In a rectangle with length L, width w, and diagonal d, the relationship d^2 = L^2 + w^2 holds. What...
GMAT Advanced Math : (Adv_Math) Questions
In a rectangle with length L, width w, and diagonal d, the relationship \(\mathrm{d^2 = L^2 + w^2}\) holds. What is L in terms of d and w?
\(\mathrm{L} = \sqrt{\mathrm{d}^2 + \mathrm{w}^2}\)
\(\mathrm{L} = \sqrt{\mathrm{d}^2 - \mathrm{w}^2}\)
\(\mathrm{L} = \mathrm{d} - \mathrm{w}\)
\(\mathrm{L} = \mathrm{d}^2 - \mathrm{w}^2\)
1. INFER the solution strategy
- We have the equation \(\mathrm{d^2 = L^2 + w^2}\) and need to find \(\mathrm{L}\)
- To isolate \(\mathrm{L}\), we need to first isolate \(\mathrm{L^2}\) by moving \(\mathrm{w^2}\) to the other side
- Then we'll take the square root to find \(\mathrm{L}\)
2. SIMPLIFY by isolating L²
- Starting equation: \(\mathrm{d^2 = L^2 + w^2}\)
- Subtract \(\mathrm{w^2}\) from both sides: \(\mathrm{d^2 - w^2 = L^2}\)
3. SIMPLIFY by taking the square root
- From \(\mathrm{L^2 = d^2 - w^2}\), take the square root of both sides
- \(\mathrm{L = \sqrt{d^2 - w^2}}\)
- Since \(\mathrm{L}\) represents length, we take the positive square root
Answer: B
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY skill: Students make algebraic sign errors and add \(\mathrm{w^2}\) instead of subtracting it.
Starting with \(\mathrm{d^2 = L^2 + w^2}\), they incorrectly think they need to "move \(\mathrm{w^2}\) to the other side" and write \(\mathrm{d^2 + w^2 = L^2}\). Taking the square root gives \(\mathrm{L = \sqrt{d^2 + w^2}}\).
This leads them to select Choice A (\(\mathrm{\sqrt{d^2 + w^2}}\))
Second Most Common Error:
Incomplete SIMPLIFY execution: Students correctly isolate \(\mathrm{L^2 = d^2 - w^2}\) but forget to take the square root as the final step.
They stop at \(\mathrm{L^2 = d^2 - w^2}\) and select this as their answer, thinking they've solved for \(\mathrm{L}\) when they've only solved for \(\mathrm{L^2}\).
This leads them to select Choice D (\(\mathrm{d^2 - w^2}\))
The Bottom Line:
This problem tests careful algebraic manipulation. The key insight is recognizing that solving for \(\mathrm{L}\) is a two-step process: first isolate \(\mathrm{L^2}\) by subtracting \(\mathrm{w^2}\), then complete the solution by taking the square root.
\(\mathrm{L} = \sqrt{\mathrm{d}^2 + \mathrm{w}^2}\)
\(\mathrm{L} = \sqrt{\mathrm{d}^2 - \mathrm{w}^2}\)
\(\mathrm{L} = \mathrm{d} - \mathrm{w}\)
\(\mathrm{L} = \mathrm{d}^2 - \mathrm{w}^2\)