Answer
The length of each side of a figure can be either $ x=50,y=20$ or $ x=46,y=28$.
Work Step by Step
Consider the area of the two adjoining square fields as $2900$ square feet, which is enclosed with $240$ feet of fencing. The perimeter of the field is the sum of all sides of adjoining squares as shown below:
$ x+x+\left( x+y \right)+y+y+\left( x-y \right)=4x+2y $.
Hence, the equation for the perimeter of the figure is:
$4x+2y=240$ (I)
Now, the area of the field is ${{x}^{2}}+{{y}^{2}}$, where x and y are the sides in the figure.
Hence, the area of two squares can be written as:
${{x}^{2}}+{{y}^{2}}=2900$ (II)
From equation (I), obtain $ y=120-2x $; and put in equation (II):
$\begin{align}
& {{x}^{2}}+{{\left( 120-2x \right)}^{2}}=2900 \\
& {{x}^{2}}+{{\left( 120 \right)}^{2}}+{{\left( 2x \right)}^{2}}-2\left( 120 \right)\left( 2x \right)=2900 \\
& {{x}^{2}}+14400+4{{x}^{2}}-480x=2900 \\
& 5{{x}^{2}}-480x+14400-2900=0
\end{align}$
Now, use the quadratic equation formula:
$\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where, $ a=5,\,\,b=-480,\,\,c=11,500$ ; put the value in the quadratic equation formula:
$\begin{align}
& \frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\frac{-\left( -480 \right)\pm \sqrt{{{\left( -480 \right)}^{2}}-4\left( 5 \right)\left( 11,500 \right)}}{2\left( 5 \right)} \\
& =\frac{480\pm \sqrt{{{\left( -480 \right)}^{2}}-4\left( 5 \right)\left( 11,500 \right)}}{2\left( 5 \right)} \\
& =\frac{480\pm \sqrt{230,400-230,000}}{10} \\
& =\frac{480\pm \sqrt{400}}{10}
\end{align}$
$=\frac{480\pm 20}{10}$
Now
Case 1:
$\begin{align}
& \frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}=\frac{480+20}{10} \\
& =\frac{500}{10} \\
& =50
\end{align}$
Case 2:
$\begin{align}
& \frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}=\frac{480-20}{10} \\
& =\frac{460}{10} \\
& =46
\end{align}$
Now, put $ x=50,46$ in $ y=120-2x $, to obtain the value of y.
Case 3:
When $ x=50$:
$\begin{align}
& y=120-2\left( 50 \right) \\
& =120-100 \\
& =20
\end{align}$
Case 4:
When $ x=46$:
$\begin{align}
& y=120-2\left( 46 \right) \\
& =120-92 \\
& =28
\end{align}$
So, there are two values for length:
Case 5:
When $ x=50,y=20$:
In that case, the length of each side will be as shown below:
$\begin{align}
& \text{AB}=x+y \\
& =50+20 \\
& =70
\end{align}$
$\begin{align}
& \text{BC}=x \\
& =50 \\
& \text{CD}=x \\
& =50
\end{align}$
$\begin{align}
& \text{DE}=x-y \\
& =50-20 \\
& =30
\end{align}$
$\begin{align}
& \text{EF}=y \\
& =20 \\
& \text{AF}=y \\
& =20
\end{align}$
Case 6:
When $ x=46,y=28$:
In that case, the length of each side will be as shown below:
$\begin{align}
& \text{AB}=x+y \\
& =46+28 \\
& =74
\end{align}$
$\begin{align}
& \text{BC}=x \\
& =46 \\
& \text{CD}=x \\
& =46
\end{align}$
$\begin{align}
& \text{DE}=x-y \\
& =46-28 \\
& =18
\end{align}$
$\begin{align}
& \text{EF}=y \\
& =28 \\
& \text{AF}=y \\
& =28
\end{align}$
Thus, the length of each side of the figure can be either $ x=50,y=20$ or $ x=46,y=28$.
