1. Joined
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    05 Oct '08 19:55
    Find the area of a quadrilateral whose vertices have coordinates (-4, -6),

    (-3, 5), (4, 7), and (5, -9).
  2. Fichtekränzi
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    05 Oct '08 20:55
    why should I?
  3. Joined
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    05 Oct '08 22:25
    Originally posted by afx
    why should I?
    If you honestly need a reason then don't. I'm just looking for a little help on my math homework. Not that you could help me anyways :]
  4. Subscriberjoe shmo
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    05 Oct '08 22:352 edits
    Originally posted by RatskcorWagoner
    If you honestly need a reason then don't. I'm just looking for a little help on my math homework. Not that you could help me anyways :]
    are you looking for an analyitical answer?, because I did it graphically

    here is what I get for the Area

    77 + 49/2 + 11/2 + 192/18 + 225/32 + 77/96 = 125.5

    like i said, probably not what your looking for?
  5. Joined
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    06 Oct '08 00:31
    Originally posted by joe shmo
    are you looking for an analyitical answer?, because I did it graphically

    here is what I get for the Area

    77 + 49/2 + 11/2 + 192/18 + 225/32 + 77/96 = 125.5

    like i said, probably not what your looking for?
    I'm not really sure what i am looking for but I know that 125.5 is incorrect.
    any other ideas?
  6. Subscriberjoe shmo
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    06 Oct '08 00:451 edit
    Originally posted by RatskcorWagoner
    I'm not really sure what i am looking for but I know that 125.5 is incorrect.
    any other ideas?
    nope...can you point out the flaw in my method? In case your wondering what I did, I plotted the points, connected the lines, and divided the area into a rectangle and several triangles, summing the areas

    What is the title of the Math course?

    here is a link with some information:http://mathworld.wolfram.com/Quadrilateral.html
  7. Joined
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    06 Oct '08 00:54
    hmm well it is pre calc so i don't know it seems like the logical thing to do..
  8. Subscriberjoe shmo
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    06 Oct '08 01:05
    Originally posted by RatskcorWagoner
    hmm well it is pre calc so i don't know it seems like the logical thing to do..
    so, what is the area? Was I even close?
  9. Joined
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    06 Oct '08 01:08
    If I knew i wouldn't be asking anyone. haha
    i will find out the answer tomorrow and will post it.
    any other ideas?
  10. Subscriberjoe shmo
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    06 Oct '08 01:19
    Originally posted by RatskcorWagoner
    If I knew i wouldn't be asking anyone. haha
    i will find out the answer tomorrow and will post it.
    any other ideas?
    but how do you know it is not 125.5?
  11. Joined
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    06 Oct '08 02:13
    I broke it into 2 triangles.

    One with sides of sqrt(122), sqrt(53), sqrt(233)...
    One with sides of sqrt(257), sqrt(53), sqrt(260).

    Using the formula A = 1/4 * sqrt( (A^2 + B^2 + C^2)^2 - 2*(A^4 + B^4 + C^4) ), I get the following..

    Triangle 1
    Area = 1/4 * Sqrt[(122 + 53 + 233)^2 - 2*(122^2 + 53^2 + 233^2)]
    Area = 1/4 * Sqrt[408^2 - 2*(14,884 + 2,809 + 54,289)]
    Area = 1/4 * Sqrt(166,464 - 2*71,982)
    Area = 1/4 * Sqrt(22,500) = 150 / 4 = 37.5

    Triangle 2
    Area = 1/4 * Sqrt[(257 + 53 + 260)^2 - 2*(257^2 + 53^2 + 260^2)]
    Area = 1/4 * Sqrt[570^2 - 2*(66,049 + 2,809 + 67,600)]
    Area = 1/4 * Sqrt(324,900 - 2*136,458)
    Area = 1/4 * Sqrt(51,984) = 228 / 4 = 57

    Total Area = 37.5 + 57 = 94.5
  12. Subscriberjoe shmo
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    06 Oct '08 02:36
    Originally posted by geepamoogle
    I broke it into 2 triangles.

    One with sides of sqrt(122), sqrt(53), sqrt(233)...
    One with sides of sqrt(257), sqrt(53), sqrt(260).

    Using the formula A = 1/4 * sqrt( (A^2 + B^2 + C^2)^2 - 2*(A^4 + B^4 + C^4) ), I get the following..

    Triangle 1
    Area = 1/4 * Sqrt[(122 + 53 + 233)^2 - 2*(122^2 + 53^2 + 233^2)]
    Area = 1/4 * Sqrt[408^2 - 2*(14,884 ...[text shortened]... (324,900 - 2*136,458)
    Area = 1/4 * Sqrt(51,984) = 228 / 4 = 57

    Total Area = 37.5 + 57 = 94.5
    that looks to technical to be false, so you da person geepamoogle.
  13. Joined
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    07 Oct '08 00:58
    Pick's Theorem? Although I'll need a piece of graph paper for that.
  14. Standard memberO Artem O
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    07 Oct '08 02:12
    so whats the answer?
  15. Subscribersonhouse
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    07 Oct '08 03:42
    Originally posted by joe shmo
    are you looking for an analyitical answer?, because I did it graphically

    here is what I get for the Area

    77 + 49/2 + 11/2 + 192/18 + 225/32 + 77/96 = 125.5

    like i said, probably not what your looking for?
    Are we assuming the first # to be the X axis, and second # to be Y?
    I got the 77, but I got 8+5.5+7+12+77 ,minus two small areas, which comes out to 109 and change.
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