\nFind the volume of the following three – dimensional object composed of two right rectangular prisms.
\n
![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-6-Lesson-26-Example-Answer-Key-1.png\")
\nAnswer:
\nVolume of object = Volume of top prism + Volume of bottom prism
\nVolume of top prism: Volume of bottom prism:
\nVolumetop prism<\/sub> = (4 m)(5 m)(5 m)
\n= 100 m3<\/sup><\/p>\n
Volumebottom prism<\/sub> = (10 m)(10 m)(3 m) There are different ways the volume of a composite figure may be calculated. If the figure is like the one shown in Example 1, where the figure can be decomposed into separate prisms and it would be impossible for the prisms to share any one dimension, the individual volumes of the decomposed prisms can be determined and then summed. If, however, the figure is similar to the figure in Exercise 1, there are two possible strategies. In Exercise 1, the figure can be decomposed into two individual prisms, but a dimension is shared between the two prisms\u2014in this case, the height. Instead of calculating the volume of each prism and then taking the sum, we can calculate the area of the entire base by decomposing it into shapes we know and then multiplying the area of the base by the height.<\/p>\n Example 2. Volume of small prism: The volume of the object is 20 \\(\\frac{1}{4}\\)cm3<\/sup>.<\/p>\n Strategy 2 Volume of object: The volume of the object is 20 \\(\\frac{1}{4}\\)cm3<\/sup>.<\/p>\n \u2192 Write a numeric expression to represent the volume of the figure in Strategy 1. \u2192 Write a numeric expression to represent the volume of the figure in Strategy 2. \u2192 How do the numeric expressions represent the problem differently? \u2192 What property allows us to show that these representations are equivalent? Example 3. b. Possible answer: Two pieces with dimensions 2.5 ft. \u2219 2 ft. \u2219 3 in. and four pieces with dimensions 2 ft. \u2219 1.25 ft. \u2219 3 in.<\/p>\n c. V1<\/sub> = 2(2.5 ft. \u2219 2 ft. \u2219 0.25 ft.) = 2.5 ft3<\/sup> d. (2.5 ft. \u2219 2 ft. \u2219 1.75 ft.) – (2 ft. \u2219 1.5 ft. \u2219 1.25 ft.) = 8.75 ft3<\/sup> – 3.75 ft3<\/sup> = 5 ft3<\/sup><\/p>\n Exercise 1. Area of Basefront prism<\/sub> = (6 in.)(2 in.) Exercise 2. Volumetriangular prism<\/sub> = Bh = (\\(\\frac{1}{2}\\) lw)h Strategy 2 Volume of object: \u2192 Write a numeric expression to represent the volume of the figure in Strategy 1. \u2192 Write a numeric expression to represent the volume of the figure in Strategy 2. \u2192 How do the numeric expressions represent the problem differently? \u2192 What property allows us to show that these representations are equivalent? Question 1. Volume of middle prism: Question 2. Volumelarge cube<\/sub> = (\\(\\frac{3}{2}\\) cm)3<\/sup> V = \\(\\frac{1}{8}\\) cm3<\/sup> + \\(\\frac{27}{8}\\) cm3<\/sup> The total volume of the object is 3\\(\\frac{1}{2}\\) cm3<\/sup>.<\/p>\n Question 3. Figure 2 b. In Figure 1, the prism is treated as two triangular prisms joined together. The volume of each triangular prism is found and then doubled, whereas in Figure 2, the prism has a base in the shape of a rhombus, and the volume is found by calculating the area of the rhomboid base and then multiplying by the height.<\/p>\n Question 4. Question 5. Question 6. Question 7. Question 8. Question 1. Engage NY Eureka Math 7th Grade Module 6 Lesson 26 Answer Key Eureka Math Grade 7 Module 6 Lesson 26 Example Answer Key Example 1. Find the volume of the following three – dimensional object composed of two right rectangular prisms. Answer: Volume of object = Volume of top prism + Volume of bottom prism … 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\/73711"}],"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=73711"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/73711\/revisions"}],"predecessor-version":[{"id":137941,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/73711\/revisions\/137941"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=73711"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=73711"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=73711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n= 300 m3<\/sup>
\nThe volume of the object is 100 m3<\/sup> + 300 m3<\/sup> = 400 m3<\/sup>.<\/p>\n
\nFind the volume of the right prism shown in the diagram whose base is the region between two right triangles. Use two different strategies.
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-7th-Grade-Module-6-Lesson-26-Example-Answer-Key-2.png\")
\nAnswer:
\nStrategy 1
\nThe volume of the right prism is equal to the difference of the volumes of the two triangular prisms.
