A right circular cylinder has a volume of 377 cubic centimeters. The area of the base of the cylinder is...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A right circular cylinder has a volume of \(377\) cubic centimeters. The area of the base of the cylinder is \(13\) square centimeters. What is the height, in centimeters, of the cylinder?
1. TRANSLATE the problem information
- Given information:
- Volume of cylinder = 377 cubic centimeters
- Area of base = 13 square centimeters
- Need to find: height in centimeters
2. INFER the approach
- Since we have volume and base area but need height, we should use the cylinder volume formula
- Key insight: The base area is already given as 13, so we don't need to find the radius separately
- We can substitute the base area directly into the volume formula
3. TRANSLATE the volume formula and substitute
- Volume formula: \(\mathrm{V} = \pi r^2 h\)
- Since base area = \(\pi r^2 = 13\), we can write: \(\mathrm{V} = (\mathrm{base\ area}) \times h\)
- Substituting known values: \(377 = 13h\)
4. SIMPLIFY to find the height
- Divide both sides by 13: \(h = 377 \div 13\)
- \(h = 29\)
Answer: 29 centimeters
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize that base area = \(\pi r^2\), so they attempt to solve for the radius first unnecessarily.
They might set up: \(13 = \pi r^2\), solve for r, then substitute back into \(\mathrm{V} = \pi r^2 h\). While this approach works, it's more complex and creates opportunities for calculation errors. Some students get lost in this longer process and make mistakes or run out of time.
Second Most Common Error:
Poor SIMPLIFY execution: Students set up the equation \(377 = 13h\) correctly but make arithmetic errors in the division.
Common mistakes include \(377 \div 13 = 27\) or \(377 \div 13 = 31\), leading them to incorrect final answers. This happens when students rush through the division or don't double-check their arithmetic.
The Bottom Line:
The key insight is recognizing that you already have everything you need - the base area can be treated as a single unit (13) rather than breaking it down into \(\pi r^2\). Students who miss this connection make the problem much harder than it needs to be.