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 (vertices on circumference)
  • Diagonal = 2 × shortest side
  • Area = 1,089√3 square units
  • Need to find: diameter of circle

2. INFER the key geometric relationship

• Since the rectangle is inscribed in the circle with all vertices on the circumference, the diagonal of the rectangle equals the diameter of the circle
• This means: diameter = diagonal length

3. TRANSLATE the side relationship

• Let \(\mathrm{s}\) = length of shortest side
• Then diagonal = \(\mathrm{2s}\) (given relationship)
• Let \(\mathrm{b}\) = length of longest side (unknown)

4. INFER how to find the longest side

• The diagonal, shortest side, and longest side form a right triangle
• We can use the Pythagorean theorem: \(\mathrm{s^2 + b^2 = (diagonal)^2}\)

5. SIMPLIFY using Pythagorean theorem

• \(\mathrm{s^2 + b^2 = (2s)^2}\)
• \(\mathrm{s^2 + b^2 = 4s^2}\)
• Subtract \(\mathrm{s^2}\) from both sides: \(\mathrm{b^2 = 3s^2}\)
• Take square root: \(\mathrm{b = s\sqrt{3}}\)

6. TRANSLATE the area information

• Area of rectangle = shortest side × longest side
• \(\mathrm{s \times s\sqrt{3} = 1{,}089\sqrt{3}}\)
• \(\mathrm{s^2\sqrt{3} = 1{,}089\sqrt{3}}\)

7. SIMPLIFY to find s

• Divide both sides by \(\mathrm{\sqrt{3}}\): \(\mathrm{s^2 = 1{,}089}\)
• Take square root: \(\mathrm{s = 33}\) (use calculator if needed)

8. INFER the final answer

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

Answer: 66

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the diagonal of an inscribed rectangle equals the diameter of the circle. Instead, they might try to work with radius relationships or get confused about which measurement represents the circle's size.

This fundamental misunderstanding prevents them from setting up the correct equation and leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{s^2 + b^2 = (2s)^2}\) but make algebraic errors when solving for \(\mathrm{b}\). They might incorrectly get \(\mathrm{b = 2s}\) or \(\mathrm{b = s}\), rather than \(\mathrm{b = s\sqrt{3}}\).

This leads them to calculate the wrong area relationship, getting an incorrect value for \(\mathrm{s}\) and subsequently an incorrect diameter.

The Bottom Line:

This problem requires recognizing the special property of inscribed rectangles (diagonal = diameter) and then carefully working through the Pythagorean relationship. Students who miss either the geometric insight or make algebraic errors will struggle to reach the correct answer.