Eureka Math Grade 7 Module 3 Lesson 22 Answer Key

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:
The surface area of the pyramid consists of one square base and four lateral triangular faces.
B = s²
B = (6 cm)²
B = 36 cm²
The pyramid’s base area is 36 cm².

LA = 4(\(\frac{1}{2}\) bh)
LA = 4 ∙ \(\frac{1}{2}\) (6 cm ∙ 7 cm)
LA = 2(6 cm ∙ 7 cm)
LA = 2(42 cm² )
LA = 84 cm²
The pyramid’s lateral area is 84 cm².
SA = LA + B
SA = 84 cm² + 36 cm²
SA = 120 cm²
The surface area of the pyramid is 120 cm².

Example 2: Using Cubes
There 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.

Answer:
The 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 .
SA = 46 ∙ A_square
SA = 46 ∙ \(\frac{1}{16}\) in²
SA = \(\frac{46}{16}\) in²
SA = 2 \(\frac{14}{16}\) in²
SA = 2 \(\frac{7}{8}\) in² The surface are of the solid is 2 \(\frac{7}{8}\) in².

Example 3.
Find the total surface area of the wooden jewelry box. The sides and bottom of the box are all \(\frac{1}{4}\) inch thick.
What are the faces that make up this box?

Answer:
The 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.

How does this box compare to other objects that you have found the surface area of?
Answer:
The 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.

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.

SA = LA + 2B
LA = P ∙ h
LA = 32 in. ∙ 4 in.
LA = 128 in²

B = lw
B = 10 in. ∙ 6 in.
B = 60 in²
The lateral area is 128 in².
The base area is 60 in².

SA = LA + 2B
SA = 128 in² + 2(60 in²)
SA = 128 in² + 120 in²
SA = 248 in²
The surface area of the larger prism is 248 in².

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.
SA = LA + 1B
LA = P ∙ h
LA = 2(9 \(\frac{1}{2}\) in. + 5 \(\frac{1}{2}\) in.) ∙ 3 \(\frac{3}{4}\) in.
LA = 2(14 in. + 1 in.) ∙ 3 \(\frac{3}{4}\) in.
LA = 2(15 in.) ∙ 3 \(\frac{3}{4}\) in.
LA = 30 in. ∙ 3 \(\frac{3}{4}\) in.
LA = 90 in² + \(\frac{90}{4}\) in²
LA = 90 in² + 22 \(\frac{1}{2}\) in²
LA = 112 \(\frac{1}{2}\) in²
The lateral area is 112 \(\frac{1}{2}\) in².

Surface Area of the Box
SA_box = SA + LA
SA_box = 248 in² + 112 \(\frac{1}{2}\) in²
SA_box = 360 \(\frac{1}{2}\) in²
The total surface area of the box is 360 \(\frac{1}{2}\) in².

Eureka Math Grade 7 Module 3 Lesson 22 Exercise Answer Key

Opening Exercise
What 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.

Answer:
There 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
A = 18 ∙ 4 units² = 72 units².
The composite figure does represent the net of a three-dimensional figure. The figure is shown below.

Eureka Math Grade 7 Module 3 Lesson 22 Problem Set Answer Key

Question 1.
For each of the following nets, draw (or describe) the solid represented by the net and find its surface area.
a. The equilateral triangles are exact copies.

Answer:
The net represents a triangular pyramid where the three lateral faces are identical to each other and the triangular base.
SA = 4B since the faces are all the same size and shape.

SA = 4B
SA = 4(35 \(\frac{1}{10}\) mm² )
SA = 140 mm² + \(\frac{4}{10}\) mm²
SA = 140 \(\frac{2}{5}\) mm²

B = \(\frac{1}{2}\) bh
B = \(\frac{1}{2}\) ∙ 9 mm ∙ 7 \(\frac{4}{5}\) mm
B = \(\frac{9}{2}\) mm ∙ 7 \(\frac{4}{5}\) mm
B = \(\frac{63}{2}\) mm² + \(\frac{36}{10}\) mm²
B = \(\frac{315}{10}\) mm² + \(\frac{36}{10}\) mm²
B = \(\frac{351}{10}\) mm² The surface area of the triangular pyramid is 140 \(\frac{2}{5}\) mm².
B = 35 \(\frac{1}{10}\) mm²

b.

