{"id":77850,"date":"2022-06-13T21:31:36","date_gmt":"2022-06-13T16:01:36","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=77850"},"modified":"2022-05-10T14:14:15","modified_gmt":"2022-05-10T08:44:15","slug":"eureka-math-grade-7-module-3-lesson-22","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-7-module-3-lesson-22\/","title":{"rendered":"Eureka Math Grade 7 Module 3 Lesson 22 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 7th Grade Module 3 Lesson 22 Answer Key<\/h2>\n

Eureka Math Grade 7 Module 3 Lesson 22 Example Answer Key<\/h3>\n

Example 1.
\nThe pyramid in the picture has a square base, and its lateral faces are triangles that are exact copies of one another. Find the surface area of the pyramid.
\n\"Engage
\nAnswer:
\nThe surface area of the pyramid consists of one square base and four lateral triangular faces.
\nB = s2<\/sup>
\nB = (6 cm)2<\/sup>
\nB = 36 cm2<\/sup>
\nThe pyramid\u2019s base area is 36 cm2<\/sup>.<\/p>\n

LA = 4(\\(\\frac{1}{2}\\) bh)
\nLA = 4 \u2219 \\(\\frac{1}{2}\\) (6 cm \u2219 7 cm)
\nLA = 2(6 cm \u2219 7 cm)
\nLA = 2(42 cm2<\/sup> )
\nLA = 84 cm2<\/sup>
\nThe pyramid\u2019s lateral area is 84 cm2<\/sup>.
\nSA = LA + B
\nSA = 84 cm2<\/sup> + 36 cm2<\/sup>
\nSA = 120 cm2<\/sup>
\nThe surface area of the pyramid is 120 cm2<\/sup>.<\/p>\n

Example 2: Using Cubes
\nThere are 13 cubes glued together forming the solid in the diagram. The edges of each cube are \\(\\frac{1}{4}\\) inch in length. Find the surface area of the solid.
\n\"Engage
\nAnswer:
\nThe surface area of the solid consists of 46 square faces, all having side lengths of \\(\\frac{1}{4}\\) inch. The area of a square having sides of length \\(\\frac{1}{4}\\) inch is \\(\\frac{1}{16}\\) in2<\/sup> .
\nSA = 46 \u2219 Asquare<\/sub>
\nSA = 46 \u2219 \\(\\frac{1}{16}\\) in2<\/sup>
\nSA = \\(\\frac{46}{16}\\) in2<\/sup>
\nSA = 2 \\(\\frac{14}{16}\\) in2<\/sup>
\nSA = 2 \\(\\frac{7}{8}\\) in2<\/sup> The surface are of the solid is 2 \\(\\frac{7}{8}\\) in2<\/sup>.<\/p>\n

Example 3.
\nFind the total surface area of the wooden jewelry box. The sides and bottom of the box are all \\(\\frac{1}{4}\\) inch thick.
\nWhat are the faces that make up this box?
\n\"Engage
\nAnswer:
\nThe box has a rectangular bottom, rectangular lateral faces, and a rectangular top that has a smaller rectangle removed from it. There are also rectangular faces that make up the inner lateral faces and the inner bottom of the box.<\/p>\n

How does this box compare to other objects that you have found the surface area of?
\nAnswer:
\nThe box is a rectangular prism with a smaller rectangular prism removed from its inside. The total surface area is equal to the surface area of the larger right rectangular prism plus the lateral area of the smaller right rectangular prism.<\/p>\n

Large Prism: The surface area of the large right rectangular prism makes up the outside faces of the box, the rim of the box, and the inside bottom face of the box.<\/p>\n

SA = LA + 2B
\nLA = P \u2219 h
\nLA = 32 in. \u2219 4 in.
\nLA = 128 in2<\/sup><\/p>\n

B = lw
\nB = 10 in. \u2219 6 in.
\nB = 60 in2<\/sup>
\nThe lateral area is 128 in2<\/sup>.
\nThe base area is 60 in2<\/sup>.<\/p>\n

