\na. Calculate the surface area of the rectangular prism.
\n
![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-6-Lesson-23-Example-Answer-Key-1.png\")
\nb. Imagine that a piece of the rectangular prism is removed. Determine the surface area of both pieces.
\n
![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-6-Lesson-23-Example-Answer-Key-2.png\")
\nc. How is the surface area in part (a) related to the surface area in part (b)?
\nAnswer:
\na. Surface area:
\n2(3 in. \u00d7 6 in.) + 2(3 in. \u00d7 12 in.) + 2(6 in. \u00d7 12 in.)
\n= 2(18 in2<\/sup>) + 2(36 in2<\/sup>) + 2(72 in2<\/sup> )
\n= 36 in2<\/sup> + 72 in2<\/sup> + 144 in2<\/sup>
\n= 252 in2<\/sup><\/p>\n
b. ![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-6-Lesson-23-Example-Answer-Key-3.png\")
\n\u2192 How did you determine the surface area of the shape on the left?
\n\u2192 I was able to calculate the area of the sides that are rectangles using length times width. For the two bases that are C – shaped, I used the area of the original top and bottom and subtracted the piece that was taken off.<\/p>\n
c. If I add the surface area of both figures, I will get more than the surface area of the original shape. Opening Exercise Surface area: \u2192 Explain the process you used to determine the surface area of the pyramid. \u2192 The surface area of a pyramid is the union of its base region and all its lateral faces. \u2192 Would this method work for any prism or pyramid? Determine the surface area of the right prisms. Exercise 2. Exercise 3. Area for front: 7 ft. \u00d7 3 ft. = 21 ft2<\/sup><\/p>\n Area of corner cutout: (2 ft. \u00d7 3 ft.) + (1 ft. \u00d7 3 ft.) = 9 ft2<\/sup> Surface area: 124 ft2<\/sup> + 24 ft2<\/sup> + 21 ft2<\/sup> + 9 ft2<\/sup> + 18 ft2<\/sup> + 24 ft2<\/sup> = 220 ft2<\/sup><\/p>\n Exercise 4. Surface area: 110 m2<\/sup> + 300 m2<\/sup> = 410 m2<\/sup><\/p>\n Exercise 5. Determine the surface area of the figures. Question 2. Area of sides: 2(8 ft. \u00d7 2 ft.) = 32 ft2<\/sup><\/p>\n Area of top and bottom: 2(9 ft. \u00d7 8 ft.) – 2(4 ft. \u00d7 3 ft.) Question 3. Surface area: 272 in2<\/sup> + 720 in2<\/sup> = 992 in2<\/sup><\/p>\n Question 4. Area that can be seen from left and right: \\(\\frac{1}{2}\\)bh Question 5. Surface area: 120 cm2<\/sup> + 200 cm2<\/sup> + 375 cm2<\/sup> + 425 cm2<\/sup> = 1,120 cm2<\/sup><\/p>\n Determine and explain how to find the surface area of the following right prisms. Question 2. Engage NY Eureka Math 7th Grade Module 6 Lesson 23 Answer Key Eureka Math Grade 7 Module 6 Lesson 23 Example Answer Key Example 1. a. Calculate the surface area of the rectangular prism. b. Imagine that a piece of the rectangular prism is removed. Determine the surface area of both pieces. c. How is … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72692"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=72692"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72692\/revisions"}],"predecessor-version":[{"id":92186,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72692\/revisions\/92186"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=72692"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=72692"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=72692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n234 in2<\/sup> + 90 in2<\/sup> = 324 in2<\/sup>
\n324 in2<\/sup> – 252 in2<\/sup> = 72 in2<\/sup>
\n72 in2<\/sup> is twice the area of the region where the two pieces fit together.
