Chapter 12.6, Problem 6ST

To determine

To find: the length and width of a rectangle.

Answer to Problem 6ST

The length of a rectangle is $50m$ and the width of a rectangle is $36m$ .

Explanation of Solution

Given information:

The area of a rectangle $1800m^2$ and the perimeter of a rectangle is $172m$ .

Formula:

Area of rectangle $=l*b$ where $l$ is the length and $b$ is the width of the rectangle.

Perimeter of a rectangle $=2*\left(l+b\right)$ where $l$ is the length and $b$ is the width of the rectangle.

  ${\left(a-b\right)}^2={\left(a+b\right)}^2-4*a*b.$

Calculation:

Consider the length of rectangle be $l$ and the width be $b$ .

As mentioned above that the area of a rectangle $1800m^2$ and the perimeter of a rectangle is $172m$ .

Therefore $l*b=1800m^2$ and $2*\left(l+b\right)=172\Rightarrow \left(l+b\right)=\mathrm{86.......}\left(i\right)$ .

Recall that ${\left(a-b\right)}^2={\left(a+b\right)}^2-4*a*b\Rightarrow {\left(l-b\right)}^2={\left(l+b\right)}^2-4*l*b$ where $l$ is the length and $b$ is the width of the rectangle.

Therefore

  $\begin{array}{l}{\left(l-b\right)}^2={\left(86\right)}^2-4*1800\Rightarrow {\left(l-b\right)}^2=7396-7200=196\\ \Rightarrow \left(l-b\right)=\mathrm{14.....}\left(ii\right)\end{array}$

Adding $\left(i\right)\&\left(ii\right)$ we have $2l=100\Rightarrow l=50.$

Again we substitute the $l=50$ in $\left(i\right)$ .

We have $b=86-50=36$ .

Therefore the length of a rectangle is $50m$ and the width of a rectangle is $36m$ .