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dariyana

A triangle has the following measurements. What is a possible length for the third side? AB = 5, BC = 7, CA = ? 2 12 5 13

  • one year ago
  • one year ago

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  1. dariyana
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    @Wired

    • one year ago
  2. dariyana
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    i think it B

    • one year ago
  3. lgbasallote
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    *greater

    • one year ago
  4. Wired
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    Thanks @lgbasallote , brain fart :) Any two sides when added have to be greater than the third side. Therefore, AB + BC > CA If it was B then AB + AC = CA

    • one year ago
  5. Wired
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    If they were less than or equal to the third side, you wouldn't have a triangle.

    • one year ago
  6. dariyana
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    so it would be 13

    • one year ago
  7. eyust707
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    |dw:1341624363071:dw|

    • one year ago
  8. eyust707
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    i dont think 13 will work

    • one year ago
  9. eyust707
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    We could make them all the way flat but they still wouldnt be 13

    • one year ago
  10. dariyana
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    when you add the two they equal 12

    • one year ago
  11. eyust707
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    Yes so it cant be 12 either

    • one year ago
  12. eyust707
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    other wise they would be flat on top of each other

    • one year ago
  13. Wired
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    No. AB + BC > CA 7 + 5 > CA 12 > CA Possible equations: 12 > 2 12 > 5 12 > 12 12 > 13 It's not 12 or 13, because 12 isn't greater than itself, and 12 isn't greater than a large number. Now, since CA = 2 or 5: AB < BC + CA (again, two sides added up are greater than the third) 7 < 5 + 2 7 < 7 7 < 5 + 5 7 < 10 7 can't be less than itself. 7 is less than 10, so you know 5 is the answer.

    • one year ago
  14. eyust707
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    12 doesnt work: |dw:1341624504291:dw|

    • one year ago
  15. dariyana
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    lol so it 5

    • one year ago
  16. eyust707
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    YEP! its 5 Cause 2 is too small |dw:1341624584353:dw|

    • one year ago
  17. Calcmathlete
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    Here. x < 5 + 7 x + 5 > 7 x < 12 x > 2 Therefore, you know that the set of numbers possible are \(2 < x < 12\). Which of those numbers is within that range?

    • one year ago
  18. dariyana
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    ok thanks you guys

    • one year ago
  19. Calcmathlete
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    You're welcome :)

    • one year ago
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