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Thread: 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;


    $\displaystyle 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;


    $\displaystyle 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.
    $\displaystyle {\text{Let side of square be }} = a \hfill \\$

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

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

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

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

    $\displaystyle a = \frac{{16 - 2x}}
    {4} \hfill \\$

    $\displaystyle a = \frac{{8 - x}}
    {2}...............................{\text{eqn(1)}} \hfill \\$

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

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

    $\displaystyle A = a^2 + x \hfill \\$

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

    $\displaystyle A = \left( {\frac{{8 - x}}
    {2}} \right)^2 + x \hfill \\$

    $\displaystyle A = \frac{{64 - 16x + x^2 }}
    {4} + x \hfill \\$

    $\displaystyle A = 16 - 4x + \frac{{x^2 }}
    {4} + x \hfill \\$

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

    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:
    . . $\displaystyle A \;= \;16-3x + \frac{x^2}{4} $
    Code:
                x
          * - - - - - *
          |           |
        1 |           | 1
          |           |
          * - - - - - *
                x

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


    The rectangle has a perimeter of $\displaystyle 2x + 2$ meters.

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

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

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


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

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