Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY Lesson 3: The Scaling Principle for Area Student Outcomes ο§ Students understand that a similarity transformation with a scale factor of π multiplies the area of a planar region by a factor of π 2 . ο§ Students understand that if a planar region is scaled by factors of π and π in two perpendicular directions, then its area is multiplied by a factor of ππ. Lesson Notes In Lesson 3, students experiment with figures that have been dilated by different scale factors and observe the effect that the dilation has on the area of the figure (or pre-image) as compared to its image. In Topic B, the move is made from the scaling principle for area to the scaling principle for volume. This shows up in the use of the formula π = π΅β; more importantly, it is how we establish the volume formula for pyramids and cones. The scaling principle for area helps us develop the scaling principle for volume, which in turn helps us develop the volume formula for general cones (G-GMD.A.1). Classwork Exploratory Challenge (10 minutes) In the Exploratory Challenge, students determine the area of similar triangles and similar parallelograms and then compare the scale factor of the similarity transformation to the value of the ratio of the area of the image to the area of the pre-image. The goal is for students to see that the areas of similar figures are related by the square of the scale factor. It may not be necessary for students to complete all of the exercises in order to see this relationship. Monitor the class. If most students understand it, then move on to the Discussion that follows. Exploratory Challenge Complete parts (i)β(iii) of the table for each of the figures in questions (a)β(d): (i) Determine the area of the figure (preimage), (ii) determine the scaled dimensions of the figure based on the provided scale factor, and (iii) determine the area of the dilated figure. Then, answer the question that follows. In the final column of the table, find the value of the ratio of the area of the similar figure to the area of the original figure. a. b. c. d. (i) Area of Original Figure Scale Factor (ii) Dimensions of Similar Figure (iii) Area of Similar Figure ππ π ππ × π πππ π. π π ππ × π ππ π. π × π π π. π × π ππ. π ππ π Lesson 3: π π π π The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 Ratio of Areas ππ«πππ¬π’π¦π’π₯ππ« : ππ«πππ¨π«π’π π’π§ππ₯ πππ =π ππ ππ =π π. π π π = ππ π ππ. π ππ π = = π ππ π 36 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY a. Scaffolding: i. ii. iii. π π ο§ Consider dividing the class and having students complete two of the four problems, and then share their results before making the conjecture in Exploratory Challenge, part (e). (π)(π) = ππ The base of the similar triangle is π(π) or ππ, and the height of the similar triangle is π(π) or π. π π (ππ)(π) = πππ ο§ Model the process of determining the dimensions of the similar figure for the whole class or a small group. b. i. ii. iii. π π (π)(π) = π. π The base of the similar triangle is π(π) or ππ, and the height of the similar triangle is π(π) or π. π π (ππ)(π) = ππ c. i. (π)(π) = ππ ii. The base of the similar parallelogram is π ( ) or π. π, and the height of the similar parallelogram is π π π π π ( ) or π. iii. π. π(π) = π Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 37 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY d. i. (π)(π) = π ii. The base of the similar parallelogram is π ( ) or π. π, and the height of the similar parallelogram is π π π π π ( ) or π. iii. e. MP.3 & MP.8 π. π(π) = ππ. π Make a conjecture about the relationship between the areas of the original figure and the similar figure with respect to the scale factor between the figures. It seems as though the value of the ratio of the area of the similar figure to the area of the original figure is the square of the scale factor of dilation. Discussion (13 minutes) Select students to share their conjecture from Exploratory Challenge, part (e). Then formalize their observations with the Discussion below about the scaling principle of area. ο§ We have conjectured that the relationship between the area of a figure and the area of a figure similar to it is the square of the scale factor. ο§ Polygon π is the image of Polygon π under a similarity transformation with a scale factor of π. How can we show that our conjecture holds for such a polygon? Polygon π ο§ Polygon π Polygon π is the image of Polygon π under a similarity transformation with a scale factor of π. How can we show that our conjecture holds for such a polygon? οΊ We can find the area of each and compare the areas of the two figures. ο§ How can we compute the area of a polygon like this? ο§ Can any polygon be decomposed into non-overlapping triangles? οΊ οΊ ο§ We can break it up into triangles. Yes If we can prove that the relationship holds for any triangle, then we can extend the relationship to any polygon. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 38 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY THE SCALING PRINCIPLE FOR TRIANGLES: If similar triangles πΊ and π» are related by a scale factor of π, then the respective areas are related by a factor of ππ . ο§ To prove the scaling principle for triangles, consider a triangle π with base and height, π and β, respectively. Then the base and height of the image of π are ππ and πβ, respectively. Triangle π οΊ ο§ MP.3 Triangle π 1 2 1 2 1 2 The area of π is Area(π) = πβ, and the area of π is Area(π) = πππβ = ( πβ) π 2 . How could we show that the ratio of the areas of π and π is equal to π 2 ? οΊ Area(π) Area(π) = 1 2 ( πβ)π 2 1 πβ 2 = π2 Therefore, we have proved the scaling principle for triangles. ο§ Given the scaling principle for triangles, can we use that to come up with a scaling principle for any polygon? οΊ Any polygon can be subdivided into non-overlapping triangles. Since each area of a scaled triangle is π 2 times the area of its original triangle, then the sum of all the individual, scaled areas of triangles should be the area of the scaled polygon. THE SCALING PRINCIPLE FOR POLYGONS: If similar polygons π· and πΈ are related by a scale factor of π, then their respective areas are related by a factor of ππ . ο§ Imagine subdividing similar polygons π and π into non-overlapping triangles. Polygon π Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 Polygon π 39 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY ο§ Each of the lengths in polygon π is π times the corresponding lengths in polygon π. ο§ The area of polygon π is Area(π) = π1 + π2 + π3 + π4 + π5 , th where ππ is the area of the π triangle, as shown. ο§ By the scaling principle of triangles, the area of each triangle in π is π 2 times the area of each corresponding triangle in π. ο§ Then the area of polygon π is Area(π) = π 2 π1 + π 2 π2 + π 2 π3 + π 2 π4 + π 2 π5 Area(π) = π 2 (π1 + π2 + π3 + π4 + π5 ) Area(π) = π 2 (Area(π)). Polygon π ο§ Polygon π Since the same reasoning applies to any polygon, we have proven the scaling principle for polygons. Exercises 1β2 (8 minutes) Students apply the scaling principle for polygons to determine unknown areas. Exercises 1β2 1. Rectangles π¨ and π© are similar and are drawn to scale. If the area of rectangle π¨ is ππ π¦π¦π, what is the area of rectangle π©? Length scale factor: ππ ππ = ππ π = π. πππ Area scale factor: (π. πππ)π ππ«ππ(π©) = (π. πππ)π × ππ«ππ(π¨) ππ«ππ(π©) = (π. πππ)π × ππ ππ«ππ(π©) = πππ. πππ The area of rectangle π© is πππ. πππ π¦π¦π. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 40 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY 2. Figures π¬ and π are similar and are drawn to scale. If the area of figure π¬ is πππ π¦π¦π, what is the area of figure π? π. π ππ¦ = ππ π¦π¦ Length scale factor: ππ ππ = π π = π. πππ Area scale factor: (π. πππ)π ππ«ππ(π) = (π. πππ)π × ππ«ππ(π¬) ππ«ππ(π) = (π. πππ)π × πππ ππ«ππ(π) = ππ. πππ The area of figure π is ππ. πππ π¦π¦π. Discussion (7 minutes) ο§ How can you describe the scaling principle for area? Allow students to share ideas out loud before confirming with the formal principle below. THE SCALING PRINCIPLE FOR AREA: If similar figures π¨ and π© are related by a scale factor of π, then their respective areas are related by a factor of ππ . ο§ The following example shows another circumstance of scaling and its effect on area: Give students 90 seconds to discuss the following sequence of images with a partner. Then ask for an explanation of what they observe. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 41 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M3 GEOMETRY Ask follow-up questions such as the following to encourage students to articulate what they notice: ο§ Is the 1 × 1 unit square scaled in both dimensions? ο§ By what scale factor is the unit square scaled horizontally? How does the area of the resulting rectangle compare to the area of the unit square? οΊ οΊ ο§ The unit square is scaled horizontally by a factor of 3, and the area is three times as much as the area of the unit square. What is happening between the second image and the third image? οΊ ο§ No, only the length is scaled and, therefore, affects the area by only the scale factor. The horizontally scaled figure is now scaled vertically by a factor of 4. The area of the new figure is 4 times as much as the area of the second image. Notice that the directions of scaling applied to the original figure, the horizontal and vertical scaling, are perpendicular to each other. Furthermore, with respect to the first image of the unit square, the third image has 12 times the area of the unit square. How is this related to the horizontal and vertical scale factors? οΊ The area has changed by the same factor as the product of the horizontal and vertical scale factors. ο§ We generalize this circumstance: When a figure is scaled by factors π and π in two perpendicular directions, then its area is multiplied by a factor of ππ: ο§ We see this same effect when we consider a triangle with base 1 and height 1, as shown below. ο§ We can observe this same effect with non-polygonal regions. Consider a unit circle, as shown below. