\n
![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Example-Answer-Key-1.png\")
\nAnswer:
\n\u2192 What information are we given in this problem?
\nWe have been given the area of the base and the volume.<\/p>\n
\u2192 How can we use the information to determine the height?
\nWe know that the area of the base times the height gives the volume. Since we already have the volume, we can do the opposite and divide to get the height.<\/p>\n
Notice that the number for the volume is less than the number for the area.
\n\u2192 What does that tell us about the height?
\nIf the product of the area of the base and the height is less than the area, we know that the height must be less than 1.<\/p>\n
\u2192 Calculate the height by solving a one-step equation.
\nVolume = Area of the base \u00d7 height
\nV = bh
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Example-Answer-Key-2.png\")
\nThe height of the sandbox is \\(\\frac{3}{4}\\) ft.<\/p>\n
\u2192 We could also calculate the height using the equation Height = Volume \u00f7 Area of the base. Solve using this equation to determine if the height will be the same.
\nHeight = Volume \u00f7 Area of the base
\nh = V \u00f7 b
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Example-Answer-Key-3.png\")
\nThe height of the sandbox is \\(\\frac{3}{4}\\) ft.<\/p>\n
b. The sandbox was filled with sand, but after the kids played, some of the sand spilled out. Now, the sand is at a height of \\(\\frac{1}{2}\\) ft. Determine the volume of the sand in the sandbox after the children played in it.
\nAnswer:<\/p>\n
\u2192 What new information have we been given in this problem?
\nThis means that the sandbox is not totally filled. Therefore, the volume of sand used is not the same as the volume of the sandbox.<\/p>\n
\u2192 How will we determine the volume of the sand? Example 2:<\/p>\n A special-order sandbox has been created for children to use as an archeological digging area at the zoo. Determine the volume of the sandbox. \u2192 Describe this three-dimensional figure. \u2192 Or, I could think of it as a piece in front and a piece behind. \u2192 How can we determine the volume of this figure? \u2192 Does it matter which way we divide the shape when we calculate the volume? \u2192 At this point, you can divide the class in half and have each half determine the volume using one of the described methods. \u2192 Volume of the prism on the right = l w h. V = \\(\\frac{9}{4}\\) m \u00d7 \\(\\frac{13}{3}\\) m \u00d7 \\(\\frac{1}{5}\\) m<\/p>\n V = \\(\\frac{117}{60}\\) m<\/p>\n V = \\(\\frac{39}{20}\\) m3<\/sup><\/p>\n Total volume = volume of left + volume of right<\/p>\n Total volume = \\(\\frac{22}{20}\\) m3<\/sup> + \\(\\frac{39}{20}\\) m3<\/sup><\/p>\n Total volume = \\(\\frac{61}{20}\\) m3<\/sup> = 3\\(\\frac{1}{20}\\) m3<\/sup><\/p>\n If the shape is divided into a figure with a piece in front and piece behind, we have the following:<\/p>\n Volume of the back piece = l w h Volume of the front piece = l w h V = \\(\\frac{63}{60}\\) m3 = 1\\(\\frac{3}{60}\\) m = 1\\(\\frac{1}{20}\\) m3<\/sup> \u2192 What do you notice about the volumes determined in each method? Exercise 1.<\/p>\n a. The volume of the rectangular prism is \\(\\frac{36}{15}\\) yd3<\/sup>. Determine the missing measurement using a one-step equation. \\(\\frac{36}{15}\\) \u00f7 \\(\\frac{12}{15}\\) = \\(\\frac{12}{15}\\)h \u00f7 \\(\\frac{12}{15}\\)<\/p>\n \\(\\frac{36}{12}\\) = h b. The volume of the box is \\(\\frac{45}{6}\\) m3<\/sup>. Determine the area of the base using a one-step equation. Exercise 2. a. Determine how much water the tank can hold. b. Determine how much water is already in the tank. c. How much more water is needed to fill the tank? Volume needed to fill = l w h Exercise 3. a. b. Another possible solution: Question 1. Question 2. Question 3. Volume of water = l w h Question 4. Volume of back piece = \\(\\frac{3}{96}\\) m3<\/sup><\/p>\n Volume of front piece = l w h<\/p>\n Volume of front piece = (\\(\\frac{1}{4}\\) m) (\\(\\frac{1}{8}\\) m) (\\(\\frac{1}{3}\\) m) Question 5. V = (1 in.) (\\(\\frac{3}{2}\\) in.) (\\(\\frac{5}{4}\\) in.) + (3 in.) (\\(\\frac{5}{2}\\) in.) (\\(\\frac{1}{4}\\) in.) Question 6. a. Write an equation to represent the volume of the composite figure. b. Use your equation to calculate the volume of the composite figure. Question 1. Volume of water = l w h Question 2. Volume of front piece = l w h Engage NY Eureka Math Grade 6 Module 5 Lesson 14 Answer Key Eureka Math Grade 6 Module 5 Lesson 14 Example Answer Key Example 1: a. The area of the base of a sandbox is 9 ft2. The volume of the sandbox is 7 ft3. Determine the height of the sandbox. Answer: \u2192 What information … 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\/84713"}],"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=84713"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/84713\/revisions"}],"predecessor-version":[{"id":137240,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/84713\/revisions\/137240"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=84713"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=84713"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=84713"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nTo determine the volume of the sand, I use the area of the base of the sandbox, but I use the height of \\(\\frac{1}{2}\\)ft. instead of the height of the sandbox.