SA = LA + 2B
\nSA = 128 in2<\/sup> + 2(60 in2<\/sup>)
\nSA = 128 in2<\/sup> + 120 in2<\/sup>
\nSA = 248 in2<\/sup>
\nThe surface area of the larger prism is 248 in2<\/sup>.<\/p>\n

Small Prism: The smaller prism is \\(\\frac{1}{2}\\) in. smaller in length and width and \\(\\frac{1}{4}\\) in. smaller in height due to the thickness of the sides of the box.
\nSA = LA + 1B
\nLA = P \u2219 h
\nLA = 2(9 \\(\\frac{1}{2}\\) in. + 5 \\(\\frac{1}{2}\\) in.) \u2219 3 \\(\\frac{3}{4}\\) in.
\nLA = 2(14 in. + 1 in.) \u2219 3 \\(\\frac{3}{4}\\) in.
\nLA = 2(15 in.) \u2219 3 \\(\\frac{3}{4}\\) in.
\nLA = 30 in. \u2219 3 \\(\\frac{3}{4}\\) in.
\nLA = 90 in2<\/sup> + \\(\\frac{90}{4}\\) in2<\/sup>
\nLA = 90 in2<\/sup> + 22 \\(\\frac{1}{2}\\) in2<\/sup>
\nLA = 112 \\(\\frac{1}{2}\\) in2<\/sup>
\nThe lateral area is 112 \\(\\frac{1}{2}\\) in2<\/sup>.<\/p>\n

Surface Area of the Box
\nSAbox<\/sub> = SA + LA
\nSAbox<\/sub> = 248 in2<\/sup> + 112 \\(\\frac{1}{2}\\) in2<\/sup>
\nSAbox<\/sub> = 360 \\(\\frac{1}{2}\\) in2<\/sup>
\nThe total surface area of the box is 360 \\(\\frac{1}{2}\\) in2<\/sup>.<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 22 Exercise Answer Key<\/h3>\n

Opening Exercise
\nWhat is the area of the composite figure in the diagram? Is the diagram a net for a three-dimensional image? If so, sketch the image. If not, explain why.
\n\"Engage
\nAnswer:
\nThere are four unit squares in each square of the figure. There are 18 total squares that make up the figure, so the total area of the composite figure is
\nA = 18 \u2219 4 units2<\/sup> = 72 units2<\/sup>.
\nThe composite figure does represent the net of a three-dimensional figure. The figure is shown below.
\n\"Engage<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 22 Problem Set Answer Key<\/h3>\n

Question 1.
\nFor each of the following nets, draw (or describe) the solid represented by the net and find its surface area.
\na. The equilateral triangles are exact copies.
\n\"Eureka
\nAnswer:
\nThe net represents a triangular pyramid where the three lateral faces are identical to each other and the triangular base.
\nSA = 4B since the faces are all the same size and shape.<\/p>\n

SA = 4B
\nSA = 4(35 \\(\\frac{1}{10}\\) mm2<\/sup> )
\nSA = 140 mm2<\/sup> + \\(\\frac{4}{10}\\) mm2<\/sup>
\nSA = 140 \\(\\frac{2}{5}\\) mm2<\/sup><\/p>\n

B = \\(\\frac{1}{2}\\) bh
\nB = \\(\\frac{1}{2}\\) \u2219 9 mm \u2219 7 \\(\\frac{4}{5}\\) mm
\nB = \\(\\frac{9}{2}\\) mm \u2219 7 \\(\\frac{4}{5}\\) mm
\nB = \\(\\frac{63}{2}\\) mm2<\/sup> + \\(\\frac{36}{10}\\) mm2<\/sup>
\nB = \\(\\frac{315}{10}\\) mm2<\/sup> + \\(\\frac{36}{10}\\) mm2<\/sup>
\nB = \\(\\frac{351}{10}\\) mm2<\/sup> The surface area of the triangular pyramid is 140 \\(\\frac{2}{5}\\) mm2<\/sup>.
\nB = 35 \\(\\frac{1}{10}\\) mm2<\/sup><\/p>\n

b.
\n\"Eureka
\nAnswer:
\nThe net represents a square pyramid that has four identical lateral faces that are triangles. The base is a square.
\nB = s2<\/sup>
\nB = (12 in.)2<\/sup>
\nB = 144 in2<\/sup><\/p>\n