\nThere are 72 more square inches when the prisms are separated.<\/p>\nEureka Math Grade 7 Module 6 Lesson 23 Exercise Answer Key<\/h3>\n
\nCalculate the surface area of the square pyramid.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-23-Exercise-Answer-Key-1.png\")
\nAnswer:
\nArea of the square base:
\ns2<\/sup> = (8 cm)2<\/sup>
\n= 64 cm2<\/sup>
\nArea of the triangular lateral sides:
\n\\(\\frac{1}{2}\\) bh = \\(\\frac{1}{2}\\)(8 cm)(5 cm)
\n= 20 cm2<\/sup>
\nThere are four lateral sides. So, the area of all 4 triangles is 80 cm2<\/sup>.<\/p>\n
\n(8 cm \u2219 8 cm) + 4(1\/2 (8 cm \u2219 5 cm)) = 64 cm2<\/sup> + 80 cm2<\/sup> = 144 cm2<\/sup><\/p>\n
\nAnswers will vary. I drew a net to determine the area of each side and added the areas together.<\/p>\n
\n\u2192 Explain how (8 cm \u2219 8 cm) + 4(\\(\\frac{1}{2}\\)(8 cm\u22195 cm)) represents the surface area.
\nThe area of the square with side lengths of 8 cm is represented by (8 cm \u2219 8 cm), and the area of the four lateral faces with base lengths of 8 cm and heights of 5 cm is represented by 4(\\(\\frac{1}{2}\\) (8 cm\u22195 cm)).<\/p>\n
\nAnswers will vary. Calculating the area of each face, including bases, will determine the surface area even if the areas are determined in different orders, by using a formula or net, or any other method.<\/p>\n
\nExercise 1.
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-23-Exercise-Answer-Key-2.png\")
\nAnswer:
\nArea of top and bottom: 2(\\(\\frac{1}{2}\\)(15 ft.\u2006 \u00d7 8 ft.) )
\n= 15 ft. \u00d7 8 ft. = 120 ft2<\/sup>
\nArea of front: 15 ft. \u00d7 20 ft. = 300 ft2<\/sup>
\nArea that can be seen from left: 17 ft. \u00d7 20 ft. = 340 ft2<\/sup>
\nArea that can be seen from right: 8 ft. \u00d7 20 ft. = 160 ft2<\/sup>
\nSurface area: 120 ft2<\/sup> + 300 ft2<\/sup> + 340 ft2<\/sup> + 160 ft2<\/sup> = 920 ft2<\/sup><\/p>\n
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-23-Exercise-Answer-Key-3.png\")
\nAnswer:
\nArea of front and back: 2(\\(\\frac{1}{2}\\)(10 yd. + 4 yd.)4 yd.)
\n= 14 yd. \u00d7 4 yd. = 56 yd2<\/sup>
\nArea of top: 4 yd. \u00d7 15 yd. = 60 yd2<\/sup>
\nArea that can be seen from left and right: 2(5 yd. \u00d7 15 yd.)
\n= 2(75 yd2<\/sup>) = 150 yd2<\/sup>
\nArea of bottom: 10 yd. \u00d7 15 yd. = 150 yd2<\/sup>
\nSurface area: 56 yd2<\/sup> + 60 yd2<\/sup> + 150 yd2<\/sup> + 150 yd2<\/sup> = 416 yd2<\/sup><\/p>\n
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-23-Exercise-Answer-Key-4.png\")
\nAnswer:
\nArea of top and bottom: 2((8 ft. \u00d7 6 ft.) + (7 ft. \u00d7 2 ft.))
