{"id":72818,"date":"2021-03-24T19:33:45","date_gmt":"2021-03-24T14:03:45","guid":{"rendered":"https:\/\/ccssmathanswers.com\/?p=72818"},"modified":"2022-04-01T10:27:40","modified_gmt":"2022-04-01T04:57:40","slug":"eureka-math-grade-7-module-6-lesson-24","status":"publish","type":"post","link":"https:\/\/bigideasmathanswers.com\/eureka-math-grade-7-module-6-lesson-24\/","title":{"rendered":"Eureka Math Grade 7 Module 6 Lesson 24 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 6 Lesson 24 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 6 Lesson 24 Example Answer Key<\/h3>\n

Example 1.
\nDetermine the surface area of the image.
\n\"Engage
\nAnswer:
\nSurface area of top or bottom prism:
\nLateral sides = 4(12 in. \u00d7 3 in.) = 144 in2<\/sup>
\nBase face = 12 in. \u00d7 12 in. = 144 in2<\/sup>
\nBase face with hole = 12 in. \u00d7 12 in. – 4 in. \u00d7 4 in.
\n= 128 in2<\/sup>
\nThere are two of these, making up 832 in2<\/sup>.
\nSurface area of middle prism:
\nLateral sides: = 4(4 in. \u00d7 8 in.) = 128 in2<\/sup>
\nSurface area: 832 in2<\/sup> + 128 in2<\/sup> = 960 in2<\/sup><\/p>\n

\u2192 Describe the method you used to determine the surface area.
\nAnswers will vary. I determined the surface area of each prism separately and added them together. Then, I subtracted the area of the sections that were covered by another prism.<\/p>\n

\u2192 If all three prisms were separate, would the sum of their surface areas be the same as the surface area you determined in this example?
\nNo, if the prisms were separate, there would be more surfaces shown. The three separate prisms would have a greater surface area than this example. The area would be greater by the area of four 4 in. \u00d7 4 in. squares (64 in2<\/sup>).<\/p>\n

Example 2.
\na. Determine the surface area of the cube.
\n\"Engage
\nb. A square hole with a side length of 4 inches is cut through the cube. Determine the new surface area.
\n\"Engage
\nAnswer:
\na. Surface area = 6s2<\/sup>
\nSA = 6(12 in.)2<\/sup>
\nSA = 6(144 in2<\/sup>)
\nSA = 864 in2<\/sup>
\n\u2192 Explain how 6(12 in.)2<\/sup> represents the surface area of the cube.
\nThe area of one face, one square with a side length of 12 in., is (12 in.)2<\/sup>, and so a total area of all six faces is 6(12 in.)2<\/sup>.<\/p>\n

b. Area of interior lateral sides
\n= 4(12 in. \u00d7 4 in.)
\n= 192 in2<\/sup><\/p>\n

Surface area of cube with holes
\n= 6 (12 in.)2<\/sup> – 2(4 in. \u00d7 4 in.) + 4(12 in. \u00d7 4 in.)
\n= 864 in2<\/sup> – 32 in2<\/sup> + 192 in2<\/sup>
\n= 1,024 in2<\/sup>
\n\u2192 How does cutting a hole in the cube change the surface area?
\nWe have to subtract the area of the square at the surface from each end.
\nWe also have to add the area of each of the interior faces to the total surface area.<\/p>\n

\u2192 What happens to the surfaces that now show inside the cube?
\nThese are now part of the surface area.<\/p>\n

\u2192 What is the shape of the piece that was removed from the cube?
\nA rectangular prism was cut out of the cube with the following dimensions: 4 in. \u00d7 4 in. \u00d7 12 in.<\/p>\n

\u2192 How can we use this to help us determine the new total surface area?
\nWe can find the surface area of the cube and the surface area of the rectangular prism, but we will have to subtract the area of the square bases from the cube and also exclude these bases in the area of the rectangular prism.<\/p>\n

\u2192 Why is the surface area larger when holes have been cut into the cube?
\nThere are more surfaces showing now. All of the surfaces need to be included in the surface area.<\/p>\n

