Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8•3 Lesson 5: First Consequences of FTS Student Outcomes ο§ Students verify the converse of the fundamental theorem of similarity experimentally. ο§ Students apply the fundamental theorem of similarity to find the location of dilated points on the plane. Classwork Concept Development (5 minutes) Begin by having students restate (in their own words) the fundamental theorem of similarity (FTS) that they learned in the last lesson. ο§ The fundamental theorem of similarity states: Given a dilation with center π and scale factor π, then for any two points π and π in the plane so that π, π, and π are not collinear, the lines ππ and π′π′ are parallel, where π′ = π·ππππ‘πππ(π) and π′ = π·ππππ‘πππ(π), and furthermore, |π′π′| = π|ππ|. The paraphrased version from the last lesson was: FTS states that given a dilation from center π, and points π and π (points π, π, and π are not on the same line), the segments formed when π to π and π′ are connected to π′ are parallel. More surprising is the fact that the segment ππ, even though it was not dilated as points π and π were, dilates to segment π′ π′ , and the length of segment π′π′ is the length of segment ππ multiplied by the scale factor. ο§ The converse of this theorem is also true. That is, if lines ππ and π′π′ are parallel, and |π′π′| = π|ππ|, then from a center π, π′ = π·ππππ‘πππ(π), π′ = π·ππππ‘πππ(π), and |ππ′| = π|ππ| and |ππ′| = π|ππ|. The converse of a theorem begins with the conclusion and produces the hypothesis. FTS concludes that lines are parallel and the length of segment π′π′ is the length of segment ππ multiplied by the scale factor. The converse of FTS begins with the statement that the lines are parallel and the length of segment π′π′ is the length of segment ππ multiplied by the scale factor. It ends with the knowledge that the points π′ and π′ are dilations of points π and π by scale factor π, and their respective segments, ππ′ and ππ′ , have lengths that are the scale factor multiplied by the original lengths of segment ππ and segment ππ. Consider providing students with some simple examples of converses, and discuss whether or not the converses are true. For example, “If it is raining, then I have an umbrella,” and its converse, “If I have an umbrella, then it is raining.” In this case, the converse is not necessarily true. An example where the converse is true is “If we turn the faucet on, then water comes out,” and the converse is “If water comes out, then the faucet is on.” ο§ The converse of the theorem is basically the work we did in the last lesson but backward. In the last lesson, we knew about dilated points and found out that the segments between the points π, π and their dilations π′, π′ were dilated according to the same scale factor, that is, |π′π′| = π|ππ|. We also discovered that the lines β‘ . The converse states that we are given parallel containing those segments were parallel, that is, β‘ππ β₯ π′π′ β‘ , where a segment, ππ Μ Μ Μ Μ Μ Μ , in the Μ Μ Μ Μ , in one line is dilated by scale factor π to a segment, π′π′ lines, β‘ππ and π′π′ other line. With that knowledge, we can say something about the center of dilation and the relationship between segments ππ′ and ππ, as well as segments ππ′ and ππ. Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 54 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8•3 In Exercise 1 below, students are given the information in the converse of FTS, that is, Μ Μ Μ Μ ππ β₯ Μ Μ Μ Μ Μ Μ π′π′ and |π′π′| = π|ππ|. Students then verify the conclusion of the converse of FTS by measuring lengths of segments and their dilated images to make sure that they have the properties stated, that is, |ππ′| = π|ππ| and |ππ′| = π|ππ|. Consider showing the diagram below and asking students to make conjectures about the relationship between segment ππ′ and segment ππ, as well as segment ππ′ and segment ππ. Record the conjectures and possibly have the class vote on which they believe to be true. In the diagram below, the lines containing segments ππ and π′π′ are parallel, and the length of segment π′π′ is equal to the length of segment ππ multiplied by the scale factor π, dilated from center π. Exercise 1 (5 minutes) Have students verify experimentally the validity of the converse of the theorem. They can work independently or in pairs. Note that the image is to scale in the student materials. Exercise 1 Μ Μ Μ Μ Μ , |π·πΈ| = π ππ¦, and Μ Μ Μ Μ β₯ π·′πΈ′ In the diagram below, points π· and πΈ have been dilated from center πΆ by scale factor π. π·πΈ |π·′πΈ′| = ππ ππ¦. Scaffolding: If students need help getting started, have them review the activity from Lesson 4. Ask what they should do to find the center π. Students should say to draw lines through ππ′ and ππ′; the point of intersection is the center π. Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 55 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM a. Lesson 5 8•3 Determine the scale factor π. According to FTS, |π·′πΈ′| = π|π·πΈ|. Therefore, ππ = π ⋅ π, so π = π. b. Locate the center πΆ of dilation. Measure the segments to verify that |πΆπ·′| = π|πΆπ·| and |πΆπΈ′| = π|πΆπΈ|. Show your work below. Center πΆ and measurements are shown above. |πΆπ·′ | = π|πΆπ·| π=π⋅π π=π |πΆπΈ′ | = π|πΆπΈ| π=π⋅π π=π Example 1 (5 minutes) ο§ Now that we know FTS and the converse of FTS in terms of dilations, we practice using them to find the coordinates of points and dilated points on a plane. We begin simply. ο§ In the diagram, we have center π and ray ππ΄. We want to find the coordinates of point π΄′. We are given that the scale factor of dilation is π = 2. ο§ To find π΄′ , we could use a ruler or compass to measure |ππ΄|, but now that we know about FTS, we can do this another way. First, we should look for parallel lines that help us locate point π΄′. How can we use the coordinate plane to ensure parallel lines? οΊ We could use the vertical or horizontal lines of the coordinate plane to ensure lines are parallel. The coordinate plane is set up so that the lines never intersect. You could also think of those lines as translations of the π₯-axis and π¦-axis. Therefore, we are guaranteed to have parallel lines. Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 56 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM ο§ 8•3 Let’s use the π₯-axis as one of our rays. (Show the picture below.) Where should we place a point π΅ on the ray along the π₯-axis? οΊ ο§ Lesson 5 Since we are using the lines on the coordinate plane to verify parallelism, we should place point π΅ directly below point π΄ on the π₯-axis. Point π΅ should be at (5, 0). (Show the picture below.) This is beginning to look like the activity we did in Lesson 4. We know that the scale factor is π = 2. Where should we put point π΅′? οΊ It is clear that |ππ΅| = 5; therefore, |ππ΅′| = 2 ⋅ 5, so the point π΅′ should be placed at (10, 0). Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 57 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM ο§ 8•3 (Show the picture below.) Now that we know the location of π΅′, using FTS, what do we expect to be true about the lines containing segments π΄π΅ and π΄′π΅′? οΊ ο§ Lesson 5 We expect the lines containing segments π΄π΅ and π΄′π΅′ to be parallel. (Show the picture below.) Then what is the location of point π΄′? οΊ Point π΄′ is located at (10, 4). Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 58 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM ο§ 8•3 (Show the picture below.) Could point π΄′ be located anywhere else? Specifically, could π΄′ have a different π₯coordinate? Why or why not? οΊ ο§ Lesson 5 No, Point π΄′ must be at (10, 4). If it had another π₯-coordinate, then lines π΄π΅ and π΄′π΅′ would have not been dilated by the scale factor π = 2. Could point π΄′ be located at another location that has 10 as its π₯-coordinate? For example, could π΄′ be at (10, 5)? Why or why not? οΊ No, Point π΄′ must be at (10, 4). By the definition of dilation, if point π΄ is dilated from center π, then the dilated point must be on the ray ππ΄, making points π, π΄, and π΄′ collinear. If π΄′ were at (10, 5) or at any coordinate other than (10, 4), then the points π, π΄, and π΄′ would not be collinear. Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 59 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8•3 Exercise 2 (3 minutes) Students work independently to find the location of point π΄′ on the coordinate plane. Exercise 2 In the diagram below, you are given center πΆ and ray πΆπ¨. Point π¨ is dilated by a scale factor π = π. Use what you know about FTS to find the location of point π¨′. Point π¨′ must be located at (ππ, ππ). Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 60 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8•3 Example 2 (6 minutes) ο§ In the diagram, we have center π and ray ππ΄. We are given that the scale factor of dilation is π = 11 . We 7 want to find the precise coordinates of point π΄′. Based on our previous work, we know exactly how to begin. Draw a ray ππ΅ along the π₯-axis, and mark a point π΅, directly below point π΄, on the ray ππ΅ . (Show the picture below.) The question now is, how do we locate point π΅′? Think about our work from Lesson 4. ο§ In Lesson 4, we counted lines to determine the scale factor. Given the scale factor, we know that point π΅ should be exactly 7 lines from the center π and that the dilated point, π΅′, should be exactly 11 lines from the center. Therefore, π΅′ is located at (11, 0). ο§ Now that we know the location of π΅′, where do we find π΄′? οΊ Point π΄′ is at the intersection of the ray ππ΄ and the vertical line through point π΅′, which must be parallel to the line containing Μ Μ Μ Μ π΄π΅ . Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 61 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM ο§ 8•3 (Show the picture below.) Now that we know where π΄′ is, we need to find the precise coordinates of it. The π₯-coordinate is easy, but how can we find the π¦-coordinate? (Give students time to talk in pairs or small groups.) οΊ ο§ Lesson 5 The π¦-coordinate is the exact length of the segment π΄′π΅′. To find that length, we can use what we know about the length of segment π΄π΅ and the scale factor since |π΄′π΅′| = π|π΄π΅|. The length of segment π΄′π΅′ gives us the π¦-coordinate. Then |π΄′ π΅′ | = location of π΄′ is (11, Lesson 5: 66 ). 7 First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 11 66 ⋅ 6 = . That means that the 7 7 62 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8•3 Exercise 3 (4 minutes) Students work independently or in pairs to find the location of point π΄′ on the coordinate plane. Exercise 3 In the diagram below, you are given center πΆ and ray πΆπ¨. Point π¨ is dilated by a scale factor π = π . Use what you know ππ about FTS to find the location of point π¨′. The π-coordinate of π¨′ is π. The π-coordinate is equal to the length of segment π¨′π©′. Since |π¨′π©′| = π|π¨π©|, then |π¨′π©′| = π ππ ⋅π= ≈ π. π. The location of π¨′ is (π, π. π). ππ ππ Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 63 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8•3 Example 3 (8 minutes) ο§ 5 8 In the diagram below, we have center π and rays ππ΄ and ππ΅ . We are given that the scale factor is π = . We want to find the precise coordinates of points π΄′ and π΅′. ο§ Based on our previous work, we know exactly how to begin. Describe what we should do. οΊ ο§ Draw a ray ππΆ along the π₯-axis, and mark a point πΆ, directly below point π΄, on the ray ππΆ . (Show the picture below.) Based on our work in Lesson 4 and our knowledge of FTS, we know that points π΅ and π΅′ along ray ππ΅ enjoy the same properties we have been using in the last few problems. Let’s begin by finding the coordinates of π΄′. What should we do first? οΊ First, we need to place the points π΄′, π΅′, and πΆ′ on their respective rays by using the scale factor. Since the scale factor π = Lesson 5: 5 , π΄′, π΅′, and πΆ′ have an π₯-coordinate of 5. (Also, they are 5 lines from the center.) 8 First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 64 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM ο§ 8•3 (Show the picture below.) Now that we know the π₯-coordinate, let’s now find the π¦-coordinate of π΄′. What do we need to do to find the π¦-coordinate of π΄′? οΊ The π¦-coordinate of π΄′ is the length of the segment π΄′πΆ′. We can calculate that length by using what we know about segment π΄πΆ and the scale factor. 5 8 ο§ The π¦-coordinate of π΄′ is |π΄′πΆ′| = π|π΄πΆ| = ⋅ 9 = ο§ Work with a partner to find the π¦-coordinate of π΅′. οΊ 45 ≈ 5.6. Then the location of π΄′ is (5, 5.6). 8 The π¦-coordinate of π΅′ is the length of segment π΅′πΆ′. Then |π΅′πΆ′| = π|π΅πΆ| = 5 20 ⋅4= = 2.5. The 8 8 location of π΅′ is (5, 2.5). Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 65 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 5 8•3 Closing (4 minutes) Summarize, or ask students to summarize, the main points from the lesson. ο§ We experimentally verified the converse of FTS. That is, if we are given parallel lines ππ and π′π′ and know |π′π′| = π|ππ|, then we know from a center π, π′ = π·ππππ‘πππ(π), π′ = π·ππππ‘πππ(π), |ππ′| = π|ππ|, and |ππ′| = π|ππ|. In other words, if we are given parallel lines ππ and π′π′, and we know that the length of segment π′π′ is equal to the length of segment ππ multiplied by the scale factor, then we also know that the length of segment ππ′ is equal to the length of segment ππ multiplied by the scale factor and that the length of segment ππ′ is equal to the length of segment ππ multiplied by the scale factor. ο§ We know how to use FTS to find the coordinates of dilated points, even if the dilated point is not on an intersection of graph lines. Lesson Summary Converse of the fundamental theorem of similarity: If lines π·πΈ and π·′πΈ′ are parallel and |π·′πΈ′| = π|π·πΈ|, then from a center πΆ, π·′ = π«πππππππ(π·), πΈ′ = π«πππππππ(πΈ), |πΆπ·′| = π|πΆπ·|, and |πΆπΈ′| = π|πΆπΈ|. To find the coordinates of a dilated point, we must use what we know about FTS, dilation, and scale factor. Exit Ticket (5 minutes) Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 66 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Name ___________________________________________________ Lesson 5 8•3 Date____________________ Lesson 5: First Consequences of FTS Exit Ticket 6 4 In the diagram below, you are given center π and ray ππ΄. Point π΄ is dilated by a scale factor π = . Use what you know about FTS to find the location of point π΄′. Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 67 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 8•3 Exit Ticket Sample Solutions π π In the diagram below, you are given center πΆ and ray πΆπ¨. Point π¨ is dilated by a scale factor π = . Use what you know about FTS to find the location of point π¨′. The π-coordinate of π¨′ is π. The π-coordinate is equal to the length of segment π¨′π©′. Since |π¨′π©′| = π|π¨π©|, then |π¨′π©′| = π ππ ⋅π= = π. π. The location of π¨′ is (π. π, π). π π Problem Set Sample Solutions Students practice using the first consequences of FTS in terms of dilated points and their locations on the coordinate plane. 1. π π Dilate point π¨, located at (π, π) from center πΆ, by a scale factor π = . What is the precise location of point π¨′? The π-coordinate of point π¨′ is the length of segment π¨′π©′. Since |π¨′π©′| = π|π¨π©|, then |π¨′π©′| = location of point π¨′ is (π, Lesson 5: ππ ), or approximately (π, π. π). π First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 π ππ ⋅ π = . The π π 68 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 5 NYS COMMON CORE MATHEMATICS CURRICULUM 2. 8•3 π π Dilate point π¨, located at (π, π) from center πΆ, by a scale factor π = . Then, dilate point π©, located at (π, π) from π π center πΆ, by a scale factor of π = . What are the coordinates of points π¨′ and π©′ ? Explain. The π-coordinate of point π¨′ is the length of π¨′πͺ′. Since |π¨′πͺ′| = π|π¨πͺ|, then |π¨′πͺ′| = π ππ ⋅ π = . The location of π π ππ ), or approximately (π, π. π). The π-coordinate of point π©′ is the length of π©′πͺ′. Since |π©′πͺ′| = π π ππ ππ π|π©πͺ|, then |π©′πͺ′| = ⋅ π = . The location of point π©′ is (π, ), or approximately (π, π. π). π π π point π¨′ is (π, 3. Explain how you used the fundamental theorem of similarity in Problems 1 and 2. Using what I knew about scale factor, I was able to determine the placement of points π¨′ and π©′, but I did not know the actual coordinates. So, one of the ways that FTS was used was actually in terms of the converse of FTS. I had to make sure I had parallel lines. Since the lines of the coordinate plane guarantee parallel lines, I knew that |π¨′πͺ′| = π|π¨πͺ|. Then, since I knew the length of segment π¨πͺ and the scale factor, I could find the precise location of π¨′. The precise location of π©′ was found in a similar way but using |π©′πͺ′| = π|π©πͺ|. Lesson 5: First Consequences of FTS This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M3-TE-1.3.0-08.2015 69 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.