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Math Help - Modelling and problem solving

  1. #1
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    Modelling and problem solving

    I need help with the following problems:

    1.A piece of wire 12 cm long is cut into 2 pieces. One piece is used to form a square shape and the other to form a rectangle shape of which the length is twice its width. Find the length of the side of the square if the combined area of the 2 shapes is 4.25cm^2.

    2. Using a square sheet of paper:
    choose a point P on the top edge
    join P to the bottom right hand corner
    create a 45 degree angle in the top left hand corner using P
    what position of the original point P will make the area OF THE QUADRILATERAL A MAXIMUM ?

    3. A shape that has been of interest to architects and artisits over the centuries is the 'golden rectangle'. Many have thought that it gave the perfect proportions for buildings. The rectangle is such is such that if a square is drawn on one of the longer sides then the new rectangle is similar to rectangle APQD. find the value of X. (this is known as the golden ratio).

    The outer rectangle is labelled APQD. It has sides 1 (the longer side) and width of x. The square has sides x. The inner rectangle has sides 1 - x and x.



    Looking forward to hear from someone. Thanks.
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  2. #2
    Senior Member Paze's Avatar
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    Re: Modelling and problem solving

    Hi Elmidge. Let me help you with at least point 3.




    Let \phi denote 'The golden ratio'.

    \frac{a+b}{a}=\frac{a}{b}={\phi}

    Modelling and problem solving-similargoldenrectangles.svg.png

    Let's solve this algebraically.

    \frac{a}{a}+\frac{b}{a}={a}{b}=\phi\\\\ 1+\frac{b}{a}=\frac{a}{b}=\phi\\\\ 1+\frac{b}{a}=1+\frac{1}{{a}/{b}}=\phi\\\\1+\frac{1}{\phi}=\phi
    Multiply both sides by \phi to get:

    \phi^2=\phi+1\Rightarrow\phi^2-\phi-1=0

    plug into quadratic formula:

    \frac{1\pm\sqrt{1-\left(-4\right)}}{2}}\Rightarrow\frac{1\pm\sqrt{5}}{2}} \Rightarrow \phi=1,61803...

    Please let me know if any of the steps confuse you.
    Last edited by Paze; February 17th 2013 at 03:24 PM.
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