\u2192 Explain how the expression 6(12 in.)2<\/sup> – 2(4 in. \u00d7 4 in.) + 4(12 in. \u00d7 4 in.) represents the surface area of the cube with the hole.
\nFrom the total surface area of a whole (uncut) cube, 6(12 in.)2<\/sup>, the area of the bases (the cuts made to the surface of the cube) are subtracted: 6(12 in.)2<\/sup> – 2(4 in. \u00d7 4 in.). To this expression we add the area of the four lateral faces of the cutout prism, 4(12 in. \u00d7 4 in.). The complete expression then is 6(12 in.)2<\/sup> – 2(4 in. \u00d7 4 in.) + 4(12 in. \u00d7 4 in.)<\/p>\n

Example 3.
\nA right rectangular pyramid has a square base with a side length of 10 inches. The surface area of the pyramid is 260 in2<\/sup>. Find the height of the four lateral triangular faces.
\nAnswer:
\nArea of base = 10 in. \u00d7 10 in. = 100 in2<\/sup>
\nArea of the four faces = 260 in2<\/sup> – 100 in2<\/sup> = 160 in2<\/sup>
\nThe total area of the four faces is 160 in2<\/sup>.
\nTherefore, the area of each triangular face is 40 in2<\/sup>.
\nArea of lateral side = \\(\\frac{1}{2}\\)bh
\n40 in2<\/sup> = \\(\\frac{1}{2}\\)(10 in.)h
\n40 in2<\/sup> = (5 in)h
\nh = 8 in.
\nThe height of each lateral triangular face is 8 inches.<\/p>\n

\u2192 What strategies could you use to help you solve this problem?
\nI could draw a picture of the pyramid and label the sides so that I can visualize what the problem is asking me to do.
\n\"Engage
\n\u2192 What information have we been given? How can we use the information?
\nWe know the total surface area, and we know the length of the sides of the square.
\nWe can use the length of the sides of the square to give us the area of the square base.<\/p>\n

\u2192 How does the area of the base help us determine the height of the lateral triangular face?
\nFirst, we can subtract the area of the base from the total surface area in order to determine what is left for the lateral sides.
\nNext, we can divide the remaining area by 4 to get the area of just one triangular face.
\nFinally, we can work backward. We have the area of the triangle, and we know the base is 10″ in.” , so we can solve for the height.<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 24 Exercise Answer Key<\/h3>\n

Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise.
\nExercise 1.
\n\"Engage
\nAnswer:
\nTop and bottom = 2(18 m \u00d7 5 m) = 180 m2<\/sup>
\nExtra interior sides = 2(5 m \u00d7 7 m) = 70 m2<\/sup>
\nLeft and right sides = 2(5 m \u00d7 12 m) = 120 m2<\/sup>
\nFront and back sides = 2((18 m \u00d7 12 m) – (8 m \u00d7 7 m))
\n= 2(216 m2<\/sup> – 56 m2<\/sup> )
\n= 2(160 m2<\/sup> )
\n= 320 m2<\/sup>
\nSurface area = 180 m2<\/sup> + 70 m2<\/sup> + 120 m2<\/sup> + 320 m2<\/sup>
\n= 690 m2<\/sup><\/p>\n

Exercise 2.
\nIn addition to your calculation, explain how the surface area of the following figure was determined.
\n\"Engage
\nAnswer:
\nThe surface area of the prism is found by taking the sum of the areas of the trapezoidal front and back and the areas of the back of the four different – sized rectangles that make up the lateral faces.
\nArea top = 25 cm \u00d7 8 cm
\n= 200 cm2<\/sup>
\nArea bottom = 24 cm \u00d7 8 cm
\n= 192 cm2<\/sup>
\nArea sides = (23 cm \u00d7 8 cm) + (16 cm \u00d7 8 cm)
\n= 312 cm2<\/sup>
\nArea front and back = 2(\\(\\frac{1}{2}\\)(16 cm + 23 cm)(24 cm))
\n= 2(468 cm2<\/sup> )
\n= 936 cm2<\/sup>
\nSurface area = 200 cm2<\/sup> + 192 cm2<\/sup> + 312 cm2<\/sup> + 936 cm2<\/sup>
\n= 1,640 cm2<\/sup><\/p>\n

