Question:


The figure shows a cube with edge of length e.
COLUMN A: The length of diagonal AB
COLUMN B: \(\sqrt{2}e\)

Show Hint

Memorize the formulas for the diagonals of a cube: the face diagonal is \(e\sqrt{2}\) and the space diagonal is \(e\sqrt{3}\). The space diagonal is always longer than the face diagonal.
Updated On: Oct 4, 2025
  • The quantity in Column A is greater.
  • The quantity in Column B is greater.
  • The two quantities are equal.
  • The relationship cannot be determined from the information given.
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This question asks for the length of the space diagonal of a cube and compares it to the length of a face diagonal.
Step 2: Key Formula or Approach:
1. The length of a face diagonal of a cube with side \(e\) is found using the Pythagorean theorem on a square face: \(d_{face} = \sqrt{e^2 + e^2} = \sqrt{2e^2} = e\sqrt{2}\).
2. The length of the space diagonal (like AB) is found by applying the Pythagorean theorem again to a right triangle formed by an edge, a face diagonal, and the space diagonal: \(d_{space} = \sqrt{e^2 + (d_{face})^2}\). The general formula for a space diagonal of a rectangular prism with sides l, w, h is \( \sqrt{l^2 + w^2 + h^2} \). For a cube, this is \( \sqrt{e^2 + e^2 + e^2} = \sqrt{3e^2} = e\sqrt{3} \).
Step 3: Detailed Explanation:
Column A: We need to find the length of the diagonal AB, which is a space diagonal of the cube.
Using the formula for the space diagonal of a cube with edge length \(e\):
\[ \text{Length of AB} = \sqrt{e^2 + e^2 + e^2} = \sqrt{3e^2} = e\sqrt{3} \] So, the quantity in Column A is \(e\sqrt{3}\). Column B: The quantity is given as \( \sqrt{2} e \), which is \(e\sqrt{2}\). This is the length of a face diagonal.
Comparison: We are comparing \(e\sqrt{3}\) (Column A) with \(e\sqrt{2}\) (Column B). Since \(e\) is a length, it must be positive. We can divide both sides by \(e\). The comparison is now between \( \sqrt{3} \) and \( \sqrt{2} \). Since \(3>2\), we know that \( \sqrt{3}>\sqrt{2} \). Therefore, the quantity in Column A is greater.
Step 4: Final Answer:
The length of the space diagonal is \(e\sqrt{3}\), which is greater than the length of the face diagonal, \(e\sqrt{2}\).
Was this answer helpful?
0
0

Questions Asked in GRE exam

View More Questions