Answer:
The net represents a square pyramid that has four identical lateral faces that are triangles. The base is a square.
B = s²
B = (12 in.)²
B = 144 in²

LA = 2(12 in. ∙ 14 \(\frac{3}{4}\) in.)
SA = LA + B
LA = 4 ∙ \(\frac{1}{2}\) (bh)
LA = 4 ∙ \(\frac{1}{2}\) (12 in. ∙ 14 \(\frac{3}{4}\) in.)
LA = 2(168 in² + 9 in² )
LA = 336 in² + 18 in²
LA = 354 in²
SA = LA + B
SA = 354 in² + 144 in²
SA = 498 in²
The surface area of the square pyramid is 498 in².

Question 2.
Find the surface area of the following prism.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (4 cm + 6 \(\frac{1}{2}\) cm + 4 \(\frac{1}{5}\) cm + 5 \(\frac{1}{4}\) cm) ∙ 9 cm
LA = (19 cm + \(\frac{1}{2}\) cm + \(\frac{1}{5}\) cm + \(\frac{1}{4}\) cm) ∙ 9 cm
LA = (19 cm + \(\frac{10}{20}\) cm + \(\frac{4}{20}\) cm + \(\frac{5}{20}\) cm) ∙ 9 cm
LA = (19 cm + \(\frac{19}{20}\) cm) ∙ 9 cm
LA = 171 cm² + \(\frac{171}{20}\) cm²
LA = 171 cm² + 8 \(\frac{11}{20}\) cm²
LA = 179 \(\frac{11}{20}\) cm²

B = A_rectangle + A_triangle
B = (5 \(\frac{1}{4}\) cm ∙ 4 cm) + \(\frac{1}{2}\)(4 cm ∙ 1 \(\frac{1}{4}\) cm)
B = (20 cm² + 1 cm² ) + (2 cm ∙ 1 \(\frac{1}{4}\) cm)
B = 21 cm² + 2 \(\frac{1}{2}\) cm²
B = 23 \(\frac{1}{2}\) cm²

SA = LA + 2B
SA = 179 \(\frac{11}{20}\) cm² + 2(23 \(\frac{1}{2}\) cm² )
SA = 179 \(\frac{11}{20}\) cm² + 47 cm²
SA = 226 \(\frac{11}{20}\) cm²
The surface area of the prism is 226 \(\frac{11}{20}\) cm².

Question 3.
The net below is for a specific object. The measurements shown are in meters. Sketch (or describe) the object, and then find its surface area.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = 6 cm ∙ \(\frac{1}{2}\) cm
LA = 3 cm²

B = (\(\frac{1}{2}\) cm ∙ \(\frac{1}{2}\) cm) + (\(\frac{1}{2}\) cm ∙ 1 cm) + (\(\frac{1}{2}\) cm ∙ 1 \(\frac{1}{2}\) cm)
B = (\(\frac{1}{4}\) cm²) + (\(\frac{1}{2}\) cm² ) + (\(\frac{3}{4}\) cm² )
B = (\(\frac{1}{4}\)cm²) + (\(\frac{2}{4}\) cm²) + (\(\frac{3}{4}\) cm²)
B = \(\frac{6}{4}\) cm²
B = 1 \(\frac{1}{2}\) cm²

SA = LA + 2B
SA = 3 cm² + 2(1 \(\frac{1}{2}\) cm² )
SA = 3 cm² + 3 cm²
SA = 6 cm²

The surface area of the object is 6 cm².

Question 4.
In the diagram, there are 14 cubes glued together to form a solid. Each cube has a volume of \(\frac{1}{8}\) in³. Find the surface area of the solid.

Answer:
The volume of a cube is s³, and 1/8 in³ is the same as (\(\frac{1}{2}\) in.)³, so the cubes have edges that are \(\frac{1}{2}\) in. long. The cube faces have area s², or (\(\frac{1}{2}\) in.)², or \(\frac{1}{4}\) in². There are 42 cube faces that make up the surface of the solid.
SA = \(\frac{1}{4}\) in² ∙ 42
SA = 10 \(\frac{1}{2}\) in²
The surface area of the solid is 10 \(\frac{1}{2}\) in².

