{"id":81543,"date":"2022-06-08T19:44:51","date_gmt":"2022-06-08T14:14:51","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=81543"},"modified":"2022-05-10T14:18:33","modified_gmt":"2022-05-10T08:48:33","slug":"eureka-math-grade-6-module-5-lesson-5","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-6-module-5-lesson-5\/","title":{"rendered":"Eureka Math Grade 6 Module 5 Lesson 5 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Grade 6 Module 5 Lesson 5 Answer Key<\/h2>\n

Eureka Math Grade 6 Module 5 Lesson 5 Exercise Answer Key<\/h3>\n

Opening Exercise:<\/p>\n

Here is an aerial view of a woodlot.<\/p>\n

\"Eureka<\/p>\n

Question 1.
\nIf AB = 10 units, FE = 8 units, AF = 6 units, and DE = 7 units, find the lengths of the other two sides.
\nDC =
\nBC =
\nAnswer:
\nDC = 2 units
\nBC = 13 units<\/p>\n

Question 2.
\nIf DC = 10 units, FE = 30 units, AF = 28 units, and BC = 54 units, find the lengths of the other two sides.
\nAB =
\nDE =
\nAnswer:
\nAB = 40 units
\nDE = 26 units<\/p>\n

Eureka Math Grade 6 Module 5 Lesson 5 Example Answer Key<\/h3>\n

Example 1: Decomposing Polygons into Rectangles<\/p>\n

The Intermediate School is producing a play that needs a special stage built. A diagram of the stage is shown below (not to scale).<\/p>\n

Question a.
\nOn the first diagram, divide the stage into three rectangles using two horizontal lines. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage.
\nAnswer:
\nDimensions: 2 m by 4m, 2 m by 4 m, and 7 m by 5 m
\nArea: 2 m \u00d7 4m = 8 m2<\/sup>, 2 m \u00d7 4 m = 8 m2<\/sup>, 7 m \u00d7 5 m = 35 m2<\/sup>
\nTotal: 8 m2<\/sup> + 8 m2<\/sup> + 35 m2<\/sup> = 51 m2<\/sup><\/p>\n

Question b.
\nOn the second diagram, divide the stage into three rectangles using two vertical lines. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage.
\nAnswer:
\nDimensions: 2 m by 9 m, 2 m by 9 m, and 3 m by 5 m
\nArea: 2 m \u00d7 9 m = 18 m \u00d7 2 m \u00d7 9 m = 18 m2<\/sup>, 3 m \u00d7 5m = 15 m2<\/sup>
\nTotal: 51 m2<\/sup><\/p>\n

Question c.
\nOn the third diagram, divide the stage into three rectangles using one horizontal line and one vertical line. Find the dimensions of these rectangles, and calculate the area of each. Then, find the total area of the stage.
\n\"Eureka
\nAnswer:
\nDimensions: 2 m by 9 m, 2 m by 4 m, and 5 m by 5 m
\nArea: 2 m \u00d7 9 m = 18 m2<\/sup>, 2 m \u00d7 4 m = 8 m2<\/sup>, 5 m \u00d7 5 m = 25 m2<\/sup>
\nTotal: 51 m2<\/sup><\/p>\n

Question d.
\nThink of this as a large rectangle with a piece removed.
\n\"Eureka<\/p>\n

i. What are the dimensions of the large rectangle and the small rectangle?
\nAnswer:
\nDimensions: 9 m by 7 m and 3 m by 4 m<\/p>\n

ii. What are the areas of the two rectangles?
\nAnswer:
\nArea: 9 m \u00d7 7 m = 63 m2<\/sup>, 3 m \u00d7 4 m = 12 m2<\/sup><\/p>\n

iii. What operation Is needed to find the area of the original figure?
\nAnswer:
\nSubtraction<\/p>\n

iv. What is the difference in area between the two rectangles?
\nAnswer:
\n63 m2<\/sup> – 12 m2<\/sup> = 51 m2<\/sup><\/p>\n

v. What do you notice about your answers to (a), (b), (c), and (d)? 7 m
\nAnswer:
\nThe area is the same.<\/p>\n

vi. Why do you think this is true?
\nAnswer:
\nNo matter how we decompose the figure, the total area is the sum of its ports. Even if we take the area around the figure and subtract the part that is not included, the area of the figure remains the same, 51 m2<\/sup>.<\/p>\n

