Chapter 12.6, Problem 1P

To determine

To find: length and width of a rectangular park.

Answer to Problem 1P

The length of a rectangular park is $\frac{5+34.27i}{2}$ and the width of the rectangular park is $\frac{5-34.27i}{2}.$

Explanation of Solution

Given information:

The width of a park is $5m$ shorter than its length and the area of rectangular park is $300m^2$ .

Formula:

Area of rectangle is $length*width$ .

The quadratic equation formula $l=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a,b,c$ is any constant.

Calculation:

Consider that the length of a rectangular park be $l$ .

Recall that area of rectangle is $length*width$ .

Since the width of a park is $5m$ shorter than its length.

Therefore we assume length of a rectangular park be $l$ .

Therefore width of a park is $5-l$ .

Therefore area of rectangular park will be $300=l*\left(5-l\right)$ .

Therefore on further $300=l*\left(5-l\right)\Rightarrow 5l-l^2=300\Rightarrow l^2-5l+300=0.$

Recall that the quadratic equation is $l=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=1,b=-5,c=300$ .

Therefore $l=\frac{5\pm \sqrt{{\left(-5\right)}^2-4*1*300}}{2}\Rightarrow l=\frac{5\pm \sqrt{25-1200}}{2}\Rightarrow l=\frac{5\pm \sqrt{-1175}}{2}\Rightarrow l=\frac{5\pm 34.27i}{2}\Rightarrow l=\frac{5+34.27i}{2},\frac{5-34.27i}{2}.$

We will assume length as $\frac{5+34.27i}{2}$ then width will be $5-\frac{5+34.27i}{2}=\frac{5-34.27i}{2}$ and vice versa.

Therefore the length of a rectangular park is $\frac{5+34.27i}{2}$ and the width of the rectangular park is $\frac{5-34.27i}{2}.$