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Math Help - Area and Perimeter

  1. #1
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    Exclamation Area and Perimeter

    Hello,
    I have 2 word problems that I think are unsolvable :

    1. A garden plot 4m by 12m has one side along a fence. The area of the garden is to be doubled by digging a border of uniform width on the three other sides. What should the width of the border be?

    2. Vanessa built a rectangular pen for her dogs. She used an outside wall of the garage for one of the pen sides. She had to buy 20m of fencing in order to build the other sides of the pen. Find the dimensions of the pen if the area is 48m^2 (squared).

    ?!?!?
    PLEASE HELP
    Thank you

    EDIT: I have worked at these for about an hour and a half I don't understand a word of this (measurement wise)...
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by doomdesire34 View Post
    Hello,
    I have 2 word problems that I think are unsolvable :

    1. A garden plot 4m by 12m has one side along a fence. The area of the garden is to be doubled by digging a border of uniform width on the three other sides. What should the width of the border be?
    L = 4
    W = 12
    A = L*W = 4*12 = 48

    We want double the area, which is:
    2A = 2*48 = 96

    Now, we want to change the length and width to double the area, however, we will be increasing the length and width by the same amount. We'll call this amount the "width" which we'll let equal x.
    L1 = L + x
    W1 = W + x
    A1 = L1*W1 = (L + x)(W + x) = L*W + Lx + Wx + x^2

    We want A1 to equal 2*A, so, if A1 = 96, L = 4, W = 12, then the equation becomes:
    96 = 48 + 4x + 12x + x^2
    x^2 + 16x - 48 = 0
    x = (-16 +/- sqrt(448))/2 = -8 +/- 4sqrt(7)

    But since the length must be positive, we have that:
    x = 4sqrt(7) - 8 = 2.583

    2. Vanessa built a rectangular pen for her dogs. She used an outside wall of the garage for one of the pen sides. She had to buy 20m of fencing in order to build the other sides of the pen. Find the dimensions of the pen if the area is 48m^2 (squared).

    P = W + 2L = 20 --> W = 20 - 2L
    A = W*L = 48 --> 48 = (20 - 2L)L = 20L - 2L^2

    2L^2 - 20L + 48 = 0
    2(L^2 - 10L + 24) = 0
    2(L - 12)(L + 2) = 0
    L = 12, -2

    Since length must be positive, we have that:
    L = 12
    W = 20 - 2(12) = -4

    Did I make a mistake? I'll come back and check my work in a second.
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  3. #3
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    Smile Thank You!

    Thank you very much. I greatly appreciate the help!
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Maybe my brain isn't functioning today but I can't figure out what I did wrong (and I am pretty sure I did something wrong).
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    2L^2 - 20L + 48 = 0
    2(L^2 - 10L + 24) = 0
    2(L - 12)(L + 2) = 0
    L = 12, -2
    Oops. I factored wrong .

    2(L^2 - 10L + 24) = 0
    2(L - 4)(L - 6) = 0
    L = 4, 6

    W = 20 - 2L

    W = 20 - 2(4) = 12
    W = 20 - 2(6) = 8

    4*12 = 48
    6*8 = 48

    We get two answers:
    L = 4 AND W = 12
    L = 6 AND W = 8
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