Answer
$15$ m $\times$ $8$ m
Work Step by Step
We know the following:
Perimeter = $46$ cm
Area = $120$ cm$^2$
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.
(1) $46 = (2x) + (2y)$
(2) $120 = xy$
We can then make $x$ the subject in one of the equations:
(2) $120 = xy$
$x=\frac{120}{y}$
We can then substitute this expression for $x$ into the first equation:
$46 = (2x) + (2y)$
$46 = (2)(\frac{120}{y}) + (2y)$
$46 = \frac{240}{y} + 2y$ (Make denominators the same)
$46 = \frac{240}{y} + \frac{2y^2}{y}$
$46 = \frac{240+2y^2}{y}$ (Multiply each side by y)
$46y = 240+2y^2$ (Subtract 46y from each side to make the equation =0)
$2y^2-46y+240=0$
We can then solve this equation by factoring:
$2y^2-30y-16y+240=0$
$2y(y-15)-16(y-15)=0$
$(2y-16)(y-15) = 0$
We can then make each bracket = 0 to find the solutions for $y$.
$(2y-16)=0$ or $(y-15) = 0$
$y=8$ or $y=15$
Now that we have both possible solutions for $y$, we can substitute them into our equation for $x$ to find both possible solutions for $x$.
$x=\frac{120}{8}$ or $x=\frac{120}{15}$
$x=15$ or $x=8$
We are getting the same combination for $x$ and $y$ in each case, which is $15$ m and $8$ m, so the dimensions of the rectangle are $15$ m by $8$ m.
