<meta http-equiv="refresh" content="1; url=/nojavascript/"> Areas of Similar Polygons | CK-12 Foundation
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Second Edition Go to the latest version.

10.3: Areas of Similar Polygons

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

  • Understand the relationship between the scale factor of similar polygons and their areas.
  • Apply scale factors to solve problems about areas of similar polygons.
  • Use scale models or scale drawings.


We’ll begin with a quick review of some important features of similar polygons. You remember that we studied similar figures rather extensively in Chapter 7. There you learned about scale factors and perimeters of similar polygons. In this section we’ll take similar figures one step farther. We’ll see that the areas of similar figures have a very specific relationship to the scale factor—but it’s just a bit tricky! We wrap up the section with some thoughts on why living things are the “right” size, and what geometry has to do with that!

Review - Scale Factors and Perimeter

Example 1

The diagram below shows two rhombi.

a. Are the rhombi similar? How do you know?


  • The sides are parallel, so the corresponding angles are congruent.
  • Using the Pythagorean Theorem, we can see that each side of the smaller rhombus has a length of 10, and each side of the larger rhombus has a length of 15.
  • So the lengths of the sides are proportional.
  • Polygons with congruent corresponding angles and proportional sides are similar.

b. What is the scale factor relating the rhombi?

The scale factor relating the smaller rhombus to the larger one is \frac{15}{10} = \frac{3}{2}=1.5.

c. What is the perimeter of each rhombus?


  • \mathrm{Perimeter\ of\ smaller\ rhombus} = 4 \times 10 = 40
  • \mathrm{Perimeter\ of\ larger\ rhombus} = 4 \times 15 = 60

d. What is the ratio of the perimeters?

\frac{60}{40} = \frac {3}{2}=1.5

e. What is the area of each rhombus?

\text{Area of smaller rhombus} & = \frac{d_1d_2}{2}=\frac {{12}\times{16}}{2}= 96 \\\text{Area of larger rhombus} & = \frac{d_1d_2}{2}=\frac {{18}\times{24}}{2}= 216

What do you notice in this example? The perimeters have the same ratio as the scale factor.

But what about the areas? The ratio of the areas is certainly not the same as the scale factor. If it were, the area of the larger rhombus would be 96 \times 1.5 = 144, but the area of the larger rhombus is actually 216.

What IS the ratio of the areas?

The ratio of the areas is \frac{216}{96} = \frac {9}{6}=2.25. Notice that \frac{9}{4}=\left ( \frac{3}{2} \right)^2 or in decimal, 2.25 = (1.5)^2.

So at least in this case we see that the ratio of the areas is the square of the scale factor.

Scale Factors and Areas

What happened in Example 1 is no accident. In fact, this is the basic relationship for the areas of similar polygons.

Areas of Similar Polygons

If the scale factor relating the sides of two similar polygons is k, then the area of the larger polygon is k^2 times the area of the smaller polygon. In symbols let the area of the smaller polygon be A_1 and the area of the larger polygon be A_2. Then:

A_2 = k^2 A_1

Think about the area of a polygon. Imagine that you look at a square with an area of exactly 1 \;\mathrm{square\ unit}. Of course, the sides of the square are 1 \;\mathrm{unit} of length long. Now think about another polygon that is similar to the first one with a scale factor of k. Every 1-by-1 square in the first polygon has a matching k-by-k square in the second polygon, and the area of each of these k-by-k squares is k^2. Extending this reasoning, every 1 \;\mathrm{square\ unit} of area in the first polygon has a corresponding k^2 units of area in the second polygon. So the total area of the second polygon is k^2 times the area of the first polygon.

Warning: In solving problems it’s easy to forget that you do not always use just the scale factor. Use the scale factor in problems about lengths. But use the square of the scale factor in problems about area!

Example 2

Wu and Tomi are painting murals on rectangular walls. The length and width of Tomi’s wall are 3 times the length and width of Wu’s wall.

a. The total length of the border of Tomi’s wall is 120 \;\mathrm{feet}. What is the total length of the border of Wu’s wall?

