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anonymous

  • 4 years ago

Geometry and Ratio problem!!! Help!!!

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  1. anonymous
    • 4 years ago
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    The figure ABCDEF below is a regular hexagon, and point P lies on side AB, with AP=3 cm and PB=1cm. Line PE meets AD at Q. What is the ratio of the area of quadrilateral AQEF to the area of hexagon ABCDEF? Express your answer as a common fraction.

  2. anonymous
    • 4 years ago
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    |dw:1327787358619:dw|

  3. Hero
    • 4 years ago
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    All you do is find the areas and put them in fraction form

  4. anonymous
    • 4 years ago
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    how do I find the area of AQEF?

  5. Hero
    • 4 years ago
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    It's not that difficult

  6. anonymous
    • 4 years ago
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    Can you give me some hints? Please.

  7. Hero
    • 4 years ago
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    |dw:1327787517063:dw|

  8. Hero
    • 4 years ago
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    There's one hint

  9. Hero
    • 4 years ago
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    Ready for the next hint?

  10. anonymous
    • 4 years ago
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    yes

  11. Hero
    • 4 years ago
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    Ready for the next one?

  12. anonymous
    • 4 years ago
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    Wait, so all the side lengths are 3? Isn't it 4?

  13. Hero
    • 4 years ago
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    oh okay, I didn't read that in-depth. This is different

  14. anonymous
    • 4 years ago
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    ok

  15. Hero
    • 4 years ago
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    |dw:1327787748238:dw|

  16. anonymous
    • 4 years ago
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    Yes

  17. Hero
    • 4 years ago
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    You have to find the area of the hexagon first. There's no way to avoid that

  18. Hero
    • 4 years ago
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    Good luck with this

  19. anonymous
    • 4 years ago
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    Area of the hexagon = |dw:1327788115227:dw|

  20. Hero
    • 4 years ago
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    I don't have a solid approach to this. Sorry bud

  21. Hero
    • 4 years ago
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    I can take a look at it later maybe.

  22. anonymous
    • 4 years ago
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    Oh, ok. Thanks for trying though.

  23. Hero
    • 4 years ago
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    AD splits the hex into two halves. That might help.

  24. anonymous
    • 4 years ago
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    ok, I will keep trying.

  25. Hero
    • 4 years ago
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    |dw:1327788413768:dw|

  26. Hero
    • 4 years ago
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    |dw:1327788548241:dw|

  27. anonymous
    • 4 years ago
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    Area of hexagon equals 24sqrt{3}, so ABCD equals 12sqrt{3} and ADEF equals 12 sqrt{3} as well.

  28. anonymous
    • 4 years ago
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    find the heigh of AE, calcuate the area of triangle APE and AFE...then you have an answer i guess

  29. anonymous
    • 4 years ago
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    the height of AE is 4sqrt{3}, APE is 6sqrt{3}. But how do I find AFE and AQP then?

  30. anonymous
    • 4 years ago
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    |dw:1327789175987:dw|

  31. anonymous
    • 4 years ago
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    What about AQE?

  32. anonymous
    • 4 years ago
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    I made an mistake, AEF should be 4sqrt{3}.

  33. Hero
    • 4 years ago
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    asnaseer, moneybird, and mertsj should be able to figure this out.

  34. asnaseer
    • 4 years ago
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    I think matematika and yociyoci are doing well so far...

  35. anonymous
    • 4 years ago
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    im not sure....but isn't there formula for calulating triangle if you know one side and angels?maybe with help of this you can find out area of this one

  36. anonymous
    • 4 years ago
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    But I don't know that formula, is there another approach?

  37. anonymous
    • 4 years ago
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    You can find the height of triangle EDQ

  38. anonymous
    • 4 years ago
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    triangle EDQ and triangle PAQ are similar

  39. anonymous
    • 4 years ago
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    so the height of EDQ is|dw:1327790090893:dw| ?

  40. anonymous
    • 4 years ago
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    AD splits the hexagon in halves so quadrilateral AQEF = quadrilateral AFED - triangle FDQ

  41. asnaseer
    • 4 years ago
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    you mean - EDQ moneybird

  42. anonymous
    • 4 years ago
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    is triangle EDQ eqauls to 32sqrt{3}/7 ?

  43. anonymous
    • 4 years ago
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    ya |dw:1327790174473:dw| From midpoint of ED to midpoint of AB is 4 sqrt(3)

  44. anonymous
    • 4 years ago
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    now use ratio to find the height of triangle EDQ and triangle APQ

  45. anonymous
    • 4 years ago
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    \[\frac{4}{h} = \frac{3}{4\sqrt{3}-h}\]

  46. anonymous
    • 4 years ago
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    I don't quite understand what you just did. Can you clarify, please?

  47. anonymous
    • 4 years ago
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    Which part?

  48. anonymous
    • 4 years ago
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    the diagram

  49. anonymous
    • 4 years ago
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    when you cut the hexagon into six equal pieces, you take out one of the piece, which is a triangle

  50. anonymous
    • 4 years ago
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    ok

  51. anonymous
    • 4 years ago
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    |dw:1327790925432:dw|so the area of the hexagon is 24 sqrt(3) ratio is

  52. anonymous
    • 4 years ago
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    ok, i understand. thank you very much, guys!!!! I appreciated!!!

  53. asnaseer
    • 4 years ago
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    very well explained moneybird

  54. anonymous
    • 4 years ago
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    ty

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