Question:A square has a perimeter of 968 centimeters. The four vertices of the square lie on a circumscribed circle. The...

1. TRANSLATE the problem information

• Given information:
  • Square perimeter = \(\mathrm{968}\) centimeters
  • Radius of circumscribed circle = \(\mathrm{k\sqrt{2}}\) centimeters
  • Need to find the value of k

2. TRANSLATE to find the side length

• Since perimeter = 4 × side length:
  \(\mathrm{968 = 4 \times side\ length}\)
  \(\mathrm{Side\ length = 968 \div 4 = 242}\) centimeters

3. INFER the geometric relationship

• Key insight: When a square is inscribed in a circle (vertices on the circle), the diagonal of the square equals the diameter of the circle
• This means: \(\mathrm{diagonal = 2 \times radius}\)

4. INFER and calculate the diagonal

• For any square with side length s, diagonal = \(\mathrm{s\sqrt{2}}\) (from Pythagorean theorem)
• Our diagonal = \(\mathrm{242\sqrt{2}}\) centimeters

5. SIMPLIFY to find the actual radius

• Since diagonal = diameter = 2 × radius:
  \(\mathrm{242\sqrt{2} = 2 \times radius}\)
  \(\mathrm{Radius = (242\sqrt{2}) \div 2 = 121\sqrt{2}}\) centimeters

6. SIMPLIFY to solve for k

• We know the radius equals \(\mathrm{k\sqrt{2}}\):
  \(\mathrm{k\sqrt{2} = 121\sqrt{2}}\)
  Divide both sides by \(\mathrm{\sqrt{2}}\): \(\mathrm{k = 121}\)

Answer: 121

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the relationship between the square's diagonal and the circle's diameter. They may try to use other formulas or relationships that don't apply here, such as treating the side of the square as the radius or trying to use area relationships. This leads to confusion and guessing because they can't establish the crucial geometric connection needed to proceed.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly find that the radius is \(\mathrm{121\sqrt{2}}\) but struggle with the final step. They might set up \(\mathrm{121\sqrt{2} = k\sqrt{2}}\) but then make algebraic errors, such as thinking \(\mathrm{k = 121\sqrt{2} \div \sqrt{2} = 121\sqrt{2}}\) or getting confused about how to handle the \(\mathrm{\sqrt{2}}\) terms. This may lead them to select incorrect numerical values or abandon the systematic solution.

The Bottom Line:

This problem requires students to bridge geometry (inscribed squares) with algebra (solving for k). The key breakthrough is recognizing that "circumscribed circle" means the diagonal becomes the diameter—without this insight, students get stuck on the very first strategic step.