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 42 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY ο§ Keep in mind that scale factors may have values between 0 and 1; had that been the case in the above examples, we could have seen reduced figures as opposed to enlarged ones. ο§ Our work in upcoming lessons is devoted to examining the effect that dilation has on three-dimensional figures. Closing (2 minutes) Ask students to summarize the key points of the lesson. Additionally, consider asking students the following questions independently in writing, to a partner, or to the whole class. ο§ If the scale factor between two similar figures is 1.2, what is the scale factor of their respective areas? οΊ ο§ If the scale factor between two similar figures is οΊ ο§ The scale factor of the respective areas is 1.44. 1 , what is the scale factor of their respective areas? 2 1 The scale factor of the respective areas is . 4 Explain why the scaling principle for triangles is necessary to generalize to the scaling principle for polygonal regions. οΊ Each polygonal region is composed of a finite number of non-overlapping triangles. If we know the scaling principle for triangles, and polygonal regions are comprised of triangles, then we know that what we observed for scaled triangles applies to polygonal regions in general. Lesson Summary THE SCALING PRINCIPLE FOR TRIANGLES: If similar triangles πΊ and π» are related by a scale factor of π, then the respective areas are related by a factor of ππ . THE SCALING PRINCIPLE FOR POLYGONS: If similar polygons π· and πΈ are related by a scale factor of π, then their respective areas are related by a factor of ππ . THE SCALING PRINCIPLE FOR AREA: If similar figures π¨ and π© are related by a scale factor of π, then their respective areas are related by a factor of ππ . Exit Ticket (5 minutes) Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 43 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY Name Date Lesson 3: The Scaling Principle for Area Exit Ticket Μ Μ Μ Μ and Μ Μ Μ Μ In the following figure, π΄πΈ π΅π· are segments. a. β³ π΄π΅πΆ and β³ πΆπ·πΈ are similar. How do we know this? b. What is the scale factor of the similarity transformation that takes β³ π΄π΅πΆ to β³ πΆπ·πΈ? c. What is the value of the ratio of the area of β³ π΄π΅πΆ to the area of β³ πΆπ·πΈ? Explain how you know. d. If the area of β³ π΄π΅πΆ is 30 cm2 , what is the area of β³ πΆπ·πΈ? Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 44 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY Exit Ticket Sample Solutions Μ Μ Μ Μ and π©π« Μ Μ Μ Μ Μ are segments. In the following figure, π¨π¬ a. β³ π¨π©πͺ and β³ πͺπ«π¬ are similar. How do we know this? The triangles are similar by the AA criterion. b. What is the scale factor of the similarity transformation that takes β³ π¨π©πͺ to β³ πͺπ«π¬? π= c. π ππ What is the value of the ratio of the area of β³ π¨π©πͺ to the area of β³ πͺπ«π¬? Explain how you know. ππ = ( d. ππ π π ) , or by the scaling principle for triangles. ππ πππ If the area of β³ π¨π©πͺ is ππ ππ¦π, what is the approximate area of β³ πͺπ«π¬? ππ«ππ(β³ πͺπ«π¬) = ππ × ππ ππ¦π β π ππ¦π πππ Problem Set Sample Solutions 1. A rectangle has an area of ππ. Fill in the table below by answering the questions that follow. Part of the first row has been completed for you. π π π π π π Original Dimensions Original Area Scaled Dimensions Scaled Area Scaled Area Original Area Area ratio in terms of the scale factor ππ × π ππ π×π ππ π×π ππ π π π π π π π π π π π π π π π π π π π =( ) π π π π π =( ) π π π π π =( ) π π π π π =( ) π π π π π π π π π =( ) π π π × ππ π π × ππ π ππ ππ π× π π π ×π π π π× π π × ππ π π × ππ π a. List five unique sets of dimensions of your choice for a rectangle with an area of ππ, and enter them in column 1. b. If the given rectangle is dilated from a vertex with a scale factor of , what are the dimensions of the images π π of each of your rectangles? Enter the scaled dimensions in column 3. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 45 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY c. What are the areas of the images of your rectangles? Enter the areas in column 4. d. How do the areas of the images of your rectangles compare to the area of the original rectangle? Write the value of each ratio in simplest form in column 5. e. Write the values of the ratios of area entered in column 5 in terms of the scale factor . Enter these values in π π column 6. f. If the areas of two unique rectangles are the same, π, and both figures are dilated by the same scale factor π, what can we conclude about the areas of the dilated images? The areas of the dilated images would both be ππ π and thus equal. 