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Thread: Word problem help.

  1. #1

    Thread Starter
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    Talking Word problem help.

    A rectangular garden measures 5ft by 7ft and is surrounded by a uniform path x ft wide. If the total area of garden plus path is 63 sq ft, find the width of the path.

    Im lost on this problem. Any help would be great.

    Area of the garden should be (5 * 7) which gives 35 ft. 63 - 35 = 28. So the area of the path should equal 28 ft?

    Im confused.
    Last edited by Dilenger4; Aug 29th, 2003 at 01:12 PM.

  2. #2
    So Unbanned DiGiTaIErRoR's Avatar
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    Code:
    +++++++
    +-----+
    +-----+
    +-----+
    +-----+
    +-----+
    +-----+
    +-----+
    +++++++
    
    + = Path
    - = Garden
    9*7 = 63 so a 9*7 square is around the garden... it'd seem.

    So the path is 1 foot across. Area of path, 28 sq. ft..

  3. #3

    Thread Starter
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    This is what i come up with.

    63 = (5 + 2y)(7 + 2y)
    63 = 35 + 10y + 14y + 4y
    63 = 35 + 24y + 4y
    63 - 35 + 24y + 4y
    28/28 = 28y/28
    y = 1

    Path = 1ft
    Last edited by Dilenger4; Aug 29th, 2003 at 01:13 PM.

  4. #4
    Frenzied Member nishantp's Avatar
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    Originally posted by Dilenger4
    This is what i come up with.

    63 = (5 + 2y)(7 + 2y)
    63 = 35 + 10y + 14y + 4y
    63 = 35 + 24y + 4y
    63 - 35 + 24y + 4y
    28/28 = 28y/28
    y = 1

    Path = 1ft
    That works for the wrong reason. The part I put in Bold should be 4y2, which makes the whole thing a quadratic.

    I didn't show all the steps but the algebraic solution is
    Code:
    63=(5+y)(7+y)
    0 = y2 + 12y - 28
    The roots are 2 and -14. The -14 is discarded.

    Since the path adds 2 feet to both dimensions, but goes all the way around, the width of the path is 2/2=1ft.



    PS - I deleted my earlier post because I made a mistake.
    You just proved that sig advertisements work.

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