Answer
Length: 12 feet
Width: 9 feet
Work Step by Step
Let's note:
$l$=the length of the rectangle
$w$=the width of the rectangle
We can write the system, using the area and the diagonal of the rectangle:
$\begin{cases}
lw=108\\
l^2+w^2=15^2
\end{cases}$
We will use the substitution method. Solve Equation 1 for $w$ and substitute the expression of $w$ in Equation 2 to eliminate $w$ and determine $l$:
$\begin{cases}
w=\dfrac{108}{l}\\
l^2+\left(\dfrac{108}{l}\right)^2=225
\end{cases}$
$l^2+\dfrac{11664}{l^2}=225$
$l^4+116644=225l^2$
$l^4-225l^2+11664=0$
$l^4-81l^2-144l^2+11664=0$
$l^2(l^2-81)-144(l^2-81)=0$
$(l^2-81)(l^2-144)=0$
$(l-9)(l+9)(l-12)(l+12)=0$
$l-9=0\Rightarrow l_1=9$
$l+9=0\Rightarrow l_2=-9$
$l-12=0\Rightarrow l_3=12$
$l+12=0\Rightarrow l_4=-12$
Substitute each of the values of $l$ in Equation 1 to determine $w$:
$lw=48$
$l_1=9\Rightarrow 9w=108\Rightarrow w_1=\dfrac{108}{9}\Rightarrow w_1=12$
$l_2=-9\Rightarrow -9w=108\Rightarrow w_2=\dfrac{108}{-9}\Rightarrow w_2=-12$
$l_3=12\Rightarrow 12w=108\Rightarrow w_3=\dfrac{108}{12}\Rightarrow w_3=9$
$l_4=-12\Rightarrow -12w=108\Rightarrow w_4=\dfrac{108}{-12}\Rightarrow w_4=-9$
The system's solutions are:
$(9,12),(-9,-12),(12,9),(-12,-9)$
As $l\geq w$, $l, w$ positive, the solution is:
$l=12$ feet
$w=9$ feet