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  1. #1
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    geometry help

    If the sides of a square are decreased by 2cm, the area is decreased by 36m^2. What are the dimensions of the original square?
    So A(x-36cm^2)=Lenght 2(x-2cm + Width 2(x-2cm) Is this the right set up?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lizard4 View Post
    If the sides of a square are decreased by 2cm, the area is decreased by 36m^2. What are the dimensions of the original square?
    So A(x-36cm^2)=Lenght 2(x-2cm + Width 2(x-2cm) Is this the right set up?
    Let x be the side length of the original square
    Let A be the area of the original square

    Then we have: x^2 = A ............(1)

    When the side-length is decreased by 2, the area decreases by 36, so we also have:

    (x - 2)^2 = A - 36

    \Rightarrow x^2 - 4x + 4 = A - 36

    \Rightarrow x^2 - 4x + 40 = A ..............(2)

    Now equate both formulas for A and solve for x, we get:

    x^2 - 4x + 40 = x^2

    And i think you can take it from there
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  3. #3
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    why seperate the length and the width, its a square so take advantage of the fact that all the sides are the same length.
    In this scenario you have two equations,
    s^2 = A (the original square)
    (s-2)^2 = A-36 (the new square)
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