## 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 the surface area in part (a) related to the surface area in part (b)?

Answer:

a. Surface area:

2(3 in. × 6 in.) + 2(3 in. × 12 in.) + 2(6 in. × 12 in.)

= 2(18 in^{2}) + 2(36 in^{2}) + 2(72 in^{2} )

= 36 in^{2} + 72 in^{2} + 144 in^{2}

= 252 in^{2}

b.

→ How did you determine the surface area of the shape on the left?

→ 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.

c. If I add the surface area of both figures, I will get more than the surface area of the original shape.

234 in^{2} + 90 in^{2} = 324 in^{2}

324 in^{2} – 252 in^{2} = 72 in^{2}

72 in^{2} is twice the area of the region where the two pieces fit together.

There are 72 more square inches when the prisms are separated.

### Eureka Math Grade 7 Module 6 Lesson 23 Exercise Answer Key

Opening Exercise

Calculate the surface area of the square pyramid.

Answer:

Area of the square base:

s^{2} = (8 cm)^{2}

= 64 cm^{2}

Area of the triangular lateral sides:

\(\frac{1}{2}\) bh = \(\frac{1}{2}\)(8 cm)(5 cm)

= 20 cm^{2}

There are four lateral sides. So, the area of all 4 triangles is 80 cm^{2}.

Surface area:

(8 cm ∙ 8 cm) + 4(1/2 (8 cm ∙ 5 cm)) = 64 cm^{2} + 80 cm^{2} = 144 cm^{2}

→ Explain the process you used to determine the surface area of the pyramid.

Answers will vary. I drew a net to determine the area of each side and added the areas together.

→ The surface area of a pyramid is the union of its base region and all its lateral faces.

→ Explain how (8 cm ∙ 8 cm) + 4(\(\frac{1}{2}\)(8 cm∙5 cm)) represents the surface area.

The area of the square with side lengths of 8 cm is represented by (8 cm ∙ 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∙5 cm)).

→ Would this method work for any prism or pyramid?

Answers 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.

Determine the surface area of the right prisms.

Exercise 1.

Answer:

Area of top and bottom: 2(\(\frac{1}{2}\)(15 ft. × 8 ft.) )

= 15 ft. × 8 ft. = 120 ft^{2}

Area of front: 15 ft. × 20 ft. = 300 ft^{2}

Area that can be seen from left: 17 ft. × 20 ft. = 340 ft^{2}

Area that can be seen from right: 8 ft. × 20 ft. = 160 ft^{2}

Surface area: 120 ft^{2} + 300 ft^{2} + 340 ft^{2} + 160 ft^{2} = 920 ft^{2}

Exercise 2.

Answer:

Area of front and back: 2(\(\frac{1}{2}\)(10 yd. + 4 yd.)4 yd.)

= 14 yd. × 4 yd. = 56 yd^{2}

Area of top: 4 yd. × 15 yd. = 60 yd^{2}

Area that can be seen from left and right: 2(5 yd. × 15 yd.)

= 2(75 yd^{2}) = 150 yd^{2}

Area of bottom: 10 yd. × 15 yd. = 150 yd^{2}

Surface area: 56 yd^{2} + 60 yd^{2} + 150 yd^{2} + 150 yd^{2} = 416 yd^{2}

Exercise 3.

Answer:

Area of top and bottom: 2((8 ft. × 6 ft.) + (7 ft. × 2 ft.))

= 2(48 ft^{2} + 14 ft^{2} )

= 2(62 ft^{2} ) = 124 ft^{2}

Area for back: 8 ft. × 3 ft. = 24 ft^{2}

Area for front: 7 ft. × 3 ft. = 21 ft^{2}

Area of corner cutout: (2 ft. × 3 ft.) + (1 ft. × 3 ft.) = 9 ft^{2}

Area of right side: 6 ft. × 3 ft. = 18 ft^{2}

Area of left side: 8 ft. × 3 ft. = 24 ft^{2}

Surface area: 124 ft^{2} + 24 ft^{2} + 21 ft^{2} + 9 ft^{2} + 18 ft^{2} + 24 ft^{2} = 220 ft^{2}

Exercise 4.

Answer:

Surface area of top prism:

Area of top: 4 m × 5 m = 20 m^{2}

Area of front and back sides: 2(4 m × 5 m) = 40 m^{2}

Area of left and right sides: 2(5 m × 5 m) = 50 m^{2}

Total surface area of top prism: 20 m^{2} + 40 m^{2} + 50 m^{2} = 110 m^{2}

Surface area of bottom prism:

Area of top: 10 m × 10 m – 20 m^{2} = 80 m^{2}

Area of bottom: 10 m × 10 m = 100 m^{2}

Area of front and back sides: 2(10 m × 3 m) = 60 m^{2}

Area of left and right sides: 2(10 m × 3 m) = 60 m^{2}

Total surface area of bottom prism: 80 m^{2} + 100 m^{2} + 60 m^{2} + 60 m^{2}

= 300 m^{2}

Surface area: 110 m^{2} + 300 m^{2} = 410 m^{2}

Exercise 5.