\n= 2(48 ft2<\/sup> + 14 ft2<\/sup> )
\n= 2(62 ft2<\/sup> ) = 124 ft2<\/sup>
\nArea for back: 8 ft. \u00d7 3 ft. = 24 ft2<\/sup><\/p>\n
\nArea of right side: 6 ft. \u00d7 3 ft. = 18 ft2<\/sup>
\nArea of left side: 8 ft. \u00d7 3 ft. = 24 ft2<\/sup><\/p>\n
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-23-Exercise-Answer-Key-5.png\")
\nAnswer:
\nSurface area of top prism:
\nArea of top: 4 m \u00d7 5 m = 20 m2<\/sup>
\nArea of front and back sides: 2(4 m \u00d7 5 m) = 40 m2<\/sup>
\nArea of left and right sides: 2(5 m \u00d7 5 m) = 50 m2<\/sup>
\nTotal surface area of top prism: 20 m2<\/sup> + 40 m2<\/sup> + 50 m2<\/sup> = 110 m2<\/sup>
\nSurface area of bottom prism:
\nArea of top: 10 m \u00d7 10 m – 20 m2<\/sup> = 80 m2<\/sup>
\nArea of bottom: 10 m \u00d7 10 m = 100 m2<\/sup>
\nArea of front and back sides: 2(10 m \u00d7 3 m) = 60 m2<\/sup>
\nArea of left and right sides: 2(10 m \u00d7 3 m) = 60 m2<\/sup>
\nTotal surface area of bottom prism: 80 m2<\/sup> + 100 m2<\/sup> + 60 m2<\/sup> + 60 m2<\/sup>
\n= 300 m2<\/sup><\/p>\n
\n![\"Engage](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-23-Exercise-Answer-Key-6.png\")
\nAnswer:
\nArea of top and bottom faces: 2(10 in. \u00d7 2 in.) + 2(6 in. \u00d7 2 in.)
\n= 40 in2<\/sup> + 24 in2<\/sup>
\n= 64 in2<\/sup>
\nArea of lateral faces: 2(9 in. \u00d7 2 in.) + 2(6 in. \u00d7 9 in.) + 2(9 in. \u00d7 10 in.)
\n= 36 in2<\/sup> + 108 in2<\/sup> + 180 in2<\/sup>
\n= 324 in2<\/sup>
\nSurface area:
\n64 in2<\/sup> + 324 in2<\/sup> = 388 in2<\/sup><\/p>\nEureka Math Grade 7 Module 6 Lesson 23 Problem Set Answer Key<\/h3>\n
\nQuestion 1.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-23-Problem-Set-Answer-Key-1.png\")
\nAnswer:
\nArea of top and bottom: 2(9 cm \u00d7 4 cm) = 72 cm2<\/sup>
\nArea of left and right sides: 2(4 cm \u00d7 9 cm) = 72 cm2<\/sup>
\nArea of front and back: 2(9 cm \u00d7 4 cm) + 2(4.5 cm \u00d7 5 cm)
\n= 117 cm2<\/sup>
\nSurface area: 72 cm2<\/sup> + 72 cm2<\/sup> + 117 cm2<\/sup> = 261 cm2<\/sup><\/p>\n
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-23-Problem-Set-Answer-Key-2.png\")
\nAnswer:
\nArea of front and back: 2(9 ft. \u00d7 2 ft.) = 36 ft2<\/sup><\/p>\n
\n= 144 ft2<\/sup> – 24 ft2<\/sup>
\n= 120 ft2<\/sup>
\nArea of interior sides: 2(4 ft. \u00d7 2 ft.) = 16 ft2<\/sup>
\nSurface area: 36 ft2<\/sup> + 32 ft2<\/sup> + 120 ft2<\/sup> + 16 ft2<\/sup> = 204 ft2<\/sup><\/p>\n
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-23-Problem-Set-Answer-Key-3.png\")
\nAnswer:
\nSurface area of top prism:
\nArea of top: 8 in. \u00d7 6 in. = 48 in2<\/sup>
\nArea of front and back sides: 2(6 in. \u00d7 8 in.) = 96 in2<\/sup>
\nArea of left and right sides: 2(8 in. \u00d7 8 in.) = 128 in2<\/sup>
\nTotal surface area of top prism: 48 in2<\/sup> + 96 in2<\/sup> + 128 in2<\/sup> = 272 in2<\/sup>
\nSurface area of bottom prism:
\nArea of top: 16 in. \u00d7 16 in. – 48 in2<\/sup> = 208 in2<\/sup>
\nArea of bottom: 16 in. \u00d7 16 in. = 256 in2<\/sup>
\nArea of front and back sides: 2(16 in. \u00d7 4 in.) = 128 in2<\/sup>
\nArea of left and right sides: 2(16 in. \u00d7 4 in.) = 128 in2<\/sup>
\nTotal surface area of bottom prism: 208 in2<\/sup> + 256 in2<\/sup> + 128 in2<\/sup> + 128 in2<\/sup>
\n= 720 in2<\/sup><\/p>\n
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-23-Problem-Set-Answer-Key-4.png\")
\nAnswer:
\nArea of the rectangle base: 14 ft. \u00d7 30 ft. = 420 ft2<\/sup>
\nArea of the triangular lateral sides:
\nArea of front and back: \\(\\frac{1}{2}\\)bh
\n= 2(\\(\\frac{1}{2}\\)(14 ft.)(24 ft.))