\"Eureka<\/p>\n

Area of Rectangle 1: b . h
\n2m . 4m = 8m2<\/sup>
\nArea of Rectangle 2: b . h
\n2 m 4m = 8 m2<\/sup>
\nArea of Rectangle 3: b . h
\n7m 5m 35 m2<\/sup>
\nArea of Polygon: 8m2<\/sup> + 8m2<\/sup> + 35 m2<\/sup> = 51 m2<\/sup><\/p>\n

\"Eureka<\/p>\n

Area of Rectangle 1: b . h
\n9m . 2m = 18 m2<\/sup>
\nArea of Rectangle 2: b . h
\n9m . 2m =18 m2<\/sup>
\nArea of Rectangle 3: b . h
\n3m . 5m = 15 m2<\/sup>
\nArea of Polygon: 18 m2<\/sup> + 18 m2<\/sup> + 15 m2<\/sup> = 51 m2<\/sup><\/p>\n

\"Eureka<\/p>\n

Area of Rectangle: b . h
\n9m . 7 m = 63 m2<\/sup>
\nArea of Missing Rectangle 2: b . h
\n3m . 4m = 12 m2<\/sup>
\nArea of Polygon: 63 m2<\/sup> – 12 m2<\/sup> = 51 m2<\/sup><\/p>\n

Example 2: Decomposing Polygons into Rectangles and Triangles<\/p>\n

Parallelogram ABCD is part of a large solar power collector. The base measures 6 m and the height is 4 m.
\n\"Eureka<\/p>\n

Question a.
\nDraw a diagonal from A to C. Find the area of both triangles ABC and ACD.
\nAnswer:
\n\"Eureka<\/p>\n

Triangle ABC
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (6 m) (4 m)
\nA = 12 m2<\/sup>.<\/p>\n

Triangle ACD
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (6m)(4m)
\nA = 12 m2<\/sup><\/p>\n

Question b.
\nFind the area of both triangles ABD and BCD. Then find the area of the trapezoid.
\nAnswer:
\n\"Eureka<\/p>\n

Triangle ABD
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (8m)(5m)
\nA = 20 m2<\/sup><\/p>\n

Triangle BCD
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (4 m) (5 m)
\nA = 10 m2<\/sup><\/p>\n

A = 20 m2<\/sup> + 10 m2<\/sup> = 30 m2<\/sup><\/p>\n

Question c.
\nHow else could we find this area?
\nAnswer:
\nWe could consider the rectangle that surrounds the trapezoid. Find the area of that rectangle, and then subtract the area of both right triangles.
\n\"Eureka<\/p>\n

Area of Rectangle;
\nA = bh
\nA = (8 m) (5 m)
\nA = 40 m2<\/sup><\/p>\n

Triangle 1
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (3 m) (5 m)
\nA = 7.5 m2<\/sup><\/p>\n

Triangle 2
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) (1 m) (5 m)
\nA = 2.5 m2<\/sup><\/p>\n

A = 40 m2<\/sup> – 7.5 m2<\/sup> – 2.5 m2<\/sup> = 30 m2<\/sup> (or)<\/p>\n

A = 40 m2<\/sup> – (7.5 m2<\/sup> + 2.5 m2<\/sup>) = 30 m2<\/sup><\/p>\n

Eureka Math Grade 6 Module 5 Lesson 5 Problem Set Answer Key<\/h3>\n

Question 1.
\nIf AB = 20 units, FE = 12 units, AF = 9 units, and DE = 12 units, find the length of the other two sides. Then, find the area of the irregular polygon.
\n\"Eureka
\nAnswer:
\nCD = 8 units, BC = 21 units, Area = 276 square units.<\/p>\n

Question 2.
\nIf DC = 1.9 cm, FE = 5.6 cm, AF = 4.8 cm, and BC = 10.9 cm, find the length of the other two sides. Then, find the area of the irregular polygon.
\n\"Eureka
\nAnswer:
\nAB = 7.5 cm, DE = 6.1 cm, Area = 47.59 cm2<\/sup><\/p>\n

Question 3.
\nDetermine the area of the trapezoid below. The trapezoid is not drawn to scale.
\n\"Eureka
\nAnswer:
\nArea of Triangle 1
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) \u00d7 22 m \u00d7 13 m
\nA = 198 m2<\/sup><\/p>\n

Area of Triangle 2
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) \u00d7 3 m \u00d7 18 m
\nA = 27 m2<\/sup><\/p>\n

Area of Trapezoid = Area of Triangle 1 + Area of Triangle 2
\nArea = 198 m2<\/sup> + 27 m2<\/sup> = 225 m2<\/sup><\/p>\n

