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TheForbiddenFollower

  • 3 years ago

if you have a triangle, and one side is 3 inch, other is 4. what is the largest and smallest possible lengths for the 3rd side?

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  1. LogicalApple
    • 3 years ago
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    Well you know the 3rd side cannot be larger than 3 + 4 from the triangle inequality s < 3 + 4 s < 7 So 7 is an upper bound. The length can be as close to 7 as you desire but it cannot be 7. As for the smallest length, consider what happens as the third side gets smaller and smaller. It would appear that s approaches 4 - 3 = 1. In fact, when you overlap the two sides, that leaves a distance of 1. This 1 serves as our lower bound. I.e., the third length can approach 1 but never reach it.

  2. LogicalApple
    • 3 years ago
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    I should also state that the 3rd side cannot be equal to 3 + 4 either***

  3. TheForbiddenFollower
    • 3 years ago
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    so the largest is 7 and smalled is 1? i also have to do 2in and 3in. so that would be 5 and1?

  4. LogicalApple
    • 3 years ago
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    They are boundary points. The third side can be 6.9999999999999999, so long as it is less than 7. There is no 'maximum' length as you can always pick a side closer and closer to 7. 7 is a limit that the side can never reach.

  5. LogicalApple
    • 3 years ago
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    Unless you consider a triangle with two sides overlapping, but then it wouldn't be a triangle anymore.

  6. TheForbiddenFollower
    • 3 years ago
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    ok, dude, you are awesome :) so you get a medal and a fan :) congratz, lol

  7. LogicalApple
    • 3 years ago
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    For similar reasons, the boundaries for the other triangle would be a lower limit of 1 and an upper limit of 5. But these limits can never be reached.

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