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6.8: Similarity and Dilations

Created by: CK-12

Introduction

The Log Cabin

Sherri decided to do her project on a log cabin. Mrs. Patterson suggested that she choose a time period to work with because log cabins have been being built for a long time. Years ago, they were quite small, but today, log cabins can be designer homes too.

“Mrs. Patterson, I am going to focus on a log cabin from the 1800’s,” Sherri said taking out a book that she found at the library on log cabins.

“That is a great idea. What was the average size of a log cabin in 1800?” Mrs. Patterson asked.

“What do you mean?”

“I mean the square footage. How many feet wide was the average house and how many feet long?” Mrs. Patterson explained.

“Oh, I get it. The average house was 20 \times 40 \ feet. So the average was 800 square feet,” Sherri said.

“Terrific, now make sure that your plan shows that,” Mrs. Patterson said walking away.

Sherri is puzzled. She knows that the shape of the log cabin is a rectangle given the length and width. She knows the area of the house. To create a plan, she will need to create a dilation. Sherri decides that she will use a scale factor of \frac{1}{16}. Given this information, what will the dimensions be of her house plan?

Use this lesson to learn about dilations and figure out the dimensions of the house by the end of the lesson.

What You Will Learn

In this lesson, you will learn how to demonstrate the following skills.

  • Recognize similar polygons as having congruent corresponding angles and proportional corresponding side lengths.
  • Use similar triangles to find unknown lengths or distances using indirect measurement.
  • Recognize dilation transformations (stretch or shrink) and scale factors.
  • Use coordinate notation to describe dilations of given figures and their resulting images.

Teaching Time

I. Recognize Similar Polygons as Having Congruent Corresponding Angles and Proportional Corresponding Side Lengths

We have spent a lot of time identifying and working with congruent figures. Remember that congruent means exactly the same, having the same size and shape. Sometimes, a figure will have the same shape, but not the same size. It will be either smaller or larger than the original figure. When this happens, we say that the two figures are similar. Similar figures have the same shape, but not the same size.

In this lesson you will learn all about how to work with similarity in figures. Think about it for a minute, if a figure has the same shape, but not the same size, then there is still a relationship between the two figures. The relationship is created based on the shape being the same.

That is a good question.

Let’s start by thinking about angles. With similar figures, each angle of one figure in a similar pair corresponds and is congruent to an angle in the other. For instance, the top point of one triangle corresponds to the top point of the other triangle in a similar pair. We call these corresponding parts.

Notice that the angles match in these two triangles. The shape of the triangles is the same, and you can see that the angles do match.

What about the side lengths?

The sides in similar pairs also correspond to each other (such as the base of each triangle), but they are not congruent; they are proportional. We can determine whether figures are similar to each other by comparing their corresponding parts. Corresponding parts are especially helpful when one figure is rotated so that it is not clear which angles and sides correspond to which in the other figure.

Now let’s look at the corresponding side lengths. In the first rectangle, the short side is 4 and the long side is 8. We know that opposite sides of a rectangle are congruent, so we don’t need to worry about writing measurements on the other two sides. We can compare the measurements in the first rectangle with the ones in the second rectangle. In the second rectangle, the short side is 2 and the long side is 4.

Let’s write a proportion to compare the corresponding sides.

\frac{short \ side}{short \ side} &= \frac{long \ side}{long \  side}\\\frac{4}{2} &= \frac{8}{4}

You can see that these two ratios form a proportion. You can use this information to prove whether or not two figures are similar as well. Remember, the angle measures must be the same, and the side lengths must be proportional.

Write these notes on similar figures down in your notebooks.

II. Use Similar Triangles to Find Unknown Lengths or Distances Using Indirect Measurement

Now that you understand how to identify whether or not two figures are similar, we can look similar triangles. Similar triangles are very useful because we can use them to figure out measurements. Many years ago, this is how people used to figure out the measurements for things that were too high or big to measure. They used indirect measurement. Indirect measurement uses similar triangles and proportions to figure out lengths or distances.

But first, let’s think about similar triangles.

Similar triangles have the same properties as other similar figures. The angle measures are the same and the corresponding side lengths are proportional. Let’s look at this diagram to understand this.

Now we can compare the angles and corresponding side lengths. Let’s begin with the angles.

\angle A & \cong \angle D\\\angle B & \cong \angle E\\\angle C & \cong \angle F

Next, we can look at the corresponding side lengths. In the diagram, we haven’t been given any measurements, but we can use the lowercase letters to show which sides correspond.

\frac{a}{d}=\frac{b}{e}=\frac{c}{f}

This shows that the side lengths form a ratio and that each of these is proportional to the other.

We can use this information when problem solving for missing side lengths.

That is a good question. First, we would have to know some of the side lengths. Let’s assign some lengths to the sides in the diagram above.

a &= 12\\b &= ?\\c &= 3\\d &= 4\\e &= 3\\f &= 1

Now we can take these given measures and substitute them into the proportion that we wrote earlier. Notice that we don’t have the measure of side b, so we will need to solve for that missing measurement.