LA = 2(12 in. \u2219 14 \\(\\frac{3}{4}\\) in.)
\nSA = LA + B
\nLA = 4 \u2219 \\(\\frac{1}{2}\\) (bh)
\nLA = 4 \u2219 \\(\\frac{1}{2}\\) (12 in. \u2219 14 \\(\\frac{3}{4}\\) in.)
\nLA = 2(168 in2<\/sup> + 9 in2<\/sup> )
\nLA = 336 in2<\/sup> + 18 in2<\/sup>
\nLA = 354 in2<\/sup>
\nSA = LA + B
\nSA = 354 in2<\/sup> + 144 in2<\/sup>
\nSA = 498 in2<\/sup>
\nThe surface area of the square pyramid is 498 in2<\/sup>.<\/p>\n

Question 2.
\nFind the surface area of the following prism.
\n\"Eureka
\nAnswer:
\nSA = LA + 2B
\nLA = P \u2219 h
\nLA = (4 cm + 6 \\(\\frac{1}{2}\\) cm + 4 \\(\\frac{1}{5}\\) cm + 5 \\(\\frac{1}{4}\\) cm) \u2219 9 cm
\nLA = (19 cm + \\(\\frac{1}{2}\\) cm + \\(\\frac{1}{5}\\) cm + \\(\\frac{1}{4}\\) cm) \u2219 9 cm
\nLA = (19 cm + \\(\\frac{10}{20}\\) cm + \\(\\frac{4}{20}\\) cm + \\(\\frac{5}{20}\\) cm) \u2219 9 cm
\nLA = (19 cm + \\(\\frac{19}{20}\\) cm) \u2219 9 cm
\nLA = 171 cm2<\/sup> + \\(\\frac{171}{20}\\) cm2<\/sup>
\nLA = 171 cm2<\/sup> + 8 \\(\\frac{11}{20}\\) cm2<\/sup>
\nLA = 179 \\(\\frac{11}{20}\\) cm2<\/sup><\/p>\n

B = Arectangle<\/sub> + Atriangle<\/sub>
\nB = (5 \\(\\frac{1}{4}\\) cm \u2219 4 cm) + \\(\\frac{1}{2}\\)(4 cm \u2219 1 \\(\\frac{1}{4}\\) cm)
\nB = (20 cm2<\/sup> + 1 cm2<\/sup> ) + (2 cm \u2219 1 \\(\\frac{1}{4}\\) cm)
\nB = 21 cm2<\/sup> + 2 \\(\\frac{1}{2}\\) cm2<\/sup>
\nB = 23 \\(\\frac{1}{2}\\) cm2<\/sup><\/p>\n

SA = LA + 2B
\nSA = 179 \\(\\frac{11}{20}\\) cm2<\/sup> + 2(23 \\(\\frac{1}{2}\\) cm2<\/sup> )
\nSA = 179 \\(\\frac{11}{20}\\) cm2<\/sup> + 47 cm2<\/sup>
\nSA = 226 \\(\\frac{11}{20}\\) cm2<\/sup>
\nThe surface area of the prism is 226 \\(\\frac{11}{20}\\) cm2<\/sup>.<\/p>\n

Question 3.
\nThe net below is for a specific object. The measurements shown are in meters. Sketch (or describe) the object, and then find its surface area.
\n\"Eureka
\nAnswer:
\nSA = LA + 2B
\nLA = P \u2219 h
\nLA = 6 cm \u2219 \\(\\frac{1}{2}\\) cm
\nLA = 3 cm2<\/sup><\/p>\n

