Which one of the following sets of three lengths can form the sides of a triangle under the triangle inequality rule?

Introduction / Context:
This question tests your knowledge of the triangle inequality theorem, a fundamental rule in geometry that determines whether three given lengths can form a triangle. The theorem states that in any triangle, the sum of the lengths of any two sides must be strictly greater than the length of the remaining side. Aspirants for competitive exams must be very comfortable applying this rule quickly to check potential side combinations, since it appears frequently in aptitude tests and school level questions.

Given Data / Assumptions:
• Four different sets of three positive lengths are given as options.
• We need to identify which set can represent the three sides of a triangle.
• All lengths are in centimetres and are assumed to be exact side lengths, not approximate or ranges.

Concept / Approach:
The triangle inequality theorem has three conditions for each triple of side lengths a, b and c (with c being the largest):
• a + b > c,
• a + c > b,
• b + c > a.
If any one of these inequalities fails (that is, if the sum is less than or equal to the remaining side), then the three lengths cannot form a valid triangle. To solve this question efficiently, you can quickly sort or identify the largest side in each option, then check whether the sum of the other two sides is strictly greater than that largest side.

Step-by-Step Solution:
Step 1: Consider option (a): 9 cm, 6 cm, 2 cm. The largest side is 9 cm. Check 6 + 2 = 8, which is less than 9, so this set cannot form a triangle. Step 2: Consider option (b): 11 cm, 3 cm, 12 cm. The largest side is 12 cm. Check 11 + 3 = 14, which is greater than 12. For completeness, 11 + 12 > 3 and 3 + 12 > 11 are obviously true. So this set satisfies the triangle inequality and can form a triangle. Step 3: Consider option (c): 3 cm, 5 cm, 8 cm. The largest side is 8 cm. Check 3 + 5 = 8, which is equal to 8, not strictly greater. This represents a degenerate case, not a proper triangle, so the triangle inequality fails. Step 4: Consider option (d): 5 cm, 7 cm, 13 cm. The largest side is 13 cm. Check 5 + 7 = 12, which is less than 13, so this set cannot form a triangle. Step 5: Conclude that only option (b) satisfies all the triangle inequality conditions.

Verification / Alternative check:
For option (b) with sides 11 cm, 3 cm and 12 cm, we can quickly see that all conditions are met: 11 + 3 = 14 > 12, 11 + 12 = 23 > 3, and 3 + 12 = 15 > 11. This ensures that a non degenerate triangle exists. For the other options, at least one sum of two sides is less than or equal to the third side, preventing the formation of a proper triangle. Thus, the triangle inequality is applied correctly and unambiguously identifies the only valid choice.

Why Other Options Are Wrong:
In option (a), 6 + 2 = 8, which is less than 9, so a side would be too long to meet the others. In option (c), 3 + 5 = 8 exactly, which would make the three points collinear, leading to a degenerate triangle with zero area, which is not acceptable for most exam definitions of a triangle. In option (d), 5 + 7 = 12, which is less than 13, making one side longer than the sum of the other two, again impossible for a triangle. Therefore, all these violate the strict triangle inequality rule.

Common Pitfalls:
A common mistake is to treat the condition a + b ≥ c as sufficient and to allow equality, but equality leads only to a degenerate triangle. Exams usually expect strict inequalities. Another pitfall is to check only the sum of the two smallest sides without carefully identifying the largest side, or to forget to verify the other two inequalities even though they are usually satisfied once the largest side check passes. Developing the habit of quickly spotting the largest side and checking the sum of the remaining two against it will make triangle inequality problems much faster and more reliable to solve.

Final Answer:
The only valid combination that can form the sides of a triangle is 11 cm, 3 cm, 12 cm (option b).