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Math Help - Calculate the area

  1. #1
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    Calculate the area

    Can somone help me solve this equation. I've already tried with
    (x-1)=x/2, but I failed.


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  2. #2
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    Quote Originally Posted by kevin3000 View Post
    Can somone help me solve this equation. I've already tried with
    (x-1)=x/2, but I failed.


    I would let x represent the length of the square.

    Therefore the base and height of the triangle are x + 1.


    Since the areas are the same...

    x^2 = \frac{1}{2}(x + 1)^2

    x^2 = \frac{1}{2}x^2 + x + \frac{1}{2}

    \frac{1}{2}x^2 - x - \frac{1}{2} = 0

    x^2 - 2x - 1 = 0

    x^2 - 2x + (-1)^2 - (-1)^2 - 1 = 0

    (x - 1)^2 - 2 = 0

    (x - 1)^2 = 2

    x - 1 = \pm \sqrt{2}

    x = 1 \pm \sqrt{2}.


    Since you obviously can not have a negative length, the length of the square is

    1 + \sqrt{2} cm.


    Thus the area is

    A= (1 + \sqrt{2})^2

     = 1 + 2\sqrt{2} + 2

     = 3 + 2\sqrt{2}\,\textrm{cm}^2.
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  3. #3
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    Hello, Kevin!

    Can somone help me solve this equation?
    I've already tried with : . (x-1)^2 \:=\:\frac{x^2}{2} . . . but I failed.

    Your equation is correct . . . It must be your algebra.

    We have: . (x-1)^2 \:=\:\frac{x^2}{2}

    Multiply by 2: . 2(x-1)^2 \:=\:x^2 \quad\Rightarrow\quad 2x^2 - 4x + 2 \:=\:x^2 \quad\Rightarrow\quad x^2 - 4x + 2 \:=\:0

    Quadratic Formula: . x \;=\;\frac{\text{-}(\text{-}4) \pm \sqrt{(\text{-}4)^2 - 4(1)(2)}}{2(1)} \;=\;\frac{4 \pm\sqrt{8}}{2} \:=\:\frac{4 \pm2\sqrt{2}}{2} \:=\:2 \pm \sqrt{2}


    The side of the triangle is: . 2 + \sqrt{2} cm.

    The side of the square is: . 1 + \sqrt{2} cm.


    The area of the square is: . (1 + \sqrt{2})^2 \;=\;1 + 2\sqrt{2} + 2 \;=\;\boxed{3 + 2\sqrt{2}\,\text{ cm}^2}

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  4. #4
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    Ok, thanks guys.
    I understand better now
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