The volume of a right circular cone is 96pi cubic meters. The radius of the base of the cone is...

1. TRANSLATE the problem information

• Given information:
  • Volume: \(\mathrm{V = 96π}\) cubic meters
  • Base radius: \(\mathrm{r = 6}\) meters
  • Unknown: height h

2. INFER the approach needed

• Since we have volume and radius but need height, we should use the cone volume formula
• The formula \(\mathrm{V = \frac{1}{3}πr^2h}\) contains all three quantities we're working with
• Strategy: Substitute known values and solve for h

3. SIMPLIFY through algebraic manipulation

• Start with the cone volume formula: \(\mathrm{V = \frac{1}{3}πr^2h}\)
• Substitute the known values: \(\mathrm{96π = \frac{1}{3}π(6)^2h}\)
• Calculate \(\mathrm{(6)^2}\): \(\mathrm{96π = \frac{1}{3}π(36)h}\)
• Simplify \(\mathrm{\frac{1}{3} × 36}\): \(\mathrm{96π = 12πh}\)
• Divide both sides by 12π: \(\mathrm{h = \frac{96π}{12π} = 8}\)

Answer: 8 meters

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make arithmetic errors in the algebraic steps, particularly when calculating \(\mathrm{(6)^2 = 36}\) or when simplifying \(\mathrm{\frac{1}{3} × 36 = 12}\).

For example, they might calculate \(\mathrm{\frac{1}{3} × 36 = 18}\) instead of 12, leading to the equation \(\mathrm{96π = 18πh}\), which gives \(\mathrm{h = \frac{96}{18} ≈ 5.33}\). This causes confusion since the answer doesn't come out to a clean integer, leading to guessing.

Second Most Common Error:

Missing conceptual knowledge: Students don't remember the cone volume formula and may confuse it with the cylinder formula \(\mathrm{V = πr^2h}\).

Using \(\mathrm{V = πr^2h}\) instead: \(\mathrm{96π = π(6)^2h}\) → \(\mathrm{96π = 36πh}\) → \(\mathrm{h = \frac{96}{36} ≈ 2.67}\). This leads to confusion and guessing since this answer also doesn't match typical answer choices.

The Bottom Line:

This problem requires precise recall of the cone volume formula and careful arithmetic. The fractional coefficient \(\mathrm{\frac{1}{3}}\) in the formula creates multiple opportunities for calculation errors, making systematic algebraic manipulation the key to success.