Kaitlyn is looking at a map and wants to find the area of the state of Nevada, United States. The map shows a scale of \begin{align*}{\frac{1}{2}}^{{\prime}{\prime}}\end{align*} to 20 miles. If the distance on the map is wide and long, what is the area, in square miles of Nevada?

In this concept, you will learn to read and interpret maps involving distance and area by using scale measurement.

### Distance and Area of Maps

A map is another type of scale drawing of a region. Maps can be very detailed or very simple, showing only points of interest and distances. You can read a map just like any other scale drawing—by using the scale.

Let’s look at an example.

On the map below, the straight-line distance between San Francisco and San Diego is 3 inches. What is the actual straight-line distance between San Francisco and San Diego?

First, set up a proportion. The scale in the drawing says that \begin{align*}0.5 \ \text{inches} = 75 \ \text{miles}\end{align*}, therefore the proportion is:

\begin{align*}\frac{0.5 \ \text{inches}}{75 \ \text{miles}}\end{align*}

Next, write the second ratio. You know the scale length is 3 inches. The unknown length is \begin{align*}x\end{align*}.

\begin{align*}\frac{0.5 \ \text{inches}}{75 \ \text{miles}} = \frac{3 \ \text{inches}}{x \ \text{miles}}\end{align*}

Then, cross-multiply.

\begin{align*}\begin{array}{rcl} \frac{0.5}{75} &=& \frac{3}{x}\\ 0.5x &=& 75\times3 \\ 0.5x &=& 225 \end{array}\end{align*}Then, divide both sides by 0.5 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl}
0.5x &=& 225 \\
\frac{0.5x}{0.5} &=& \frac{225}{0.5} \\
x &=& 450
\end{array}\end{align*}

The answer is 450.

The straight-line distance from San Francisco to San Diego is 450 miles.

Note: The straight-line distance is also known as “as the crow flies.” If you were actually traveling from San Francisco to San Diego, it would be farther than 450 miles, since you would need to drive on highways that are not necessarily in a straight line.

You can also use a scale to find the area of a space or region. First, you need to figure out the length and width then we can complete any necessary calculations. Sometimes, you will have two different distances or areas that you are working to compare. When this happens, you can use proportions to compare the differences and similarities.

Let’s look at an example.

Marta has a square with a side length of 4 inches. She has a similar square with dimensions that are twice the first square. How does the area of the larger square compare to the area of the smaller square?

First, find the dimensions of the larger square. The problem states that the dimensions are twice the first square. You can use this information to figure out the scale factor, and this means they are scaled up by a factor of 2. The side length of the larger square is:

\begin{align*}4 \ \text{inches}\times 2 = 8 \ \text{inches}\end{align*}.

Next, find the area of both squares and compare.

\begin{align*}\begin{array}{rcl} && \text{Area of smaller square} \qquad \ \ \ \text{Area of larger square}\\ && \text{A = lw} \qquad \qquad \qquad \qquad \ \quad \text{A = lw} \\ && \text{A = (4 inches)(4 inches)} \qquad \text{A = (8 inches)(8 inches)} \\ && \text{A} = 16 \ \text{in}^2 \qquad \qquad \qquad \ \ \ \quad \text{A} =64 \ \text{in}^2 \end{array}\end{align*}

Then, compare the two areas. You want to know how the area of the larger square compares to the area of the smaller square. Write a ratio comparing the two areas.

\begin{align*}\frac{64 \ \text{in}^2}{16 \ \text{in}^2} = 4\end{align*}

The answer is 4.

The area of the larger square is 4 times larger than the area of the smaller square.

This leads to a rule when comparing areas of similar figures. The ratio of areas of similar figures is the square of the scale factor.

### Examples

#### Example 1

Earlier, you were given a problem about Kaitlyn and her interest in Nevada.

First, set up a proportion. The scale in the drawing says that \begin{align*}\frac{1}{2} \text{ inch} = 20 \ \text{miles}\end{align*}, therefore the proportion is:

\begin{align*}\frac{0.5\ \text{inch}}{20 \ \text{miles}}\end{align*}

Next, write the ratio representing the length. You know the scale length is 8 inches. The unknown length is \begin{align*}x\end{align*}.

\begin{align*}\frac{0.5 \ \text{inch}}{20 \ \text{miles}} = \frac{8 \ \text{inches}}{x \ \text{miles}}\end{align*}

Then, cross-multiply to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl}
\frac{0.5}{20} &=& \frac{8}{x}\\
0.5x &=& 20\times8\\
0.5x&=& 160
\end{array}\end{align*}

Then, divide both sides by 0.5 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 0.5x &=& 160\\ \frac{0.5x}{0.5} &=& \frac{160}{0.5}\\ x&=& 320 \end{array}\end{align*}

The answer is 320.

The width of Nevada is 320 miles.

