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anonymous

  • one year ago

∆ABC has vertices A(0, 0), B(2, 4), and C(4, 2). ∆RST has vertices R(0, 3), S(-1, 5), and T(-2, 4). Prove that ∆ABC ∼ ∆RST. (Hint: Graph ∆ABC and ∆RST in the coordinate plane.)

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  1. dm91
    • one year ago
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    @steve816 he can help you i think.

  2. steve816
    • one year ago
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    Well, do you know how to plot points on the coordinate plane? All you do is draw the points and connect the dots.

  3. dm91
    • one year ago
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    |dw:1443590087569:dw|

  4. anonymous
    • one year ago
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    omg thank you

  5. steve816
    • one year ago
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    No problem :)

  6. anonymous
    • one year ago
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    so how do i do this problem?

  7. anonymous
    • one year ago
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    @steve816

  8. dm91
    • one year ago
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    ok so as steve put it do you have a graph?

  9. anonymous
    • one year ago
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    ive already graphed the points but i dont know how to prove it similar

  10. steve816
    • one year ago
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    If the triangles are the same size once you graphed, then they are similar.

  11. anonymous
    • one year ago
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    i thought you had to do the distance formula?

  12. anonymous
    • one year ago
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    they werent the same size

  13. steve816
    • one year ago
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    Yes, you can use the distance formula

  14. steve816
    • one year ago
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    For each of the sides

  15. anonymous
    • one year ago
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    okay how can i do it with the distance formula?

  16. steve816
    • one year ago
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    Plug in your (x, y) coordinates into this formula: http://2.bp.blogspot.com/-GMrbM3fg4NU/VHljne7DleI/AAAAAAAABec/3emecyWG4ew/s1600/Slide1.jpg

  17. anonymous
    • one year ago
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    ok give me a second

  18. anonymous
    • one year ago
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    wait which numbers do i put into it?

  19. anonymous
    • one year ago
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    i did AB and got 20 is that correct?

  20. anonymous
    • one year ago
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    ?

  21. anonymous
    • one year ago
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    @steve816 is that correct?

  22. anonymous
    • one year ago
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    hello?

  23. anonymous
    • one year ago
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    i need help fast can someone help?

  24. dm91
    • one year ago
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    ok now what?

  25. anonymous
    • one year ago
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    is that correct?

  26. anonymous
    • one year ago
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    i got the squareroot of 20 for AB

  27. dm91
    • one year ago
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    yes

  28. anonymous
    • one year ago
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    okay thanks

  29. dm91
    • one year ago
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    yep

  30. anonymous
    • one year ago
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    for rs i got the sqaureroot of 5 is that correct?

  31. dm91
    • one year ago
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    yes

  32. anonymous
    • one year ago
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    and for CT i got the sqareroot of 40 is that correct?

  33. dm91
    • one year ago
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    yes

  34. anonymous
    • one year ago
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    ok what do i do now?

  35. dm91
    • one year ago
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    no that one would have to be up to Steve because he gave you the square root idea but i might be able to get you someone else that can help because Steve is helping other pp at the time.

  36. anonymous
    • one year ago
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    okay thanks!

  37. dm91
    • one year ago
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    @dan815

  38. dm91
    • one year ago
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    yep

  39. dm91
    • one year ago
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    @Koikkara

  40. Koikkara
    • one year ago
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    Hai, how can i help you ?

  41. anonymous
    • one year ago
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    i was doing the distance formula for the problem and dont know what to do next

  42. Koikkara
    • one year ago
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    |dw:1443591908295:dw| According to your question..

  43. anonymous
    • one year ago
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    yes

  44. anonymous
    • one year ago
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    i did Ab and got the squareroot of 20 then i did RS and got the sqareroot of 5 then i did CT and got the squareroot of 40

  45. Koikkara
    • one year ago
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    hmm... better, this user @dan815 can help you more clearly.

  46. zpupster
    • one year ago
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    this is cheating a bit becasue this program gives the lentgh of the sides but you need to use the SSS postulate and this is easier because 2 side length are the same \[\frac{ BC}{ EF} = \frac{ AB }{DF} \]

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