{"id":77173,"date":"2022-06-13T21:31:07","date_gmt":"2022-06-13T16:01:07","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=77173"},"modified":"2022-05-10T14:14:16","modified_gmt":"2022-05-10T08:44:16","slug":"eureka-math-grade-7-module-4-lesson-14","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-7-module-4-lesson-14\/","title":{"rendered":"Eureka Math Grade 7 Module 4 Lesson 14 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 4 Lesson 14 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 4 Lesson 14 Example Answer Key<\/h3>\n

Example 1.
\nThe distance around the entire small boat is 28.4 units. The larger figure is a scale drawing of the smaller drawing of the boat. State the scale factor as a percent, and then use the scale factor to find the distance around the scale drawing.
\n\"Engage
\nAnswer:
\nScale factor:
\nHorizontal distance of the smaller boat: 8 units Vertical sail distance of smaller boat: 6 units
\nHorizontal distance of the larger boat: 22 units Vertical sail distance of larger boat: 16.5 units
\nScale factor: Quantity = Percent \u00d7 Whole
\nSmaller boat is the whole.
\nTotal Distance:
\nDistance around smaller boat = 28.4 units
\nDistance around larger boat = 28.4(275%) = 28.4(2.75) = 78.1
\nThe distance around the larger boat is 78.1 units.
\nLength in larger = Percent \u00d7 Length in smaller
\n22 = P \u00d7 8
\n\\(\\frac{22}{8}\\) = 2.75 = 275%<\/p>\n

Length in larger = Percent \u00d7 Length in smaller
\n16.5 = P \u00d7 6
\n\\(\\frac{16.5}{6}\\) = 2.75 = 275%<\/p>\n

Example 2: Time to Garden
\n\"Engage
\nSherry designed her garden as shown in the diagram above. The distance between any two consecutive vertical grid lines is 1 foot, and the distance between any two consecutive horizontal grid lines is also 1 foot. Therefore, each grid square has an area of one square foot. After designing the garden, Sherry decided to actually build the garden 75% of the size represented in the diagram.
\na. What are the outside dimensions shown in the blueprint?
\nAnswer:
\nBlueprint dimensions: Length: 26 boxes = 26 ft.
\nWidth: 12 boxes = 12 ft.<\/p>\n

b. What will the overall dimensions be in the actual garden? Write an equation to find the dimensions. How does the problem relate to the scale factor?
\nAnswer:
\nActual garden dimensions (75% of blueprint): 19.5 ft. \u00d7 9 ft.
\nLength: (26 ft.)(0.75) = 19.5 ft.
\nWidth: (12 ft.)(0.75) = 9 ft.
\nSince the scale factor was given as 75%, each dimension of the actual garden should be 75% of the original corresponding dimension. The actual length of the garden,19.5 ft., is 75% of 26 ft., and the actual width of the garden, 9 ft., is 75% of 12 ft.<\/p>\n

c. If Sherry plans to use a wire fence to divide each section of the garden, how much fence does she need?
\nAnswer:
\nDimensions of the blueprint:
\n\"Engage
\nTotal amount of wire needed for the blueprint:
\n26(4)+12(2)+4.5(4)+14 = 160
\nThe amount of wire needed is 160 ft.
\nNew dimensions of actual garden:
\nLength: 19.5 ft. (from part (b))
\nWidth: 9 ft. (from part (b))
\nInside borders: 4.5(0.75) = 3.375; 3.375 ft.
\n14(0.75) = 10.5; 10.5 ft.
\nThe dimensions of the inside borders are 3.375 ft. by 10.5 ft.<\/p>\n

Total wire with new dimensions:
\n19.5(4)+9(2)+3.375(4)+10.5 = 120
\nOR
\n160(0.75) = 120
\nTotal wire with new dimensions is 120 ft.
\nSimpler way: 75% of 160 ft. is 120 ft.<\/p>\n

d. If the fence costs $3.25 per foot plus 7% sales tax, how much would the fence cost in total?
\nAnswer:
\n3.25(120) = 390
\n390(1.07) = 417.30
\nThe total cost is $417.30.<\/p>\n