This is a question about lengths, so you use the scale factor itself. All the sides of Tomi’s wall are 3 times the length of the corresponding side of Wu’s wall, so the perimeter of Tomi’s wall is also 3 times the perimeter of Wu’s wall.

The total length of the border (perimeter) of Wu’s wall is \frac{120}{3} = 40 \;\mathrm{feet}.

b. Wu can cover his wall with 6 quarts of paint. How many quarts of paint will Tomi need to cover her wall?

This question is about area, since the area determines the amount of paint needed to cover the walls. The ratio of the amounts of paint is the same as the ratio of the areas (which is the square of the scale factor). Let x be the amount of paint that Tomi needs.

\frac{x}{6} & = k^2=3^2=9 \\x & = 6 \times 9 = 54

Tomi would need 54 quarts of paint.

Summary of Length and Area Relationships for Similar Polygons

If two similar polygons are related by a scale factor of k, then:

  • Length: The lengths of any corresponding parts have the same ratio, k. Note that this applies to sides, *Area: The ratio of the areas is k^2. Note that this applies to areas, and any aspect of an object that

Note: You might be able to make a pretty good guess about the volumes of similar solid (3-D) figures. You’ll see more about that in Chapter 11.

Scale Drawings and Scale Models

One important application of similar figures is the use of scale drawings and scale models. These are two-dimensional (scale drawings) or three-dimensional (scale models) representations of real objects. The drawing or model is similar to the actual object.

Scale drawings and models are widely used in design, construction, manufacturing, and many other fields. Sometimes a scale is shown, such as “1\;\mathrm{inch}= 5\;\mathrm{miles}” on a map. Other times the scale may be calculated, if necessary, from information about the object being modeled.

Example 3

Jake has a map for a bike tour. The scale is 1\;\mathrm{inch} = 5\;\mathrm{miles}. He estimated that two scenic places on the tour were about 3\frac{1}{2}\;\mathrm{inches} apart on the map. How far apart are these places in reality?

Each inch on the map represents a distance of 5\;\mathrm{miles}. The places are about 3\frac{1}{2}\times 5 =17.5\;\mathrm{miles} apart.

Example 4

Cristy’s design team built a model of a spacecraft to be built. Their model has a scale of 1:24. The actual spacecraft will be 180\;\mathrm{feet} long. How long should the model be?

Let x be the length of the model.

\frac{1}{24}& =\frac {x}{180} \\24x & = 180 \\x & = 7.5

The model should be 7.5 \;\mathrm{feet} long.

Example 5

Tasha is making models of several buildings for her senior project. The models are all made with the same scale. She has started the chart below.

a. What is the scale of the models?

1250 \div 20 = 62.5

The scale is 1\;\mathrm{inch} = 62.5\;\mathrm{feet}.

b. Complete the chart below.

Building Actual height (feet) Model height (inches)

Sears Tower


? 23.2

Empire State Building

(New York City)

1250 20

Columbia Center


930 ?

Sears Tower: 23.2 \times 62.5 = 1450 . It is 1450\;\mathrm{feet} high.

Columbia Center: \mathrm{Let}\ x = \mathrm{the\ model\ height}.

\frac{1250}{20}& =\frac {930}{x} \\1250x & = 20 \times 930 \\x & =\frac{{20}\times{930}}{1250}\thickapprox 14.9

The model should be about 14.9\;\mathrm{inches} high.

Why There Are No 12-Foot-Tall Giants

Why are there no 12-\mathrm{foot}-tall giants? One explanation for this is a matter of similar figures.

Let’s suppose that there is a 12-\;\mathrm{foot}-tall human. Compare this giant (?) to a 6-\;\mathrm{foot}-tall person. Now let’s apply some facts about similar figures.

The scale factor relating these two hypothetical people is \frac{12}{6}=2 . Here are some consequences of this scale factor.

  • All linear dimensions of the giant would be 2 times the corresponding dimensions of the real person. This includes height, bone length, etc.
  • All area measures of the giant would be 2^2 = 4 times the corresponding area measures of the real person. This includes respiration (breathing) and metabolism (converting nutrients to usable materials and energy) rates, because these processes take place along surfaces in the lungs, intestines, etc. This also includes the strength of bones, which depends on the cross-section area of the bone.
  • All volume measures of the giant would be 2^3 = 8 times the corresponding volume measures of the real person. (You’ll learn why in Chapter 11.) The volume of an organism generally determines its weight and mass.

