The area of an equilateral triangle is 48sqrt(3) square inches. The three vertices of the triangle lie on a circle....

1. TRANSLATE the problem information

• Given information:
  • Equilateral triangle area = \(48\sqrt{3}\) square inches
  • Triangle vertices lie on a circle (circumscribed)
  • Circle circumference = \(k\pi\) inches
  • Need to find k
• What this tells us: We need to work from triangle area → side length → circle radius → circumference

2. INFER the solution strategy

• To find k in the circumference \(k\pi\), we need the actual circumference
• To get circumference, we need the circle's radius
• The circle's radius relates to the triangle's side length
• We can get the side length from the given area
• Strategy: Area → Side length → Radius → Circumference → k

3. SIMPLIFY to find the triangle's side length

• Use equilateral triangle area formula: \(\mathrm{A} = \frac{\mathrm{s}^2\sqrt{3}}{4}\)
• Set up equation: \(48\sqrt{3} = \frac{\mathrm{s}^2\sqrt{3}}{4}\)
• Divide both sides by \(\sqrt{3}\): \(48 = \frac{\mathrm{s}^2}{4}\)
• Multiply both sides by 4: \(\mathrm{s}^2 = 192\)
• Take square root: \(\mathrm{s} = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}\) inches

4. INFER the relationship between triangle and circle

• For an equilateral triangle inscribed in a circle, the circumradius formula is \(\mathrm{R} = \frac{\mathrm{s}}{\sqrt{3}}\)
• This comes from the geometry of equilateral triangles and their circumscribed circles

5. SIMPLIFY to find the circle's radius

• \(\mathrm{R} = \frac{\mathrm{s}}{\sqrt{3}} = \frac{8\sqrt{3}}{\sqrt{3}} = 8\) inches

6. SIMPLIFY to find k

• Circumference = \(2\pi\mathrm{R} = 2\pi(8) = 16\pi\)
• Since the problem states circumference = \(k\pi\):
• \(k\pi = 16\pi\)
• Therefore \(k = 16\)

Answer: 16

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Circumradius formula for equilateral triangle

Many students know the area formula for equilateral triangles, but don't recall or never learned that for an equilateral triangle inscribed in a circle, \(\mathrm{R} = \frac{\mathrm{s}}{\sqrt{3}}\). Without this key relationship, they get stuck after finding the side length and resort to guessing. This leads to confusion and random answer selection.

Second Most Common Error:

Weak SIMPLIFY execution: Algebraic manipulation errors

Students might correctly set up \(48\sqrt{3} = \frac{\mathrm{s}^2\sqrt{3}}{4}\) but make errors when dividing by \(\sqrt{3}\) or simplifying \(\sqrt{192}\). For instance, they might incorrectly calculate \(\sqrt{192}\) as something other than \(8\sqrt{3}\), or make errors in the fraction \(\frac{8\sqrt{3}}{\sqrt{3}}\), leading to an incorrect radius and therefore wrong value of k.

The Bottom Line:

This problem tests whether students can work systematically through a multi-step geometric relationship. The key insight is recognizing that you must work backwards from the given area through the geometric relationships to reach the final answer - and knowing the specific circumradius formula for equilateral triangles is crucial.