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anonymous

  • one year ago

The figure shows three right triangles. Triangles JKM, KLM, and JLK are similar.

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  1. anonymous
    • one year ago
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    Theorem: If two triangles are similar, the corresponding sides are in proportion.

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    Using the given theorem, which two statements help to prove that if segment JL is x, then x2 = 100? A.Segment JL • segment JM = 64 Segment JL • segment LM = 48 B. Segment JL • segment JM = 48 Segment JL • segment LM = 36 C. Segment JL • segment JM = 64 Segment JL • segment LM = 36 D.Segment JL • segment JM = 36 Segment JL • segment LM = 64

  4. anonymous
    • one year ago
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    anyone know how to do this?

  5. anonymous
    • one year ago
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    do you know? @Calcmathlete

  6. anonymous
    • one year ago
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    I know how to do the problem, but it's worded very strangely to me. It would make much more sense for x to be JK, KL, or KM. Give me a second, and I'll help you.

  7. anonymous
    • one year ago
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    Ok. Have you done work with geometric means in these types of problems before?

  8. anonymous
    • one year ago
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    no its actually a pretest so i havent done this before

  9. anonymous
    • one year ago
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    but if x^2 is 100 then x must be 10

  10. anonymous
    • one year ago
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    JL is 10 i mean

  11. anonymous
    • one year ago
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    i thought it was x2 not x^2

  12. anonymous
    • one year ago
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    Whenever you copy and paste, an exponent, it usually pastes it as x2 or something. @Venice-Gribbin

  13. anonymous
    • one year ago
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    sorry hehe its x squared

  14. anonymous
    • one year ago
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    i see.

  15. anonymous
    • one year ago
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    Alright. Before we start, you may have noticed that you could've used Pythagorean Theorem to verify that JL = 10. Did you notice that?

  16. anonymous
    • one year ago
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    yeah 36 plus 64=100

  17. anonymous
    • one year ago
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    Ok. That's something that happens a lot in geometry. The key is that you need to use the given, which is the theorem they gave you. So, looking at similarity. The way the triangles are written are very significant. The order in which the letters appear tell you which sides are proportionate to what. So, if they were written as (JKM, KLM, and JLK) as you said, JL is proportionate to KL and JK. So, you could say that:\[\frac{JL}{KL}=\frac{JK}{KM}\]Are you following me so far?

  18. anonymous
    • one year ago
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    yes so can we plug in their values into this proportion?

  19. anonymous
    • one year ago
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    \[\frac{ 10 }{ 8 }=\frac{ 6 }{ ? }\]

  20. anonymous
    • one year ago
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    If you have them, yes. Now, with this type of problem, you can only work with the values they gave you, even though it's obvious that JL does indeed equal 10. So, the only values you can actually plug in are JK and KL. The above proportion won't help you here. That was just for example. Hold on, and I'll help you get the right equation.

  21. anonymous
    • one year ago
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    Lol @Gjallerhorn your name is perfect man

  22. anonymous
    • one year ago
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    fellow destiny player?

  23. anonymous
    • one year ago
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    Yeah I play it on xbox one. But before we continue this convo I want you to succeed in your math lol so ill wait till you get the answer

  24. anonymous
    • one year ago
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    anyone who is familiar with Feizel crux's masterpiece knows their "Destiny..."

  25. anonymous
    • one year ago
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    lolwut

  26. anonymous
    • one year ago
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    I'll tell you how I got this specific proportion afterwards, but the one that will help you is\[\frac{JM}{JK}=\frac{JK}{JL}\]This can be simplified to \[JM\times JL=JK^2\]Now, plug in the values you have.

  27. anonymous
    • one year ago
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    into the proportion or the equation under it?

  28. anonymous
    • one year ago
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    The equation under it. I got the bottom equation by cross multiplying. The only values they gave you were JK and KL.

  29. anonymous
    • one year ago
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    JMxJL=6^2 or 36?

  30. anonymous
    • one year ago
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    Yes. That gives you the answer you need because none of the other answers have that as part of the answer. If you wanted to completely do the question, the other proportion you would use is:\[\frac{JL}{KL}=\frac{KL}{LM}\]which simplifies to \[JL\times LM=KL^2\]

  31. anonymous
    • one year ago
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    ok lemme plug in

  32. anonymous
    • one year ago
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    JLxLM=64 or 8^2

  33. anonymous
    • one year ago
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    with those two values my answer would be D. correct?

  34. anonymous
    • one year ago
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    Yes. That's correct. I'll write how I got those specific proportions in a second.

  35. anonymous
    • one year ago
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    Now, the way I got those proportions is something called geometric means. A geometric mean between two numbers a and b is defined as \[Geometric ~Mean=\sqrt{ab}\]or you can also say \[ (Geometric~Mean)^2=a\times b\]Whenever you have a situation like this where you have the three similar right triangles with 2 of them within the bigger one, there are three relationships of "geometric means" that are extremely helpful (they are just similar triangles, so you can derive them later too). |dw:1436552274512:dw|

  36. anonymous
    • one year ago
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    One of the geometric mean relationships is that BD is the geometric mean of AD and DC. In other words, \[\frac{AD}{BD}=\frac{BD}{DC}\]or\[AD\times DC=BD^2\]which is a geometric mean. The other two are that BC is the geometric mean of CD and AC and that AB is the geometric mean of AD and AC.

  37. anonymous
    • one year ago
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    I remember it with the pneumonic "ALL".|dw:1436552524856:dw| That's the 'A' part of it.

  38. anonymous
    • one year ago
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    These are the other two 'L' |dw:1436552558243:dw| |dw:1436552572987:dw| Sorry that the explanation is so long, just thought it'd be good if you knew where I got the information from.

  39. anonymous
    • one year ago
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    wow thanks so much!!!!

  40. anonymous
    • one year ago
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    No problem. If you need me to explain anything else, let me know.

  41. anonymous
    • one year ago
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    Also, keep in mind that they need to be right triangles for these things to work because otherwise, they wouldn't be similar triangles.

  42. anonymous
    • one year ago
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    im definitely fanning you.

  43. anonymous
    • one year ago
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    thanks for the guidance and explanation!

  44. anonymous
    • one year ago
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    No problem. I'm not online very often tbh, but when I am, I try to help as many people as possible.

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