Chapter 8, Problem 3PFA

a.

To determine

To write an expression that represents the perimeter and area of the garden.

Answer to Problem 3PFA

Perimeter: $16+4w$

Area: $8w+w^2$

Explanation of Solution

Given:

Length of the garden is 8 feet more than the width.

Perimeter is 108 feet.

Concept used:

  $\begin{array}{l}\text{Perimeter of rectangle}=2(l+b)\\ \text{Area of rectangle}=l\times b\end{array}$

Calculation:

Let x be the length of the rectangular garden and w be the width of the rectangular garden.

Given that the length of the garden is 8 feet more than the width.

Therefore,

  $x=8+w$

  $\begin{array}{c}\text{Perimeter of rectangle}=2((8+w)+w)\\ =2(8+2w)\\ \text{Perimeter of rectangle}=16+4w\end{array}$

  $\begin{array}{c}\text{Area of rectangle}=(8+w)\times w\\ \text{Area of rectangle}=8w+w^2\end{array}$

Conclusion:

Therefore, the perimeter of rectangular garden is $16+4w$ and the area of the rectangular garden is $8w+w^2$ .

b.

To determine

To calculate the width of the rectangular garden.

Answer to Problem 3PFA

23 feet

Explanation of Solution

Given:

  $\text{Perimeter of rectangle}=108\text{feet}$

From subpart (a), Perimeter of the rectangular garden is $16+4w.$

Formula used:

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

Calculation:

Substituting the length of the rectangle and width of the rectangle,

  $\begin{array}{l}\text{Perimeter of rectangle}=2(l+b)\\ \Rightarrow 108=2((8+w)+w)\\ \Rightarrow 108=2(8+2w)\\ \Rightarrow 108=16+4w\\ \Rightarrow 108-16=4w\\ \Rightarrow 92=4w\\ \Rightarrow w=23\end{array}$

Conclusion:

Therefore, the width of the rectangular garden is 23 feet.

c.

To determine

To explain the mathematical practise used in solving the problem.

Answer to Problem 3PFA

Substitution

Explanation of Solution

The mathematical practise used in solving the problem is substitution.

First, an equation is formed for the length of the rectangular garden, later the length which is interms of width is substituted to find the remaining values.