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Math Help - Completely stumped by a picture frame >_<

  1. #1
    Newbie Monocerotis's Avatar
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    Completely stumped by a picture frame >_<

    So here is the question:

    A picture that measures 10cm by 8cm is to be surrounded by a mat before being framed. The width of the mat is to be the same on all sides of the picture. The area of the mat is to equal the area of the picture. What is the width of the mat, to the nearest tenth of a centimeter?

    I can't figure out how to begin working this one out. Can anyone point me along in the right direction ? If required the question is from a unit we are doing on complex numbers and quadratic functions. Maybe I'm just reading it wrong but if the area of the mat is equal to the area of the picture, wouldn't the width of the mat then be either 10, or 8 centimeters ?
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Monocerotis View Post
    So here is the question:

    A picture that measures 10cm by 8cm is to be surrounded by a mat before being framed. The width of the mat is to be the same on all sides of the picture. The area of the mat is to equal the area of the picture. What is the width of the mat, to the nearest tenth of a centimeter?

    I can't figure out how to begin working this one out. Can anyone point me along in the right direction ? If required the question is from a unit we are doing on complex numbers and quadratic functions. Maybe I'm just reading it wrong but if the area of the mat is equal to the area of the picture, wouldn't the width of the mat then be either 10, or 8 centimeters ?
    Let x denote the width of the mat. Then the entire area (picture + mat) is calculated by:

    (10+2x)(8+2x)=2 \cdot (10 \cdot 8)

    Solve for x.
    Spoiler:
    I've got x = 1.8
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  3. #3
    Newbie Monocerotis's Avatar
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    Quote Originally Posted by earboth View Post
    Let x denote the width of the mat. Then the entire area (picture + mat) is calculated by:

    (10+2x)(8+2x)=2 \cdot (10 \cdot 8)

    Solve for x.
    Spoiler:
    I've got x = 1.8
    Thanks for the help earboth, but would you mind explaining to me how you formulated that function.

    For some reason I'm just really confused by this word problem, it would be a huge help to me.
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  4. #4
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    Hello, Monocerotis!

    A picture that measures 10 cm 8 cm is to be surrounded by a mat.
    The width of the mat is to be the same on all sides of the picture.
    The area of the mat is to equal the area of the picture.
    What is the width of the mat, to the nearest tenth of a centimeter?

    I can't figure out how to begin working this one out.
    Start by making a sketch . . .

    Code:
          : x : - 8 - : x :
        - * - - - - - - - * - -
        x |               |   :
        - |   * - - - *   |   :
        : |   |       |   |   :
        : |   |       |   |   :
       10 |   |10     |   | 2x+10
        : |   |       |   |   :
        : |   |   8   |   |   :
        - |   * - - - *   |   :
        x |               |   :
        - * - - - - - - - * - -
          : - - 2x+8  - - :

    The dimensions of the picture is: . 8\times10
    . . Its area is: . 8\cdot10 \:=\:80\text{ cm}^2

    The dimensions of the frame is: . (2x+8) \times (2x+10)
    . . Its area is: . (2x+8)(2x+10)


    \text{Area of mat} \;=\;\text{Area of frame} - \text{Area of picture}
    n . . . . . . . = \;(2x+8)(2x+10) - 80
    n . . . . . . . = \;4x ^2 + 36x

    We are told: . \text{Area of mat} \:=\:\text{Area of picture}
    . . . . . . . . . . . 4x^2 + 36x \;\;=\quad 80

    We have: . 4x^2 + 36x - 80 \:=\:0 \quad\Rightarrow\quad x^2 + 9x - 20 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{-9\pm\sqrt{161}}{2}

    Therefore: . x \:=\:\frac{-9 +\sqrt{161}}{2} \;\approx\;1.8\text{ cm}

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