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Math Help - halving a rectangle

  1. #1
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    halving a rectangle

    I'm trying to find an expression, equation.

    Given rectangle x by y, what is the single value, a, by which I can reduce

    each side to halve the area.

    xy / 2 = (x-a)(y-a) solve for a
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by gowan View Post
    I'm trying to find an expression, equation.

    Given rectangle x by y, what is the single value, a, by which I can reduce

    each side to halve the area.

    xy / 2 = (x-a)(y-a) solve for a
    Looks like you've got it. Can you not solve for a?
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  3. #3
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    I'm brand new to this forum. I'm guessing that this is how I respond to a question.
    No I can't solve for a in a way that would permit me to find 'a' given x and y.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by gowan View Post
    I'm brand new to this forum. I'm guessing that this is how I respond to a question.
    No I can't solve for a in a way that would permit me to find 'a' given x and y.
    \frac{xy}{2}=xy-ax-ay+a^2

    -\frac{xy}{2}=a^2-a(x+y)

    Note that the coeficient of the linear term is -(x+y). So, can you complete the square?
    Last edited by VonNemo19; January 10th 2010 at 06:17 PM.
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  5. #5
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    Hello, gowan!

    Given rectangle x by y, what is the single value, a,
    by which I can reduce each side to halve the area?

    . . (x-a)(y-a)\:=\:\frac{xy}{2}\qquad \hdots Solve for a.

    We have: . xy - ax - ay + a^2 \:=\:\frac{xy}{2} \quad\Rightarrow\quad 2xy -2ax - 2ay + 2a^2 \:=\:xy

    . . which simplifies to:. . 2a^2 - 2(x+y)a + xy \:=\:0 . . . a quadratic in a


    Quadratic Formula: . a \;=\;\frac{2(x+y) \pm\sqrt{4(x+y)^2 - 8xy}}{4} \;=\;\frac{(x+y) \pm\sqrt{x^2+y^2}}{2}

    The answer is the smaller of the two values: . a \;=\;\frac{(x+y)-\sqrt{x^2+y^2}}{2}

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  6. #6
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    So, once again, we have "to solve a problem, don't think at all, just memorize a formula."
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