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Math Help - help me find dimensions of rectangle and triangles

  1. #1
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    help me find dimensions of rectangle and triangles

    a rectangular pen is subdivided by two dividers that meet at the midpoint P of the base of the rectangle. ( the picture is a rectangle, in which there is a line segment coming from each upper corner of the rectangle, coming together at midpoint P) each divider is 10 m long, and the perimeter of the pen is 30 min. what are the dimensions of the rectangle?
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    Quote Originally Posted by polakio92 View Post
    a rectangular pen is subdivided by two dividers that meet at the midpoint P of the base of the rectangle. ( the picture is a rectangle, in which there is a line segment coming from each upper corner of the rectangle, coming together at midpoint P) each divider is 10 m long, and the perimeter of the pen is 30 min. what are the dimensions of the rectangle?
    Let a and b denote the sides of the rectangle. Then you get 2 equations:

    2a+2b=30~\implies~ b=15-a ....... [1]

    \left(\frac12 a\right)^2+b^2=100 ....... [2]

    Plug in the term of b from [1] into [2]:

    \left(\frac12 a\right)^2+(15-a)^2=100 and solve for a:

    a=12\pm2\sqrt{11}~\implies~b=3\mp 2\sqrt{11}

    The negative value of b isn't very plausible so there is only one solution left.
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  3. #3
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    Quote Originally Posted by earboth View Post
    Let a and b denote the sides of the rectangle. Then you get 2 equations:

    2a+2b=30~\implies~ b=15-a ....... [1]

    \left(\frac12 a\right)^2+b^2=100 ....... [2]

    Plug in the term of b from [1] into [2]:

    \left(\frac12 a\right)^2+(15-a)^2=100 and solve for a:

    a=12\pm2\sqrt{11}~\implies~b=3\mp 2\sqrt{11}

    The negative value of b isn't very plausible so there is only one solution left.

    for some reason i can't get a to end in the square root of 11. am i doing something wrong?
    Last edited by polakio92; March 30th 2008 at 05:59 AM. Reason: spelling change
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    yes, it is not working for some reason. is there any further explanation?
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  5. #5
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    Quote Originally Posted by polakio92 View Post
    for some reason i can't get a to end in the square root of 11. am i doing something wrong?
    Do you mean that you can't get the result? If so:

    \left(\frac12 a\right)^2+(15-a)^2=100 ~\iff~ \frac14 a^2 + 225 -30a + a^2 = 100 ~\iff~ \frac54 a^2 - 30a + 125 = 0

    Now apply the formula to solve this quadratic equation:

    a = \frac{30 \pm \sqrt{900 - 4 \cdot  \frac54 \cdot 125}}{2 \cdot \frac54} =  \frac{30 \pm \sqrt{900 - 625}}{ \frac52} = \frac{ 2 \cdot 30 \pm 2 \cdot 5 \cdot  \sqrt{11}}{5} which will yield the result I posted in my previous reply.
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