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Math Help - Fractions of area - the pained pane

  1. #1
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    Talking Fractions of area - the pained pane

    Last question i need help with:

    THE PAINED PANE:
    A triangular piece of glass (labeled XYC) is cut from a rectangular pane (labled ABCD)

    What fraction of the area of ABCD is XYC?

    What I know:
    Points X and Y are the midpoints of the two sides of the rectangular pane.


    The diagram/picture: http://www.imagemule.com/uploads/07E8gE.jpg
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  2. #2
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    Talking

    From your diagram...

    Area of Triangle XBY is:

    Area \ XBY = \frac{1}{2}.\frac{1}{2}x.\frac{1}{2}y

     Area \ XBY = \frac{1}{8}xy

    Area of Trapezium AXCD

     Area \ AXCD = \frac{1}{2}(y+\frac{1}{2}y)x

     Area \ AXCD = \frac{3}{4}xy

    That means that the total area that HASN'T been cut out is...

     AreaXBY + Area AXCD=\frac{7}{8}xy

    The area of the WHOLE rectangle is Total \ Area=xy

    That means that the fractional area of triangle cut out is...

    xy-\frac{7}{8}xy=\frac{1}{8}xy

    That means that the area of XYC is \frac{1}{8} of Area ABCD



    This is the first time I've helped someone so I hope its correct!
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  3. #3
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    maybe u misread me but the triangle is XYC.

    Ps. i really appreciate your help thankies
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  4. #4
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    That's the one I worked out! =P


    Well by subtracting the other two areas around it
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  5. #5
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    ok.. but it seems confusing Your answer is correct and I know that but I don't know how u got it. can you explain plz? thankies
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  6. #6
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    Hello, anthmoo!

    Let L \,= \,AB \,= \,CD, the length of the rectangle.
    Let W \,= \,AD \,= \,BC, the width of the rectangle.


    Tip your head to the left and look at \Delta XYC.

    . . The base is: CY \,=\,\frac{1}{2}BC \,=\,\frac{1}{2}W
    . . Its height is: XB \,=\,\frac{1}{2}AB\,=\,\frac{1}{2}L

    The area of \Delta XYC is: . \frac{1}{2}(CY)(XB)\:=\:\frac{1}{2}\left(\frac{1}{  2}W\right)\left(\frac{1}{2}L\right) \:=\:\frac{1}{8}(LW)


    The area of the triangle is one-eighth of the area of the rectangle.

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  7. #7
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    Quote Originally Posted by puppy_wish View Post
    ok.. but it seems confusing Your answer is correct and I know that but I don't know how u got it.
    I was scared of this...ok:

    Work through my solution step by step by looking at the diagram...

    Step 1: I found the area of triangle XBY (the triangle surrounded by the points X, B and Y). This area is \frac{1}{8} the area of the whole thing (this is not XYC but the areas are awkardly the same...).

    Step 2: I found the area of the trapezium AXCD (you see I'm finding the area of everything EXCEPT the area of XYC). This turned out to be \frac{3}{4} the area of the whole thing. This can be said as \frac{6}{8} the area of the whole thing, to make it easier later.

    Step 3: I added the two areas together which is \frac{1}{8} + \frac{6}{8} = \frac{7}{8} . That is the area of EVERYTHING but the pane of glass we are cutting out.

    Step 4: We can now work out the area of pane of glass by \frac{8}{8} - \frac{7}{8} = \frac{1}{8}. That means that fraction area of the pain of glass is \frac{1}{8} that of the rectangular area of the pane of glass.
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  8. #8
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    THANK YOU SO MUCH BOTH OF YOU! This really helped me. THANK YOU AGAIN, you guys really rok!!
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