A sphere has a surface area equal to the area of a certain square. If the sphere has a radius...

1. TRANSLATE the problem setup

• Given information:
  • Sphere \(\mathrm{radius = 3}\)
  • Sphere's surface area equals a square's area
  • Need to find: side length of the square

• What this tells us: We need to find the sphere's surface area first, then use that to find the square's side length.

2. INFER the solution approach

• Since the surface areas are equal, we can set up an equation
• Strategy: Calculate sphere's surface area → set equal to square's area → solve for side

3. Find the sphere's surface area

• Use the surface area formula: \(\mathrm{SA = 4\pi r^2}\)
• Substitute \(\mathrm{r = 3}\):
  \(\mathrm{SA = 4\pi(3)^2}\)
  \(\mathrm{= 4\pi(9)}\)
  \(\mathrm{= 36\pi}\)

4. TRANSLATE the equal areas condition

• The sphere's surface area equals the square's area
• So: \(\mathrm{s^2 = 36\pi}\) (where s is the square's side length)

5. INFER how to find the side length

• To find s from \(\mathrm{s^2 = 36\pi}\), take the square root of both sides
• \(\mathrm{s = \sqrt{36\pi}}\)

6. SIMPLIFY the radical expression

• Use the property \(\mathrm{\sqrt{ab} = \sqrt{a} \times \sqrt{b}}\)
• \(\mathrm{s = \sqrt{36\pi}}\)
  \(\mathrm{= \sqrt{36} \times \sqrt{\pi}}\)
  \(\mathrm{= 6\sqrt{\pi}}\)

Answer: \(\mathrm{6\sqrt{\pi}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may misinterpret "surface area equal to area of square" and try to set the radius equal to the side length, or confuse which geometric formulas to use.

Instead of setting \(\mathrm{4\pi r^2 = s^2}\), they might write something like \(\mathrm{4\pi r = s^2}\) or even \(\mathrm{r = s}\). This fundamental misreading of the problem setup makes it impossible to arrive at the correct answer, leading to confusion and guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{s^2 = 36\pi}\) but struggle with simplifying \(\mathrm{\sqrt{36\pi}}\).

They might leave their answer as \(\mathrm{\sqrt{36\pi}}\) instead of recognizing they can separate it into \(\mathrm{\sqrt{36} \times \sqrt{\pi} = 6\sqrt{\pi}}\). This could lead them to select Choice \(\mathrm{(C) (6\pi)}\) if they incorrectly think \(\mathrm{\sqrt{36\pi} = 6\pi}\), or get stuck and guess randomly.

The Bottom Line:

This problem tests whether students can connect different geometric formulas through an equality relationship. Success depends on careful translation of the problem statement and systematic algebraic manipulation of square roots.