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Homework Help: Area of irregular square

  1. Dec 10, 2011 #1

    How do you solve for the area of irregular square. What's the formula? For example. A square has the following 4 sides:

    side a: 11.83 meters
    side b: 38.74 meters
    side c: 12.00 meters
    side d: 36.02 meters

    What is the total area? Thanks.
  2. jcsd
  3. Dec 10, 2011 #2

    Simon Bridge

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    Well, it is not a square, and you need to know the angles.

    I guess it's trying to be a rectangle if a is opposite c, and the angles are as close as possible to right-angles.
    Then the shape will cover the maximum possible area for the sides - this what you mean?
    Or do you mean any old tetragon?
    Last edited: Dec 10, 2011
  4. Dec 11, 2011 #3
    The area of an irregular quadrilateral is

    [tex]A= \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cdot\cos^2{\frac{\alpha +\gamma}{2}}}[/tex]

    where a,b,c,d are the sides. s is the semi-perimeter and [itex]\alpha[/itex] and [itex]\gamma[/itex] are any two opposite angles.
  5. Dec 11, 2011 #4

    Simon Bridge

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    @Blandongstein: awesome first post, welcome to PF.
    Unfortunately we are not supplied with any angles ... so more information is needed from stglyde.

    I was intrigued by the description as a "irregular square" ... another common formulation is to inscribe the tetragon/quadrilateral inside a circle for example. If we know the constraints on how squashed the shape can be, we can answer the question.
  6. Dec 11, 2011 #5
    I just want to get the approximate area and I think it is easy by simply multiplying 12 x 37 or 444 so I'm satisfied. Thanks for the help.
  7. Dec 11, 2011 #6

    Simon Bridge

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    There you go you see - not enough information was supplied.
    The area must be pretty close to rectangular for that approximation to work.
    But if, say, the angle between side a and side b is small, then a better approximation would be for a triangle. See why you got such complicated answers?

    Oh well. Good luck.
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