Remember Jeremy and his grandpa from the Scale Distances or Dimensions Concept? Well, they used the actual bookcase to create a design using a unit scale. Now Jeremy is going to figure out the actual size of a different bookcase using a drawing. Take a look.

Jeremy has a drawing of the bookcase which shows that the bookcase is \begin{align*}5.5\end{align*}

Given this scale, what is the actual height the bookcase?

**This Concept will show you how to use unit scale to figure out actual dimensions. We will look at this problem again at the end of the Concept.**

### Guidance

**Sometimes, you will have a scale drawing, map or model to work with first. You won’t be given the actual dimensions. Instead, you will have to use the unit scale that accompanies the scale model, drawing or map to figure out the actual dimensions.**

This often happens with maps. You look at a map and try to figure out how far it is from one city to another. The scale in the bottom of the map can help you with this. If the scale says that 1” = 100 miles and the map indicates that there is 4 inches between one city and the next, then you can say that the actual distance between the two cities is 400 miles.

Hector made this scale model of the Statue of Liberty. The scale height of his model, from the base to the torch, is 4.65 centimeters. Find the actual height of the Statue of Liberty.

Write the unit scale as a ratio.

\begin{align*}\frac{centimeters}{meters} = \frac{1}{10}\end{align*}

The scale height of the model is 4.65 centimeters. Use \begin{align*}h\end{align*}

\begin{align*}\frac{centimeters}{meters} = \frac{4.65}{h}\end{align*}

Set up and solve proportions to find the actual height, \begin{align*}h\end{align*}

\begin{align*}\frac{1}{10} &= \frac{4.65}{h}\\
10 \cdot 4.65 &= 1 \cdot h\\
46.5 &= h\end{align*}

**The actual height of the Statue of Liberty, from the base to the torch, is 46.5 meters.**

**So far, we have considered scale drawings and models that are used to represent very large objects. Scale drawings or models can also help us represent very small objects.**

A beetle that Amelia observed was too small to draw at its actual size. So, she made this scale drawing.

In her drawing, the length of the beetle was 4.8 cm. What was the length of the actual beetle?

Write the unit scale, 0.4 cm = 1 mm, as a ratio.

\begin{align*}\frac{centimeters}{millimeters} = \frac{0.4}{1}\end{align*}

In the scale drawing, the beetle has a length of 4.8 centimeters. Use \begin{align*}m\end{align*}

\begin{align*}\frac{centimeters}{millimeters} = \frac{4.8}{m}\end{align*}

Set up and solve a proportion to find the actual length, \begin{align*}m\end{align*}

\begin{align*}\frac{0.4}{1} &= \frac{4.8}{m}\\
1 \cdot 4.8 &= 0.4 \cdot m\\
4.8 &= 0.4m\\
\frac{4.8}{0.4} &= \frac{0.4m}{0.4}\\
12 &= m\end{align*}

**The actual length of the beetle Amelia observed was 12 millimeters.**

Now it's your turn. Use unit scale 1" = 200 miles to find the actual dimensions of distance on the map.

#### Example A

500 miles

**Solution: 2.5 inches**

#### Example B

600 miles

**Solution: 3 inches**

#### Example C

3000 miles

**Solution: 15 inches**

Here is the original problem once again.

Jeremy has a drawing of the bookcase which shows that the bookcase is \begin{align*}5.5\end{align*}

Given this scale, what is the actual height the bookcase?

To solve this problem, we can write a proportion and solve for the missing value.

\begin{align*}\frac{2}{1} = \frac{x}{5.5}\end{align*}

Now cross multiply and solve.

\begin{align*}x = 11\end{align*}

**The actual height of the bookcase is 11 feet.**

### Vocabulary

- Ratio
- a comparison between two quantities. Ratios can be written in fraction form, with a colon or by using the word “to”.

- Proportion
- when two ratios are equal or equivalent, they form a proportion.

- Scale Drawing
- a smaller drawing that is used to represent a larger, life-size building or model.

- Unit Scale
- a measurement meant to represent the actual measurements of a larger, life-size building, map or other item. For example 1” = 2 feet would be a unit scale.

### Guided Practice

Here is one for you to try on your own.