Exercise 3.
\n\"Engage
\nAnswer:
\nSurface area of one of the prisms on the sides:
\nArea of front and back = 2(2 in. \u00d7 14 in.)
\n= 56 in2<\/sup>
\nArea of top and bottom = 2(2 in. \u00d7 10 in.)
\n= 40 in2<\/sup>
\nArea of side = 14 in. \u00d7 10 in. = 140 in2<\/sup>
\nArea of side with hole = 14 in. \u00d7 10 in. – 3 in. \u00d7 3 in.
\n= 131 in2<\/sup><\/p>\n

There are two such rectangular prisms, so the surface area of both is 734 in2<\/sup>.
\nSurface area of middle prism:
\nArea of front and back = 2(3 in. \u00d7 12<\/sup>in.) = 72 in2<\/sup>
\nArea of sides = 2(3 in. \u00d7 12<\/sup> in.) = 72 in2<\/sup>
\nSurface area of middle prism = 72 in2<\/sup> + 72 in2<\/sup> = 144 in2<\/sup>
\nThe total surface area of the figure is 734 in2<\/sup> + 144 in2<\/sup> = 878 in2<\/sup>.<\/p>\n

Exercise 4.
\nIn addition to your calculation, explain how the surface area was determined.
\n\"Engage
\nAnswer:
\nThe surface area of the prism is found by taking the area of the base of the rectangular prism and the area of its four lateral faces and adding it to the area of the four lateral faces of the pyramid.
\nArea of base = 9 ft. \u00d7 9 ft.
\n= 81 ft2<\/sup>
\nArea of rectangular sides = 4(9 ft. \u00d7 5 ft.)
\n= 180 ft2<\/sup>
\nArea of triangular sides = 4(\\(\\frac{1}{2}\\)(9 ft.)(6 ft.))
\n= 108 ft2<\/sup>
\nSurface area = 81 ft2<\/sup> + 180 ft2<\/sup> + 108 ft2<\/sup>
\n= 369 ft2<\/sup><\/p>\n

Exercise 5.
\nA hexagonal prism has the following base and has a height of 8 units. Determine the surface area of the prism.
\n\"Engage
\nAnswer:
\nArea of bases = 2(48 + 6 + 20 + 10) = 168
\nArea of 4 unit sides = 4(5 \u00d7 8)
\n= 160
\nArea of other sides = (4 \u00d7 8) + (12 \u00d7 8)
\n= 128
\nSurface area = 168 + 160 + 128
\n= 456
\nThe surface area of the hexagonal prism is 456 units2<\/sup>.<\/p>\n

Exercise 6.
\nDetermine the surface area of each figure.
\na.
\n\"Engage
\nb. A cube with a square hole with 3 m side lengths has been cut through the cube.
\n\"Engage
\nc. A second square hole with 3 m side lengths has been cut through the cube.
\n\"Engage
\nAnswer:
\na. SA = 6s2<\/sup>
\n= 6(9 m)2<\/sup>
\n= 6(81m2<\/sup>)
\n= 486 m2<\/sup><\/p>\n

b. Lateral sides of the hole = 4(9 m \u00d7 3 m) = 108 m2<\/sup>
\nSurface area of cube with holes = 486 m2<\/sup> – 2(3 m \u00d7 3 m) + 108 m2<\/sup>
\n= 576 m2<\/sup><\/p>\n

c. Surface area = 576 m2<\/sup> – 4(3 m \u00d7 3 m) + 2(4(3 m \u00d7 3 m))
\n= 612 m2<\/sup><\/p>\n

Exercise 7.
\nThe figure below shows 28 cubes with an edge length of 1 unit. Determine the surface area.
\n\"Engage
\nAnswer:
\nArea top and bottom = 24 units2<\/sup>
\nArea sides = 18 units2<\/sup>
\nArea front and back = 28 units2<\/sup>
\nSurface area
\n= 24 units2<\/sup> + 18 units 2<\/sup> + 28 units2<\/sup>
\n= 70 units2<\/sup><\/p>\n

