The base of a right prism is a right triangle with legs of lengths 6 units and 8 units. The...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
The base of a right prism is a right triangle with legs of lengths \(6\) units and \(8\) units. The volume of the prism is \(2{,}880\) cubic units. What is the height, in units, of the prism?
- \(60\)
- \(80\)
- \(120\)
- \(240\)
1. TRANSLATE the problem information
- Given information:
- Base is a right triangle with legs 6 units and 8 units
- Volume of prism is 2,880 cubic units
- Need to find the height of the prism
2. INFER the solution approach
- To find the height using V = Base Area × Height, we first need the base area
- Since the base is a right triangle, we can calculate its area using the legs
3. Calculate the base area
- Area of right triangle = \(\frac{1}{2} \times \mathrm{leg_1} \times \mathrm{leg_2}\)
- Area = \(\frac{1}{2} \times 6 \times 8 = 24\) square units
4. SIMPLIFY to find the height
- Using Volume = Base Area × Height:
- \(2{,}880 = 24 \times \mathrm{h}\)
- \(\mathrm{h} = 2{,}880 \div 24 = 120\) units
Answer: C (120)
Why Students Usually Falter on This Problem
Most Common Error Path:
Conceptual confusion about triangle measurements: Students may confuse the triangle's legs (6 and 8) with the base and height of the entire prism, trying to use 6 or 8 as the prism height directly.
They might calculate: \(\mathrm{Volume} \div 6 = 2{,}880 \div 6 = 480\) or \(\mathrm{Volume} \div 8 = 2{,}880 \div 8 = 360\), leading to confusion since neither appears in the answer choices. This leads to guessing.
Second Most Common Error:
Missing the triangle area formula: Students may forget to use \(\frac{1}{2}\) in the right triangle area calculation, computing base area as \(6 \times 8 = 48\) instead of 24.
This gives them \(\mathrm{h} = 2{,}880 \div 48 = 60\), leading them to select Choice A (60).
The Bottom Line:
This problem requires clear understanding that the 'base' of the prism is the triangular face, and its area must be calculated before using the volume formula.