\nVolume = Area of the base \u00d7 height
\nVolume = 9\\(\\frac{1}{2}\\) ft2<\/sup> \u00d7 \\(\\frac{1}{2}\\) ft.
\nVolume = \\(\\frac{19}{2}\\) ft2<\/sup> \u00d7 \\(\\frac{1}{2}\\) ft.
\nVolume = \\(\\frac{19}{4}\\) ft3<\/sup>
\nVolume = 4\\(\\frac{3}{4}\\) ft3<\/sup>
\nThe volume of the sand is 4\\(\\frac{3}{4}\\) ft3<\/sup>.<\/p>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Example-Answer-Key-4.png\")
\nAnswer:<\/p>\n
\nThis figure looks like two rectangular prisms that have been placed together to form one large prism.
\nI could think of it as a piece on the left and a piece on the right.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Example-Answer-Key-5.png\")
<\/p>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Example-Answer-Key-6.png\")
<\/p>\n
\nWe can find the volume of each piece and then add the volumes together to get the volume of the entire figure.<\/p>\n
\nAnswers will vary.<\/p>\n
\nIf the shape is divided into a figure on the left and a figure on the right, we would have the following:
\nVolume of prism on the left = l w h.
\nV = 2\\(\\frac{3}{4}\\) m \u00d7 2 m \u00d7 \\(\\frac{1}{5}\\) m
\nV = \\(\\frac{11}{4}\\) m \u00d7 2 m \u00d7 \\(\\frac{1}{5}\\) m
\nV = \\(\\frac{22}{20}\\) m3<\/sup><\/p>\n
\nV = 2\\(\\frac{1}{4}\\) m \u00d7 4\\(\\frac{1}{3}\\) m \u00d7 \\(\\frac{1}{5}\\) m<\/p>\n
\nV = 5 m \u00d7 2 m \u00d7 \\(\\frac{1}{5}\\) m
\nV = 2m3<\/sup><\/p>\n
\nV = 2\\(\\frac{1}{4}\\) m \u00d7 2\\(\\frac{1}{3}\\) m \u00d7 \\(\\frac{1}{5}\\) m
\nV = \\(\\frac{9}{4}\\) m \u00d7 \\(\\frac{7}{3}\\) m \u00d7 \\(\\frac{1}{5}\\) m<\/p>\n
\nTotal volume = volume of back + volume of front
\nTotal volume = 2 m3<\/sup> + 1\\(\\frac{1}{20}\\) m3<\/sup>
\nTotal volume = 3\\(\\frac{1}{20}\\) m3<\/sup><\/p>\n
\nThe volume calculated with each method is the same. It does not matter how we break up the shape. We still get the same volume.<\/p>\nEureka Math Grade 6 Module 5 Lesson 14 Exercise Answer Key<\/h3>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-7.png\")
\nAnswer:
\nV = bh
\n\\(\\frac{36}{15}\\) = \\(\\frac{4}{5}\\) h<\/p>\n
\n3 = h
\nThe height is 3 yd.<\/p>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-8.png\")
\nAnswer:
\nV = bh
\n\\(\\frac{45}{6}\\) = b(\\(\\frac{2}{9}\\))
\n\\(\\frac{45}{6}\\) \u00f7 \\(\\frac{27}{6}\\) = b(\\(\\frac{27}{6}\\)) \u00f7 \\(\\frac{7}{6}\\)
\n\\(\\frac{45}{27}\\) = b
\n\\(\\frac{5}{3}\\) = b
\nThe area of the base is \\(\\frac{45}{6}\\) m2<\/sup>.<\/p>\n
\nMarissa\u2019s fish tank needs to be filled with more water.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-9.png\")
<\/p>\n
