Question 3 of 15intermediate🔍 AnalyzeLong Answer4 marks

For each set of lengths seen so far, you might have noticed that in at least two of the comparisons, the direct length was less than the sum of the other two (if not, check again!). For example, for the set of lengths 10 cm, 15 cm and 30 cm, there are two comparisons where this happens: 10 < 15 + 30, 15 < 10 + 30. But this doesn't happen for the third length: 30 > 10 + 15. Will this always happen? That is, for any set of lengths, will there be at least two comparisons where the direct length is less than the sum of the other two? Explore for different sets of lengths. Further, for a given set of lengths, is it possible to identify which lengths will immediately be less than the sum of the other two, without calculations? [Hint: Consider the direct lengths in the increasing order.]

Correct Answer

Yes, for any set of three lengths, there will always be at least two comparisons where the direct length is less than the sum of the other two.

Reasoning: When we compare the two smaller lengths with the largest length, the smaller lengths will always be less than the sum of the other two (since the sum includes the largest length).

Example with lengths a, b, c where a ≤\leq b ≤\leq c: a < b + c (always true since b + c ≥\geq c ≥\geq a) b < a + c (always true since a + c ≥\geq c ≥\geq b) The only comparison that can fail is whether c < a + b.

Conclusion: ∴ Without calculations, the two smaller lengths will always be less than the sum of the other two. ∴ Only the largest length needs to be checked against the sum of the two smaller lengths.

Exercise: Figure it Out (Section 2) | Q: 3 | (Page Page 9)
For More Understanding

Explanation

The chapter guides students to discover this pattern. When lengths are arranged in increasing order (a ≤\leq b ≤\leq c), the two smaller lengths (a and b) will always satisfy the inequality because their sum with the largest length will always exceed them. The only critical comparison is whether the largest length (c) is less than the sum of the two smaller lengths (a + b). This is why we only need to check if the longest side is less than the sum of the other two.

Solution Steps

  1. Arrange the three lengths in increasing order: a ≤\leq b ≤\leq c

  2. Compare a with (b + c): Since b + c ≥\geq c ≥\geq a, this inequality is always true

  3. Compare b with (a + c): Since a + c ≥\geq c ≥\geq b, this inequality is always true

  4. Compare c with (a + b): This is the only comparison that can fail

  5. Conclusion: The two smaller lengths always satisfy the inequality; only the largest length needs to be checked

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