An equilateral triangle is inscribed in a circle. The radius of the circle is 18 inches. What is the perimeter,...

1. VISUALIZE the problem setup

• Given information:
  • Equilateral triangle inscribed in a circle
  • Circle radius = 18 inches
  • Need to find triangle perimeter

• VISUALIZE the geometry by drawing the circle with center O and the inscribed triangle ABC, where all vertices A, B, C lie on the circle.

2. INFER the key geometric relationships

• Since it's an equilateral triangle inscribed in a circle, the center O is equidistant from all three vertices
• The central angles are equal: \(\mathrm{360°/3 = 120°}\) each
• We need to find the side length first, then multiply by 3

3. INFER the solution strategy

• To find side length, focus on one side (like AB) and the triangle it forms with the center
• Drop a perpendicular from center O to side AB, meeting at point M
• This creates two congruent right triangles, each being a 30-60-90 triangle

4. SIMPLIFY using trigonometry

• In right triangle OMA:
  • OA = 18 (the radius)
  • Angle AOM = \(\mathrm{60°}\) (half of the \(\mathrm{120°}\) central angle)
  • \(\mathrm{AM = 18 \times sin(60°)}\)
    \(\mathrm{= 18 \times (\sqrt{3}/2)}\)
    \(\mathrm{= 9\sqrt{3}}\)

5. SIMPLIFY to find the full side length

• Since M is the midpoint of AB:
• Side length \(\mathrm{AB = 2 \times AM}\)
  \(\mathrm{= 2 \times 9\sqrt{3}}\)
  \(\mathrm{= 18\sqrt{3}}\) inches

6. SIMPLIFY to find the perimeter

• \(\mathrm{Perimeter = 3 \times (side\ length)}\)
  \(\mathrm{= 3 \times 18\sqrt{3}}\)
  \(\mathrm{= 54\sqrt{3}}\) inches

Answer: D) \(\mathrm{54\sqrt{3}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak VISUALIZE skill: Students struggle to set up the auxiliary construction (dropping the perpendicular from center to side) needed to create workable right triangles.

Without this visualization, they may try to work directly with the \(\mathrm{120°}\) central angle or attempt to use formulas they don't fully understand. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the 30-60-90 triangle setup but make calculation errors, such as finding \(\mathrm{AM = 9\sqrt{3}}\) but forgetting that this is only half the side length.

This leads them to calculate perimeter as \(\mathrm{3 \times 9\sqrt{3} = 27\sqrt{3}}\), causing them to select Choice B (\(\mathrm{27\sqrt{3}}\)).

The Bottom Line:

This problem requires strong spatial visualization to construct the right auxiliary lines, combined with solid trigonometric calculation skills. The key insight is recognizing that dropping a perpendicular creates the familiar 30-60-90 triangles that make the problem solvable.