Points Q and R lie on a circle with center P. The radius of this circle is 9 inches. Triangle...

1. TRANSLATE the problem information

• Given information:
  • Points Q and R lie on a circle with center P
  • Circle radius = 9 inches
  • Triangle PQR perimeter = 31 inches
  • Find length of side QR

2. INFER the key geometric relationship

• Since P is the center of the circle and both Q and R lie ON the circle, this means:
  • PQ is a radius = 9 inches
  • PR is a radius = 9 inches
• This is the crucial insight - two sides of triangle PQR are already known!

3. TRANSLATE the perimeter condition into an equation

• Perimeter = sum of all three sides
• \(\mathrm{PQ + PR + QR = 31\ inches}\)
• Substituting known values: \(\mathrm{9 + 9 + QR = 31}\)

4. SIMPLIFY the algebra to find QR

\(\mathrm{18 + QR = 31}\)

\(\mathrm{QR = 31 - 18 = 13\ inches}\)

Answer: B. 13

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that PQ and PR are both radii of the circle, so they equal 9 inches each.

Without this key insight, students might think they don't have enough information to solve the problem, or they might try to use more complex geometric relationships (like the Pythagorean theorem) that aren't needed here. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Conceptual confusion about radius vs. diameter: Students might mistakenly think that since the "radius is 9 inches," the sides PQ and PR are each 18 inches (confusing radius with diameter).

This would lead to: \(\mathrm{18 + 18 + QR = 31}\), giving \(\mathrm{QR = -5}\), which is impossible for a length. This causes them to get stuck and randomly select an answer.

The Bottom Line:

This problem tests whether students can connect the abstract concept of "center and points on a circle" to the concrete fact that "these create segments equal to the radius." Once that connection is made, it becomes a simple algebra problem.