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anonymous

  • 4 years ago

Need help with Geometry and Ratio Question!! Please Help!!

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  1. anonymous
    • 4 years ago
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    The base of a triangular piece of paper ABC is 12 cm long. The paper is folded down over the base, with crease DE parallel to the base of the paper. The area of the triangle that projects below the base is 16% that of the area of the triangle ABC. What is the length of DE, in cm.

  2. anonymous
    • 4 years ago
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    |dw:1327792772445:dw|

  3. anonymous
    • 4 years ago
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    |dw:1327793675334:dw| ok so we actually are interested in the area of the triangle CFG too, since that is the one that projects below the base.

  4. anonymous
    • 4 years ago
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    Yes

  5. anonymous
    • 4 years ago
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    What is the connection between those triangles?

  6. anonymous
    • 4 years ago
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    These are similar triangles. There exists a scale factor s>1 such that \[ABC = s^2CFG.\] Another fact about s is that \[FGs = AB = 12\] Therefore \[s = \frac{12}{FG}.\] There also exists another scale factor, r>1 such that \[ABC=r^2CDE\] and obviously another fact about r is that \[rDE = AB = 12\] Therefore \[r=\frac{12}{DE}\] We are also told that the area of the triangle that projects below the base is 16% that of the area of the triangle ABC. So we get \[0.16ABC = CFG.\] What I think we should do is solve for r solve for s and then solve for DE.

  7. anonymous
    • 4 years ago
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    i understand that they are similar but I don't quite understand why is (s^2)(CFG)=ABC. Why does s have to be squared?

  8. anonymous
    • 4 years ago
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    is it some sort of formula?

  9. anonymous
    • 4 years ago
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    \[0.16s^2CFG = CFG \Longrightarrow s=\sqrt{\frac{1}{0.16}} = 2.5\] Also, \[s=\frac{12}{FG} \Longrightarrow FG = 4.8\]

  10. anonymous
    • 4 years ago
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    well similar lengths are scaled by a factor, and similar areas are scaled by the square of that factor

  11. anonymous
    • 4 years ago
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    ok so we have found s, maybe try to find r next

  12. anonymous
    • 4 years ago
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    oh, i see.

  13. anonymous
    • 4 years ago
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    so does it work the same for r too?

  14. anonymous
    • 4 years ago
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    |dw:1327795417846:dw|

  15. anonymous
    • 4 years ago
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    r is a little more tricky...

  16. anonymous
    • 4 years ago
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    because we don't know what percent is CDE of ABC

  17. anonymous
    • 4 years ago
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    actually no... \[(0.4)^2 ABh_1 = FGh_2\] So it must be that \[0.4h_1 = h_2\]

  18. anonymous
    • 4 years ago
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    yes, but what about CED?

  19. anonymous
    • 4 years ago
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    Yeah, we need to try to figure out something relating s and CDE so we can use \[ABC = r^2CDE\]

  20. anonymous
    • 4 years ago
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    oh, ok

  21. anonymous
    • 4 years ago
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    I'm going to think about it for a bit lol

  22. anonymous
    • 4 years ago
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    ok,thx

  23. phi
    • 4 years ago
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    If you want a hint. I hope this is not too cryptic

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  24. anonymous
    • 4 years ago
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    Can you clarify ? |dw:1327797362393:dw|

  25. anonymous
    • 4 years ago
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    why is it 2/5?

  26. phi
    • 4 years ago
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    0.4 you figured that out up top. It is the ratio of altitudes of 2 similar triangles. When the areas are in the ratio of 0.16/1 the altitudes are in the ratio sqrt(.16)/sqrt(1)= 0.4= 2/5

  27. anonymous
    • 4 years ago
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    oh yeah. Thank you.

  28. anonymous
    • 4 years ago
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    yeah phi that's excellent, 0.4 is the scale factor for the heights of the triangles I had above: \[0.4h_1 = h_2\] and yeah r is the ratio of the heights of CDE and ABC. Thanks phi

  29. anonymous
    • 4 years ago
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    Thank you, guys!!!

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