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barnoldwrl

  • 3 years ago

Is it possible for a triangle to have sides with the given lengths: 5in, 8in, & 15in?

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  1. barnoldwrl
    • 3 years ago
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    @hba

  2. UnkleRhaukus
    • 3 years ago
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    |dw:1358502234947:dw|

  3. barnoldwrl
    • 3 years ago
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    ??

  4. UnkleRhaukus
    • 3 years ago
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    the sides wont all meet

  5. barnoldwrl
    • 3 years ago
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    How about 10cm,12cm,20cm? @UnkleRhaukus

  6. PhoenixFire
    • 3 years ago
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    Three things must be true for the sides: Two sides added together must be larger than the third side. a+b>c, a+c>b, b+c>a So, for your first one: a=5, b=8, c=15 5+8=13 > 15 <---- false 8+15=23 > 5 <---- true 5+15=20 > 8 <---- true Because there is one false, the sides are invalid.

  7. barnoldwrl
    • 3 years ago
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    How about the second one... im not sure howt o do the formula though.. @PhoenixFire

  8. UnkleRhaukus
    • 3 years ago
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    Try construction

  9. PhoenixFire
    • 3 years ago
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    a=10 b=12 c=20 Check the three inequalities I gave you and see if any result in a false. If so, the triangle is invalid. If they are all true, then it is a valid triangle.

  10. barnoldwrl
    • 3 years ago
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    It isn't possible. thats what i got.

  11. barnoldwrl
    • 3 years ago
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    @PhoenixFire

  12. PhoenixFire
    • 3 years ago
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    10+12=22 > 20 <---- true 10+20=30 > 12 <---- true 12+20=32 > 10 <---- true The triangle is VALID.

  13. barnoldwrl
    • 3 years ago
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    I dont understand though.

  14. barnoldwrl
    • 3 years ago
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    Nevermind! i understand it now.. it has to be greater then the number. thats the part i wasn't understanding. i was doing it the opposite way thats why it through me off . thank you so much!! :D

  15. PhoenixFire
    • 3 years ago
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    haha. No problems.

  16. barnoldwrl
    • 3 years ago
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    I have 2 more questions that i need help on, do you mind giving me a hand? @PhoenixFire

  17. PhoenixFire
    • 3 years ago
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    sure.

  18. barnoldwrl
    • 3 years ago
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    Triangle PQR has medians QM and PN that intersect at Z. if Zm=4, what is QZ and QM. @PhoenixFire

  19. PhoenixFire
    • 3 years ago
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    |dw:1358503491884:dw| Supposedly ZM is half the length of QZ: |dw:1358503968318:dw| So if ZM=4 then QZ=8 which means QM=ZM+QZ=4+8=12 @UnkleRhaukus If you can validate this. I'm not a triangle expert.

  20. PhoenixFire
    • 3 years ago
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    It's a property of medians and centroids I believe. I'm not 100% certain on this.

  21. barnoldwrl
    • 3 years ago
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    Thats what i was thinking too. that it was half of that. so whats QZ and QM then?

  22. PhoenixFire
    • 3 years ago
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    I gave the answer in my post. ZM=Centroid to midpoint=a=4 QZ=vertex to centroid=2a=8 QM=vertex to midpoint=2a+a=12

  23. barnoldwrl
    • 3 years ago
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    QZ=8 and QM=12 is this right?

  24. PhoenixFire
    • 3 years ago
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    Yes

  25. barnoldwrl
    • 3 years ago
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    okay i have one more.... this one is hard i think lol

  26. barnoldwrl
    • 3 years ago
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    |dw:1358428925101:dw| @PhoenixFire

  27. PhoenixFire
    • 3 years ago
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    find X?

  28. barnoldwrl
    • 3 years ago
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    Yes, sorry.

  29. barnoldwrl
    • 3 years ago
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    you there? @phoenixfire

  30. PhoenixFire
    • 3 years ago
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    Thinking. Sorry. :P

  31. barnoldwrl
    • 3 years ago
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    Its okay.

  32. PhoenixFire
    • 3 years ago
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    I'm guessing the 2x line goes between midpoints of those sides?

  33. barnoldwrl
    • 3 years ago
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    Yeah

  34. PhoenixFire
    • 3 years ago
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    So what I've gathered is that the line connecting two midpoints is half the size of the third side. \[2x={1 \over 2}(x+12)\] So, solve for x I guess. \[4x=x+12\]\[3x=12\]\[x=4\]

  35. barnoldwrl
    • 3 years ago
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    Thank you!!!!

  36. PhoenixFire
    • 3 years ago
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    No problem. :)

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