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Math Help - Algebra help. (word problem)

  1. #1
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    Algebra help. (word problem)

    I am really having difficulty with this question, I just dont seem to get it.

    A piece of wire which is 18 meters long is cut into two parts. The first part is bent to form the four sides of a square. The second part is bent to form the fours sides of a rectangle. The breadth of the rectangle is one meter. And its length is x meters. If the sum of the areas of the square and rectangle is A square meters show that;


     A = 16-3x + \frac{x^2}{4}

    I don't really know how to get to this formula, it does not say that the the wire is cut into 2 equal parts, so i can't assume that can i ?

    I understand were the number 16 comes from and that's about it.
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  2. #2
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    Quote Originally Posted by hana_102 View Post
    I am really having difficulty with this question, I just dont seem to get it.

    A piece of wire which is 18 meters long is cut into two parts. The first part is bent to form the four sides of a square. The second part is bent to form the fours sides of a rectangle. The breadth of the rectangle is one meter. And its length is x meters. If the sum of the areas of the square and rectangle is A square meters show that;


     A = 16-3x + \frac{x^2}{4}

    I don't really know how to get to this formula, it does not say that the the wire is cut into 2 equal parts, so i can't assume that can i ?

    I understand were the number 16 comes from and that's about it.
     {\text{Let side of square be }} = a \hfill \\

     {\text{Perimeter of square  +  perimeter of rectangle  =  18}} \hfill \\

    4a + 2\left( {1 + x} \right) = 18 \hfill \\

     4a + 2 + 2x = 18 \hfill \\

    4a = 18 - 2 - 2x \hfill \\

    a = \frac{{16 - 2x}}<br />
{4} \hfill \\

    a = \frac{{8 - x}}<br />
{2}...............................{\text{eqn(1)}} \hfill \\

    {\text{A =  area of square  +  area of rectangle}} \hfill \\

    A = \left( {a^2 } \right) + \left( {1 \times x} \right) \hfill \\

    A = a^2  + x \hfill \\

    {\text{Put the value of }}a{\text{ from eqn (1)}} \hfill \\

     A = \left( {\frac{{8 - x}}<br />
{2}} \right)^2  + x \hfill \\

    A = \frac{{64 - 16x + x^2 }}<br />
{4} + x \hfill \\

    A = 16 - 4x + \frac{{x^2 }}<br />
{4} + x \hfill \\

      A = 16 - 3x + \frac{{x^2 }}<br />
{4} \hfill \\ <br />

    Did you get it now??? if not, please ask.
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  3. #3
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    Thank you I understand it now, that's a very good explanation, but I would have never thought of using the perimeters to work it out in the first place.
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  4. #4
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    Hello, hana_102!

    We just have to baby-talk our way through it . . .


    A piece of wire which is 18 meters long is cut into two parts.
    The first part is bent to form the four sides of a square.
    The second part is bent to form the fours sides of a rectangle.
    The breadth of the rectangle is one meter. And its length is x meters.

    Show that the total area of the square and rectangle is:
    . . A \;= \;16-3x + \frac{x^2}{4}
    Code:
                x
          * - - - - - *
          |           |
        1 |           | 1
          |           |
          * - - - - - *
                x

    The rectangle has an area of: . x\cdot1 \:=\:\boxed{x\text{ m}^2}


    The rectangle has a perimeter of 2x + 2 meters.

    This leaves: . 18 - (2x+2) \:=\:16-2x meters for the square.

    The side of the square is: . \frac{16-2x}{4} \:=\:\frac{8-x}{2} meters.

    The area of the square is: . \left(\frac{8-x}{2}\right)^2 \;=\;\frac{64-16x + x^2}{4} \;=\;\boxed{16 - 4x + \frac{x^2}{4}\text{ m}^2}


    Therefore, the total area is: . A \;=\;x + \left(16 - 4x + \frac{x^2}{4}\right) \;=\;{\color{blue}16 - 3x + \frac{x^2}{4}}

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