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Area of a Quadrilateral

  1. Apr 4, 2016 #1
    • Originally posted in a non-homework forum section
    Hi,
    I got a question from a book to find the area of a Quadrilateral. I divided the quadrilateral into two triangles but answer is not correct. Some body please guide me.

    I am uploading my work in a attached file.

    Zulfi. Book Page 315.jpg
     
  2. jcsd
  3. Apr 4, 2016 #2

    ProfuselyQuarky

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    For ADB, did you write ##A=\frac {1}{2}(6)(4)^2##?

    And is triangle DBC a right triangle?
     
    Last edited: Apr 4, 2016
  4. Apr 4, 2016 #3

    ProfuselyQuarky

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    Oh, and this should also be in the homework forum, not general math.
     
  5. Apr 4, 2016 #4

    Ssnow

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    For ADB is correct. For the second isn't simply ##\frac{1}{2}6\times 6##, hint: calculate ##BD## and ##DC## (that are not ##6##) ...
     
  6. Apr 4, 2016 #5

    ProfuselyQuarky

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    distance formula :)
     
  7. Apr 4, 2016 #6
    Hi,
    Thanks for replies. This is not a hw. Its exam preparation. But i would put my future questions related to book in a hw forum. I think you judge hws with the usage of books?
    BD= sqrt(sqr( 5-1) + sqr(1-7))
    = sqrt(16 + 36)
    =2 *Sqrt(13)
    Ans=??? (I cant use calculator in exam)
    DC = sqrt(sqr(11-5) + sqr(5-1))
    = sqrt(36 + 16)
    = 2*sqrt(13)

    area of DBC = 1/2 (2*sqrt(13)) * ( 2 *sqrt(13))
    Ans = 26

    Area od ABCD = 26 + 12 =38.

    This is the correct answer. Thanks.

    Zulfi.
     
  8. Apr 4, 2016 #7

    SammyS

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    How do you know that ΔDBC is a right triangle ?
     
  9. Apr 4, 2016 #8
    Hi,
    No I am not saying that DBC is a right angle triangle. That's why i think they have asked me to use distance formula. However from fig it looks that ABD is a right angle triangle, so i am not using distance formula here and calculating the area directly using 1/2 alt * base formula.

    I also have a question:
    for triangle DBC why are we using sides DB & DC? why cant we take sides DB & BC??
    Some body please guide me.

    Zulfi.
     
  10. Apr 4, 2016 #9

    SammyS

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    What are you using as a formula to calculate the area of a triangle ?

    The altitude is the perpendicular distance from the base to the opposite vertex.
     
    Last edited: Apr 4, 2016
  11. Apr 4, 2016 #10

    Mark44

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    Thread moved.

    This forum section is for homework and coursework, which includes problems found in textbooks.
     
  12. Apr 4, 2016 #11

    ProfuselyQuarky

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    That is why I asked whether DBC was a right triangle or not. If DBC is not a right triangle, then you can not you ##A=\frac {1}{2}bh## to find the area. Instead, you have to use Heron’s theorem which states ##A=\sqrt {(s-a)(s-b)(s-c)}## where ##s=\frac {1}{2}(a+b+c)## and variables ##a##, ##b##, and ##c## are the sides of the triangle.
     
  13. Apr 4, 2016 #12

    SammyS

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    Are you giving the solution, or are you trying to help OP.

    Heron's theorem is not needed here.
     
  14. Apr 4, 2016 #13

    ProfuselyQuarky

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    I was just helping the OP overall. Nothing to do with the specific problem :smile:
     
  15. Apr 5, 2016 #14

    Samy_A

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    One easy way to solve this kind of exercises is a variation on the "box method", illustrated here for the area of a triangle.

    Add the points E(11,7) and F(11,1) to your graph, and the area of ABCD can be easily computed:

    quadri.jpg

    Or decompose your quadrilateral as follows:

    quadri2.jpg
     
  16. Apr 5, 2016 #15

    LCKurtz

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    @Samy_A Nice pictures. What software did you use for them?
     
  17. Apr 6, 2016 #16

    Samy_A

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    Geogebra
     
  18. Apr 10, 2016 #17

    micromass

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