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Thread: The rectangular playground quardratic equations question

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Question The rectangular playground quardratic equations question

    The question goes like this:

    ''A rectangular playground has a perimeter of $\displaystyle 60m$ and an area of $\displaystyle 216m^2$. Find the dimensions of the playground.''

    What I did is this (perimeter):

    Let the length be $\displaystyle x$ meters and the breadth be $\displaystyle y$ metres.
    From the given information,
    $\displaystyle 2x + 2y = 60$

    $\displaystyle 2x +2y -60 = 0$

    That was when I got stuck in this status quo...
    It's common sense that I couldn't solve that equation.

    Now if I were to take another way, which is the equation of the area of the playground:

    $\displaystyle
    (x^2)(y^2)= 216$

    $\displaystyle x^2y^2 -216 = 0$

    I still couldn't solve this...

    Can anyone help me to solve this question? Thank you so much for taking your time!
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  2. #2
    Senior Member Educated's Avatar
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    You are correct in the first part of the equation:

    $\displaystyle 2x + 2y = 60m$

    But your second equation is wrong. The area of a rectangle is base * height... so where did you get the x^2 and y^2 from?
    The equation for the area should be:

    $\displaystyle xy = 216m^2$

    Now that we have 2 equations with 2 variables, we can use simultaneous equation to solve for x and y, the dimensions of the playground.

    First equation: $\displaystyle xy = 216m^2$ which if we rewrite in the form of y= , we get $\displaystyle y = \frac{216}{x}$

    Second equation: $\displaystyle 2x + 2y = 60m$


    Now we substitute from the first equation of y = 216/x into the second equation whenever y occurs. This leaves us with:

    $\displaystyle 2x + 2(\frac{216}{x}) = 60$

    $\displaystyle 2x + \frac{432}{x} = 60$

    Multiply everything by x to make it easier to solve:
    $\displaystyle 2x^2 -60x + 432 = 0$

    Now solve the equation for x, using the quadratic formula or by factorising. You should get 2 answers, and those are the dimensions of the playground.
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  3. #3
    MHF Contributor

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    Or, from 2x+ 2y= 60, x+ y= 30 so that y= 30- x. Putting that into xy= 316 gives the quadratic equation $\displaystyle x(30- x)= 30x- x^2= 316$ or $\displaystyle x^2- 30x+ 316= 0$. That is, of course, the same as $\displaystyle 2x^2- 60x+ 432= 0$ divided by 2.

    By the way, area is always measured in "square units" where "units" is whatever a distance is measured in. If x and y are measured in meters, for example, $\displaystyle x^2y^2$ would be $\displaystyle (m^2)(m^2)= m^4$, while $\displaystyle xy$ gives the correct $\displaystyle (meters)(meters)= meters^2$.
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