\frac{12}{4}=\frac{b}{3}=\frac{3}{1}

Next, we can use two of the three ratios to solve the proportions. We have three ratios, but we don’t need all three because two equal ratios form a proportion. This means that we only need to work with two ratios to solve for the value of b.

\frac{12}{4}=\frac{b}{3}

Now we can cross - multiply and solve the proportion.

4b &= 36\\b &= 9

The value of b is 9.

We can also use this kind of indirect measurement to figure out lengths and distances. To do this, we will rely on what we have already learned about similar figures and proportions.

Example

A person is five feet tall and casts a shadow of 2 feet. A tower casts a shadow that is 10 feet long. What is the height of the tower?

This may seem like a very challenging problem to solve, however if you think about people and shadows as they are related to triangles it becomes much easier. Look at this picture.

Notice that the person and the shadow form two sides of a triangle and we can draw an imaginary line from the head of the person to the tip of the shadow. Shadows are a way of working with triangles and indirect measurement. In fact, you will often hear these types of problems referred to as shadow problems.

To solve this one, let’s figure out how to use similar triangles to figure out the height of the tower. First, think about what is being compared. We are comparing the height of the person with the length of the shadow. That is the first ratio.

\frac{person}{shadow}=\frac{5 \ ft}{2 \ ft}

Next, we look at the tower. We don’t know the height of the tower that is our variable. We do know the length of the shadow. Here is our second ratio.

\frac{tower}{shadow}=\frac{x}{10 \ ft}

We can say that these two triangles are similar and that similar triangles are proportional. Therefore, these two ratios form a proportion. Let’s write them as a proportion.

\frac{5 \ ft}{2 \ ft}=\frac{x}{10 \ ft}

Now we can cross - multiply and solve the proportion.

2x &= 50 \ ft\\x &= 25 \ ft

The tower is 25 feet tall.

The key to working with indirect measurement is to always be clear about what is being compared. You write your ratios and then form a proportion and solve for the missing length or distance.

Write some notes about indirect measurement down in your notebook.

III. Recognize Dilation Transformations (stretch or shrink) and Scale Factors

We have talked about transformations of many different kinds. We can also stretch or shrink a figure to create a new one. This is called a dilation. A dilation is a transformation created by a scale factor. We can create a dilation that is smaller or larger than the original figure. Either way, a similar figure is created through a dilation.

Let’s think about scale factors for a minute. The scale factor is the ratio that determines the proportional relationship between the sides of similar figures. For the pairs of sides to be proportional to each other, they must have the same scale factor. In other words, similar figures have congruent angles and sides with the same scale factor. A scale factor of two means that each side of the larger figure is exactly twice as long as the corresponding side in the smaller figure.

When we compare the corresponding sides of a figure, we can figure out the scale factor of that figure.

Example

A figure has a side length of 3 feet. What would be the corresponding side length of the next figure is the scale factor was 4?

Let’s think about this. We know the length of one of the sides of the first figure and we know the scale factor. To figure out the new length, we can multiply the scale factor times the first length.

3 \times 4 = 12

The length of the corresponding side of the second figure is 12 feet.

When we have a figure that is larger than the original, we have a scale factor that is greater than one. If we have a figure that is smaller than the original, then we have a scale factor that is less than one or a fraction.

Example

A figure has a side length of 5 meters. What would be the corresponding side length of the new figure if the scale factor is \frac{1}{2}?

To figure this out, we have to take the given length of the first figure and divide it in half. This will give us the corresponding length of the second figure.

5 \left(\frac{1}{2}\right) = 2.5

The length of the corresponding side will be 2.5 meters.

IV. Use Coordinate Notation to Describe Dilations of Given Figures and Their Resulting Images

Now that you understand dilations, we can look at how to work with them on the coordinate plane. Once again, we will be using coordinate notation to describe the different dilations that are created on the coordinate plane.

Let’s look at this figure and then see how we can graph the dilation of it.

Graphing dilations of geometric figures is actually fairly easy to do when we know the scale factor. We simply multiply both coordinates for each vertex by the scale factor to produce new coordinates.

Suppose we want to make an enlargement of the rectangle above using a scale factor of 3. We need to multiply each coordinate by 3.

& (-2,-3) \ (-2,3) \ (2,3) \ (2,-3)\\& \times 3\\& (-6,-9) \ (-6,9) \ (6,9) \ (6,-9)

Now we can graph it on the coordinate plane.

We can create a reduction too. We create a reduction by dividing each coordinate by the scale factor. This will give us the new measurements of the figure.

Example

Graph a reduction of the following figure if the scale factor is 2.

Notice that each of the original coordinates were divided by two to create the coordinates of the reduction.

