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Math Help - Find Area of Rectangle

  1. #1
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    Find Area of Rectangle

    The perimeter and the diagonal of a rectangle are 18m and 5m respectively. Find its area.
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  2. #2
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    Hello, magentarita!

    Is there a typo in the problem?
    As stated, it has no solution.


    The perimeter and the diagonal of a rectangle are 14m and 5m respectively.
    Find its area.
    Code:
          * - - - - - *
          |        *  |
          |     *5    | W
          |  *        |
          * - - - - - *
                L

    The perimeter is 14: . 2L + 2W = 14 \quad\Rightarrow\quad L + W \:=\:7 .[1]

    The diagonal is 5: . L^2 + W^2 \:=\:5^2 .[2] . . . (Pythagorus)


    From [1], we have: . W \:=\:7-L

    Substitute into [2]: . L^2 + (7-L)^2 \:=\:25 \quad\Rightarrow\quad l^2-7L + 12 :=\:0

    Factor: . (L-3)(L-4) \:=\:0 \quad\Rightarrow\quad L \;=\;3,\:4 \quad\Rightarrow\quad W \;=\;4,\:3


    The area is: . 3 \times 4 \:=\:12\text{ m}^2

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  3. #3
    MHF Contributor
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    I got..........

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Is there a typo in the problem?
    As stated, it has no solution.

    Code:
          * - - - - - *
          |        *  |
          |     *5    | W
          |  *        |
          * - - - - - *
                L
    The perimeter is 14: . 2L + 2W = 14 \quad\Rightarrow\quad L + W \:=\:7 .[1]

    The diagonal is 5: . L^2 + W^2 \:=\:5^2 .[2] . . . (Pythagorus)


    From [1], we have: . W \:=\:7-L

    Substitute into [2]: . L^2 + (7-L)^2 \:=\:25 \quad\Rightarrow\quad l^2-7L + 12 :=\:0

    Factor: . (L-3)(L-4) \:=\:0 \quad\Rightarrow\quad L \;=\;3,\:4 \quad\Rightarrow\quad W \;=\;4,\:3


    The area is: . 3 \times 4 \:=\:12\text{ m}^2
    I understood the fact that the diagonal is 5m and that length and width both equal x but got stuck after that.
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