B = (\\(\\frac{1}{2}\\) cm \u2219 \\(\\frac{1}{2}\\) cm) + (\\(\\frac{1}{2}\\) cm \u2219 1 cm) + (\\(\\frac{1}{2}\\) cm \u2219 1 \\(\\frac{1}{2}\\) cm)
\nB = (\\(\\frac{1}{4}\\) cm2<\/sup>) + (\\(\\frac{1}{2}\\) cm2<\/sup> ) + (\\(\\frac{3}{4}\\) cm2<\/sup> )
\nB = (\\(\\frac{1}{4}\\)cm2<\/sup>) + (\\(\\frac{2}{4}\\) cm2<\/sup>) + (\\(\\frac{3}{4}\\) cm2<\/sup>)
\nB = \\(\\frac{6}{4}\\) cm2<\/sup>
\nB = 1 \\(\\frac{1}{2}\\) cm2<\/sup><\/p>\n

SA = LA + 2B
\nSA = 3 cm2<\/sup> + 2(1 \\(\\frac{1}{2}\\) cm2<\/sup> )
\nSA = 3 cm2<\/sup> + 3 cm2<\/sup>
\nSA = 6 cm2<\/sup><\/p>\n

The surface area of the object is 6 cm2<\/sup>.<\/p>\n

Question 4.
\nIn the diagram, there are 14 cubes glued together to form a solid. Each cube has a volume of \\(\\frac{1}{8}\\) in3<\/sup>. Find the surface area of the solid.
\n\"Eureka
\nAnswer:
\nThe volume of a cube is s3<\/sup>, and 1\/8 in3<\/sup> is the same as (\\(\\frac{1}{2}\\) in.)3<\/sup>, so the cubes have edges that are \\(\\frac{1}{2}\\) in. long. The cube faces have area s2<\/sup>, or (\\(\\frac{1}{2}\\) in.)2<\/sup>, or \\(\\frac{1}{4}\\) in2<\/sup>. There are 42 cube faces that make up the surface of the solid.
\nSA = \\(\\frac{1}{4}\\) in2<\/sup> \u2219 42
\nSA = 10 \\(\\frac{1}{2}\\) in2<\/sup>
\nThe surface area of the solid is 10 \\(\\frac{1}{2}\\) in2<\/sup>.<\/p>\n

Question 5.
\nThe nets below represent three solids. Sketch (or describe) each solid, and find its surface area.
\na.
\n\"Eureka
\nAnswer:
\nSA = LA + 2B
\nLA = P \u2219 h
\nLA = 12 \u2219 3
\nLA = 36 cm2<\/sup><\/p>\n

B = s2<\/sup>
\nB = (3 cm)2<\/sup>
\nB = 9 cm2<\/sup><\/p>\n

SA =
\n36 cm2<\/sup> + 2(9 cm2<\/sup>)
\nSA = 36 cm2<\/sup> + 18 cm2<\/sup>
\nSA = 54 cm2<\/sup>
\n\"Eureka<\/p>\n

