The sum of the lengths of all 12 edges of a cube is a, where a is a positive constant....

1. TRANSLATE the problem information

• Given information:
  • The sum of lengths of all 12 edges of a cube equals a
  • a is a positive constant
  • Need to find surface area in terms of a

• What this tells us: We need to work backwards from total edge length to find the cube's dimensions, then calculate surface area.

2. INFER the relationship between edges and side length

• A cube has 12 edges, and all edges of a cube are equal in length
• If we call the side length s, then: \(12\mathrm{s} = \mathrm{a}\)
• Solving for the side length: \(\mathrm{s} = \frac{\mathrm{a}}{12}\)
• Strategy: Use this side length to find surface area

3. INFER the surface area calculation

• A cube has 6 faces, each being a square with area \(\mathrm{s}^2\)
• Surface area formula: \(\mathrm{SA} = 6\mathrm{s}^2\)
• Substitute our expression for s: \(\mathrm{SA} = 6\left(\frac{\mathrm{a}}{12}\right)^2\)

4. SIMPLIFY the algebraic expression

• \(\mathrm{SA} = 6\left(\frac{\mathrm{a}}{12}\right)^2\)
• \(\mathrm{SA} = 6 \times \frac{\mathrm{a}^2}{144}\)
• \(\mathrm{SA} = \frac{6\mathrm{a}^2}{144}\)
• \(\mathrm{SA} = \frac{\mathrm{a}^2}{24}\)

Answer: B (\(\frac{\mathrm{a}^2}{24}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students incorrectly count the number of edges in a cube, thinking it has 8 edges (confusing edges with vertices) or 6 edges (confusing edges with faces).

If they think a cube has 8 edges, they set up \(8\mathrm{s} = \mathrm{a}\), leading to \(\mathrm{s} = \frac{\mathrm{a}}{8}\), then \(\mathrm{SA} = 6\left(\frac{\mathrm{a}}{8}\right)^2 = \frac{6\mathrm{a}^2}{64} = \frac{3\mathrm{a}^2}{32}\). This doesn't match any answer choice, leading to confusion and guessing.

If they think it has 6 edges, they get \(\mathrm{s} = \frac{\mathrm{a}}{6}\), then \(\mathrm{SA} = 6\left(\frac{\mathrm{a}}{6}\right)^2 = \frac{6\mathrm{a}^2}{36} = \frac{\mathrm{a}^2}{6}\), leading them to select Choice E (\(\frac{\mathrm{a}^2}{6}\)).

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{SA} = 6\left(\frac{\mathrm{a}}{12}\right)^2\) but make arithmetic errors in the simplification.

A common mistake is: \(\mathrm{SA} = 6\left(\frac{\mathrm{a}}{12}\right)^2 = \frac{6\mathrm{a}^2}{12} = \frac{\mathrm{a}^2}{2}\) (forgetting to square the denominator). Since \(\frac{\mathrm{a}^2}{2}\) isn't an option, they might pick the closest-looking choice or guess.

The Bottom Line:

Success requires both spatial visualization (correctly identifying a cube's 12 edges) and careful algebraic manipulation. Students who rush through either the setup or the simplification often select incorrect answers that result from systematic computational errors.