Geometry Module 2, Topic A, Lesson 4: Student Version
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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 4 M2 GEOMETRY Lesson 4: Comparing the Ratio Method with the Parallel Method Classwork Today, our goal is to show that the parallel method and the ratio method are equivalent; that is, given a figure in the plane and a scale factor π > 0, the scale drawing produced by the parallel method is congruent to the scale drawing produced by the ratio method. We start with two easy exercises about the areas of two triangles whose bases lie on the same line, which helps show that the two methods are equivalent. Opening Exercise a. Suppose two triangles, β³ π΄π΅πΆ and β³ π΄π΅π·, share the same base π΄π΅ such that points πΆ and π· lie on a line parallel to β‘π΄π΅ . Show that their areas are equal, that is, Area(β³ π΄π΅πΆ) = Area(β³ π΄π΅π·). (Hint: Why are the altitudes of each triangle equal in length?) b. Suppose two triangles have different-length bases, π΄π΅ and π΄π΅′, that lie on the same line. Furthermore, suppose they both have the same vertex πΆ opposite these bases. Show that the value of the ratio of their areas is equal to the value of the ratio of the lengths of their bases, that is, Area(β³ π΄π΅πΆ) π΄π΅ = . ′ Area(β³ π΄π΅ πΆ) π΄π΅′ Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.24 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 4 M2 GEOMETRY Discussion To show that the parallel and ratio methods are equivalent, we need only look at one of the simplest versions of a scale drawing: scaling segments. First, we need to show that the scale drawing of a segment generated by the parallel method is the same segment that the ratio method would have generated and vice versa. That is, The parallel method βΉ The ratio method, and The ratio method βΉ The parallel method. The first implication above can be stated as the following theorem: PARALLEL βΉ RATIO THEOREM: Given π΄π΅ and point π not on β‘π΄π΅ , construct a scale drawing of π΄π΅ with scale factor π > 0 using the parallel method: Let π΄′ = π·π,π (π΄), and β be the line parallel to β‘π΄π΅ that passes through π΄′. Let π΅′ be the point where ππ΅ intersects β. Then π΅′ is the same point found by the ratio method, that is, π΅′ = π·π,π (π΅). PROOF: We prove the case when π > 1; the case when 0 < π < 1 is the same but with a different picture. Construct two line segments π΅π΄′ and π΄π΅′ to form two triangles β³ π΅π΄π΅′ and β³ π΅π΄π΄′ , labeled as π1 and π2 , respectively, in the picture below. Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.25 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M2 GEOMETRY The areas of these two triangles are equal, Area(π1 ) = Area(π2 ), by Exercise 1. Why? Label β³ ππ΄π΅ by π0 . Then Area(β³ ππ΄′ π΅) = Area(β³ ππ΅′ π΄) because areas add: Area(β³ ππ΄′ π΅) = Area(π0 ) + Area(π2 ) = Area(π0 ) + Area(π1 ) = Area(β³ ππ΅′ π΄). Next, we apply Exercise 2 to two sets of triangles: (1) π0 and β³ ππ΄′ π΅ and (2) π0 and β³ ππ΅′π΄. (2) π»π and β³ πΆπ©′π¨ with β‘ ′ bases on πΆπ© (1) π»π and β³ πΆπ¨′ π© with bases on β‘πΆπ¨′ Therefore, Area(β³ ππ΄′ π΅) ππ΄′ = , and Area(π0 ) ππ΄ Area(β³ ππ΅′ π΄) ππ΅′ = . Area(π0 ) ππ΅ Since Area(β³ ππ΄′ π΅) = Area(β³ ππ΅′ π΄), we can equate the fractions: dilating ππ΄ to ππ΄′ , we know that ππ΄′ ππ΄ = π; therefore, ππ΅′ ππ΅ ππ΄′ ππ΄ = ππ΅′ ππ΅ . Since π is the scale factor used in = π, or ππ΅′ = π ⋅ ππ΅. This last equality implies that π΅′ is the dilation of π΅ from π by scale factor π, which is what we wanted to prove. Next, we prove the reverse implication to show that both methods are equivalent to each other. Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.26 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M2 GEOMETRY RATIO βΉ PARALLEL THEOREM: Given π΄π΅ and point π not on β‘π΄π΅ , construct a scale drawing π΄′ π΅′ of π΄π΅ with scale factor π > 0 using the ratio method (i.e., find π΄′ = π·π,π (π΄) and π΅′ = π·π,π (π΅), and draw π΄′ π΅′ ). Then π΅′ is the same as the point found using the parallel method. PROOF: Since both the ratio method and the parallel method start with the same first step of setting π΄′ = π·π,π (π΄), the only difference between the two methods is in how the second point is found. If we use the parallel method, we construct the line β parallel to β‘π΄π΅ that passes through π΄′ and label the point where β intersects ππ΅ by πΆ. Then π΅′ is the same as the point found using the parallel method if we can show that πΆ = π΅′ . C πΆπ¨ πΆπ¨ π ⋅ πΆπ¨ π ⋅ πΆπ¨ The ratio method The parallel method By the parallel βΉ ratio theorem, we know that πΆ = π·π,π (π΅), that is, that πΆ is the point on ππ΅ such that ππΆ = π ⋅ ππ΅. But π΅′ is also the point on ππ΅ such that ππ΅′ = π ⋅ ππ΅. Hence, they must be the same point. The fact that the ratio and parallel methods are equivalent is often stated as the triangle side splitter theorem. To understand the triangle side splitter theorem, we need a definition: B D SIDE SPLITTER: A line segment πΆπ· is said to split the sides of β³ ππ΄π΅ proportionally if πΆ is a point on Μ Μ Μ Μ ππ΄, π· is a point on Μ Μ Μ Μ ππ΅ , and (or equivalently, ππΆ ππ΄ = ππ· ππ΅ ππ΄ ππΆ = ππ΅ ππ· ). We call line segment πΆπ· a side splitter. O C A TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a triangle proportionally if and only if it is parallel to the third side. Restatement of the triangle side splitter theorem: Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.27 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M2 GEOMETRY Lesson Summary THE TRIANGLE SIDE SPLITTER THEOREM: A line segment splits two sides of a triangle proportionally if and only if it is parallel to the third side. Problem Set 1. 2. Use the diagram to answer each part below. a. Measure the segments in the figure below to verify that the proportion is true. ππ΄′ ππ΅′ = ππ΄ ππ΅ b. Is the proportion c. Is the proportion ππ΄ ππ΄′ π΄π΄′ ππ΄′ = = ππ΅ ππ΅ ′ π΅π΅ ′ ππ΅ ′ also true? Explain algebraically. also true? Explain algebraically. Given the diagram below, π΄π΅ = 30, line β is parallel to Μ Μ Μ Μ π΄π΅ , and the distance from Μ Μ Μ Μ π΄π΅ to β is 25. Locate point πΆ on line β such that β³ π΄π΅πΆ has the greatest area. Defend your answer. l 3. Μ Μ Μ Μ and Μ Μ Μ Μ Given β³ πππ, ππ ππ are partitioned into equal-length segments by the endpoints of the dashed segments as shown. What can be concluded about the diagram? Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.28 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M2 GEOMETRY 4. Given the diagram, π΄πΆ = 12, π΄π΅ = 6, π΅πΈ = 4, π∠π΄πΆπ΅ = π₯°, and π∠π· = π₯°, find πΆπ·. 5. What conclusions can be drawn from the diagram shown to the right? Explain. 6. Parallelogram πππ π is shown. Two triangles are formed by a diagonal within the parallelogram. Identify those triangles, and explain why they are guaranteed to have the same areas. Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.29 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. M2 Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM GEOMETRY 7. In the diagram to the right, π»πΌ = 36 and πΊπ½ = 42. If the ratio of the areas of the triangles is Area βπΊπ»πΌ Area βπ½π»πΌ 5 = , find 9 π½π», πΊπ», πΊπΌ, and π½πΌ. Lesson 4: Comparing the Ratio Method with the Parallel Method This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015 S.30 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
