Chapter 1, Problem 16P

(a)

To determine

The conversion factor between ${\text{ft}}^{\text{2}}{\text{and yard}}^{\text{2}}$ .

Answer to Problem 16P

The conversion factor is $1{\text{ft}}^2=0.111{\text{yard}}^2$

Explanation of Solution

Given:

The units are ${\text{ft}}^{\text{2}}{\text{and yard}}^{\text{2}}$

Formula used:

1 feet = 0.3333 yards

Calculation:

Since 1 feet = 0.3333 yards

So,

  $\begin{array}{l}\text{1ft}=\text{0}\text{.333yard}\\ {\text{1ft}}^{\text{2}}=\text{0}{\text{.333}}^{\text{2}}{\text{yard}}^{\text{2}}\\ {\text{1ft}}^{\text{2}}=\text{0}{\text{.111yard}}^{\text{2}}\end{array}$

Conclusion:

Thus, the conversion factor is $1{\text{ft}}^2=0.111{\text{yard}}^2$

(b)

To determine

The conversion factor between ${\text{m}}^{\text{2}}{\text{and ft}}^{\text{2}}$ .

Answer to Problem 16P

The conversion factor is

  $1{\text{m}}^2=10.76{\text{ft}}^2$

Explanation of Solution

Given:

The units are ${\text{ft}}^{\text{2}}{\text{and m}}^{\text{2}}$

Formula used:

1 m = 3.28 ft

Calculation:

Since 1 m = 3.28 ft

So,

  $\begin{array}{l}\text{1m}=3.\text{28ft}\\ {\text{1m}}^{\text{2}}=\text{3}{\text{.28}}^{\text{2}}{\text{ft}}^{\text{2}}\\ {\text{1m}}^{\text{2}}=\text{10}{\text{.76ft}}^{\text{2}}\end{array}$

Conclusion:

Thus, the conversion factor is

  $1{\text{m}}^2=10.76{\text{ft}}^2$