b.
\n\"Eureka
\nAnswer:
\nSA = 3Asquare<\/sub> + 3Art triangle<\/sub> + Aequ triangle<\/sub>
\nAsquare<\/sub> = s2<\/sup>
\nAsquare<\/sub> = (3 cm)2<\/sup>
\nAsquare<\/sub> = 9 cm2<\/sup>
\nArt triangle<\/sub> = \\(\\frac{1}{2}\\) bh
\nArt triangle<\/sub> = \\(\\frac{1}{2}\\) \u2219 3 cm \u2219 3 cm
\nArt triangle<\/sub> = \\(\\frac{9}{2}\\)
\nArt triangle<\/sub> = 4 \\(\\frac{1}{2}\\) cm2<\/sup>
\nAequ triangle<\/sub> = \\(\\frac{1}{2}\\) bh
\nAequ triangle<\/sub> = \\(\\frac{1}{2}\\) \u2219 (4 \\(\\frac{1}{5}\\) cm) \u2219 (3 \\(\\frac{7}{10}\\) cm)
\nAequ triangle<\/sub> = 2 \\(\\frac{1}{10}\\) cm \u2219 3 \\(\\frac{7}{10}\\) cm
\nAequ triangle<\/sub> = \\(\\frac{21}{10}\\) cm \u2219 \\(\\frac{37}{10}\\) cm
\nAequ triangle<\/sub> = 7 \\(\\frac{77}{100}\\) cm2<\/sup>
\nAequ triangle<\/sub> = 7 \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 3(9 cm2<\/sup> ) + 3(4 \\(\\frac{1}{2}\\) cm2<\/sup> ) + 7 \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 27 cm2<\/sup> + (12 + \\(\\frac{3}{2}\\)) cm2<\/sup> + 7 \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 47 cm2<\/sup> + \\(\\frac{1}{2}\\) cm2<\/sup> + \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 47 cm2<\/sup> + \\(\\frac{50}{100}\\) cm2<\/sup> + \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 47 cm2<\/sup> + \\(\\frac{127}{100}\\) cm2<\/sup>
\nSA = 47 cm2<\/sup> + 1 cm2<\/sup> + \\(\\frac{27}{100}\\) cm2<\/sup>
\nSA = 48 \\(\\frac{27}{100}\\) cm2<\/sup>
\n\"Eureka<\/p>\n

c.
\n\"Eureka
\nAnswer:
\nSA = 3Art triangle<\/sub> + Aequ triangle<\/sub>
\nSA = 3(4 \\(\\frac{1}{2}\\) ) cm2<\/sup> + 7 \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 12 cm2<\/sup> + \\(\\frac{3}{2}\\) cm2<\/sup> + 7 cm2<\/sup> + \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 20 cm2<\/sup> + \\(\\frac{1}{2}\\) cm2<\/sup> + \\(\\frac{77}{100}\\) cm2<\/sup>
\nSA = 20 cm2<\/sup> + 1 cm2<\/sup> + \\(\\frac{27}{100}\\) cm2<\/sup>
\nSA = 21 \\(\\frac{27}{100}\\) cm2<\/sup>
\n\"Eureka<\/p>\n

d. How are figures (b) and (c) related to figure (a)?
\nAnswer:
\nIf the equilateral triangular faces of figures (b) and (c) were matched together, they would form the cube in part (a).<\/p>\n

Question 6.
\nFind the surface area of the solid shown in the diagram. The solid is a right triangular prism (with right triangular bases) with a smaller right triangular prism removed from it.
\n\"Eureka
\nAnswer:
\nSA = LA + 2B
\nLA = P \u2219 h
\nLA = (4 in. + 4 in. + 5 \\(\\frac{13}{20}\\) in.) \u2219 2 in.
\nLA = (13 \\(\\frac{13}{20}\\) in.) \u2219 2 in.
\nLA = 26 in2<\/sup> + \\(\\frac{13}{10}\\) in2<\/sup>
\nLA = 26 in2<\/sup> + 1 in2<\/sup> + \\(\\frac{3}{10}\\) in2<\/sup>
\nLA = 27 \\(\\frac{3}{10}\\) in2<\/sup>
\nThe \\(\\frac{1}{4}\\)in. by 4 \\(\\frac{19}{20}\\) in. rectangle has to be taken away from the lateral area:
\nA = lw
\nA = 4 \\(\\frac{19}{20}\\) in \u2219 \\(\\frac{1}{4}\\) in
\nA = 1 in2<\/sup> + 19\/80 in2<\/sup>
\nA = 1 19\/80 in2<\/sup><\/p>\n

