Chapter 9.4, Problem 33E

(a)

To determine

The length of the rectangle

Answer to Problem 33E

  $l=2w$

Explanation of Solution

Given:

Area = 20 sq. meters

Length is two times width

Calculation:

Let the width be w

The length can be written as

  $l=2w$

(b)

To determine

The equation for the area of the rectangle

Answer to Problem 33E

  $20=2w^2$

Explanation of Solution

Given:

Area = 20 sq. meters

Length is two times width

Calculation:

Let the width be w

The length can be written as

  $l=2w$

The area of the rectangle will be

  $\begin{array}{l}A=l\times w\\ A=2w\times w\\ 20=2w^2\end{array}$

(c)

To determine

The width of the rectangle

Answer to Problem 33E

  $w=1.4\text{m}$

Explanation of Solution

Given:

Area = 20 sq. meters

Length is two times width

Calculation:

Let the width be w

The length can be written as

  $l=2w$

The area of the rectangle will be

  $\begin{array}{l}A=l\times w\\ A=2w\times w\\ 20=2w^2\\ w=\sqrt{\frac{20}{10}}\\ w=\sqrt{2}\\ w=1.4\text{m}\end{array}$

(d)

To determine

The length of the rectangle

Answer to Problem 33E

  $l=2.8\text{m}$

Explanation of Solution

Given:

Area = 20 sq. meters

Length is two times width

Calculation:

Let the width be w

The length can be written as

  $l=2w$

The area of the rectangle will be

  $\begin{array}{l}A=l\times w\\ A=2w\times w\\ 20=2w^2\\ w=\sqrt{\frac{20}{10}}\\ w=\sqrt{2}\\ w=1.4\text{m}\\ l=2w\\ l=2\left(1.4\right)\\ l=2.8\text{m}\end{array}$