Question 3 - Figure it out - Page 159 - Chapter 7 Class 7 - A tale of three Intersecting Lines (Ganit Prakash)

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Question 3 For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values could also be chosen): (a) 1, 100 (b) 5, 5 (c) 3, 7 See if you can describe all possible lengths of the third side in each case, so that a triangle exists with those sidelengths. For example, in case (a), all numbers strictly between 99 and 101 would be possible. Let’s first do this in theory Let two sides given be a, b and 3rd side we want to find be x Now, applying Triangle inequality on this Triangle inequality says Sum of two sides > 3rd side Since we don’t know which side is bigger, we try all three cases Case 1: a + b > x Case 2: a + x > b Case 3: b + x > a From Case 1 a + b > x x < a + b Thus, 3rd side must be less than the sum of both sides From Case 2 a + x > b x > b – a Thus, 3rd side must be greater than the difference of both sides From Case 3 b + x > a x > a – b Thus, 3rd side must be greater than the difference of both sides Thus, our rules are Third side must be less than sum of both sides Third side must be greater than difference of both sides (a) 1, 100 Sum of both sides = 100 + 1 = 101 Difference of both sides = 100 – 1 = 99 So, our third side must be between 101 and 99 Note: This is between, not equal Thus, numbers could be 99.5, 100, 100.5, 100.6, 100.7 (b) 5, 5 Sum of both sides = 5 + 5 = 10 Difference of both sides = 5 – 5 = 0 So, our third side must be between 0 and 10 Note: This is between, not equal Thus, numbers could be 1, 2, 3, 4, 5 (c) 3, 7 Sum of both sides = 3 + 7 = 10 Difference of both sides = 7 – 3 = 4 So, our third side must be between 4 and 10 Note: This is between, not equal Thus, numbers could be 5, 6, 7, 8, 9