Exercise 8.
\nThe base rectangle of a right rectangular prism is 4 ft. \u00d7 6 ft. The surface area is 288 ft2<\/sup>. Find the height. Let h be the height in feet.
\nAnswer:
\nArea of one base: 4 ft. \u00d7 6 ft. = 24 ft2<\/sup>
\nArea of two bases: 2(24 ft2<\/sup>) = 48 ft2<\/sup>
\nNumeric area of four lateral faces: 288 ft2<\/sup> – 48 ft2<\/sup> = 240 ft2<\/sup>
\nAlgebraic area of four lateral faces: 2(6h + 4h) ft2<\/sup>
\nSolve for h.
\n2(6h + 4h) = 240
\n10h = 120
\nh = 12
\nThe height is 12 feet.<\/p>\n

Eureka Math Grade 7 Module 6 Lesson 24 Problem Set Answer Key<\/h3>\n

Determine the surface area of each figure.
\nQuestion 1.
\nIn addition to the calculation of the surface area, describe how you found the surface area.
\n\"Eureka
\nAnswer:
\nArea of top = 28 cm \u00d7 13 cm
\n= 364 cm2<\/sup>
\nArea of bottom = 28 cm \u00d7 12 cm
\n= 336 cm2<\/sup>
\nArea of left and right sides = 28 cm \u00d7 20 cm + 15 cm \u00d7 28 cm
\n= 980 cm2<\/sup>
\nArea of front and back sides = 2(12 cm \u00d7 15 cm) + \\(\\frac{1}{2}\\)(5 cm \u00d7 12 cm)
\n= 2(180 cm2<\/sup> + 30 cm2<\/sup> )
\n= 2(210 cm2<\/sup> )
\n= 420 cm2<\/sup>
\nSurface area = 364 cm2<\/sup> + 336 cm2<\/sup> + 980 cm2<\/sup> + 420 cm2<\/sup>
\n= 2,100 cm2<\/sup>
\nSplit the area of the two trapezoidal bases into triangles and rectangles, take the sum of the areas, and then add the areas of the four different – sized rectangles that make up the lateral faces.<\/p>\n

Question 2.
\n\"Eureka
\nAnswer:
\nArea of front and back = 2(18.4 in. \u00d7 8.6 in.) = 316.48 in2<\/sup>
\nArea of sides = 2(8.6 in. \u00d7 24 in.) = 412.8 in2<\/sup>
\nArea of top and bottom = 2(18.4 in. \u00d7 24 in.) – (4.8 in. \u00d7 2.2 in.)
\n= 2(441.6 in2<\/sup> – 10.56 in2<\/sup>)
\n= 2(431.04 in2<\/sup>)
\n= 862.08 in2<\/sup>
\nSurface area = 316.48 in2<\/sup> + 412.8 in2<\/sup> + 862.08 in2<\/sup>
\n= 1,591.36 in2<\/sup><\/p>\n

Question 3.
\n\"Eureka
\nAnswer:
\nArea of front and back = 2(\\(\\frac{1}{2}\\)(32 m + 16 m)15 m)
\n= 720 m2<\/sup>
\nArea of top = 16 m \u00d7 36 m
\n= 576 m2<\/sup>
\nArea of left and right sides = 2(17 m \u00d7 36 m)
\n= 2(612 m2<\/sup> )
\n= 1,224 m2<\/sup>
\nArea of bottom = 32 m \u00d7 36 m
\n= 1,152 m2<\/sup>
\nSurface area = 720 m2<\/sup> + 1,152 m2<\/sup> + 1,224 m2<\/sup> + 576 m2<\/sup>
\n= 3,672 m2<\/sup><\/p>\n