Then, write the ratio representing the length. You know the scale length is 12.25 inches. The unknown length is \begin{align*}x\end{align*}.

\begin{align*}\frac{0.5\ \text{inch}}{20 \ \text{miles}} = \frac{12.25 \ \text{inches}}{x \ \text{miles}}\end{align*}

Then, cross-multiply to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl}
\frac{0.5}{20} &=& \frac{12.25}{x}\\
0.5x &=& 20\times12.25\\
0.5x&=& 245
\end{array}\end{align*}

Then, divide both sides by 0.5 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 0.5x &=& 245\\ \frac{0.5x}{0.5} &=& \frac{245}{0.5}\\ x&=& 490 \end{array}\end{align*}

The answer is 490.

The length of Nevada is 490 miles.

Finally, find the area of Nevada.

\begin{align*}\begin{array}{rcl} && \text{A = lw} \\ && A = (320 \ \text{miles})(490 \ \text{miles}) \\ && A = 156,800 \ \text{square miles} \end{array}\end{align*}

The answer is 156,800.

The area of Nevada is

.#### Example 2

If the scale is \begin{align*}0.5\ \text{inches} = 100 \ \text{miles}\end{align*}, how many inches is 500 miles?

First, set up the proportion to solve.

\begin{align*} \frac{0.5 \ \text{inches}}{100 \ \text{miles}}= \frac{x \ \text{inches}}{500 \ \text{miles}}\end{align*}

Next, cross multiply.

\begin{align*}\begin{array}{rcl} \frac{0.5}{100}&=& \frac{x}{500}\\ 100x &=& 0.5\times500\\ 100x &=& 250 \end{array} \end{align*}Then, divide by 100 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 100x &=& 250\\ \frac{100x}{100} &=& \frac{250}{100}\\ x&=& 2.5 \end{array} \end{align*}The answer is 2.5.

The distance on the map will be 2.5 inches.

#### Example 3

If \begin{align*}3^{\prime \prime}\end{align*}.

, find each actual distance with a scale measurement ofFirst, set up a proportion. The scale in the drawing says that \begin{align*}1\ \text{inch} = 2000 \ \text{miles}\end{align*}, therefore the proportion is:

\begin{align*}\frac{1\ \text{inch}}{2000 \ \text{miles}}\end{align*}

Next, write the second ratio. You know the scale length is 3 inches. The unknown length is \begin{align*}x\end{align*}.

\begin{align*}\frac{1\ \text{inch}}{2000 \ \text{miles}} = \frac{3 \ \text{inches}}{x \ \text{miles}}\end{align*}

Then, cross-multiply to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{1}{2000} &=& \frac{3}{x}\\ 1x &=& 2000\times3\\ x&=& 6000 \end{array}\end{align*}

The answer is 6000.

The actual distance is 6000 miles.

#### Example 4

If

, find each actual distance with a scale measurement of .First, set up a proportion. The scale in the drawing says that \begin{align*}1\ \text{inch} = 2000 \ \text{miles}\end{align*}, therefore the proportion is:

\begin{align*}\frac{1\ \text{inch}}{2000 \ \text{miles}}\end{align*}

Next, write the second ratio. You know the scale length is \begin{align*}0.5^{\prime \prime}\end{align*} inches. The unknown length is \begin{align*}x\end{align*}.

\begin{align*}\frac{1\ \text{inch}}{2000 \ \text{miles}} = \frac{0.5\ \text{inches}}{x\ \text{miles}}\end{align*}

Then, cross-multiply to solve for .

\begin{align*}\begin{array}{rcl} \frac{1}{2000} &=& \frac{0.5}{x}\\ 1x &=& 2000\times0.5\\ x&=& 1000 \end{array} \end{align*}The answer is 1000.

The actual distance is 1000 miles.

#### Example 5

If \begin{align*}{\frac{1}{4}}^{{\prime}{\prime}}\end{align*}.

, find each actual distance with a scale measurement ofFirst, set up a proportion. The scale in the drawing says that \begin{align*}1\text{ inch} = 2000 \text{ miles}\end{align*}, therefore the proportion is:

\begin{align*}\frac{1\ \text{inch}}{2000 \ \text{miles}}\end{align*}

Next, write the second ratio. You know the scale length is \begin{align*}x\end{align*}.

inches. The unknown length is\begin{align*}\frac{1\ \text{inch}}{2000 \ \text{miles}} = \frac{0.25 \ \text{inches}}{x \ \text{miles}}\end{align*}

Then, cross-multiply to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{1}{2000} &=& \frac{0.25}{x}\\ 1x &=& 2000\times0.25\\ x&=& 500 \end{array} \end{align*}The answer is 500.

The actual distance is 500 miles.

### Review

Using the scale \begin{align*}1 \ \text{inch} = 5.5 \ \text{miles}\end{align*}, figure out the number of inches needed to map each number of miles. Use proportions to figure out your answers.

1. 16.5 miles

2. 11 miles

3. 27.5 miles

4. 8.25 miles

5. 33 miles

6. 60.5 miles

7. 13.75 miles

Using the scale

, figure out the number of actual miles represented by each scale measurement.8. \begin{align*}1^{\prime \prime}\end{align*}

9.

10. \begin{align*}3^{\prime \prime}\end{align*}

11. \begin{align*}{\frac{1}{4}}^{{\prime}{\prime}}\end{align*}

12.

13.

14. \begin{align*}2{\frac{1}{2}}^{{\prime}{\prime}}\end{align*}

15.

16.