What is the length of one side of a square that has the same perimeter as a circle with radius...

1. TRANSLATE the problem information

• Given information:
  • Circle has radius 2
  • Square has the same perimeter as this circle
  • Need to find: side length of the square

• What this tells us: We need to find the circle's perimeter first, then use it to determine the square's side length

2. INFER the solution approach

• Key insight: "Same perimeter" means we can set the two perimeter expressions equal
• Strategy: Find circle circumference → Set equal to square perimeter → Solve for side length

3. TRANSLATE and calculate the circle's circumference

• Circumference = \(\mathrm{2πr}\) = \(\mathrm{2π(2)}\) = \(\mathrm{4π}\)

4. INFER the equation setup

• Since perimeters are equal: Square perimeter = Circle circumference
• Square perimeter formula: \(\mathrm{4s}\) (where \(\mathrm{s}\) = side length)
• Equation: \(\mathrm{4s = 4π}\)

5. SIMPLIFY to find the side length

• Divide both sides by 4: \(\mathrm{s = \frac{4π}{4} = π}\)

Answer: \(\mathrm{π}\) (Choice C)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students confuse radius with circumference or use diameter instead of radius in the circumference formula.

Some students see "radius 2" and think the circumference is just 2, forgetting the \(\mathrm{π}\) completely. Others might use \(\mathrm{C = πd}\) and substitute 2 for \(\mathrm{d}\) instead of recognizing that \(\mathrm{d = 2r = 4}\). This leads to getting circumference = \(\mathrm{2π}\) instead of \(\mathrm{4π}\), which would give them \(\mathrm{s = \frac{2π}{4} = \frac{π}{2}}\).

This may lead them to select Choice A (\(\mathrm{\frac{π}{2}}\)).

Second Most Common Error:

Inadequate SIMPLIFY execution: Students set up the equation correctly as \(\mathrm{4s = 4π}\) but make algebraic errors when solving.

The most common mistake is forgetting to divide the \(\mathrm{π}\) by 4, leaving \(\mathrm{s = 4π}\) instead of \(\mathrm{s = π}\). Or sometimes students get confused about which number to divide by which.

This may lead them to select Choice D (\(\mathrm{2π}\)) or causes confusion and guessing.

The Bottom Line:

This problem tests whether students can systematically work with two different perimeter formulas and connect them through an equation. The key is recognizing that "same perimeter" creates an algebraic relationship, not just asking for separate calculations.