Question 5.
The nets below represent three solids. Sketch (or describe) each solid, and find its surface area.
a.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = 12 ∙ 3
LA = 36 cm²

B = s²
B = (3 cm)²
B = 9 cm²

SA =
36 cm² + 2(9 cm²)
SA = 36 cm² + 18 cm²
SA = 54 cm²

b.

Answer:
SA = 3A_square + 3A_{rt triangle} + A_{equ triangle}
A_square = s²
A_square = (3 cm)²
A_square = 9 cm²
A_{rt triangle} = \(\frac{1}{2}\) bh
A_{rt triangle} = \(\frac{1}{2}\) ∙ 3 cm ∙ 3 cm
A_{rt triangle} = \(\frac{9}{2}\)
A_{rt triangle} = 4 \(\frac{1}{2}\) cm²
A_{equ triangle} = \(\frac{1}{2}\) bh
A_{equ triangle} = \(\frac{1}{2}\) ∙ (4 \(\frac{1}{5}\) cm) ∙ (3 \(\frac{7}{10}\) cm)
A_{equ triangle} = 2 \(\frac{1}{10}\) cm ∙ 3 \(\frac{7}{10}\) cm
A_{equ triangle} = \(\frac{21}{10}\) cm ∙ \(\frac{37}{10}\) cm
A_{equ triangle} = 7 \(\frac{77}{100}\) cm²
A_{equ triangle} = 7 \(\frac{77}{100}\) cm²
SA = 3(9 cm² ) + 3(4 \(\frac{1}{2}\) cm² ) + 7 \(\frac{77}{100}\) cm²
SA = 27 cm² + (12 + \(\frac{3}{2}\)) cm² + 7 \(\frac{77}{100}\) cm²
SA = 47 cm² + \(\frac{1}{2}\) cm² + \(\frac{77}{100}\) cm²
SA = 47 cm² + \(\frac{50}{100}\) cm² + \(\frac{77}{100}\) cm²
SA = 47 cm² + \(\frac{127}{100}\) cm²
SA = 47 cm² + 1 cm² + \(\frac{27}{100}\) cm²
SA = 48 \(\frac{27}{100}\) cm²

c.

Answer:
SA = 3A_{rt triangle} + A_{equ triangle}
SA = 3(4 \(\frac{1}{2}\) ) cm² + 7 \(\frac{77}{100}\) cm²
SA = 12 cm² + \(\frac{3}{2}\) cm² + 7 cm² + \(\frac{77}{100}\) cm²
SA = 20 cm² + \(\frac{1}{2}\) cm² + \(\frac{77}{100}\) cm²
SA = 20 cm² + 1 cm² + \(\frac{27}{100}\) cm²
SA = 21 \(\frac{27}{100}\) cm²

d. How are figures (b) and (c) related to figure (a)?
Answer:
If the equilateral triangular faces of figures (b) and (c) were matched together, they would form the cube in part (a).

Question 6.
Find 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.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (4 in. + 4 in. + 5 \(\frac{13}{20}\) in.) ∙ 2 in.
LA = (13 \(\frac{13}{20}\) in.) ∙ 2 in.
LA = 26 in² + \(\frac{13}{10}\) in²
LA = 26 in² + 1 in² + \(\frac{3}{10}\) in²
LA = 27 \(\frac{3}{10}\) in²
The \(\frac{1}{4}\)in. by 4 \(\frac{19}{20}\) in. rectangle has to be taken away from the lateral area:
A = lw
A = 4 \(\frac{19}{20}\) in ∙ \(\frac{1}{4}\) in
A = 1 in² + 19/80 in²
A = 1 19/80 in²

LA = 27 \(\frac{3}{10}\) in²-1 \(\frac{19}{80}\) in²
LA = 27 \(\frac{24}{80}\) in²-1 \(\frac{19}{80}\) in²
LA = 26 \(\frac{5}{80}\) in²
LA = 26 \(\frac{1}{16}\) in²

