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Area of quadrilateral

  1. Jun 12, 2010 #1
    Is it true that the area of quadrilateral in general is the product of its diagonal divided by 2 ? Does this include rhombus or parallellogram ?
     
  2. jcsd
  3. Jun 12, 2010 #2

    Mark44

    Staff: Mentor

    Did you mean "product of its diagonals divided by 2"? If so, this isn't true.

    As a counterexample, consider a rectangle whose width is w and length l. The length of the diagonal is sqrt(w^2 + l^2). The product of the diagonals is w^2 + l^2, and half that is (1/2)(w^2 + l^2) != lw.

    If that's not what you meant, what did you mean?
     
  4. Jun 12, 2010 #3
    thanks Mark, yes thats what i meant.

    I came across this question asking to find the area of quadrilateral and i have the values of its diagonals. I got the answer coincidentally by multiplying its diagonals and halved it.

    What's the correct formula ?
     
  5. Jun 12, 2010 #4
    no this is not true.
     
  6. Jun 13, 2010 #5
    Refer to this quick diagram:

    http://yfrog.com/afpf4j

    OK so you know the values of [itex]A[/itex] and [itex]B[/itex] right, and you need to find the area of the shape.

    Obviously the area is given by [itex]X \times Y[/itex].

    Also the length of diagonal [itex]A[/itex], which is equal to the diagonal length [itex]B[/itex], is obviously given by [itex]A=B=\sqrt{X^{2}+Y^{2}}[/itex]

    If you follow your method:

    [tex]\frac{A\times B}{2}=\frac{(\sqrt{X^{2}+Y^{2}})\times (\sqrt{X^{2}+Y^{2}})}{2}=\frac{X^{2}+Y^{2}}{2}[/tex]

    Hence can see:

    [tex]\frac{X^{2}+Y^{2}}{2}\neq XY[/tex]

    Although like you said you can 'accidently' get the correct answer, if for example [itex]X=Y=2[/itex] but it's not true in general.

    Hope that helps :smile:
     
  7. Jun 14, 2010 #6

    thanks ! So that method only works for quadrilaterals with all equal sides ie squares and rhombus.
     
  8. Jun 14, 2010 #7
    I Believe that to be the case.

    If the sides are equal:

    [tex]
    \frac{A\times A}{2}=\frac{(\sqrt{X^{2}+X^{2}})\times (\sqrt{X^{2}+X^{2}})}{2}=\frac{2X^{2}}{2}=X^{2}
    [/tex]

    Hence clearly:

    [tex]X^{2}=X\times X [/tex]

    So the method would work. But surely it's much easier to just use [itex]Area=X\times Y[/itex] :smile:
     
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