A cone has a base diameter of 8 inches and a height of 12 inches. What is the volume, in...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A cone has a base diameter of \(8\) inches and a height of \(12\) inches. What is the volume, in cubic inches, of the cone?
- \(64\pi\)
- \(192\pi\)
- \(256\pi\)
- \(768\pi\)
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 = \(\mathrm{8 \div 2 = 4}\) 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 (\(\mathrm{64\pi}\))
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:
\(\mathrm{V = \frac{1}{3}\pi(8)^2(12)}\)
\(\mathrm{= \frac{1}{3}\pi(64)(12)}\)
\(\mathrm{= \frac{1}{3}(768)\pi = 256\pi}\)
This may lead them to select Choice C (\(\mathrm{256\pi}\)).
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 \(\mathrm{16 \times 12}\) as something other than 192
This may lead them to select Choice B (\(\mathrm{192\pi}\)) or Choice D (\(\mathrm{768\pi}\)).
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.