Two bases of the larger and smaller triangular prisms must be added:
SA = 26 \(\frac{1}{16}\) in² + 2(3 \(\frac{1}{2}\) in ∙ \(\frac{1}{4}\) in) + 2(\(\frac{1}{2}\) ∙ 4 in ∙ 4 in)
SA = 26 \(\frac{1}{16}\) in² + 2 ∙ \(\frac{1}{4}\) in ∙ 3 \(\frac{1}{2}\) in + 16 in²
SA = 26 \(\frac{1}{16}\) in² + \(\frac{1}{2}\) in ∙ 3 \(\frac{1}{2}\) in + 16 in²
SA = 26 \(\frac{1}{16}\) in² + (\(\frac{3}{2}\) in² + \(\frac{1}{4}\) in² ) + 16 in²
SA = 26 \(\frac{1}{16}\) in² + 1 in² + \(\frac{8}{16}\) in² + \(\frac{4}{16}\) in² + 16 in²
SA = 43 \(\frac{13}{16}\) in²
The surface area of the solid is 43 \(\frac{13}{16}\) in².

Question 7.
The 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.

Answer:
Exterior surfaces of the cube (SA₁):
SA₁ = 6(1 m)²-6(\(\frac{1}{2}\) m)²
SA₁ = 6(1 m² )-6(\(\frac{1}{4}\) m²)
SA₁ = 6 m²–\(\frac{6}{4}\) m²
SA₁ = 6 m²-(1 \(\frac{1}{2}\) m²)
SA₁ = 4 \(\frac{1}{2}\) m²
Just 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.
SA₂ = 6(LA)
SA₂ = 6(2 m ∙ \(\frac{1}{4}\) m)
SA₂ = 6 ∙ \(\frac{1}{2}\) m²
SA₂ = 3 m²

The total surface area of the remaining solid is the sum of these two areas:
SA_T = SA₁ + SA₂.
SA_T = 4 \(\frac{1}{2}\) m² + 3 m²
SA_T = 7 \(\frac{1}{2}\) m²
The surface area of the remaining solid is 7 \(\frac{1}{2}\) m².

Eureka Math Grade 7 Module 3 Lesson 22 Exit Ticket Answer Key

Question 1.
The 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.

Answer:
SA = LA + 1B
LA = 6 ∙ \(\frac{1}{2}\) (bh)
LA = 6 ∙ \(\frac{1}{2}\) (5 ft. ∙ 15 ft.)
LA = 3 ∙ 75 ft²
LA = 225 ft²

B = A_rectangle + 2A_triangle
B = (8 ft. ∙ 5 ft.) + 2 ∙ \(\frac{1}{2}\) (8 ft. ∙ 3 ft.)
B = 40 ft² + (8 ft. ∙ 3 ft.)
B = 40 ft² + 24 ft²

B = 64 ft²
SA = LA + 1B
SA = 225 ft² + 64 ft²
SA = 289 ft²
The surface area of the pyramid is 289 ft².

Question 2.
Six 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?

Answer:
There 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 “inside” of the front).
A = s²
A = (1 \(\frac{1}{2}\) in.)²
A = (1 \(\frac{1}{2}\) in.)(1 \(\frac{1}{2}\) in.)
A = 1 \(\frac{1}{2}\) in.(1 in. + \(\frac{1}{2}\) in.)
A = (1 \(\frac{1}{2}\) in. ∙ 1 in.) + (1 \(\frac{1}{2}\) in. ∙ \(\frac{1}{2}\) in.)
A = 1 \(\frac{1}{2}\) in² + \(\frac{3}{4}\) in²
A = 1 \(\frac{2}{4}\) in² + \(\frac{3}{4}\) in²
A = 1 \(\frac{5}{4}\) in² = 2 \(\frac{1}{4}\) in²

SA = 26 ∙ (2 \(\frac{1}{4}\) in²)
SA = 52 in² + \(\frac{26}{4}\) in²
SA = 52 in² + 6 in² + \(\frac{1}{2}\) in²
SA = 58 \(\frac{1}{2}\) in²
The surface area of the solid is 58 \(\frac{1}{2}\) in².