Question 4.
\nDetermine the surface area after two square holes with a side length of 2 m are cut through the solid figure composed of two rectangular prisms.
\n\"Eureka
\nAnswer:
\nSurface area of top prism before the hole is drilled:
\nArea of top = 4 m \u00d7 5 m
\n= 20 m2<\/sup>
\nArea of front and back = 2(4 m \u00d7 5 m)
\n= 40 m2<\/sup>
\nArea of sides = 2(5 m \u00d7 5 m)
\n= 50 m2<\/sup>
\nSurface area of bottom prism before the hole is drilled:
\nArea of top = 10 m \u00d7 10 m – 20 m2<\/sup>
\n= 80 m2<\/sup>
\nArea of bottom = 10 m \u00d7 10 m
\n= 100 m2<\/sup>
\nArea of front and back = 2(10 m \u00d7 3 m)
\n= 60 m2<\/sup>
\nArea of sides = 2(10 m \u00d7 3 m)
\n= 60 m2<\/sup><\/p>\n

Surface area of interiors:
\nArea of interiors = 4(2 m \u00d7 4 m) + 4(2 m \u00d7 3 m)
\n= 56 m2<\/sup>
\nSurface Area = 110 m2<\/sup> + 300 m2<\/sup> + 56 m2<\/sup> – 16 m2<\/sup>
\n= 450 m2<\/sup><\/p>\n

Question 5.
\nThe base of a right prism is shown below. Determine the surface area if the height of the prism is 10 cm. Explain how you determined the surface area.
\n\"Eureka
\nAnswer:
\nTake the sum of the areas of the two bases made up of two right triangles, and add to it the sum of the areas of the lateral faces made up of rectangles of different sizes.
\nArea of sides = (20 cm \u00d7 10 cm) + (15 cm \u00d7 10 cm) + (24 cm \u00d7 10 cm) + (7 cm \u00d7 10 cm)
\n= 200 cm2<\/sup> + 150 cm2<\/sup> + 240 cm2<\/sup> + 70 cm2<\/sup>
\n= 660 cm2<\/sup>
\nArea of bases = 2(\\(\\frac{1}{2}\\)(7 cm \u00d7 24 cm) + 1\/2 (20 cm \u00d7 15 cm))
\n= (7 cm \u00d7 24 cm) + (20 cm \u00d7 15 cm)
\n= 168 cm2<\/sup> + 300 cm2<\/sup>
\n= 468 cm2<\/sup>
\nSurface area = 660 cm2<\/sup> + 468 cm2<\/sup>
\n= 1,128 cm2<\/sup><\/p>\n

Eureka Math Grade 7 Module 6 Lesson 24 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nDetermine the surface area of the right rectangular prism after the two square holes have been cut. Explain how you determined the surface area.
\n\"Eureka
\nAnswer:
\nArea of top and bottom = 2(15 cm \u00d7 6 cm)
\n= 180 cm2<\/sup>
\nArea of sides = 2(6 cm \u00d7 8 cm)
\n= 96 cm2<\/sup>
\nArea of front and back = 2(15 cm \u00d7 8 cm) – 4(5 cm \u00d7 5 cm)
\n= 140 cm2<\/sup>
\nArea inside = 8(5 cm \u00d7 6 cm)
\n= 240 cm2<\/sup>
\nSurface area = 180 cm2<\/sup> + 96 cm2<\/sup> + 140 cm2<\/sup> + 240 cm2<\/sup>
\n= 656 cm2<\/sup>
\nTake the sum of the areas of the four lateral faces and the two bases of the main rectangular prism, and subtract the areas of the four square cuts from the area of the front and back of the main rectangular prism. Finally, add the lateral faces of the prisms that were cut out of the main prism.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 6 Lesson 24 Answer Key Eureka Math Grade 7 Module 6 Lesson 24 Example Answer Key Example 1. Determine the surface area of the image. Answer: Surface area of top or bottom prism: Lateral sides = 4(12 in. \u00d7 3 in.) = 144 in2 Base face = 12 … 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\/72818"}],"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=72818"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72818\/revisions"}],"predecessor-version":[{"id":92185,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/72818\/revisions\/92185"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=72818"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=72818"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=72818"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}