What kinds of problems do we see for our giant? Here are two severe ones.

  1. The giant would have bones that are 4 times as strong, but those bones have to carry a body weight that is 8 times as much. The bones would not be up to the task. In fact it appears that the giant’s own weight would be able to break its bones.
  2. The giant would have 8 times the weight, number of cells, etc. of the real person, but only 4 times as much ability to supply the oxygen, nutrition, and energy needed.

Conclusion: There are no 12-\;\mathrm{foot}-giants, and some of the reasons are nothing more, or less, than the geometry of similar figures.

For further reading: On Being the Right Size, by J. B. S. Haldane, also available at http://irl.cs.ucla.edu/papers/right-size.html

Lesson Summary

In his lesson we focused on one main point: The areas of similar polygons have a ratio that is the square of the scale factor. We also used ideas about similar figures to analyze scale drawings and scale models, which are actually similar representations of actual objects.

Points to Consider

You have now learned quite a bit about the lengths of sides and areas of polygons. Next we’ll build on knowledge about polygons to come to a conclusion about the “perimeter” of the “ultimate polygon,” which is the circle.

Suppose we constructed regular polygons that are all inscribed in the same circle.

  • Think about polygons that have more and more sides.
  • How would the perimeter of the polygons change as the number of sides increases?

The answers to these questions will lead us to an understanding of the formula for the circumference (perimeter) of a circle.

Review Questions

The figure below is made from small congruent equilateral triangles.

4 congruent small triangles fit together to make a bigger, similar triangle.

  1. What is the scale factor of the large and small triangles?
  2. If the area of the large triangle is 20\;\mathrm{square\ units}, what is the area of a small triangle? The smallest squares in the diagram below are congruent.
  3. What is the scale factor of the shaded square and the largest square?
  4. If the area of the shaded square is 50\;\mathrm{square\ units}, what is the area of he largest square?
  5. Frank drew two equilateral triangles. Each side of one triangle is 2.5 times as long as a side of the other triangle. The perimeter of the smaller triangle is 40\;\mathrm{cm}. What is the perimeter of the larger triangle? In the diagram below, \overline{M N}: \overline{P Q}.
  6. What is the scale factor of the small triangle and the large triangle?
  7. If the perimeter of the large triangle is 42, what is the perimeter of the small triangle?
  8. If the area of the small triangle is A, write an expression for the area of the large triangle.
  9. If the area of the small triangle is K, write an expression for the area of the trapezoid.
  10. The area of one square on a game board is exactly twice the area of another square. Each side of the larger square is 50\;\mathrm{mm} long. How long is each side of the smaller square?
  11. The distance from Charleston to Morgantown is 160\;\mathrm{miles}. The distance from Fairmont to Elkins is 75\;\mathrm{miles}. Charleston and Morgantown are 5\;\mathrm{inches} apart on a map. How far apart are Fairmont and Elkins on the same map?

Marlee is making models of historic locomotives (train engines). She uses the same scale for all of her models.

  • The S1 locomotive was 140 \;\mathrm{feet} long. The model is 8.75 \;\mathrm{inches} long.
  • The 520 Class locomotive was 87\;\mathrm{feet} long.
  1. What is the scale of Marlee’s models?
  2. How long is the model of the 520 Class locomotive?

Review Answers

  1. 2
  2. 5
  3. \frac{4}{9} or 4: 9
  4. 112.5
  5. 100\;\mathrm{cm}
  6. \frac{2}{3}
  7. 28
  8. \frac{9}{4}A or \frac{9A}{4}
  9. \frac{5}{4}K or \frac{5K}{4}
  10. 35.4\;\mathrm{mm}
  11. 2.3\;\mathrm{inches}
  12. 1\;\mathrm{inch}= 16\;\mathrm{feet} or equivalent
  13. 5.4\;\mathrm{inches}

Image Attributions

Files can only be attached to the latest version of None


Please wait...
Please wait...
Image Detail
Sizes: Medium | Original

Original text