2. Find the ratio of the areas of each pair of similar figures. The lengths of corresponding line segments are shown. a. The scale factor that takes the smaller pentagon to the larger π pentagon is . The area of the larger pentagon is equal to the π π π π area of the smaller pentagon times ( ) = ππ . Therefore, the π ratio of the area of the smaller pentagon to the larger pentagon is π: ππ. b. The scale factor that takes the larger region to the smaller π region is . The area of the smaller region is equal to the area π π π π π of the larger region times ( ) or . Therefore, the ratio of the π area of the larger region to the smaller region is π: π. c. π The scale factor that takes the small star to the large star is . π The area of the large star is equal to the area of the small star π π π times ( ) = ππ . Therefore, the ratio of the area of the small ππ star to the area of the large star is ππ: ππ. 3. In β³ π¨π©πͺ, line segment π«π¬ connects two sides of the triangle and is parallel to line segment π©πͺ. If the area of β³ π¨π©πͺ is ππ and π©πͺ = ππ«π¬, find the area of β³ π¨π«π¬. The smaller triangle is similar to the larger triangle with a scale factor of π π π triangle is ( ) = π . So, the area of the small π π the area of the large triangle. π π π ππ«ππ(π¨π«π¬) = (ππ) ππ«ππ(π¨π«π¬) = π The area of β³ π¨π«π¬ is π square units. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 46 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY 4. The small star has an area of π. The large star is obtained from the small star by stretching by a factor of π in the horizontal direction and by a factor of π in the vertical direction. Find the area of the large star. The area of a figure that is scaled in perpendicular directions is equal to the area of the original figure times the product of the scale factors for each direction. The large star therefore has an area equal to the original star times the product π β π. ππ«ππ = π β π β π ππ«ππ = ππ The area of the large star is ππ square units. 5. A piece of carpet has an area of ππ π²ππ. How many square inches will this be on a scale drawing that has π π’π§. represent π π²π.? One square yard will be represented by one square inch. So, ππ square yards will be represented by ππ square inches. 6. An isosceles trapezoid has base lengths of ππ π’π§. and ππ π’π§. If the area of the larger shaded triangle is ππ π’π§π , find the area of the smaller shaded triangle. The triangles must be similar by AA criterion, so the smaller triangle is the result of a similarity transformation of the larger triangle including a dilation with a scale factor of ππ ππ π = . By the scaling π principle for area, the area of the smaller triangle must be equal to the area of the larger triangle times the square of the scale factor used: π π π ππ«ππ(π¬π¦ππ₯π₯ ππ«π’ππ§π π₯π) = ( ) β ππ«ππ(π₯ππ«π π ππ«π’ππ§π π₯π) π π ππ«ππ(π¬π¦ππ₯π₯ ππ«π’ππ§π π₯π) = (ππ) ππ«ππ(π¬π¦ππ₯π₯ ππ«π’ππ§π π₯π) = ππ The area of the smaller triangle with base ππ π’π§. is ππ π’π§π . 7. Μ Μ Μ Μ . The lengths of certain Triangle π¨π©πΆ has a line segment π¨β²π©β² connecting two of its sides so that Μ Μ Μ Μ Μ Μ π¨β² π©β² β₯ π¨π© segments are given. Find the ratio of the area of triangle πΆπ¨β²π©β² to the area of the quadrilateral π¨π©π©β²π¨β². O 4 3 B' A' 8 6 B A β³ πΆπ¨β² π©β² ~ β³ πΆπ¨π©. The area of β³ πΆπ¨β² π©β² is So, the area of the quadrilateral is π π π π π π π π of the area of the area of β³ πΆπ¨π© because ( π π π π π ) =( ) = . π+π π π of the area of β³ πΆπ¨π©. The ratio of the area of β³ πΆπ¨β² π©β² to the area of the quadrilateral π¨π©π©β² π¨β² is : , or π: π. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 47 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM M3 GEOMETRY 8. A square region πΊ is scaled parallel to one side by a scale factor of π, π β π, and is scaled in a perpendicular direction by a scale factor of one-third of π to yield its image πΊβ². What is the ratio of the area of πΊ to the area of πΊβ²? π Let the sides of square πΊ be π. Therefore, the resulting scaled image would have lengths ππ and ππ. Then the area of square πΊ would be ππ , and the area of πΊβ² would be π π π π ππ(ππ) or π π (ππ)πor π π π ππ π π . π π The ratio of areas of πΊ to πΊβ² is then ππ : ππ ππ or π: ππ or π: ππ . 9. Figure π»β² is the image of figure π» that has been scaled horizontally by a scale factor of π and vertically by a scale π factor of . If the area of π»β² is ππ square units, what is the area of figure π»? π π β π β ππ«ππ(π») π π ππ = ππ«ππ(π») π ππ«ππ(π»β² ) = π β ππ = ππ«ππ(π») π ππ = ππ«ππ(π») The area of π» is ππ square units. 10. What is the effect on the area of a rectangle if β¦ a. Its height is doubled and base left unchanged? The area would double. b. Its base and height are both doubled? The area would quadruple. c. Its base is doubled and height cut in half? The area would remain unchanged. Lesson 3: The Scaling Principle for Area This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M3-TE-1.3.0-08.2015 48 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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