A square is inscribed in a circle. The radius of the circle is (20sqrt(2))/2 inches. What is the side length,...

1. TRANSLATE the problem information

• Given information:
  • Square is inscribed in a circle
  • Radius of circle = \(\frac{20\sqrt{2}}{2}\) inches
  • Need to find: side length of the square

2. INFER the key geometric relationship

• Key insight: When a square is inscribed in a circle, the diagonal of the square is exactly equal to the diameter of the circle
• This means we need to find the diameter first, then use it to find the square's side length

3. SIMPLIFY to find the diameter

• Diameter = \(2 \times \text{radius}\)
• Diameter = \(2 \times \frac{20\sqrt{2}}{2}\)
• \(= 20\sqrt{2}\) inches
• So the diagonal of our square is \(20\sqrt{2}\) inches

4. INFER the relationship between square's side and diagonal

• In any square, the diagonal and side are related by the Pythagorean theorem
• If side length = \(s\), then diagonal = \(s\sqrt{2}\)
• This comes from the 45-45-90 triangle formed when we draw the diagonal

5. SIMPLIFY to solve for the side length

• We know: diagonal = \(s\sqrt{2} = 20\sqrt{2}\)
• Dividing both sides by \(\sqrt{2}\):
• \(s = \frac{20\sqrt{2}}{\sqrt{2}}\)
• \(= 20\)
• Therefore, side length = \(20\) inches

Answer: A. 20

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Not recognizing that the diagonal of the inscribed square equals the diameter of the circle.

Without this key insight, students might try to work directly with the radius or get confused about how the square and circle relate to each other. They may attempt to use the radius as if it were the side length or try to apply formulas that don't apply to this situation. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Making errors when working with the radical expressions.

Students might incorrectly calculate the diameter as just \(\frac{20\sqrt{2}}{2}\) instead of doubling it to get \(20\sqrt{2}\), or they might make errors when dividing \(20\sqrt{2}\) by \(\sqrt{2}\). For example, they might think \(\frac{20\sqrt{2}}{\sqrt{2}} = \frac{20\sqrt{2}}{2}\), leading them to select Choice B (\(\frac{20\sqrt{2}}{2}\)).

The Bottom Line:

This problem tests whether students can connect geometric relationships (inscribed figures) with algebraic manipulation (radical expressions). The key breakthrough is recognizing that inscription creates a specific relationship between the square's diagonal and the circle's diameter.