Answer
$20$ m $\times$ $60$ m
Work Step by Step
We know the following:
Perimeter = $160$ m
Area = $1200 m^2$
We know the following equations as well for a rectangle:
Perimeter = (2 $\times$ Length) + (2 $\times$ Width)
Area = Length $\times$ Width
We can make the width of the rectangle = $y$ and the length of the rectangle = $x$. Then, substitute these variables into the formulas.
$160 = (2x) + (2y)$
$1200 = xy$
We can then make $x$ the subject in one of the equations:
$1200 = xy$
$x=\frac{1200}{y}$
We can then substitute this expression for $x$ into the first equation:
$160 = (2x) + (2y)$
$160 = (2)(\frac{1200}{y}) + (2y)$
$160 = \frac{2400}{y} + 2y$ (Make denominators the same)
$160 = \frac{2400}{y} + \frac{2y^2}{y}$
$160 = \frac{2400+2y^2}{y}$ (Multiply each side by y)
$160y = 2400+2y^2$ (Subtract 160y from each side to make the equation =0)
$2y^2-160y+2400=0$
We can then solve this equation by factoring:
$2y^2-120y-40y+2400=0$
$2y(y-60)-40(y-60)=0$
$(2y-40)(y-60) = 0$
We can then make each bracket = 0 to find the solutions for $y$.
$(2y-40)=0$ or $(y-60) = 0$
$y=20$ or $y=60$
Now that we have both possible solutions for $y$, we can substitute them into our equation for $x$ to find both possible solutions for $x$.
$x=\frac{1200}{20}$ or $x=\frac{1200}{60}$
$x=60$ or $x=20$
We are getting the same combination for $x$ and $y$ in each case, which is $20$ m and $60$ m, so the dimensions of the rectangle are $20$ m by $60$ m.
