Precalculus (6th Edition) Blitzer

The length of each side of a figure can be either $x=50,y=20$ or $x=46,y=28$.
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 $x=46,y=28$.