# Thread: The rectangular playground quardratic equations question

1. ## 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!

2. 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.

3. 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 $\displaystyle (meters)(meters)= meters^2$.