Chapter 8, Problem 9PFA

To determine

To write an expression that represents the perimeter of the room.

Answer to Problem 9PFA

  $4(x+1)$

Explanation of Solution

Given:

  $\text{Area of a rectangular room}=x^2+2x-224$

  $\text{Length of the rectangular room}=x+16$

Formula used:

  $\text{Area of a rectangle}=l\times b$

  $\text{Perimeter of a rectangle}=2(l+b)$

Calculation:

Given that the area of the rectangular room is $x^2+2x-224$ and length of the room is $x+16.$

  $\begin{array}{l}\text{Area of a rectangle}=l\times b\\ \Rightarrow x^2+2x-224=(x+16)\times b\\ \Rightarrow b=\frac{x^2+2x-224}{x+16}\end{array}$

Dividing the above expression,

  $x+16\begin{array}{c}\hfill x-14\\ \hfill \overline{)\begin{array}{l}{x}^2+2x-224\\ \frac{x^2+16x}{\begin{array}{l}-14x-224\\ \frac{-14x-224}{0}\end{array}}\end{array}}\end{array}$

  $\begin{array}{l}\Rightarrow b=\frac{x^2+2x-224}{x+16}\\ \Rightarrow b=x-14\end{array}$

Therefore, the width of the rectangular room is $x-14.$

  $\begin{array}{c}\text{Perimeter of a rectangle}=2(l+b)\\ =2((x+16)+(x-14))\\ =2(x+16+x-14)\\ =2(2x+2)\\ =4x+4\\ =4(x+1)\end{array}$

Conclusion:

Therefore, the perimeter of the rectangular room is $4(x+1).$