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Math Help - Express in terms of X

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    Express in terms of X

    A piece of wire x m long is cut into two pieces, the length of the first piece being 14 m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.

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    Word Problem

    Hello,

    Have a look here for a similar starting : http://www.mathhelpforum.com/math-he...entiation.html, which should give you ideas as to how you can express this in terms of x.
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  3. #3
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    Hello, gearshifter!

    A tricky one . . . it requires baby-talking our way through it.


    A piece of wire x m long is cut into two pieces.
    The length of the first piece is 14 m and is bent into a circle.
    The other is bent into a rectangle with length twice the width.
    Give an expression for the total area A enclosed in the two shapes in terms of x.
    Code:
          :    14     P   x-14    :
          * - - - - - * - - - - - *
          A         x             B
    The wire is: . AB \:=\:x

    It is cut at point P\!:\;\;AP = 14,\;\;PB \:=\:x-14


    Length AP = 14 is bent into a circle.
    The circumference of a circle is: . C \:=\:2\pi r
    Since the circumference is 14, we have: . 2\pi r \:=\:14 \quad\Rightarrow\quad r \:=\:\tfrac{7}{\pi}
    The area of a circle is: . A \:=\:\pi r^2
    . . So we have: . A_c \:=\:\pi\left(\tfrac{7}{\pi}\right)^2 \quad\Rightarrow\quad\boxed{ A_c\:=\:\frac{49}{\pi}}


    The other piece, PB = x-14, is bent into a rectangle.
    . . Its length is twice its width.
    Code:
                2W
          * - - - - - - *
          |             |
        W |             | W
          |             |
          * - - - - - - *
                2W
    Its perimeter is x-14.

    We have: . 6W \:=\:x-14 \quad\Rightarrow\quad W \:=\:\frac{x-14}{6}

    And the length is: . L \:=\:2W \:=\:\frac{x-14}{3}

    The area of the rectangle is: . A \:=\:L\cdot W\quad\Rightarrow\quad\boxed{ A_r \:=\:\frac{(x-14)^2}{18}}


    Therefore, the total area is: . \boxed{{\color{blue}A \;=\;\frac{49}{\pi} + \frac{(x-14)^2}{18}}} square meters.

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