(2, 4) \div 2 &= (1, 2)\\(8, -4) \div 2 &= (4, -2)\\(-6, -2) \div 2 &= (-3, -1)

Now let’s look back at the problem from the introduction and see how similar figures can help us to solve this problem.  

Real-Life Example Completed

The Log Cabin

Here is the original problem once again. Reread it and then solve the problem for the plan dimensions.

Sherri decided to do her project on a log cabin. Mrs. Patterson suggested that she choose a time period to work with because log cabins have been being built for a long time. Years ago, they were quite small, but today, log cabins can be designer homes too.

“Mrs. Patterson, I am going to focus on a log cabin from the 1800’s,” Sherri said taking out a book that she found at the library on log cabins.

“That is a great idea. What was the average size of a log cabin in 1800?” Mrs. Patterson asked.

“What do you mean?”

“I mean the square footage. How many feet wide was the average house and how many feet long?” Mrs. Patterson explained.

“Oh, I get it. The average house was 20 \times 40 \ feet. So the average was 800 square feet,” Sherri said.

“Terrific, now make sure that your plan shows that,” Mrs. Patterson said walking away.

Sherri is puzzled. She knows that the shape of the log cabin is a rectangle given the length and width. She knows the area of the house. To create a plan, she will need to create a dilation. Sherri decides that she will use a scale factor of \frac{1}{16}. Given this information, what will the dimensions be of her house plan?

Now solve for the plan dimensions.

Solution to Real – Life Example

To solve this problem, we begin with the actual dimensions of the log cabin. The log cabin has real – world dimensions of 20 \times 40 \ feet.

Sherri is using a scale factor of \frac{1}{16}. That means that the dilation will be a reduction. We divide both dimensions by 16.

20 \div 16 &= 1.25 \ ft.\\40 \div 16 &= 2.5 \ ft

The dimensions of Sherri’s plan will be 1.25 \ ft \ wide \times 2.5 \ ft \ long.

Vocabulary

Here are the vocabulary words that are found in this lesson.

Congruent
having the same size, shape and measure.
Similar
having the same shape, but not the same size. Angle measures are the same and side lengths are proportional.
Proportional
the side lengths create ratios that form a proportion.
Indirect Measurement
using similar triangles to figure out challenging distances or lengths.
Dilation
to reduce or enlarge a figure according to a scale factor.
Scale Factor
the ratio that compares the lengths of corresponding sides to each other. That comparison is the scale factor.

Time to Practice

Directions: Identify whether or not each pair of triangles is similar based on the ratios of their sides.

  1. Triangle A has side lengths of 2, 4, and 6. Triangle B has side lengths of 6, 12 and 24. Are these triangles similar?
  2. Triangle C has side lengths of 4, 5, and 10. Triangle B has side lengths of 2, 2.5 and 5. Are these two triangles similar?
  3. Triangle D has side lengths of 5, 8, and 12. Triangle B has side lengths of 10, 16 and 24. Are these two triangles similar?
  4. Triangle A has side lengths of 10, 12, and 14. Triangle B has side lengths of 5, 7 and 9. Are these two triangles similar?
  5. Triangle B has side lengths of 8, 14, and 20. Triangle C has side lengths of 4, 7 and 10. Are these two triangles similar?
  6. Triangle E has side lengths of 20, 11 and 8. Triangle F has side lengths of 10, 5.5 and 5. Are these two triangles similar?
  7. Triangle G has side lengths of 6, 8 and 12. Triangle H has side lengths of 18, 24 and 36. Are these two triangles similar?
  8. Triangle I has side lengths of 8, 12, and 16. Triangle J has side lengths of 4, 8 and 10. Are these two triangles similar?

Directions: Find the missing length by looking at each series of ratios. The top value represents the side lengths of the first similar triangle. The bottom value represents the side lengths of the second similar triangle.

  1. \frac{1}{2}=\frac{3}{6}=\frac{9}{x}
  2. \frac{3}{6}=\frac{6}{12}=\frac{10}{x}
  3. \frac{4}{2}=\frac{10}{x}=\frac{12}{6}
  4. \frac{6}{2}=\frac{9}{x}=\frac{12}{4}
  5. \frac{5}{10}=\frac{10}{20}=\frac{15}{x}
  6. \frac{12}{6}=\frac{20}{10}=\frac{15}{x}
  7. \frac{16}{x}=\frac{20}{5}=\frac{24}{6}

Directions: Use each scale factor to determine the new dimensions of each figure.

  1. A triangle with side measures of 4, 5, 9 and a scale factor of 2.
  2. A triangle with side measures of 8, 10, 14 and a scale factor of 3.
  3. A triangle with side measures of 2, 4, 6 and a scale factor of 2.
  4. A quadrilateral with side measures of 4, 6, 8, 10 and a scale factor of \frac{1}{2}
  5. A quadrilateral with side measures of 9, 12, 18, 24 and a scale factor of \frac{1}{3}

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CK.MAT.ENG.SE.1.Middle-School-Math-Grade-8.6.8

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