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Math Help - Pentagon.

  1. #1
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    Pentagon.



    a || b

    Construct a 4 angled figure ( could be Square, trapezoid, deltoid or whatever ) whose Surface will be equal to the surface of the pentagon.
    Last edited by mr fantastic; March 27th 2011 at 12:14 PM. Reason: Re-titled.
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  2. #2
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    Isn't this a pentagon?

    I guess you can set the angle E or C to be 180 degrees, and you will have a quadrilateral with the same surface as the figure.
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  3. #3
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    Crap yeah, Pentagon. it is a pentagon.

    I edited it.

    can u copy my figure and show how is it done cuz I tried as you said and it looks like a pentagon again.
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  4. #4
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    Ok. If we make the angle E 180 degrees, we get this:

    Pentagon.-quadrilateral1.gif

    And If we make the angle C 180 degrees, we get this:

    Pentagon.-quadrilateral2.gif

    Both of these are quadrilaterals, as in the first case E went from being a corner to becoming a point on the side AD, and in the second case the same thing happened to C.
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  5. #5
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    Quote Originally Posted by sturmgewehr View Post

    Construct a 4 angled figure ( could be Square, trapezoid, deltoid or whatever ) whose Surface will be equal to the surface of the pentagon.
    I'll show you how to get whatever

    Use shearing to get a quadrilateral:

    \Delta(ECD) \cong \Delta(ECD')~\implies~(ABCD') is a quadrilateral.

    Use shearing again to get a triangle:

    (\overline{D'B}) \parallel (\overline{CC'}) therefore: \Delta(D'BC) \cong \Delta(D'BC')

    The area of the pentagon didn't change therefore the area of \Delta(AC'D') = (ABCDE)
    Attached Thumbnails Attached Thumbnails Pentagon.-doppelscherung.png  
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  6. #6
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    Quote Originally Posted by scounged View Post
    Ok. If we make the angle E 180 degrees, we get this:

    Click image for larger version. 

Name:	quadrilateral1.gif 
Views:	15 
Size:	7.3 KB 
ID:	21279

    And If we make the angle C 180 degrees, we get this:

    Click image for larger version. 

Name:	quadrilateral2.gif 
Views:	12 
Size:	7.3 KB 
ID:	21280

    Both of these are quadrilaterals, as in the first case E went from being a corner to becoming a point on the side AD, and in the second case the same thing happened to C.



    Smart move.

    But I how do we prove that the surface of rectangle EQE1 = the Surface of the Rectangle QDC




    Quote Originally Posted by earboth View Post
    I'll show you how to get whatever

    Use shearing to get a quadrilateral:

    \Delta(ECD) \cong \Delta(ECD')~\implies~(ABCD') is a quadrilateral.

    Use shearing again to get a triangle:

    (\overline{D'B}) \parallel (\overline{CC'}) therefore: \Delta(D'BC) \cong \Delta(D'BC')

    The area of the pentagon didn't change therefore the area of \Delta(AC'D') = (ABCDE)
    Ok maybe I am too stupid to understand this but as far as I know we need 3 elements in common for 2 Triangles to be equal as u stated ( Triangle ECD = Triangle ECD` ), as far as I can see those 2 triangles have only one thing in common and that is EC base.
    Last edited by sturmgewehr; March 27th 2011 at 09:32 AM.
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  7. #7
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    Code:
            E
    
                                 E                F
    
    D              A             D                A
    
    
    
    
    
    C              B             C                B
    The way question is worded does not prevent using an isosceles
    triangle on top of a square, then made into a rectangle.
    EXAMPLE:
    Square ABCD sides = 6, AE = DE = 5 ; so area ABCDE = 36 + 12 = 48.
    Rectangle becomes a 8 by 6 (ADEF being 6 by 2), so area 48.
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  8. #8
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    Quote Originally Posted by sturmgewehr View Post
    Smart move.

    But I how do we prove that the surface of rectangle EQE1 = the Surface of the Rectangle QDC
    I don't think we have to do that.
    Instead, we can do this:



    \mbox{Here the height, h, is the same for both}~\triangle {ECE_{1}}~\mbox{and}~\triangle {ECD}.

    As both triangles have the same base and the same height, they will also have the same area of

    \displaystyle{\frac{h(\mid{EC}\mid)}{2}}
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  9. #9
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    Quote Originally Posted by scounged View Post
    I don't think we have to do that.
    Instead, we can do this:



    \mbox{Here the height, h, is the same for both}~\triangle {ECE_{1}}~\mbox{and}~\triangle {ECD}.

    As both triangles have the same base and the same height, they will also have the same area of

    \displaystyle{\frac{h(\mid{EC}\mid)}{2}}
    this is very good answer, simple and precise.

    thank you very much
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  10. #10
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    No problem.
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