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Running by the pool

  1. Mar 18, 2008 #1
    The length of a rectangular swimming pool is twice its width. The pool is surrounded by a cement walk 4 ft wide. If the area of the walk is 748 square feet, determine the dimensions of the pool.

    A= the total area bounded by the outer edge of the surrounding walk
    W= the area of the walk
    P= the area of the pool
    x= the length of the short side of the pool

    A=P+W
    P=x2x=[tex]2x^{2}[/tex]
    W=748
    A=(x+8)(2x+8)=[tex]2x^{2}+24x+64[/tex]=[tex]2x^{2}+748
    [/tex]
    solving for x gives x=28.5 feet and 2x=57 feet
    The answer sheet says the dimensions are 30 and 60 feet.
    Where am I going wrong?
    Thanks for any replies.
     
  2. jcsd
  3. Mar 18, 2008 #2
    Well, if you plug 30 and 60 back in, you get 784 so maybe it's a typo.

    Or they rounded.
     
  4. Mar 18, 2008 #3

    symbolipoint

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    Did you draw and label a picture?
     
  5. Mar 18, 2008 #4
    Yes. I just don't know how to reproduce it in the post.
     
  6. Mar 18, 2008 #5

    symbolipoint

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    Good. I believe you and see now that some of what you showed indicated that you used a picture. I'm working on the exercise right now. Someone will probably write a response before I finish. So far, I'm looking at A of pool equals A of wholeRectangle minus A of walkway. I obtained 2x^2 = (x+8)(2x+8) - 748, and I'm not finished.
     
  7. Mar 18, 2008 #6

    symbolipoint

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    bacon, in fact, I found the same results as you did: x=28.5, 2x=57
     
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