\nVolume of object = Volumelarge prism<\/sub> – Volumesmall prism<\/sub>
\nVolume of large prism:
\nVolumelarge prism<\/sub> = (\\(\\frac{1}{2}\\) \u2219 3 cm \u2219 4 cm)4 \\(\\frac{1}{2}\\)cm
\n= 27 cm3<\/sup><\/p>\n
\nVolumesmall prism<\/sub> = (\\(\\frac{1}{2}\\) \u2219 1 \\(\\frac{1}{2}\\)cm \u2219 2 cm) 4\\(\\frac{1}{2}\\)cm
\n= 6 \\(\\frac{3}{4}\\)cm3<\/sup><\/p>\n
\nThe volume of a right prism is equal to the area of its base times its height. The base is the region between two right triangles.
\nVolume of object = Bh
\nB = Arealarge triangle<\/sub> – Areasmall triangle<\/sub>
\nArealarge triangle<\/sub> = \\(\\frac{1}{2}\\) \u2219 3 cm \u2219 4 cm = 6 cm2<\/sup>
\nAreasmall triangle<\/sub> = \\(\\frac{1}{2}\\) \u2219 1 1\\(\\frac{1}{2}\\)cm \u2219 2 cm = 1 \\(\\frac{1}{2}\\)cm2<\/sup>
\nB = 6 cm2<\/sup> – 1 \\(\\frac{1}{2}\\)cm2<\/sup> = 4 \\(\\frac{1}{2}\\)cm2<\/sup><\/p>\n
\nVolumeobject<\/sub> = Bh
\n= (4 \\(\\frac{1}{2}\\)cm2<\/sup> \u2219 4 \\(\\frac{1}{2}\\)cm)
\n= 20 \\(\\frac{1}{4}\\)cm3<\/sup><\/p>\n
\n(\\(\\frac{1}{2}\\) \u2219 3 cm \u2219 4 cm)4 \\(\\frac{1}{2}\\)cm – (\\(\\frac{1}{2}\\) \u2219 1 \\(\\frac{1}{2}\\)cm \u2219 2 cm)4 \\(\\frac{1}{2}\\)cm<\/p>\n
\n4 \\(\\frac{1}{2}\\)cm<\/p>\n
\nThe first expression is appropriate to use when the volume of the smaller prism is being subtracted away from the volume of the larger prism, whereas the second expression is used when the area of the base of the composite figure is found and then multiplied by the height to determine the volume.<\/p>\n
\nThe distributive property<\/p>\n
\nA box with a length of 2 ft., a width of 1.5 ft., and a height of 1.25 ft. contains fragile electronic equipment that is packed inside a larger box with three inches of styrofoam cushioning material on each side (above, below, left side, right side, front, and back).
\na. Give the dimensions of the larger box.
\nb. Design styrofoam right rectangular prisms that could be placed around the box to provide the cushioning (i.e., give the dimensions and how many of each size are needed).
\nc. Find the volume of the styrofoam cushioning material by adding the volumes of the right rectangular prisms in the previous question.
\nd. Find the volume of the styrofoam cushioning material by computing the difference between the volume of the larger box and the volume of the smaller box.
\nAnswer:
\na. Length 2.5 ft., width 2 ft., and height 1.75 ft.<\/p>\n
\nV2<\/sub> = 4(2 ft. \u2219 1.25 ft. \u2219 0.25 ft.) = 2.5 ft3<\/sup>
\nV1<\/sub> + V2<\/sub> = 2.5 ft3<\/sup> + 2.5 ft3<\/sup> = 5 ft3<\/sup><\/p>\nEureka Math Grade 7 Module 6 Lesson 26 Exercise Answer Key<\/h3>\n
\nFind the volume of the following three – dimensional figure composed of two right rectangular prisms.
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-26-Exercise-Answer-Key-1.png\")
\nAnswer:
\nArea of Baseback prism<\/sub> = (2 in.)( 10 in.)
\n= 20 in.2<\/sup><\/p>\n
\n= 12 in2<\/sup>
\nArea of Base = 20 in2<\/sup> + 12 in2<\/sup> = 32 in2<\/sup>
\nThe volume of the object is (9 in.)(32 in2<\/sup> ) = 288 in3<\/sup>.<\/p>\n
\nThe right trapezoidal prism is composed of a right rectangular prism joined with a right triangular prism. Find the volume of the right trapezoidal prism shown in the diagram using two different strategies.
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Engage-NY-Math-Grade-7-Module-6-Lesson-26-Exercise-Answer-Key-2.png\")
\nAnswer:
\nStrategy 1
\nThe volume of the trapezoidal prism is equal to the sum of the volumes of the rectangular and triangular prisms.