\nAnswer:
\nVolume of entire tank = l w h
\nV = (\\(\\frac{3}{4}\\) m) (\\(\\frac{1}{4}\\) m) (\\(\\frac{3}{5}\\) m)
\nV = \\(\\frac{9}{80}\\) m3<\/sup>.<\/p>\n
\nAnswer:
\nVolume of water in the tank = l w h
\nV = (\\(\\frac{3}{4}\\) m) (\\(\\frac{1}{4}\\) m) (\\(\\frac{3}{8}\\) m)
\nV = \\(\\frac{9}{128}\\) m3<\/sup>.<\/p>\n
\nAnswer:
\nHeight of empty part of tank:
\nh = \\(\\frac{3}{5}\\) m – \\(\\frac{3}{8}\\) m
\n= \\(\\frac{24}{40}\\) m – \\(\\frac{15}{40}\\) m
\n= \\(\\frac{9}{40}\\) m<\/p>\n
\nV = (\\(\\frac{3}{4}\\) m) (\\(\\frac{1}{4}\\) m) (\\(\\frac{9}{40}\\) m)
\nV = \\(\\frac{27}{640}\\) m3<\/sup><\/p>\n
\nDetermine the volume of composite figures.<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-10.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-11.png\")
<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-12.png\")
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-13.png\")
<\/p>\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-14.png\")
<\/p>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exercise-Answer-Key-15.png\")
<\/p>\nEureka Math Grade 6 Module 5 Lesson 14 Problem Set Answer Key<\/h3>\n
\nThe volume of a rectangular prism is \\(\\frac{21}{12}\\) ft3<\/sup>, the height of the prism is \\(\\frac{3}{4}\\) ft. Determine the area of the base.
\nAnswer:
\nV = bh
\n\\(\\frac{21}{12}\\) = b (\\(\\frac{3}{4}\\))
\n\\(\\frac{21}{9}\\) = b
\nThe area of the base is \\(\\frac{21}{9}\\) ft2<\/sup> OR \\(\\frac{7}{3}\\) ft2<\/sup>.<\/p>\n
\nThe volume of a rectangular prism is \\(\\frac{10}{21}\\) ft3<\/sup>. The area of the base is \\(\\frac{2}{3}\\) ft2. Determine the height of the rectangular prism.
\nAnswer:
\nHeight = Volume \u00f7 Area of the base
\nHelght = \\(\\frac{10}{21}\\) ft3<\/sup> \u00f7 \\(\\frac{2}{3}\\) ft2<\/sup>
\nHeight = \\(\\frac{10}{21}\\) ft3<\/sup> \u00f7 \\(\\frac{14}{21}\\) ft2<\/sup>
\nHeight = \\(\\frac{10}{14}\\) ft. OR \\(\\frac{5}{7}\\) ft.<\/p>\n
\nDetermine the volume of the space in the tank that still needs to be tilled with water if the water is ft. deep.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Problem-Set-Answer-Key-16.png\")
\nAnswer:
\nVolume of tank = l w h
\nVolume of tank = (5 ft.) (1\\(\\frac{2}{3}\\) ft.) (2 ft.)
\nVolume of tank = \\(\\frac{50}{3}\\) ft3<\/sup>
\nVolume to be filled = \\(\\frac{50}{3}\\) ft3<\/sup> – \\(\\frac{25}{9}\\) ft3<\/sup>
\nVolume to be filled = \\(\\frac{150}{9}\\) ft3<\/sup> – \\(\\frac{25}{9}\\) ft3<\/sup>
\nVolume to be filled = \\(\\frac{125}{9}\\) ft3<\/sup><\/p>\n
\nVoIumeofwater = (5 ft.) (1\\(\\frac{2}{3}\\) ft.) (\\(\\frac{1}{3}\\) ft.)