LA = 27 \\(\\frac{3}{10}\\) in2<\/sup>-1 \\(\\frac{19}{80}\\) in2<\/sup>
\nLA = 27 \\(\\frac{24}{80}\\) in2<\/sup>-1 \\(\\frac{19}{80}\\) in2<\/sup>
\nLA = 26 \\(\\frac{5}{80}\\) in2<\/sup>
\nLA = 26 \\(\\frac{1}{16}\\) in2<\/sup><\/p>\n

Two bases of the larger and smaller triangular prisms must be added:
\nSA = 26 \\(\\frac{1}{16}\\) in2<\/sup> + 2(3 \\(\\frac{1}{2}\\) in \u2219 \\(\\frac{1}{4}\\) in) + 2(\\(\\frac{1}{2}\\) \u2219 4 in \u2219 4 in)
\nSA = 26 \\(\\frac{1}{16}\\) in2<\/sup> + 2 \u2219 \\(\\frac{1}{4}\\) in \u2219 3 \\(\\frac{1}{2}\\) in + 16 in2<\/sup>
\nSA = 26 \\(\\frac{1}{16}\\) in2<\/sup> + \\(\\frac{1}{2}\\) in \u2219 3 \\(\\frac{1}{2}\\) in + 16 in2<\/sup>
\nSA = 26 \\(\\frac{1}{16}\\) in2<\/sup> + (\\(\\frac{3}{2}\\) in2<\/sup> + \\(\\frac{1}{4}\\) in2<\/sup> ) + 16 in2<\/sup>
\nSA = 26 \\(\\frac{1}{16}\\) in2<\/sup> + 1 in2<\/sup> + \\(\\frac{8}{16}\\) in2<\/sup> + \\(\\frac{4}{16}\\) in2<\/sup> + 16 in2<\/sup>
\nSA = 43 \\(\\frac{13}{16}\\) in2<\/sup>
\nThe surface area of the solid is 43 \\(\\frac{13}{16}\\) in2<\/sup>.<\/p>\n

Question 7.
\nThe diagram shows a cubic meter that has had three square holes punched completely through the cube on three perpendicular axes. Find the surface area of the remaining solid.
\n\"Eureka
\nAnswer:
\nExterior surfaces of the cube (SA1<\/sub>):
\nSA1<\/sub> = 6(1 m)2<\/sup>-6(\\(\\frac{1}{2}\\) m)2<\/sup>
\nSA1<\/sub> = 6(1 m2<\/sup> )-6(\\(\\frac{1}{4}\\) m2<\/sup>)
\nSA1<\/sub> = 6 m2<\/sup>–\\(\\frac{6}{4}\\) m2<\/sup>
\nSA1<\/sub> = 6 m2<\/sup>-(1 \\(\\frac{1}{2}\\) m2<\/sup>)
\nSA1<\/sub> = 4 \\(\\frac{1}{2}\\) m2<\/sup>
\nJust inside each square hole are four intermediate surfaces that can be treated as the lateral area of a rectangular prism. Each has a height of \\(\\frac{1}{4}\\) m and perimeter of \\(\\frac{1}{2}\\) m + \\(\\frac{1}{2}\\) m + \\(\\frac{1}{2}\\) m + \\(\\frac{1}{2}\\) m or 2 m.
\nSA2<\/sub> = 6(LA)
\nSA2<\/sub> = 6(2 m \u2219 \\(\\frac{1}{4}\\) m)
\nSA2<\/sub> = 6 \u2219 \\(\\frac{1}{2}\\) m2<\/sup>
\nSA2<\/sub> = 3 m2<\/sup><\/p>\n

The total surface area of the remaining solid is the sum of these two areas:
\nSAT<\/sub> = SA1<\/sub> + SA2<\/sub>.
\nSAT<\/sub> = 4 \\(\\frac{1}{2}\\) m2<\/sup> + 3 m2<\/sup>
\nSAT<\/sub> = 7 \\(\\frac{1}{2}\\) m2<\/sup>
\nThe surface area of the remaining solid is 7 \\(\\frac{1}{2}\\) m2<\/sup>.<\/p>\n

