Chapter 4, Problem 17CCSR

To determine

Find the rectangle B’s perimeter in feet.

Answer to Problem 17CCSR

Perimeter of rectangle B is $29ft$ .

Explanation of Solution

Given:

The area of rectangle A is $180f{t}^2$

Length of rectangle A is $20ft$

The area of rectangle B is $45f{t}^2$

Concept used:

1. Calculate the width of rectangle A.
2. Find the length and width of rectangle B
3. Determine the value of perimeter of B.

Formula used:

  $\text{Area}\text{of}\text{reatangle}\text{= l}\times b$

Calculation:

Consider, $\text{area}\text{of}{\text{rectangle A}}_A{\text{=180ft}}^2$

  $\begin{array}{l}{\text{A}}_A{\text{=180ft}}^{\text{2}}\\ \text{l×b}=\text{180}\\ \text{20×b}=\text{180}\\ \text{b}=\text{9}\text{ft}\end{array}$

  $\begin{array}{l}\text{Perimeter}\text{of}{\text{rectangle P}}_A=\text{2(l+b)}\\ =\text{2(20+9)}\\ =58ft\end{array}$

  $\because \text{Area}\text{of}{\text{rectangle A}}_B\text{=}\frac{1}{4}{\text{A}}_A=\frac{l}{2}\times \frac{b}{2}$

So, length and width will be half of rectangle A.

  $\begin{array}{l}\because \text{Perimeter}\text{of}{\text{rectangle P}}_B\text{=}\frac{1}{2}{\text{P}}_A\\ =29ft\end{array}$

Therefore, the perimeter of rectangle B is $29ft$ .