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Math Help - algebra problem

  1. #1
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    algebra problem

    Hi friends.
    Please can you say me what is the correct way to solve this problem? Thank you

    A rectangular area with a square at its center. If the long side of rectangle is more large than short side in 8 and the side of square is (1/10) of the short side of rectangle, the area A (area without square) is.....

    if x: short side, then A=x(x+8)-(1/10)(x)

    or x: long side, then A=(x-8)x-(1/10)(x-8)
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  2. #2
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    Re: algebra problem

    Hello, user!

    Your answers are almost right . . .


    A rectangular area with a square at its center.
    The long side of rectangle is 8 more than its short side.
    The side of square is (1/10) of the short side of rectangle.
    Find the area A (area without square).

    Let x = short side of rectange.
    Then x+8 = long side of rectange.
    . . Area of rectangle: x(x+8)

    Then \tfrac{1}{10}x = side of the square.
    . . Area of square: {\color{blue}\left(\tfrac{x}{10}\right)^2}

    Therefore: A \;=\;x(x+8) - \tfrac{x^2}{100}
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  3. #3
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    Re: algebra problem

    Thank you Soroban. But if i take x:long side of rectangle ?
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  4. #4
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    Re: algebra problem

    Hello again, user!

    Thank you, Soroban.
    But if i take x = long side of rectangle?

    Let x = long side of the rectangle.
    Then x-8 = short side of the rectangle.
    . . Area of rectangle: x(x-8)

    Then \frac{x-8}{10} = side of the square.
    . . Area of square: \left(\frac{x-8}{10}\right)^2

    Therefore: . A \;=\;x(x-8) - \left(\frac{x-8}{10}\right)^2
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  5. #5
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    Re: algebra problem

    thanks Soroban, when I obtain the both solutions, they are equals. Have a nice day.
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