Eureka Math Grade 7 Module 3 Lesson 22 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nThe right hexagonal pyramid has a hexagon base with equal-length sides. The lateral faces of the pyramid are all triangles (that are exact copies of one another) with heights of 15 ft. Find the surface area of the pyramid.
\n\"Eureka
\nAnswer:
\nSA = LA + 1B
\nLA = 6 \u2219 \\(\\frac{1}{2}\\) (bh)
\nLA = 6 \u2219 \\(\\frac{1}{2}\\) (5 ft. \u2219 15 ft.)
\nLA = 3 \u2219 75 ft2<\/sup>
\nLA = 225 ft2<\/sup><\/p>\n

B = Arectangle<\/sub> + 2Atriangle<\/sub>
\nB = (8 ft. \u2219 5 ft.) + 2 \u2219 \\(\\frac{1}{2}\\) (8 ft. \u2219 3 ft.)
\nB = 40 ft2<\/sup> + (8 ft. \u2219 3 ft.)
\nB = 40 ft2<\/sup> + 24 ft2<\/sup><\/p>\n

B = 64 ft2<\/sup>
\nSA = LA + 1B
\nSA = 225 ft2<\/sup> + 64 ft2<\/sup>
\nSA = 289 ft2<\/sup>
\nThe surface area of the pyramid is 289 ft2<\/sup>.<\/p>\n

Question 2.
\nSix cubes are glued together to form the solid shown in the diagram. If the edges of each cube measure 1 \\(\\frac{1}{2}\\) inches in length, what is the surface area of the solid?
\n\"Eureka
\nAnswer:
\nThere are 26 square cube faces showing on the surface area of the solid (5 each from the top and bottom view, 4 each from the front and back view, 3 each from the left and right side views, and 2 from the \u201cinside\u201d of the front).
\nA = s2<\/sup>
\nA = (1 \\(\\frac{1}{2}\\) in.)2<\/sup>
\nA = (1 \\(\\frac{1}{2}\\) in.)(1 \\(\\frac{1}{2}\\) in.)
\nA = 1 \\(\\frac{1}{2}\\) in.(1 in. + \\(\\frac{1}{2}\\) in.)
\nA = (1 \\(\\frac{1}{2}\\) in. \u2219 1 in.) + (1 \\(\\frac{1}{2}\\) in. \u2219 \\(\\frac{1}{2}\\) in.)
\nA = 1 \\(\\frac{1}{2}\\) in2<\/sup> + \\(\\frac{3}{4}\\) in2<\/sup>
\nA = 1 \\(\\frac{2}{4}\\) in2<\/sup> + \\(\\frac{3}{4}\\) in2<\/sup>
\nA = 1 \\(\\frac{5}{4}\\) in2<\/sup> = 2 \\(\\frac{1}{4}\\) in2<\/sup><\/p>\n

SA = 26 \u2219 (2 \\(\\frac{1}{4}\\) in2<\/sup>)
\nSA = 52 in2<\/sup> + \\(\\frac{26}{4}\\) in2<\/sup>
\nSA = 52 in2<\/sup> + 6 in2<\/sup> + \\(\\frac{1}{2}\\) in2<\/sup>
\nSA = 58 \\(\\frac{1}{2}\\) in2<\/sup>
\nThe surface area of the solid is 58 \\(\\frac{1}{2}\\) in2<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 7th Grade Module 3 Lesson 22 Answer Key Eureka Math Grade 7 Module 3 Lesson 22 Example Answer Key Example 1. The pyramid in the picture has a square base, and its lateral faces are triangles that are exact copies of one another. Find the surface area of the pyramid. Answer: … 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\/77850"}],"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=77850"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/77850\/revisions"}],"predecessor-version":[{"id":137775,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/77850\/revisions\/137775"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=77850"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=77850"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=77850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}