In a triangle XYZ, which one of the following conditions is always true for the side lengths (triangle inequality)? (Assume XY, YZ, and ZX are the lengths of the three sides.)

Difficulty: Easy

Correct Answer: XY + YZ > ZX

Explanation:


Introduction / Context:
This question tests the triangle inequality theorem, one of the most fundamental rules for triangles. A triangle is possible only if any two sides together are longer than the third side. If this is not true, the shape cannot close into a triangle; the points would lie on a straight line (degenerate) or be impossible. The theorem applies to all triangles regardless of type (scalene, isosceles, equilateral) as long as side lengths are positive.


Given Data / Assumptions:

  • XYZ is a valid triangle
  • XY, YZ, ZX are positive side lengths
  • We need a condition that is always true


Concept / Approach:
Triangle inequality requires: XY + YZ > ZX YZ + ZX > XY ZX + XY > YZ Any condition using “<” is invalid for a real triangle, and “=” represents a degenerate case (not a proper triangle).


Step-by-Step Solution:
Pick the option that matches triangle inequality form “sum of two sides is greater than the third”. Option “XY + YZ > ZX” exactly matches the theorem. Therefore, it is always true for every valid triangle XYZ.


Verification / Alternative check:
Example triangle with sides 3, 4, 5: 3 + 4 = 7 > 5, so the condition holds. If you try a non-triangle like 2, 3, 5, then 2 + 3 = 5 which is not greater, and it becomes a straight line, confirming why “>” is necessary.


Why Other Options Are Wrong:
Any “<” option contradicts triangle inequality and implies a triangle cannot be formed. The “=” option describes a degenerate case where points are collinear, not a proper triangle.


Common Pitfalls:
Using “greater than or equal to” instead of strictly greater, confusing side rules with angle rules, or accidentally interpreting expressions without the plus sign correctly.


Final Answer:
The always true condition is XY + YZ > ZX.

More Questions from Alligation or Mixture

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion