{s} = 56\):
  \(\mathrm{R} = \frac{56\sqrt{3}}{3}\)

5. SIMPLIFY to find w

• We're told the radius equals \(w\sqrt{3}\), so:
  \(w\sqrt{3} = \frac{56\sqrt{3}}{3}\)
• Divide both sides by \(\sqrt{3}\):
  \(w = \frac{56}{3}\)

Answer: 56/3

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Students don't remember or confuse the circumradius formula for an equilateral triangle

Many students know \(\mathrm{R} = \frac{\mathrm{s}}{2\sin \mathrm{A}}\) but struggle to apply it correctly to equilateral triangles, or they remember incorrect formulas like \(\mathrm{R} = \frac{\mathrm{s}}{2}\) or \(\mathrm{R} = \mathrm{s}\sqrt{3}\). This leads to getting the wrong radius value and consequently the wrong value for w.

This may lead them to answers like \(w = 56\) or \(w = \frac{112}{3}\) depending on which incorrect formula they use.

Second Most Common Error:

Weak SIMPLIFY execution: Students make algebraic mistakes when solving the area equation or manipulating the final expression

Common mistakes include forgetting to take the square root when solving \(\mathrm{s}^2 = 3136\), or incorrectly canceling the \(\sqrt{3}\) terms in \(w\sqrt{3} = \frac{56\sqrt{3}}{3}\). These calculation errors lead to wrong values for s or w.

This causes them to get stuck partway through or arrive at incorrect numerical answers.

The Bottom Line:

This problem requires students to connect three different formulas (area, circumradius, and the given form) in sequence. The key challenge is remembering the correct circumradius formula and executing the algebra cleanly through multiple steps.