We know that the length of the rectangle is 5 feet more than the width. Let x be the width of the rectangle, then its length is x+5. This can be see in the next picture
We also know that the area of the fish pond is 143 and that the area is given by
[tex]A=wl[/tex]Plugging the values we have that
[tex]143=x(x+5)[/tex]writting the equation in standard form we have that
[tex]\begin{gathered} 143=x(x+5) \\ 143=x^2+5x \\ x^2+5x-143=0 \end{gathered}[/tex]We know that any quadratic equation can be solve by
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]using it for our equation we have
[tex]\begin{gathered} x=\frac{-5\pm\sqrt[]{(5)^2-4(1)(-143)}}{2(1)} \\ =\frac{-5\pm\sqrt[]{25+572}}{2} \\ =\frac{-5\pm\sqrt[]{597}}{2} \\ \text{then} \\ x_1=\frac{-5+\sqrt[]{597}}{2}=9.72 \\ \text{and} \\ x_2=\frac{-5-\sqrt[]{597}}{2}=-14.72 \end{gathered}[/tex]As we know the quadratic equation leads to two solutions. Nevertheless the negative solution is not right in this case, since the distances have to be positive. Then x=9.72.
Once we have the value of x we can know the width a lenght
[tex]\begin{gathered} l=9.72 \\ w=14.72 \end{gathered}[/tex]And the perimeter is
[tex]\begin{gathered} P=2w+2l \\ =2(14.72)+2(9.72) \\ =48.88 \end{gathered}[/tex]The perimeter is 48.88 ft.
