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Math Help - find length of rectangle given diagonal and area

  1. #1
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    find length of rectangle given diagonal and area

    Rectangle has area=168 m^2 and diagonal of 25. Find length
    This is how tried to attempt the problem
    Area= L X W
    168 = L x W ..........(1)
    L^2 + W^2 =25^2 ............(2)
    From (1) L = 168/W...........(3)
    Substitute (3) into (2)
    (168/W)^2 +W^2 = 625
    28224/W^2 + W^2 = 625
    The problem gets complicated as I proceed
    Is this aproach correct if it is,
    Is there a convinient method
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  2. #2
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    Re: find length of rectangle given diagonal and area

    Quote Originally Posted by Bonganitedd View Post
    Rectangle has area=168 m^2 and diagonal of 25. Find length
    This is how tried to attempt the problem
    Area= L X W
    168 = L x W ..........(1)
    L^2 + W^2 =25^2 ............(2)
    From (1) L = 168/W...........(3)
    Substitute (3) into (2)
    (168/W)^2 +W^2 = 625
    28224/W^2 + W^2 = 625
    The problem gets complicated as I proceed
    Is this aproach correct if it is,
    Is there a convinient method
    Sketch a rectangle with one diagonal. laber the sides L and W and D for the diabonal. Write Pythagorean theorem, and the area = LW. Eliminate L from the the Pythagorean theorem and solve for W then find L from area formula.
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  3. #3
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    Re: find length of rectangle given diagonal and area

    Its same approach I used, I did draw a rectangle now how do eliminate L bcos the pathygras is L^2 + W^2 =25^2
    The area, LW =168
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  4. #4
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    Re: find length of rectangle given diagonal and area

    Quote Originally Posted by Bonganitedd View Post
    Rectangle has area=168 m^2 and diagonal of 25. Find length
    This is how tried to attempt the problem
    Area= L X W
    168 = L x W ..........(1)
    L^2 + W^2 =25^2 ............(2)
    From (1) L = 168/W...........(3)
    Substitute (3) into (2)
    (168/W)^2 +W^2 = 625
    28224/W^2 + W^2 = 625
    The problem gets complicated as I proceed
    Is this aproach correct if it is,
    Is there a convinient method
    Have a look at this webpage.
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  5. #5
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    Re: find length of rectangle given diagonal and area

    Hello, Bonganitedd!

    Rectangle has area=168 m^2 and diagonal of 25. Find the length.

    This is how tried to attempt the problem
    \text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]

    L^2 + W^2 \:=\:25^2\;\;[2]

    \text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625

    Is this approach correct? . Yes
    If it is, is there a convinient method?

    We have: . \frac{28,\!224}{W^2} + W^2 \:=\:625

    Multiply by W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0

    Factor: . (W^2 - 49)(W^2 - 576) \:=\:0

    \begin{array}{ccccccccc}W^2-49 \:=\:0 & \Rightarrow & W^2 \:=\:49 & \Rightarrow & W \:=\:7 \\ W^2 - 576 \:=\:0 & \Rightarrow & W^2 \:=\:576 & \Rightarrow & W \:=\:24 \end{array}


    \text{Assuming }L > W\text{, we have: }\:W\,=\,7,\;\boxed{L \,=\,24}
    Thanks from Bonganitedd
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  6. #6
    MHF Contributor

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    Re: find length of rectangle given diagonal and area

    Once you get to "[tex]W^4- 625W^2+ 28224=0 as Soroban showed, if the "fourth degree polynomial" bothers you, you can let x= W^2 so that your equation is x^2- 625x+ 28224= 0 and use whatever method you like, completing the square or the quadratic formula, to solve that quadratic equation.
    Thanks from Bonganitedd
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  7. #7
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    Re: find length of rectangle given diagonal and area

    Not taking away credit to other members who contributed, but @ Hallsofivy your advice is what I needed. Big u to MHF.
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