A cone has a base diameter of 8 inches and a height of 12 inches. What is the volume, in...

1. TRANSLATE the problem information

• Given information:
  • Base diameter: 8 inches
  • Height: 12 inches
  • Find: Volume in cubic inches

2. INFER the approach needed

• We need the volume formula for a cone: \(\mathrm{V = \frac{1}{3}\pi r^2h}\)
• The formula requires radius (r), but we're given diameter
• Strategy: Convert diameter to radius first, then apply formula

3. Convert diameter to radius

• Radius = diameter ÷ 2 = 8 ÷ 2 = 4 inches

4. SIMPLIFY by substituting into the volume formula

• \(\mathrm{V = \frac{1}{3}\pi r^2h}\)
• \(\mathrm{V = \frac{1}{3}\pi(4)^2(12)}\)
• \(\mathrm{V = \frac{1}{3}\pi(16)(12)}\)
• \(\mathrm{V = \frac{1}{3}\pi(192)}\)
• \(\mathrm{V = \frac{1}{3}(192)\pi = 64\pi}\) cubic inches

Answer: A (64π)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Using diameter directly in the formula instead of converting to radius first.

Students might substitute 8 directly for r in \(\mathrm{V = \frac{1}{3}\pi r^2h}\), getting:

\(\mathrm{V = \frac{1}{3}\pi(8)^2(12)}\)

\(\mathrm{= \frac{1}{3}\pi(64)(12)}\)

\(\mathrm{= \frac{1}{3}(768)\pi}\)

\(\mathrm{= 256\pi}\)

This may lead them to select Choice C (256π).

Second Most Common Error:

Poor SIMPLIFY execution: Making arithmetic errors in the final calculation.

Students correctly find radius = 4 and set up \(\mathrm{V = \frac{1}{3}\pi(16)(12)}\), but then calculate:

• \(\mathrm{\frac{1}{3}(16)(12) = \frac{1}{3}(192)}\) incorrectly as 192 instead of 64
• Or miscalculate 16 × 12 as something other than 192

This may lead them to select Choice B (192π) or Choice D (768π).

The Bottom Line:

This problem tests whether students can distinguish between radius and diameter in geometric formulas, and whether they can execute multi-step arithmetic accurately. The key insight is recognizing that most geometric formulas use radius, not diameter.