Example 3.
\nRace Car #2 is a scale drawing of Race Car #1. The measurement from the front of Race Car #1 to the back of Race Car #1 is 12 feet, while the measurement from the front of Race Car #2 to the back of Race Car #2 is 39 feet. If the height of Race Car #1 is 4 feet, find the scale factor, and write an equation to find the height of Race Car #2. Explain what each part of the equation represents in the situation.
\n\"Engage
\nAnswer:
\nScale Factor: The larger race car is a scale drawing of the smaller. Therefore, the smaller race car is the whole in the relationship.
\nQuantity = Percent \u00d7 Whole
\nLarger = Percent \u00d7 Smaller
\n39 = Percent \u00d7 12
\n\\(\\frac{39}{12}\\) = 3.25 = 325%
\nHeight: 4(3.25) = 13
\nThe height of Race Car #2 is 13 ft.
\nThe equation shows that the smaller height, 4 ft., multiplied by the scale factor of 3.25, equals the larger height, 13 ft.<\/p>\n

Eureka Math Grade 7 Module 4 Lesson 14 Exercise Answer Key<\/h3>\n

Exercise 1.
\nThe length of the longer path is 32.4 units. The shorter path is a scale drawing of the longer path. Find the length of the shorter path, and explain how you arrived at your answer.
\n\"Engage
\nAnswer:
\nFirst, determine the scale factor. Since the smaller path is a reduction of the original drawing, the scale factor should be less than 100%. Since the smaller path is a scale drawing of the larger, the larger path is the whole in the relationship.
\nQuantity = Percent \u00d7 Whole
\nTo determine the scale factor, compare the horizontal segments of the smaller path to the larger path.
\nSmaller = Percent \u00d7 Larger
\n2 = Percent \u00d7 6
\n\\(\\frac{2}{6}\\) = \\(\\frac{1}{3}\\) = 33 \\(\\frac{1}{3}\\)%
\nTo determine the length of the smaller path, multiply the length of the larger path by the scale factor.
\n32.4(\\(\\frac{1}{3}\\)) = 10.8
\nThe length of the shorter path is 10.8 units.<\/p>\n

Exercise 2.
\nDetermine the scale factor, and write an equation that relates the height of side A in Drawing 1 and the height of side B in Drawing 2 to the scale factor. The height of side A is 1.1 cm. Explain how the equation illustrates the relationship.
\n\"Engage
\nAnswer:
\nEquation: 1.1(scale factor) = height of side B in Drawing 2
\nFirst find the scale factor:
\nQuantity = Percent \u00d7 Whole
\nDrawing 2 = Percent \u00d7 Drawing 1
\n3.3 = Percent \u00d7 2
\n\\(\\frac{3.3}{2}\\) = 1.65 = 165%
\nEquation: (1.1)(1.65) = 1.815
\nThe height of side B in Drawing 2 is 1.815 cm.
\nOnce we determine the scale factor, we can write an equation to find the unknown height of side B in Drawing 2 by multiplying the scale factor by the corresponding height in the original drawing.<\/p>\n

Exercise 3.
\nThe length of a rectangular picture is 8 inches, and the picture is to be reduced to be 45 \\(\\frac{1}{2}\\)% of the original picture. Write an equation that relates the lengths of each picture. Explain how the equation illustrates the relationship.
\nAnswer:
\n8(0.455) = 3.64
\nThe length of the reduced picture is 3.64 in. The equation shows that the length of the reduced picture, 3.64, is equal to the original length, 8, multiplied by the scale factor, 0.455.<\/p>\n

