The perimeter of an equilateral triangle is 852 centimeters. The three vertices of the triangle lie on a circle. The...

1. TRANSLATE the problem information

• Given information:
  • Perimeter = \(852 \text{ cm}\) for an equilateral triangle
  • Triangle vertices lie on a circle (circumscribed)
  • Circle radius = \(w\sqrt{3} \text{ cm}\)
  • Need to find w
• What this tells us: We have an equilateral triangle inscribed in a circle, and we need to connect the triangle's dimensions to the circle's radius.

2. Find the side length

Since the triangle is equilateral, all three sides are equal:

Side length = \(852 \div 3 = 284 \text{ cm}\)

3. VISUALIZE the geometric setup

• Draw the equilateral triangle ABC inscribed in circle with center O
• Drop a perpendicular from center O to side AB, hitting at midpoint M
• This creates right triangle AOM, which is key to solving the problem

4. INFER the angle relationships

• In an equilateral triangle inscribed in a circle, the angle OAM = \(30°\)
• This happens because the central angle for each side is \(120°\), and when we drop to the midpoint, we get half of the inscribed angle
• Triangle AOM is a 30-60-90 right triangle

5. Set up the trigonometric relationship

In right triangle AOM:

• AM = \(284/2 = 142 \text{ cm}\) (half the side length)
• AO = \(w\sqrt{3} \text{ cm}\) (the radius)
• Angle OAM = \(30°\)

6. Apply trigonometry and SIMPLIFY

\(\cos(30°) = \frac{AM}{AO} = \frac{142}{w\sqrt{3}}\)

Since \(\cos(30°) = \frac{\sqrt{3}}{2}\):

\(\frac{\sqrt{3}}{2} = \frac{142}{w\sqrt{3}}\)

Cross multiply:

\((w\sqrt{3})(\frac{\sqrt{3}}{2}) = 142\)

\(w(\frac{3}{2}) = 142\)

\(\frac{3w}{2} = 142\)

\(3w = 284\)

\(w = \frac{284}{3}\)

To convert to decimal (use calculator): \(284 \div 3 = 94.666...\)

Answer: \(\frac{284}{3}\) (or 94.67)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak VISUALIZE skill: Students struggle to see that they need to create the right triangle by connecting the center to a vertex and dropping a perpendicular to the opposite side.

Without this geometric insight, they might try to use incorrect formulas or get stuck trying to directly relate perimeter to radius. This leads to confusion and guessing.

Second Most Common Error:

Poor INFER reasoning about angle measures: Students may not recognize that the angle OAM = \(30°\) in the constructed right triangle.

They might assume it's \(60°\) (confusing it with the triangle's internal angles) or try to use other trigonometric relationships incorrectly. This may lead them to set up wrong equations and get an incorrect value for w.

The Bottom Line:

This problem requires students to bridge geometry and trigonometry by constructing an auxiliary right triangle. The key insight is recognizing that inscribed equilateral triangles create predictable 30-60-90 triangles when you connect the center to the vertices and sides.