\n= 336 ft2<\/sup><\/p>\n
\n= 2(1\\(\\frac{1}{2}\\)(30 ft.)(20 ft.))
\n= 600 ft2<\/sup>
\nSurface area: 420 ft2<\/sup> + 336 ft2<\/sup> + 600 ft2<\/sup> = 1,356 ft2<\/sup><\/p>\n
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-23-Problem-Set-Answer-Key-5.png\")
\nAnswer:
\nArea of front and back: 2(\\(\\frac{1}{2}\\)(8 cm \u00d7 15 cm))
\n= 8 cm \u00d7 15 cm
\n= 120 cm2<\/sup>
\nArea of bottom: 8 cm \u00d7 25 cm
\n= 200 cm2<\/sup>
\nArea that can be seen from left side: 25 cm \u00d7 15 cm
\n= 375 cm2<\/sup>
\nArea that can be seen from right side: 25 cm \u00d7 17 cm
\n= 425 cm2<\/sup><\/p>\nEureka Math Grade 7 Module 6 Lesson 23 Exit Ticket Answer Key<\/h3>\n
\nQuestion 1.
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-6-Lesson-23-Exit-Ticket-Answer-Key-1.png\")
\nAnswer:
\nArea of top and bottom: 2(\\(\\frac{1}{2}\\)(12 ft. \u00d7 5 ft.))
\n= 12 ft. \u00d7 5 ft.
\n= 60 ft2<\/sup>
\nArea of front: 12 ft. \u00d7 15 ft.
\n= 180 ft2<\/sup>
\nArea seen from left: 13 ft. \u00d7 15 ft.
\n= 195 ft2<\/sup>
\nArea seen from right: 5 ft. \u00d7 15 ft.
\n= 75 ft2<\/sup>
\nSurface area: 60 ft2<\/sup> + 180 ft2<\/sup> + 195 ft2<\/sup> + 75 ft2<\/sup>
\n= 510 ft2<\/sup>
\nTo find the surface area of the triangular prism, I must sum the areas of two triangles (the bases that are equal in area) and the areas of three different – sized rectangles.<\/p>\n
\n![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-6-Lesson-23-Exit-Ticket-Answer-Key-2.png\")
\nAnswer:
\nArea of front and back: 2(10 ft. \u00d7 1 ft.) = 20 ft2<\/sup>
\nArea of sides: 2(10 ft. \u00d7 1 ft.) = 20 ft2<\/sup>
\nArea of top and bottom: 2(10 ft. \u00d7 5 ft.) + 2(4 ft. \u00d7 5 ft.)
\n= 100 ft2<\/sup> + 40 ft2<\/sup>
\n= 140 ft2<\/sup>
\nSurface area: 20 ft2<\/sup> + 20 ft2<\/sup> + 140 ft2<\/sup>
\n= 180 ft2<\/sup>
\nTo find the surface area of the prism, I must sum the composite area of the bases with rectangular areas of the sides of the prism.<\/p>\n","protected":false},"excerpt":{"rendered":"