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trantom

  • one year ago

Please help! I have this worksheet due tomorrow and I don't want answers but I don't know what to do. I tried using right triangles but there aren't any I know enough about. I'll attach a drawing of the problem in the thread. Thanks!

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  1. trantom
    • one year ago
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    |dw:1433114484012:dw|

  2. trantom
    • one year ago
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    The 11 refers to \[LT\]

  3. trantom
    • one year ago
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    |dw:1433114690021:dw|

  4. ganeshie8
    • one year ago
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    what are you trying to find

  5. trantom
    • one year ago
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    The missing lengths sorry I should have said that xD

  6. trantom
    • one year ago
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    PT, PN, LR, and RT (the lengths of all these segments)

  7. ganeshie8
    • one year ago
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    Okay what lengths are you given ?

  8. trantom
    • one year ago
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    They're labelled in the diagram

  9. ganeshie8
    • one year ago
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    diagram is a bit confusing, can you write down the given lengths

  10. trantom
    • one year ago
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    But here LM=7 MN=8 TN=10 PR=3 RM=6 LT=11

  11. trantom
    • one year ago
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    |dw:1433114943288:dw|

  12. ganeshie8
    • one year ago
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    you may use "intersecting chords theorem" to find RT

  13. ganeshie8
    • one year ago
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    |dw:1433115060081:dw|

  14. trantom
    • one year ago
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    oh ok I get that

  15. ganeshie8
    • one year ago
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    what do you get for RT ?

  16. trantom
    • one year ago
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    LR=9 RT=2

  17. trantom
    • one year ago
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    I got you man thanks so much!

  18. trantom
    • one year ago
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    Now what about PT, can I do the same thing for that?

  19. ganeshie8
    • one year ago
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    how do you know its not LR = 2 RT = 9 ?

  20. trantom
    • one year ago
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    Because the two larger ones are always on the same side of the intersection

  21. trantom
    • one year ago
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    oh but it could be the other way too.. oh actually that's weird

  22. trantom
    • one year ago
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    oh I know

  23. trantom
    • one year ago
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    A triangle with side lengths 6, 7, and 2 is impossible right? wait not that's totally possible Agh I'm confused

  24. trantom
    • one year ago
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    oh because LR goes past the center and RT doesn't start until after that

  25. ganeshie8
    • one year ago
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    maybe lets find PT first

  26. trantom
    • one year ago
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    thats a good idea xD

  27. trantom
    • one year ago
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    You could use a lot of trig to get angle LRM, then use its vertical angle PRT to find PT

  28. ganeshie8
    • one year ago
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    there is an easy formula

  29. ganeshie8
    • one year ago
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    simply use this |dw:1433115780042:dw|

  30. trantom
    • one year ago
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    so... 15x8=10(x-10) and x-10 is PT

  31. ganeshie8
    • one year ago
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    shouldnt it be 15x8=10(x\(\color{red}{+}\)10) ?

  32. trantom
    • one year ago
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    if you want x to be pt then yes, but I was having x-10 be pt anyway I got pt=12

  33. ganeshie8
    • one year ago
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    |dw:1433116168532:dw|

  34. trantom
    • one year ago
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    or x is 4 in yours

  35. ganeshie8
    • one year ago
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    NT = 10 right ?

  36. ganeshie8
    • one year ago
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    you should be getting PT=2 http://www.wolframalpha.com/input/?i=solve%208(8%2B7)%20%3D%2010(10%2Bx)&t=crmtb01

  37. trantom
    • one year ago
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    where did you get x+8 from? there's no 8 there

  38. trantom
    • one year ago
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    shouldn't it be 15*8 = 10(x+10)

  39. ganeshie8
    • one year ago
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    sry it was a typo yeah it is `15*8 = 10(x+10)`

  40. trantom
    • one year ago
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    12= x+10 x=2 yep

  41. trantom
    • one year ago
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    Ok I get it thanks!

  42. trantom
    • one year ago
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    and the reason why LR=9 and RT=2 instead of LR=2 and RT=9 is because a triangle with side lengths 2, 3, and 9 is impossible!

  43. ganeshie8
    • one year ago
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    Looks good!

  44. trantom
    • one year ago
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    Thank you so much I've been getting B+'s and A-'s on a couple tests lately I need to get back to being an A student xD

  45. ganeshie8
    • one year ago
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    I see you're very good in geometry, you deserve A+

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