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Thread: golden ratio question

  1. #1
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    golden ratio question

    A rectangle has width 1 unit and length x units.
    A square is cut off it.
    the rectangle left behind is similar to the original rectangle.
    Find the value of x in surd form.
    The value of x is called Golden ratio

    MY attempt:
    area of the original rectangle = x sq units

    area of square = 1 sq units
    are of b ( small rectangle) = x-1 sq units


    i found the golden ratio by doing this:
     x = \frac{1}{x-1}
     x(x-1)=1
     x^2 -x=1
     (x-\frac{1}{2})^2-\frac{1}{4}=1
     (x-\frac{1}{2})^2 = \frac{5}{4}
     x = \frac{1}{2}\pm\frac{\sqrt{5}}{2}

    since width and length are positive units

    thus  x = \frac{1}{2} + \frac{\sqrt{5}}{2}

    I found the answer but I read off a forum that i needed to set up the equations relating ratios involving x


    but when i tihnk about it why is  x = \frac{1}{x-1} this is the part im confused about
    in my head 3 = (6/2) thus i thought for a long time that the ratio should be  1 = \frac{x}{x-1} not  x = \frac{1}{x-1}

    Pleqase can some explain more in order to deepen my understanding thank you
    Attached Thumbnails Attached Thumbnails golden  ratio question-b1.jpg   golden  ratio question-b2.jpg  
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  2. #2
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    Re: golden ratio question

    rectangle 1 is length $x$, and width $1$

    a little thought shows that $x < 2$ and you must swap the orientation of the 2nd rectangle with respect to the first

    rectangle 2 is thus length $1$ and width $x-1$

    the two rectangles are similar so

    $\dfrac{x}{1} = \dfrac{1}{x-1}$

    $x = \dfrac {1}{x-1}$
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  3. #3
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    Re: golden ratio question

    Quote Originally Posted by romsek View Post
    rectangle 1 is length $x$, and width $1$

    a little
    thank you very much i cant add to your reputation. It does not allow me to.
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  4. #4
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    Re: golden ratio question

    Quote Originally Posted by bigmansouf View Post
    thank you very much i cant add to your reputation. It does not allow me to.
    you can always "thank" a responder by clicking the "thanks" link in the bottom right hand corner of a post.
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