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Math Help - Solve for Area of This Quadrilateral

  1. #1
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    Solve for Area of This Quadrilateral

    Here is a link to the quadrilateral:

    http://img200.imageshack.us/img200/3690/quad.png

    I am having a tough time solving for the area. I know you have to split it up into simpler shapes, but it is giving me a hard time.

    Please go through the steps for me.

    Thanks!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Brahmagupta's formula - Wikipedia, the free encyclopedia

    I tink that the above link could shed a little light on your problem. Also, BD can be found by using the law of cosines.
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  3. #3
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    Is there a way to do it using easier methods?
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  4. #4
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    Quote Originally Posted by csawyer1109 View Post
    Here is a link to the quadrilateral:

    http://img200.imageshack.us/img200/3690/quad.png

    I am having a tough time solving for the area. I know you have to split it up into simpler shapes, but it is giving me a hard time.

    Please go through the steps for me.

    Thanks!
    1. Calculate the length BD using Cosine rule (7.8465)

    2. BD is the hypotenus in the right triangle ADB. Calculate the length of AD using Pythagorean theorem (5.29562)

    3. The area of ADB is a_{ADB}=\dfrac12 \cdot AB \cdot AD

    4. The area of the triangle DBC is a_{DCB} = \dfrac12 \cdot DC \cdot BC \cdot \sin(55.2^\circ)
    Attached Thumbnails Attached Thumbnails Solve for Area of This Quadrilateral-quadlit_split.png  
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Earboth did it faster than I could, and there's no easier way than the way he describes. I will say this though, in case there was any confusion, BC(sin52.2) is another is a fancy way to describe the height of triangle BCD. So the last line he gives is simply Area of triangle BCD=(1/2)base*height.
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  6. #6
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    Can you go over #4 in more detail. I trust that your answer is correct, however I would like to fully understand it.

    Thanks earboth!


    Quote Originally Posted by earboth View Post
    1. Calculate the length BD using Cosine rule (7.8465)

    2. BD is the hypotenus in the right triangle ADB. Calculate the length of AD using Pythagorean theorem (5.29562)

    3. The area of ADB is a_{ADB}=\dfrac12 \cdot AB \cdot AD

    4. The area of the triangle DBC is a_{DCB} = \dfrac12 \cdot DC \cdot BC \cdot \sin(55.2^\circ)
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  7. #7
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    Quote Originally Posted by csawyer1109 View Post
    Here is a link to the quadrilateral:

    http://img200.imageshack.us/img200/3690/quad.png

    I am having a tough time solving for the area. I know you have to split it up into simpler shapes, but it is giving me a hard time.

    Please go through the steps for me.

    Thanks!
    use the Cosine Law to determine distance DB
     c = a^2 + b^2 - 2 a b\, \cos C
    Distance DB being the c (small cap) in the above equation.
    Then use pythagoras to deteremine distance DA.

     8.99 \times \sin 55.2 = perpendicular dist D is from line BC

    The area of a triangle is = base x height / 2

    You have two triangles: you know the base of each, and you know the height of each.
    There are other methods, but that's the way I would approach it.
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  8. #8
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    Ok I have got it now!

    Thanks for the help guys!
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