Eureka Math Grade 7 Module 4 Lesson 14 Problem Set Answer Key<\/h3>\n

Question 1.
\nThe smaller train is a scale drawing of the larger train. If the length of the tire rod connecting the three tires of the larger train, as shown below, is 36 inches, write an equation to find the length of the tire rod of the smaller train. Interpret your solution in the context of the problem.
\n\"Eureka
\nAnswer:
\nScale factor:
\nSmaller = Percent \u00d7 Larger
\n6 = Percent \u00d7 16
\n\\(\\frac{6}{16}\\) = 0.375 = 37.5%
\nTire rod of smaller train: (36)(0.375) = 13.5
\nThe length of the tire rod of the smaller train is 13.5 in.
\nSince the scale drawing is smaller than the original, the corresponding tire rod is the same percent smaller as the windows. Therefore, finding the scale factor using the windows of the trains allows us to then use the scale factor to find all other corresponding lengths.<\/p>\n

Question 2.
\nThe larger arrow is a scale drawing of the smaller arrow. If the distance around the smaller arrow is 25.66 units. What is the distance around the larger arrow? Use an equation to find the distance and interpret your solution in the context of the problem.
\n\"Eureka
\nAnswer:
\nHorizontal distance of smaller arrow: 8 units
\nHorizontal distance of larger arrow: 12 units
\nScale factor:
\nLarger = Percent \u00d7 Smaller
\n12 = Percent \u00d7 8
\n\\(\\frac{12}{8}\\) = 1.5 = 150%
\nDistance around larger arrow:
\n(25.66)(1.5) = 38.49
\nThe distance around the larger arrow is 38.49 units.
\nAn equation where the distance of the smaller arrow is multiplied by the scale factor results in the distance around the larger arrow.<\/p>\n

Question 3.
\nThe smaller drawing below is a scale drawing of the larger. The distance around the larger drawing is 39.4 units. Using an equation, find the distance around the smaller drawing.
\n\"Eureka
\nAnswer:
\nVertical distance of larger drawing: 10 units
\nVertical distance of smaller drawing: 4 units
\nScale factor:
\nSmaller = Percent \u00d7 Larger
\n4 = Percent \u00d7 10
\n\\(\\frac{4}{10}\\) = 0.4 = 40%
\nTotal distance:
\n(39.4)(0.4) = 15.76
\nThe total distance around the smaller drawing is 15.76 units.<\/p>\n

Question 4.
\nThe figure is a diagram of a model rocket and is a two-dimensional scale drawing of an actual rocket. The length of a model rocket is 2.5 feet, and the wing span is 1.25 feet. If the length of an actual rocket is 184 feet, use an equation to find the wing span of the actual rocket.
\n\"Eureka
\nAnswer:
\nLength of actual rocket: 184 ft.
\nLength of model rocket: 2.5 ft.
\nScale Factor:
\nActual = Percent \u00d7 Model
\n184 = Percent \u00d7 2.5
\n\\(\\frac{184}{2.5}\\) = 73.60 = 7,360%<\/p>\n

Wing span:
\nModel rocket wing span: 1.25 ft.
\nActual rocket wing span : (1.25)(73.60) = 92
\nThe wing span of the actual rocket is 92 ft.<\/p>\n

Eureka Math Grade 7 Module 4 Lesson 14 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nEach of the designs shown below is to be displayed in a window using strands of white lights. The smaller design requires 225 feet of lights. How many feet of lights does the enlarged design require? Support your answer by showing all work and stating the scale factor used in your solution.
\n\"Eureka
\nAnswer:
\nScale Factor:
\nBottom horizontal distance of the smaller design: 8
\nBottom horizontal distance of the larger design: 16
\nThe smaller design represents the whole since we are going from the smaller to the larger.
\nQuantity = Percent \u00d7 Whole
\nLarger = Percent \u00d7 Smaller
\n16 = Percent \u00d7 8
\n\\(\\frac{16}{8}\\) = 2 = 200%
\nNumber of feet of lights needed for the larger design:
\n225 ft.(200%) = 225 ft.(2) = 450 ft.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 4 Lesson 14 Answer Key Eureka Math Grade 7 Module 4 Lesson 14 Example Answer Key Example 1. The distance around the entire small boat is 28.4 units. The larger figure is a scale drawing of the smaller drawing of the boat. State the scale factor as a … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/77173"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=77173"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/77173\/revisions"}],"predecessor-version":[{"id":137783,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/77173\/revisions\/137783"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=77173"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=77173"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=77173"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}