\nVolume of water = \\(\\frac{25}{9}\\) ft3<\/sup><\/p>\n
\nDetermine the volume of composite figure.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Problem-Set-Answer-Key-17.png\")
\nAnswer:
\nVolume of back piece = l w h
\nVolume of back piece = (\\(\\frac{3}{4}\\) m) (\\(\\frac{1}{8}\\) m) (\\(\\frac{1}{3}\\) m)<\/p>\n
\nVolume of front piece = \\(\\frac{1}{96}\\) m3<\/sup>
\nTotal volume = \\(\\frac{3}{96}\\) m3<\/sup> + \\(\\frac{1}{96}\\) m3<\/sup> = \\(\\frac{4}{96}\\) m3<\/sup> OR \\(\\frac{1}{24}\\)m3<\/sup><\/p>\n
\nDetermine the volume of the composite figure.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Problem-Set-Answer-Key-18.png\")
\nAnswer:
\nV = (1 in.) (1\\(\\frac{1}{2}\\) in.) (1\\(\\frac{1}{4}\\) in.) + (3 in.) (2\\(\\frac{1}{2}\\) in.) (\\(\\frac{1}{4}\\) in.)<\/p>\n
\nV = \\(\\frac{15}{8}\\) in3<\/sup> + \\(\\frac{15}{8}\\) in3<\/sup>
\nV = \\(\\frac{30}{8}\\) in3<\/sup> = 3\\(\\frac{6}{8}\\) in3<\/sup> OR 3\\(\\frac{3}{4}\\) in3<\/sup><\/p>\n
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Problem-Set-Answer-Key-19.png\")
<\/p>\n
\nAnswer:
\nV = (3\\(\\frac{1}{2}\\) m \u00d7 2 m \u00d7 1\\(\\frac{1}{4}\\) m) + (3\\(\\frac{3}{4}\\) m \u00d7 2 m \u00d7 2\\(\\frac{1}{4}\\) m)<\/p>\n
\nAnswer:
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Problem-Set-Answer-Key-20.png\")
<\/p>\nEureka Math Grade 6 Module 5 Lesson 14 Exit Ticket Answer Key<\/h3>\n
\nDetermine the volume of the water that would be needed to fill the rest of the tank.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exit-Ticket-Answer-Key-21.png\")
\nAnswer:
\nVolume of tank = l w h
\nVolume of tank = (1\\(\\frac{1}{4}\\) m) (\\(\\frac{1}{2}\\) m) (\\(\\frac{3}{4}\\) m)
\nVolume of tank = \\(\\frac{15}{32}\\) m3<\/sup><\/p>\n
\nVolume of water = (1\\(\\frac{1}{4}\\) m) (\\(\\frac{1}{2}\\) m) (\\(\\frac{1}{2}\\) m)
\nVolume of waler = \\(\\frac{5}{16}\\) m3<\/sup> = \\(\\frac{10}{32}\\) m
\nRemaining water needed = \\(\\frac{15}{32}\\) m3<\/sup> – \\(\\frac{10}{32}\\) m3<\/sup> = \\(\\frac{5}{32}\\)m3<\/sup><\/p>\n
\nDetermine the volume of the composite figure.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-6-Module-5-Lesson-14-Exit-Ticket-Answer-Key-22.png\")
Answer:
\nVolume of back piece = l w h
\nVolume of back piece = (\\(\\frac{5}{8}\\) ft.) (\\(\\frac{1}{3}\\) ft.) (\\(\\frac{1}{4}\\) ft.)
\nVolume of back piece = \\(\\frac{5}{96}\\) ft3<\/sup><\/p>\n
\nVolume of front piece = (\\(\\frac{1}{4}\\) ft.) (\\(\\frac{1}{6}\\) ft.) (\\(\\frac{1}{4}\\) ft.)
\nVolume of front piece = \\(\\frac{1}{96}\\) ft3<\/sup>
\nTotal volume = \\(\\frac{5}{96}\\) ft3<\/sup> + \\(\\frac{1}{96}\\) ft3<\/sup> = \\(\\frac{6}{96}\\) ft3<\/sup> = \\(\\frac{2}{32}\\) ft3<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"