Answer:

Area of top and bottom faces: 2(10 in. × 2 in.) + 2(6 in. × 2 in.)

= 40 in^{2} + 24 in^{2}

= 64 in^{2}

Area of lateral faces: 2(9 in. × 2 in.) + 2(6 in. × 9 in.) + 2(9 in. × 10 in.)

= 36 in^{2} + 108 in^{2} + 180 in^{2}

= 324 in^{2}

Surface area:

64 in^{2} + 324 in^{2} = 388 in^{2}

### Eureka Math Grade 7 Module 6 Lesson 23 Problem Set Answer Key

Determine the surface area of the figures.

Question 1.

Answer:

Area of top and bottom: 2(9 cm × 4 cm) = 72 cm^{2}

Area of left and right sides: 2(4 cm × 9 cm) = 72 cm^{2}

Area of front and back: 2(9 cm × 4 cm) + 2(4.5 cm × 5 cm)

= 117 cm^{2}

Surface area: 72 cm^{2} + 72 cm^{2} + 117 cm^{2} = 261 cm^{2}

Question 2.

Answer:

Area of front and back: 2(9 ft. × 2 ft.) = 36 ft^{2}

Area of sides: 2(8 ft. × 2 ft.) = 32 ft^{2}

Area of top and bottom: 2(9 ft. × 8 ft.) – 2(4 ft. × 3 ft.)

= 144 ft^{2} – 24 ft^{2}

= 120 ft^{2}

Area of interior sides: 2(4 ft. × 2 ft.) = 16 ft^{2}

Surface area: 36 ft^{2} + 32 ft^{2} + 120 ft^{2} + 16 ft^{2} = 204 ft^{2}

Question 3.

Answer:

Surface area of top prism:

Area of top: 8 in. × 6 in. = 48 in^{2}

Area of front and back sides: 2(6 in. × 8 in.) = 96 in^{2}

Area of left and right sides: 2(8 in. × 8 in.) = 128 in^{2}

Total surface area of top prism: 48 in^{2} + 96 in^{2} + 128 in^{2} = 272 in^{2}

Surface area of bottom prism:

Area of top: 16 in. × 16 in. – 48 in^{2} = 208 in^{2}

Area of bottom: 16 in. × 16 in. = 256 in^{2}

Area of front and back sides: 2(16 in. × 4 in.) = 128 in^{2}

Area of left and right sides: 2(16 in. × 4 in.) = 128 in^{2}

Total surface area of bottom prism: 208 in^{2} + 256 in^{2} + 128 in^{2} + 128 in^{2}

= 720 in^{2}

Surface area: 272 in^{2} + 720 in^{2} = 992 in^{2}

Question 4.

Answer:

Area of the rectangle base: 14 ft. × 30 ft. = 420 ft^{2}

Area of the triangular lateral sides:

Area of front and back: \(\frac{1}{2}\)bh

= 2(\(\frac{1}{2}\)(14 ft.)(24 ft.))

= 336 ft^{2}

Area that can be seen from left and right: \(\frac{1}{2}\)bh

= 2(1\(\frac{1}{2}\)(30 ft.)(20 ft.))

= 600 ft^{2}

Surface area: 420 ft^{2} + 336 ft^{2} + 600 ft^{2} = 1,356 ft^{2}

Question 5.

Answer:

Area of front and back: 2(\(\frac{1}{2}\)(8 cm × 15 cm))

= 8 cm × 15 cm

= 120 cm^{2}

Area of bottom: 8 cm × 25 cm

= 200 cm^{2}

Area that can be seen from left side: 25 cm × 15 cm

= 375 cm^{2}

Area that can be seen from right side: 25 cm × 17 cm

= 425 cm^{2}

Surface area: 120 cm^{2} + 200 cm^{2} + 375 cm^{2} + 425 cm^{2} = 1,120 cm^{2}

### Eureka Math Grade 7 Module 6 Lesson 23 Exit Ticket Answer Key

Determine and explain how to find the surface area of the following right prisms.

Question 1.

Answer:

Area of top and bottom: 2(\(\frac{1}{2}\)(12 ft. × 5 ft.))

= 12 ft. × 5 ft.

= 60 ft^{2}

Area of front: 12 ft. × 15 ft.

= 180 ft^{2}

Area seen from left: 13 ft. × 15 ft.

= 195 ft^{2}

Area seen from right: 5 ft. × 15 ft.

= 75 ft^{2}

Surface area: 60 ft^{2} + 180 ft^{2} + 195 ft^{2} + 75 ft^{2}

= 510 ft^{2}

To 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.

Question 2.

Answer:

Area of front and back: 2(10 ft. × 1 ft.) = 20 ft^{2}

Area of sides: 2(10 ft. × 1 ft.) = 20 ft^{2}

Area of top and bottom: 2(10 ft. × 5 ft.) + 2(4 ft. × 5 ft.)

= 100 ft^{2} + 40 ft^{2}

= 140 ft^{2}

Surface area: 20 ft^{2} + 20 ft^{2} + 140 ft^{2}

= 180 ft^{2}

To 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.