A rectangle is inscribed in a circle, such that each vertex of the rectangle lies on the circumference of the...

1. TRANSLATE the problem information

• Given information:
  • Rectangle inscribed in circle (all vertices on circumference)
  • Diagonal = \(2 \times \text{(shortest side length)}\)
  • Area = \(1,089\sqrt{3}\) square units
  • Need to find: diameter of circle

2. INFER the key relationship

• When a rectangle is inscribed in a circle, the diagonal of the rectangle equals the diameter of the circle
• This means: \(\mathrm{diameter = diagonal = 2 \times (shortest\ side)}\)
• Strategy: Find the shortest side, then multiply by 2

3. INFER how to use the Pythagorean theorem

• Let \(\mathrm{s}\) = shortest side length
• Let \(\mathrm{b}\) = longest side length
• Diagonal = \(\mathrm{2s}\)
• Using Pythagorean theorem: \(\mathrm{s^2 + b^2 = (diagonal)^2}\)

4. SIMPLIFY to find the relationship between sides

• \(\mathrm{s^2 + b^2 = (2s)^2}\)
• \(\mathrm{s^2 + b^2 = 4s^2}\)
• \(\mathrm{b^2 = 4s^2 - s^2 = 3s^2}\)
• \(\mathrm{b = s\sqrt{3}}\)

5. TRANSLATE the area information and solve

• Area = shortest side × longest side
• Area = \(\mathrm{s \times (s\sqrt{3}) = s^2\sqrt{3}}\)
• Given: \(\mathrm{s^2\sqrt{3} = 1,089\sqrt{3}}\)
• Dividing both sides by \(\sqrt{3}\): \(\mathrm{s^2 = 1,089}\)
• Taking square root: \(\mathrm{s = 33}\)

6. APPLY CONSTRAINTS and find final answer

• Diagonal = \(\mathrm{2s = 2(33) = 66}\)
• Since \(\mathrm{diagonal = diameter}\): \(\mathrm{diameter = 66}\)

Answer: 66

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that the diagonal of an inscribed rectangle equals the circle's diameter

Students may try to work with the circle's radius or circumference instead of realizing that the rectangle's diagonal IS the diameter. Without this key insight, they get lost trying to relate the rectangle's dimensions to the circle through more complex geometric relationships.

This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Making algebraic errors when working with the Pythagorean relationship

Students correctly set up \(\mathrm{s^2 + b^2 = (2s)^2}\) but then make mistakes like:

• Incorrectly expanding \(\mathrm{(2s)^2}\) as \(\mathrm{2s^2}\) instead of \(\mathrm{4s^2}\)
• Errors in isolating \(\mathrm{b^2 = 3s^2}\)
• Mistakes when taking the square root to get \(\mathrm{b = s\sqrt{3}}\)

These calculation errors lead to wrong values for s, resulting in an incorrect diameter.

The Bottom Line:

This problem requires recognizing a key geometric property (inscribed rectangle's diagonal = circle's diameter) and then carefully executing multi-step algebra involving radicals. Students who miss either the conceptual insight or make algebraic errors will struggle to reach the correct answer.