Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’3 Lesson 9: Basic Properties of Similarity Student Outcomes ο§ Students know that similarity is both a symmetric and a transitive relation. Classwork Concept Development (4 minutes) ο§ If we say that one figure, π, is similar to another figure, πβ², that is, π~πβ², can we also say that πβ²~π? That is, is similarity symmetric? Keep in mind that there is a very specific sequence of a dilation followed by a congruence that would map π to πβ². οΊ ο§ Expect students to say yes; they would expect similarity to be symmetric. If we say that figure π is similar to another figure, π (i.e., π~π), and figure π is similar to yet another figure, π (i.e., π~π), is it true that π~π? That is, is similarity transitive? οΊ Expect students to say yes; they would expect similarity to be transitive. The Exploratory Challenges that follow are for students to get an intuitive sense that, in fact, these two statements are true. Exploratory Challenge 1 (10 minutes) Students work in pairs to complete Exploratory Challenge 1. Exploratory Challenge 1 The goal is to show that if β³ π¨π©πͺ is similar to β³ π¨β² π©β² πͺβ² , then β³ π¨β² π©β² πͺβ² is similar to β³ π¨π©πͺ. Symbolically, if β³ π¨π©πͺ βΌβ³ π¨β²π©β²πͺβ², then β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ. Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 128 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM a. 8β’3 First, determine whether or not β³ π¨π©πͺ is in fact similar to β³ π¨β² π©β² πͺβ² . (If it isnβt, then no further work needs to be done.) Use a protractor to verify that the corresponding angles are congruent and that the ratios of the corresponding sides are equal to some scale factor. The corresponding angles are congruent: β π¨ β β π¨β² , β π© β β π©β² , and β πͺ β β πͺβ², therefore |β π¨| = |β π¨β² | = ππ°, |β π©| = |β π©β² | = ππ°, and |β πͺ| = |β πͺβ² | = ππ°. The ratios of the corresponding sides are equal: b. π π = π π = π π = π. Describe the sequence of dilation followed by a congruence that proves β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² . π π To map β³ π¨π©πͺ onto β³ π¨β² π©β² πͺβ² , dilate β³ π¨π©πͺ from center πΆ by scale factor π = , noted in the diagram above by the red triangle. Then, translate the red triangle up two units and five units to the right. Next, rotate the red triangle πππ degrees around point π¨β² until π¨πͺ coincides with π¨β²πͺβ². Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 129 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM c. 8β’3 Describe the sequence of dilation followed by a congruence that proves β³ π¨β²π©β²πͺβ² βΌβ³ π¨π©πͺ. Note that in the diagram below, both axes have been compressed. To map β³ π¨β²π©β²πͺβ² onto β³ π¨π©πͺ, dilate β³ π¨β²π©β²πͺβ² from center πΆ by scale factor π = π, noted by the blue triangle in the diagram. Then, translate the blue triangle ten units to the left and two units down. Next, rotate the blue triangle πππ degrees around point π¨ until side π¨β²πͺβ² coincides with side π¨πͺ. d. Is it true that β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ? Why do you think this is so? Yes, it is true that β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ. I think it is true because when we say figures are similar, it means that they are the same figure, just a different size because one has been dilated by a scale factor. For that reason, if one figure, like β³ π¨π©πͺ, is similar to another, like β³ π¨β² π©β² πͺβ² , it must mean that β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ. However, the sequence you would use to map one of the figures onto the other is different. Concept Development (3 minutes) Ask students to share what they wrote for part (d) of Exploratory Challenge 1. Expect students to respond in a similar manner to the response for part (d). If they do not, ask questions about what similarity means, what a dilation does, and how figures are mapped onto one another. ο§ For any two figures π and πβ², if π~π β² , then πβ²~π. This is what the statement that similarity is a symmetric relation means. Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 130 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 9 8β’3 Exploratory Challenge 2 (15 minutes) Students work in pairs to complete Exploratory Challenge 2. Exploratory Challenge 2 The goal is to show that if β³ π¨π©πͺ is similar to β³ π¨β²π©β²πͺβ² and β³ π¨β²π©β²πͺβ² is similar to β³ π¨β²β²π©β²β²πͺβ²β², then β³ π¨π©πͺ is similar to β³ π¨β²β²π©β²β²πͺβ²β². Symbolically, if β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β²π©β²πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β², then β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β². Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 131 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM a. Lesson 9 8β’3 Describe the similarity that proves β³ π¨π©πͺ βΌβ³ π¨β²π©β²πͺβ². To map β³ π¨π©πͺ onto β³ π¨β²π©β²πͺβ², we need to first determine the scale factor that makes β³ π¨π©πͺ the same size as π π.π π β³ π¨β² π©β² πͺβ² . Then, = = = π. Dilate β³ π¨π©πͺ from center πΆ by scale factor π = π, shown in red in the π π.π π diagram. Then, translate the red triangle π units up. b. Describe the similarity that proves β³ π¨β²π©β²πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β². To map β³ π¨β²π©β²πͺβ² onto β³ π¨β²β²π©β²β²πͺβ²β², we need to first determine the scale factor that makes β³ π¨β²π©β²πͺβ² the same π.π π π π size as β³ π¨β²β²π©β²β²πͺβ²β². Then, = = = π. Dilate β³ π¨β²π©β²πͺβ² from center πΆ by scale factor π = , shown in π π.π π π blue in the diagram. Then, translate the blue triangle π. π units down and π units to the right. Next, rotate the blue triangle ππ degrees clockwise around point π¨β²β² until the blue triangle coincides with β³ π¨β²β²π©β²β²πͺβ²β². Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 132 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM c. Lesson 9 8β’3 Verify that, in fact, β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β² by checking corresponding angles and corresponding side lengths. Then, describe the sequence that would prove the similarity β³ π¨π©πͺ βΌβ³ π¨β²β² π©β²β² πͺβ²β² . The corresponding angles are congruent: β π¨ β β π¨β²β², β π© β β π©β²β² , and β πͺ β β πͺβ²β²; therefore, |β π¨| = |β π¨β²β² | = ππ°, |β π©| = |β π©β²β² | = πππ°, and |β πͺ| = |β πͺβ²β² | = ππ°. The ratio of the corresponding sides is equal: π.π π π = = = π. Dilate β³ π¨π©πͺ from center πΆ by scale factor π = π, shown as the pink triangle in the π.π π π diagram. Then, translate the pink triangle π units to the right. Finally, rotate the pink triangle ππ degrees clockwise around point π¨β²β² until the pink triangle coincides with β³ π¨β²β²π©β²β²πͺβ²β². d. Is it true that if β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β², then β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β²? Why do you think this is so? Yes, it is true that if β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² βΌβ³ π¨β²β² π©β²β² πͺβ²β² , then β³ π¨π©πͺ βΌβ³ π¨β²β² π©β²β² πͺβ²β² . Again, because these figures are similar, it means that they have equal angles and are made different sizes based on a specific scale factor. Since dilations map angles to angles of the same degree, it makes sense that all three figures would have the βsame shape.β Also, using the idea that similarity is a symmetric relation, the statement that β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² implies that β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ. Since we know that β³ π¨β² π©β² πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β², it is reasonable to conclude that β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β². Concept Development (3 minutes) Ask students to share what they wrote for part (d) of Exploratory Challenge 2. Expect students to respond in a similar manner to the response for part (d). If they do not, ask questions about what similarity means, what a dilation does, and how they might use what they just learned about similarity being a symmetric relation. ο§ For any three figures π, π, and π, if π~π and π~π, then π~π. This is what the statement that similarity is a transitive relation means. Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 133 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 9 8β’3 Closing (5 minutes) Summarize, or ask students to summarize, the main points from the lesson. ο§ We know that similarity is a symmetric relation. That means that if one figure is similar to another, π~πβ², then we can be sure that πβ²~π. The sequence that maps one onto the other is different, but we know that it is true. ο§ We know that similarity is a transitive relation. That means that if we are given two similar figures, π~π, and another statement about π~π, then we also know that π~π. Again, the sequence and scale factor are different to prove the similarity, but we know it is true. Lesson Summary Similarity is a symmetric relation. That means that if one figure is similar to another, πΊ~πΊβ², then we can be sure that πΊβ²~πΊ. Similarity is a transitive relation. That means that if we are given two similar figures, πΊ~π», and another statement about π»~πΌ, then we also know that πΊ~πΌ. Exit Ticket (5 minutes) Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 134 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM Name 8β’3 Date Lesson 9: Basic Properties of Similarity Exit Ticket Use the diagram below to answer Problems 1 and 2. 1. Which two triangles, if any, have similarity that is symmetric? 2. Which three triangles, if any, have similarity that is transitive? Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 135 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM 8β’3 Exit Ticket Sample Solutions Use the diagram below to answer Problems 1 and 2. 1. Which two triangles, if any, have similarity that is symmetric? β³ πΊ~ β³ πΉ and β³ πΉ~ β³ πΊ β³ πΊ~ β³ π» and β³ π»~ β³ πΊ β³ π»~ β³ πΉ and β³ πΉ~ β³ π» 2. Which three triangles, if any, have similarity that is transitive? One possible solution: Since β³ πΊ~ β³ πΉ and β³ πΉ~ β³ π», then β³ πΊ~ β³ π». Note that β³ πΌ and β³ π½ are not similar to each other or any other triangles. Therefore, they should not be in any solution. Problem Set Sample Solutions 1. Would a dilation alone be enough to show that similarity is symmetric? That is, would a dilation alone prove that if β³ π¨π©πͺ βΌ β³ π¨β²π©β²πͺβ², then β³ π¨β² π©β² πͺβ² βΌ β³ π¨π©πͺ? Consider the two examples below. a. Given β³ π¨π©πͺ βΌ β³ π¨β² π©β² πͺβ² , is a dilation enough to show that β³ π¨β² π©β² πͺβ² βΌ β³ π¨π©πͺ? Explain. For these two triangles, a dilation alone is enough to show that if β³ π¨π©πͺ βΌβ³ π¨β²π©β²πͺβ², then β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ. The reason that dilation alone is enough is because both of the triangles have been dilated from the same center. Therefore, to map one onto the other, all that would be required is a dilation. Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 136 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM b. 8β’3 Given β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² , is a dilation enough to show that β³ π¨β² π©β² πͺβ² βΌβ³ π¨π©πͺ? Explain. For these two triangles, a dilation alone is not enough to show that if β³ π¨π©πͺ βΌ β³ π¨β² π©β² πͺβ², then β³ π¨β² π©β² πͺβ² βΌ β³ π¨π©πͺ. The reason is that a dilation would just make them the same size. It would not show that you could map one of the triangles onto the other. To do that, you would need a sequence of basic rigid motions to demonstrate the congruence. c. In general, is dilation enough to prove that similarity is a symmetric relation? Explain. No, in general, a dilation alone does not prove that similarity is a symmetric relation. In some cases, like part (a), it would be enough, but because we are talking about general cases, we must consider figures that require a sequence of basic rigid motions to map one onto the other. Therefore, in general, to show that there is a symmetric relationship, we must use what we know about similar figures, a dilation followed by a congruence, as opposed to dilation alone. 2. Would a dilation alone be enough to show that similarity is transitive? That is, would a dilation alone prove that if β³ π¨π©πͺ βΌβ³ π¨β²π©β²πͺβ², and β³ π¨β² π©β² πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β², then β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β²? Consider the two examples below. a. Given β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β², is a dilation enough to show that β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β²? Explain. Yes, in this case, we could dilate by different scale factors to show that all three triangles are similar to each other. Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 137 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 9 NYS COMMON CORE MATHEMATICS CURRICULUM b. 8β’3 Given β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² βΌβ³ π¨β²β²π©β²β²πͺβ²β², is a dilation enough to show that β³ π¨π©πͺ βΌβ³ π¨β²β²π©β²β²πͺβ²β²? Explain. In this case, it would take more than just a dilation to show that all three triangles are similar to one another. Specifically, it would take a dilation followed by a congruence to prove the similarity among the three. c. In general, is dilation enough to prove that similarity is a transitive relation? Explain. In some cases, it might be enough, but the general case requires the use of dilation and a congruence. Therefore, to prove that similarity is a transitive relation, you must use both a dilation and a congruence. 3. In the diagram below, β³ π¨π©πͺ βΌβ³ π¨β² π©β² πͺβ² and β³ π¨β² π©β² πͺβ² ~ β³ π¨β²β²π©β²β²πͺβ²β². Is β³ π¨π©πͺ~ β³ π¨β²β²π©β²β²πͺβ²β²? If so, describe the dilation followed by the congruence that demonstrates the similarity. Yes, β³ π¨π©πͺ~ β³ π¨β²β²π©β²β²πͺβ²β² because similarity is transitive. Since π|π¨π©| = |π¨β²β²π©β²β²|, then π × π = π, which means π π π = . Then, a dilation from the origin by scale factor π = makes β³ π¨π©πͺ the same size as β³ π¨β²β²π©β²β²πͺβ²β². Translate π π π the dilated image of β³ π¨π©πͺ π units to the left and then reflect across line π¨β²β²π©β²β². The sequence of the dilation and π the congruence map β³ π¨π©πͺ onto β³ π¨β²β² π©β²β² πͺβ²β² , demonstrating the similarity. Lesson 9: Basic Properties of Similarity This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 138 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
© Copyright 2026 Paperzz