\nVolume of object = Volume of rectangular prism + Volume of triangular prism
\nVolume of rectangular prism: Volume of triangular prism:
\nVolumerectangular prism<\/sub> = Bh
\n= (lw)h
\n= (3 cm \u2219 2 cm) \u2219 1 \\(\\frac{1}{2}\\)cm
\n= 9 cm3<\/sup><\/p>\n
\n= (\\(\\frac{1}{2}\\) \u2219 3 cm \u2219 2\\(\\frac{1}{4}\\)cm) \u2219 1 \\(\\frac{1}{2}\\)cm
\n= 5 \\(\\frac{1}{16}\\) cm3<\/sup>
\nThe volume of the object is 9 cm3<\/sup> + 5 \\(\\frac{1}{16}\\) cm3<\/sup> = 14 \\(\\frac{1}{16}\\) cm3<\/sup>.<\/p>\n
\nThe volume of a right prism is equal to the area of its base times its height. The base consists of a rectangle and a triangle.
\nVolume of object = Bh
\nB = Arearectangle<\/sub> + Areatriangle<\/sub>
\nArearectangle<\/sub> = 3 cm \u2219 2 cm = 6 cm2<\/sup>
\nAreatriangle<\/sub> = \\(\\frac{1}{2}\\) \u2219 3 cm \u2219 2 \\(\\frac{1}{4}\\)cm = 3 \\(\\frac{3}{8}\\)cm2<\/sup>
\nB = 6 cm2<\/sup> + 3 \\(\\frac{3}{8}\\)cm2<\/sup> = 9 \\(\\frac{3}{8}\\)cm2<\/sup><\/p>\n
\nVolumeobject<\/sub> = Bh
\n= (9 \\(\\frac{3}{8}\\)cm2<\/sup> )(1\\(\\frac{1}{2}\\)cm)
\n= 14 \\(\\frac{1}{16}\\)cm3<\/sup>
\nThe volume of the object is 14\\(\\frac{1}{16}\\)cm3<\/sup>.<\/p>\n
\n(3 cm \u2219 2 cm) \u2219 1 \\(\\frac{1}{2}\\)cm + (\\(\\frac{1}{2}\\) \u2219 3 cm \u2219 2 \\(\\frac{1}{4}\\)cm) \u2219 1 \\(\\frac{1}{2}\\)cm<\/p>\n
\n(3 cm \u2219 2 cm + \\(\\frac{1}{2}\\) \u2219 3 cm \u2219 2 \\(\\frac{1}{4}\\)cm )(1 \\(\\frac{1}{2}\\)cm)<\/p>\n
\nThe first expression is appropriate to use when individual volumes of the decomposed figure are being added together, whereas the second expression is used when the area of the base of the composite figure is found and then multiplied by the height to determine the volume.<\/p>\n
\nThe distributive property.<\/p>\nEureka Math Grade 7 Module 6 Lesson 26 Problem Set Answer Key<\/h3>\n
\nFind the volume of the three – dimensional object composed of right rectangular prisms.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-26-Problem-Set-Answer-Key-1.png\")
\nAnswer:
\nVolumeobject<\/sub> = Volumetop and bottom prisms<\/sub> + Volumemiddle prism<\/sub>
\nVolume of top and bottom prisms:
\nV = 2(12 in. \u2219 12 in. \u2219 3 in.)
\n= 864 in3<\/sup><\/p>\n
\n= 4 in. \u2219 4 in. \u2219 8 in.
\n= 128 in3<\/sup>
\nThe volume of the object is 864 in3<\/sup> + 128 in3<\/sup> = 992 in3<\/sup>.<\/p>\n
\nA smaller cube is stacked on top of a larger cube. An edge of the smaller cube measures \\(\\frac{1}{2}\\) cm in length, while the larger cube has an edge length three times as long. What is the total volume of the object?
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-26-Problem-Set-Answer-Key-2.png\")
\nAnswer:
\nVolumeobject<\/sub> = Volumesmall cube<\/sub> + Volumelarge cube<\/sub>
\nVolumesmall cube<\/sub> = (\\(\\frac{1}{2}\\) cm)3<\/sup>
\n= \\(\\frac{1}{8}\\) cm3<\/sup><\/p>\n
\n= \\(\\frac{27}{8}\\) cm3<\/sup><\/p>\n
\n= 3 \\(\\frac{1}{2}\\) cm3<\/sup><\/p>\n
\nTwo students are finding the volume of a prism with a rhombus base but are provided different information regarding the prism. One student receives Figure 1, while the other receives Figure 2.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-26-Problem-Set-Answer-Key-3.png\")
\na. Find the expression that represents the volume in each case; show that the volumes are equal.
\nb. How does each calculation differ in the context of how the prism is viewed?
