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Math Help - Rectangle with given sides (in terms of x) and diagonal.

  1. #1
    Senior Member Mukilab's Avatar
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    Rectangle with given sides (in terms of x) and diagonal.

    I have a rectangle of sides x and y
    x+y=5
    The diagonal of the rectangle is 4
    Show that 2x^2-10x+9=0

    I did \sqrt{x^2+y^2}=4\longrightarrow x^2+y^2=16, x+y=10

    so i tried doing a simultaneous equation by cancelling out the y^2 first
    I did the x+y=10 times (x-y)
    10x-10y=x^2-y^2
    So I added them together
    -2x^2+10x+16=10y

    Now I'm stuck. What have I done wrong?
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  2. #2
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    Quote Originally Posted by Mukilab View Post
    I have a rectangle of sides x and y
    x+y=5
    The diagonal of the rectangle is 4
    Show that 2x^2-10x+9=0

    I did \sqrt{x^2+y^2}=4\longrightarrow x^2+y^2=16, x+y=10

    so i tried doing a simultaneous equation by cancelling out the y^2 first
    I did the x+y=10 times (x-y)
    10x-10y=x^2-y^2
    So I added them together
    -2x^2+10x+16=10y

    Now I'm stuck. What have I done wrong?
    x+y=5 or x+y=10 ?
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  3. #3
    Senior Member Mukilab's Avatar
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    sorry, x+y=5, got confused because it previously asked me to show that x+y=5 (2x+2y=10=perimeter)
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  4. #4
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    Quote Originally Posted by Mukilab View Post
    I have a rectangle of sides x and y
    x+y=5
    The diagonal of the rectangle is 4
    Show that 2x^2-10x+9=0

    I did \sqrt{x^2+y^2}=4\longrightarrow x^2+y^2=16, x+y=10

    so i tried doing a simultaneous equation by cancelling out the y^2 first
    I did the x+y=10 times (x-y)
    10x-10y=x^2-y^2
    So I added them together
    -2x^2+10x+16=10y

    Now I'm stuck. What have I done wrong?
    x^2 + y^2 = 4^2 .... (1)

    x + y = 5 => y = 5 - x .... (2)

    Substitute (2) into (1), expand, simplify and re-arrange.
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