The map below shows the distances between three towns.

a. On the map, the distance between Smithville and Frankton is \begin{align*}2 \frac{1}{4}\end{align*} inches. Find the actual straight-line distance between Smithville and Frankton.

b. On the map, the distance between Frankton and Blair is \begin{align*}1 \frac{1}{2}\end{align*} inches. Find the actual straight-line distance between Frankton and Blair.

c. How many miles closer is Frankton to Blair than to Smithville?

**Answer**

First, let’s look at the information that we have been given. Then we can use this information to solve each part of the problem. Notice that there are three parts, \begin{align*}a, b\end{align*} and \begin{align*}c\end{align*}.

The unit scale is \begin{align*}\frac{1}{4} \ in. = 2 \ mi\end{align*}. Express that ratio as a fraction. Since \begin{align*}\frac{1}{4} = 0.25\end{align*}, use 0.25 mile as the first term of the ratio.

\begin{align*}\frac{inches}{miles} = \frac{0.25}{2}\end{align*}

**Now, consider part \begin{align*}a\end{align*}.**

You know that the scale distance between Smithville and Frankton is \begin{align*}2 \frac{1}{4}\end{align*} inches. The actual distance between those two towns is unknown, so use \begin{align*}x\end{align*} to represent that distance. Write a ratio to represent this. Since \begin{align*}2 \frac{1}{4} = 2.25\end{align*}, use 2.25 as the first term.

\begin{align*}\frac{inches}{miles} = \frac{2.25}{x}\end{align*}

Set up a proportion using the unit scale and the ratio above and solve for \begin{align*}x\end{align*}.

\begin{align*}\frac{0.25}{2} &= \frac{2.25}{x}\\ 2 \cdot 2.25 &= 0.25 \cdot x\\ 4.5 &= 0.25x\\ \frac{4.5}{0.25} &= \frac{0.25x}{0.25}\\ 18 &= x\end{align*}

**The actual distance between Smithville and Frankton is 18 miles.**

**Next, consider part \begin{align*}b\end{align*}.**

You know that the scale distance between Frankton and Blair is \begin{align*}1 \frac{1}{2}\end{align*} inches. The actual distance between those two towns is unknown, so use \begin{align*}x\end{align*} to represent that distance. Write a ratio to represent this. Since \begin{align*}1 \frac{1}{2} = 1.5\end{align*}, use 1.5 as the first term.

\begin{align*}\frac{inches}{miles} = \frac{1.5}{x}\end{align*}

Set up a proportion using the unit scale and the ratio above and solve for \begin{align*}x\end{align*}.

\begin{align*}\frac{0.25}{2} &= \frac{1.5}{x}\\ 2 \cdot 1.5 &= 0.25 \cdot x\\ 3 &= 0.25x\\ \frac{3}{0.25} &= \frac{0.25x}{0.25}\\ 12 &= x\end{align*}

**The actual distance between Frankton and Blair is 12 miles.**

**Finally, consider part \begin{align*}c\end{align*}.**

We know that Smithville is 18 miles from Frankton, and that Blair is 12 miles from Frankton. So, we can subtract to find out how much closer Frankton is to Blair than to Smithville.

\begin{align*}18 - 12 = 6\end{align*}

**Frankton is 6 miles closer to Blair than to Smithville.**

### Video Review

- This is a James Sousa video on unit scale.

### Practice

Directions: Use the given scale to determine the actual distance.

Given: Scale 1” = 100 miles

1. How many miles is 2” on the map?

2. How many miles is \begin{align*}2\frac{1}{2} inch\end{align*} on the map?

3. How many miles is \begin{align*}\frac{1}{4} inch\end{align*} on the map?

4. How many miles is 10 inches on the map?

5. How many miles is 11 inches on the map?

6. How many miles is 12.5 inches on the map?

7. How many miles is \begin{align*}\frac{1}{2}\end{align*} inch on the map?

8. How many miles is \begin{align*}5 \frac{1}{4}\end{align*} inch on the map?

Given: 1 cm = 20 mi

9. How many miles is 2 cm on the map?

10. How many miles is 4 cm on the map?

11. How many miles is 8.5 cm on the map?

12. How many miles is 19 cm on the map?

13. How many miles is 13 cm on the map?

14. How many miles is \begin{align*}\frac{1}{2}\end{align*} cm on the map?

15. How many miles is \begin{align*}1 \frac{1}{2}\end{align*} cm on the map?