Chapter 1, Problem 1.113QP

(a)

Interpretation Introduction

Interpretation:

The perimeter of property is given as front boundary and back boundary were measured in meters: $\text{85}\text{m}\text{and}\text{95}\text{m}$ respectively.  The side boundaries are measured in feet: $\text{435}\text{ft}\text{and}\text{515}\text{ft}$.  The problem solving strategy has to be given which is used to determine the perimeter in kilometers.

Explanation of Solution

The problem solving strategy is conversion of units.

The meter can be converted into kilometer by the conversion factor:

  $\begin{array}{l}\text{1}\text{km}\text{=}{\text{10}}^{\text{3}}\text{m}\\ \text{conversion}\text{factor:}\\ \text{=}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\end{array}$

The feet can be converted into kilometers by a sequence of conversion by the conversion factors:

  $\begin{array}{l}\text{Conversion}\text{of}\text{feet}\text{to}\text{yard:}\\ \text{3}\text{ft}\text{=}\text{1}\text{yd}\\ \text{Conversion}\text{factor:}\\ \text{=}\frac{\text{1}\text{yd}}{\text{3}\text{ft}}\\ \text{Conversion}\text{of}\text{yard}\text{to}\text{meter:}\\ \text{1}\text{yd}\text{=}\text{0}\text{.914}\text{m}\\ \text{Conversion}\text{factor:}\\ \text{=}\frac{\text{0}\text{.914}\text{m}}{\text{1}\text{yd}}\\ \text{Conversion}\text{of}\text{meter}\text{to}\text{kilometer:}\\ \text{1}\text{km}\text{=}{\text{10}}^{\text{3}}\text{m}\\ \text{conversion}\text{factor:}\\ \text{=}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\end{array}$

The overall conversion of feet to kilometer can be given as

  $\text{=}\frac{\text{1}\text{yd}}{\text{3}\text{ft}}\text{×}\frac{\text{0}\text{.914}\text{m}}{\text{1}\text{yd}}\text{×}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}$

(b)

Interpretation Introduction

Interpretation:

The perimeter of property is given as front boundary and back boundary were measured in meters: $\text{85}\text{m}\text{and}\text{95}\text{m}$ respectively.  The side boundaries are measured in feet: $\text{435}\text{ft}\text{and}\text{515}\text{ft}$.  The perimeter has to be calculated in kilometers.

Explanation of Solution

Given, front and back boundaries in meters as $\text{85}\text{m}\text{and}\text{95}\text{m}$.  Using the conversion factor, the meter is converted into kilometer.

  $\begin{array}{l}\text{Given,}\text{85}\text{m}\text{and}\text{95}\text{m}\\ \text{85}\text{m:}\\ \text{1}\text{km}\text{=}{\text{10}}^{\text{3}}\text{m}\\ \text{=}\text{85}\text{m}\text{×}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\\ \text{=}\text{0}\text{.085}\text{km}\\ \text{95}\text{m:}\\ \text{1}\text{km}\text{=}{\text{10}}^{\text{3}}\text{m}\\ \text{=}\text{95}\text{m}\text{×}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\\ \text{=}\text{0}\text{.095}\text{km}\end{array}$

The side boundaries are given in feet: $\text{435}\text{ft}\text{and}\text{515}\text{ft}$.  Using conversion factor, feet is converted into kilometer.

  $\begin{array}{l}\text{435}\text{ft}\text{and}\text{515}\text{ft}\\ \text{435}\text{ft:}\\ \text{=}\text{435}\text{ft}\text{×}\frac{\text{1}\text{yd}}{\text{3}\text{ft}}\text{×}\frac{\text{0}\text{.914}\text{m}}{\text{1}\text{yd}}\text{×}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\\ \text{=}\text{0}\text{.132}\text{km}\\ \text{515}\text{ft:}\\ \text{=}\text{515}\text{ft}\text{×}\frac{\text{1}\text{yd}}{\text{3}\text{ft}}\text{×}\frac{\text{0}\text{.914}\text{m}}{\text{1}\text{yd}}\text{×}\frac{\text{1}\text{km}}{{\text{10}}^{\text{3}}\text{m}}\\ \text{=}\text{0}\text{.156}\text{km}\end{array}$

The perimeter in kilometers are calculated as front and back boundaries are $\text{0}\text{.085}\text{km}\text{and}\text{0}\text{.095}\text{km}$ respectively and side boundaries are $\text{0}\text{.132}\text{km}\text{and}\text{0}\text{.156}\text{km}$.