\nAnswer:
\na. Figure 1
\n2(\\(\\frac{1}{2}\\) \u2219 14.6 mm \u2219 3.6 mm) \u2219 9 mm
\n473.04 mm3<\/sup><\/p>\n
\n((8 mm \u2219 6.57 mm) \u2219 9 mm)
\n473.04 mm3<\/sup><\/p>\n
\nFind the volume of wood needed to construct the following side table composed of right rectangular prisms.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-26-Problem-Set-Answer-Key-4.png\")
\nAnswer:
\nVolume of bottom legs: V = 2(8 in. \u2219 1 in. \u2219 0.75 in.)
\n= 12 in3<\/sup>
\nVolume of vertical legs: V = 2(1 in. \u2219 9.5 in. \u2219 0.75 in.)
\n= 14.25 in3<\/sup>
\nVolume of tabletop: V = 8 in. \u2219 6 in. \u2219 1.5 in.
\n= 72 in3<\/sup>
\nThe volume of the table is
\n12 in3<\/sup> + 14.25 in3<\/sup> + 72 in3<\/sup> = 98.25 in3<\/sup>.<\/p>\n
\nA plastic die (singular for dice) for a game has an edge length of 1.5 cm. Each face of the cube has the number of cubic cutouts as its marker is supposed to indicate (i.e., the face marked 3 has 3 cutouts). What is the volume of the die?
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-26-Problem-Set-Answer-Key-5.png\")
\nAnswer:
\nNumber of cubic cutouts:
\n1 + 2 + 3 + 4 + 5 + 6 = 21
\nVolume of cutout cubes: Volume of large cube:
\nV = 21(2 mm)3<\/sup>
\nV = 168 mm3<\/sup> = 0.168 cm3<\/sup> V = (1.5 cm)3<\/sup>
\nV = 3.375 cm3<\/sup>
\nThe total volume of the die is
\n3.375 cm3<\/sup> – 0.168 cm3<\/sup> = 3.207 cm3<\/sup>.<\/p>\n
\nA wooden cube with an edge length of 6 inches has square holes (holes in the shape of right rectangular prisms) cut through the centers of each of the three sides as shown in the figure. Find the volume of the resulting solid if the square for the holes has an edge length of 1 inch.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-7th-Grade-Module-6-Lesson-26-Problem-Set-Answer-Key-6.png\")
\nAnswer:
\nThink of making the square holes between opposite sides by cutting three times: The first cut removes 6 in3<\/sup>, and the second and third cuts each remove 5 in3<\/sup>. The resulting solid has a volume of (6 in.)3<\/sup> – 6 in3<\/sup> – 5 in3<\/sup> – 5 in3<\/sup> = 200 in3<\/sup>.<\/p>\n
\nA right rectangular prism has each of its dimensions (length, width, and height) increased by 50%. By what percent is its volume increased?
\nAnswer:
\nV = l \u2219 w \u2219 h
\nV’ = 1.5l \u2219 1.5w \u2219 1.5h
\nV’ = 3.375lwh
\nThe larger volume is 337.5% of the smaller volume. The volume has increased by 237.5%.<\/p>\n
\nA solid is created by putting together right rectangular prisms. If each of the side lengths is increase by 40%, by what percent is the volume increased?
\nAnswer:
\nIf each of the side lengths is increased by 40%, then the volume of each right rectangular prism is multiplied by 1.43<\/sup> = 2.744. Since this is true for each right rectangular prism, the volume of the larger solid, V’, can be found by multiplying the volume of the smaller solid, V, by 2.744 = 274.4% (i.e., V’ = 2.744V). This is an increase of 174.4%.<\/p>\nEureka Math Grade 7 Module 6 Lesson 26 Exit Ticket Answer Key<\/h3>\n
\nA triangular prism has a rectangular prism cut out of it from one base to the opposite base, as shown in the figure. Determine the volume of the figure, provided all dimensions are in millimeters.
\nIs there any other way to determine the volume of the figure? If so, please explain.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/03\/Eureka-Math-Grade-7-Module-6-Lesson-26-Exit-Ticket-Answer-Key-1.png\")
\nAnswer:
\nPossible response:
\nVolume of the triangular prism: (\\(\\frac{1}{2}\\) \u2219 16 mm \u2219 13 mm)(14 mm) = 1,456 mm3<\/sup>
\nVolume of the rectangular prism: (6 mm \u2219 3 mm \u2219 14 mm) = 252 mm3<\/sup>
\nVolume of the composite prism: 1,456 mm3<\/sup> – 252 mm3<\/sup> = 1,204 mm3<\/sup>
\nThe calculations above subtract the volume of the cutout prism from the volume of the main prism. Another strategy would be to find the area of the base of the figure, which is the area of the triangle less the area of the rectangle, and then multiply by the height to find the volume of the prism.<\/p>\n","protected":false},"excerpt":{"rendered":"