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Math Help - Narrative Quadratic

  1. #1
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    Narrative Quadratic

    Hey folks. Been trying the one for over an hour and can't seem to get it at all, whats worse I need it for tomorow as everyone in our class was assigned a question to answer and I don't want to be the odd one out, although after seeing the questions I'm sure there will be a few odd ones out. Anyway here's the question

    A square and a rectangle have equal perimeters.

    The length of the rectangle is 11cm and the area of the square is 4cm squared larger than the area of the rectangle. Find the length of a side on the square.

    So far I've tried relating the area of the square to be x^2 and the area of the rectangle to be (x^2)-4. After checking the solution I found the two possible answers to be 9 and 13 which add to give 22 so I believed I had to insert 2 x length of the rectangle somewhere but I don't know how. Really frustrated atm and hope you can help me!
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  2. #2
    Super Member

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    Hello, fladam!

    A square and a rectangle have equal perimeters.
    The length of the rectangle is 11cm.
    The area of the square is 4cm larger than the area of the rectangle.
    Find the length of a side on the square.
    Code:
                            * - - - - *
          * - - - - - *     |         |
          |           |     |         | x
        w |    11w    |     |    x   |
          |           |     |         |
          * - - - - - *     * - - - - *
                11                x

    The perimeter of the rectangle is: 2w + 22
    The perimeter of the square is: 4x

    The perimeters are equal: . 2w + 22 \:=\:4x \quad\Rightarrow\quad w \:=\:2x-11\;\;{\color{blue}[1]}


    The area of the rectangle is: 11w
    The area of the square is: x^2

    The area of the square is 4 more than that of the rectangle: . x^2 \:=\:11w + 4\;\;{\color{blue}[2]}


    Substitute [1] into [2]: . x^2 \:=\:11(2x-11) +4\quad\Rightarrow\quad x^2 - 22x + 117 \:=\:0

    Factor: . (x-9)(x-13) \:=\:0

    And we have two solutions: . x \:=\:9,\:13


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Check


    If x = 9, substitute into [1]: . w \:=\:2(9) - 11 \:=\:7

    The rectangle is 7 11.
    Its perimeter is 36 cm.
    Its area is 77 cm.

    The square is 9 9.
    Its perimeter is 36 cm.
    Its area is: 81 cm.

    . . - - - - - - - - - -

    If x = 13, then: . w \:=\:2(13)-11 \:=\:15

    The rectangle is: 15 11.
    Its perrimer is 52 cm.
    Its area is: 165 cm.

    The square is: 13 13.
    Its perimeter is: 52 cm.
    Its area is: 169 cm.


    Both solutions check out!

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