Question 4.
\nDetermine the area of the shaded isosceles trapezoid below. The image is not drawn to scale.
\n\"Eureka
\nAnswer:
\nArea of Rectangle
\nA = bh
\nA = 18 m \u00d7 12 m
\nA = 216 m2<\/sup><\/p>\n

Area of Triangles 1 and 2
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) \u00d7 7.5 m \u00d7 12 m
\nA = 45 m2<\/sup><\/p>\n

Area of Trapezoid = Area of Rectangle – Area of Triangle 1 – Area of Triangle 2
\nA = 216 m2<\/sup> – 45 m2<\/sup> – 45 m2<\/sup> = 126 m2<\/sup><\/p>\n

Question 5.
\nHere is a sketch of a wall that needs to be painted:
\n\"Eureka<\/p>\n

a. The windows and door will not be painted. Calculate the area of the wall that will be painted.
\nAnswer:
\nWhole wall: 12 ft. \u00d7 8 ft. = 96 ft2<\/sup>
\nWindow: 2 ft. \u00d7 2 ft. = 4 ft2<\/sup>
\nThere are two identical windows, 4 ft2<\/sup> \u00d7 2 = 8 ft2<\/sup>
\nDoor: 6 ft. \u00d7 3 ft. = 18 ft2<\/sup>
\n96 ft2<\/sup> – 8 ft2<\/sup> – 18 ft2<\/sup> = 70 ft2<\/sup><\/p>\n

b. If a quart of Extra-Thick Gooey Sparkle paint covers 30 ft2<\/sup>. how many quarts must be purchased for the painting job?
\nAnswer:
\n70 \u00f7 30 = 2\\(\\frac{1}{3}\\)
\nTherefore, 3 quarts must be purchased.<\/p>\n

Question 6.
\nThe figure below shows a floor plan of a new apartment. New carpeting has been ordered, which will cover the living room and bedroom but not the kitchen or bathroom. Determine the carpeted area by composing or decomposing in two different ways, and then explain why they are equivalent.<\/p>\n

\"Eureka
\nAnswer:
\nAnswers will vary. Sample student responses are shown.
\nBedroom: 15 ft. \u00d7 25 ft. = 375 ft2<\/sup>
\nLiving room: 35 ft. \u00d7 20 ft. = 700 ft2<\/sup>
\nSum of bedroom and living room: 375 ft2<\/sup> + 700 ft2<\/sup> = 1,075 ft2<\/sup><\/p>\n

Alternatively, the whole apartment is 45 ft. \u00d7 35 ft. = 1,575 ft2<\/sup>
\nSubtracting the kitchen and bath (300 ft2<\/sup> and 200 ft2<\/sup>) still gives 1,075 ft2<\/sup>.
\nThe two areas are equivalent because they both represent the area of the living room and bedroom.<\/p>\n

Eureka Math Grade 6 Module 5 Lesson 5 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nFind the missing dimensions of the figure below, and then find the area. The figure is not drawn to scale.
\n\"Eureka
\nAnswer:
\nThe solution can be any of these below.
\n\"Eureka<\/p>\n

Question 2.
\nFind the area of the parallelogram below by decomposing it into two triangles. The figure is not drawn to scale.
\n\"Eureka
\nAnswer:
\nArea of Triangle 1
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) \u00d7 8 mi.\u00d7 10 mi.
\nA = 40 mi2<\/sup><\/p>\n

Area of Triangle 2
\nA = \\(\\frac{1}{2}\\) bh
\nA = \\(\\frac{1}{2}\\) \u00d7 8 mi. \u00d7 10 mi.
\nA = 40 mi2<\/sup><\/p>\n

Area of Parallelogram = Area of Triangle 1 + Area of Triangle 2
\nA = 40 mi2<\/sup> + 40 mi2<\/sup> = 80 mi2<\/sup>
\nThe area of the parallelogram is 80 mi2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math Grade 6 Module 5 Lesson 5 Answer Key Eureka Math Grade 6 Module 5 Lesson 5 Exercise Answer Key Opening Exercise: Here is an aerial view of a woodlot. Question 1. If AB = 10 units, FE = 8 units, AF = 6 units, and DE = 7 units, find the … 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\/81543"}],"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=81543"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/81543\/revisions"}],"predecessor-version":[{"id":137558,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/81543\/revisions\/137558"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=81543"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=81543"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=81543"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}