Lesson 24: Surface Area

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Lesson 24: Surface Area

Student Outcomes

 Students determine the surface area of three-dimensional figures, those that are composite figures and those that have missing sections.

Lesson Notes

This lesson is a continuation of Lesson 23. Students will continue to work on surface area advancing to figures with missing sections.

Classwork

Example 1 (8 minutes)

Students should solve this problem on their own.

Example 1

Determine the surface area of the image.

Surface area of top and bottom prisms:

Lateral sides: = 𝟒𝟒(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.× 𝟑𝟑 𝐢𝐢𝐢𝐢. ) = 𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏 Base face: = 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. = 𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏 Base face with hole: = 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. −𝟒𝟒 𝐢𝐢𝐢𝐢.× 𝟒𝟒 𝐢𝐢𝐢𝐢.

= 𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

There are two of these, making up 𝟏𝟏𝟑𝟑𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏. Surface area of middle prism:

Lateral sides = 𝟒𝟒(𝟒𝟒 𝐢𝐢𝐢𝐢.× 𝟏𝟏 𝐢𝐢𝐢𝐢. ) = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏 Surface area: 𝟏𝟏𝟑𝟑𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏= 𝟗𝟗𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏

 Describe the method you used to determine the surface area.

 Answers 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.

 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?

 No, 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.× 4 in. squares (64 in

²

).

Scaffolding:

As in Lesson 23, students can draw nets of the figures to help them visualize the area of the faces. They could determine the area of these without the holes first and subtract the surface area of the holes.

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MP.8

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7•6

Lesson 24

Example 2 (5 minutes)

Example 2

a. Determine the surface area of the cube.

 Explain how 6(12 in. )

²

represents the surface area of the cube.

 The area of one face, one square with a side length of 12 in., is (12 in. )

²

, and so a total area of all six faces is 6(12 in. )

²

.

b. A square hole with a side length of 𝟒𝟒 inches is drilled through the cube. Determine the new surface area.

 How does drilling a hole in the cube change the surface area?

 We have to subtract the area of the square at the surface from each end.

 What happens to the surfaces that now show inside the cube?

 These are now part of the surface area.

 What is the shape of the piece that was removed from the cube?

 A rectangular prism was drilled out of the cube with the following dimensions: 4 in.× 4 in.× 12 in.

 How can we use this to help us determine the new total surface area?

 We 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.

 Why is the surface area larger when holes have been cut into the cube?

 There are more surfaces showing now. All of the surfaces need to be included in the surface area.

Scaffolding:

As in Lesson 23, students can draw nets of the figures to help them visualize the area of the faces. They could determine the area of these without the holes first and subtract the surface area of the holes.

𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒 𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒 = 𝟗𝟗𝒔𝒔^𝟏𝟏 𝑺𝑺𝑺𝑺 = 𝟗𝟗(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢

.

⁾^𝟏𝟏

𝑺𝑺𝑺𝑺 = 𝟗𝟗(𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏) 𝑺𝑺𝑺𝑺 = 𝟏𝟏𝟗𝟗𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏

Area of interior lateral sides:

= 𝟒𝟒(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢

.

^{× 𝟒𝟒 𝐢𝐢𝐢𝐢}

.

⁾

= 𝟏𝟏𝟗𝟗𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

Surface area of cube with holes:

= 𝟗𝟗 (𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢

.

⁾^𝟏𝟏− 𝟏𝟏(𝟒𝟒 𝐢𝐢𝐢𝐢

.

× 𝟒𝟒 𝐢𝐢𝐢𝐢

.

) + 𝟒𝟒(𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢

.

× 𝟒𝟒 𝐢𝐢𝐢𝐢

.

)

= 𝟏𝟏𝟗𝟗𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏− 𝟑𝟑𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏+ 𝟏𝟏𝟗𝟗𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

= 𝟏𝟏𝟗𝟗𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏

Lesson 24: Surface Area

267 A STORY OF RATIOS

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 Explain how the expression 6(12 in. )

²

− 2(4 in.× 4 in. ) + 4(12 in.× 4 in. ) represents the surface area of the cube with the hole.

 From the total surface area of a whole (uncut) cube, 6(12 in. )

²

, the area of the bases (the cuts made to the surface of the cube) are subtracted: 6(12 in. )

²

− 2(4 in.× 4 in. ). To this expression we add the area of the four lateral faces of the cutout prism, 4(12 in.× 4 in. ). The complete expression then is 6(12 in. )

²

− 2(4 in.× 4 in. ) + 4(12 in.× 4 in. ).

Example 3 (5 minutes)

Example 3

A right rectangular pyramid has a square base with a side length of 𝟏𝟏𝟗𝟗 inches. The surface area of the pyramid is 𝟏𝟏𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏. Find the height of the four lateral triangular faces.

Area of base = 𝟏𝟏𝟗𝟗 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟗𝟗 𝐢𝐢𝐢𝐢. = 𝟏𝟏𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏 Area of the four faces = 𝟏𝟏𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏− 𝟏𝟏𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏= 𝟏𝟏𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏

The total area of the four faces is 𝟏𝟏𝟗𝟗𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏. Therefore, the area of each triangular face is 𝟒𝟒𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏. Area of lateral side = 𝟏𝟏𝟏𝟏𝒃𝒃𝒃𝒃

𝟒𝟒𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏=𝟏𝟏

𝟏𝟏(𝟏𝟏𝟗𝟗 𝐢𝐢𝐢𝐢. )𝒃𝒃 𝟒𝟒𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏= (𝟓𝟓 𝐢𝐢𝐢𝐢.)𝒃𝒃

𝒃𝒃 = 𝟏𝟏 𝐢𝐢𝐢𝐢.

The height of each lateral triangular face is 𝟏𝟏 inches.

 What strategies could you use to help you solve this problem?

 I could draw a picture of the pyramid and label the sides so that I can visualize what the problem is asking me to do.

 What information have we been given? How can we use the information?

 We know the total surface area, and we know the length of the sides of the square.

 We can use the length of the sides of the square to give us the area of the square base.

 How will the area of the base help us determine the slant height?

 First, we can subtract the area of the base from the total surface area in order to determine what is left for the lateral sides.

 Next, we can divide the remaining area by 4 to get the area of just one triangular face.

 Finally, 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.

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7•6

Lesson 24

Exercises 1–8 (20 minutes)

Students work in pairs to complete the exercises.

Exercises

Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise.

1.

2. In addition to your calculation, explain how the surface area of the following figure was determined.

The surface area of the prism is found by taking the sum of the areas of the trapezoidal front and the areas of the back of the four different-sized rectangles that make up the lateral faces.

Area top = 𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟏𝟏 𝐒𝐒𝐜𝐜

= 𝟏𝟏𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏 Area bottom = 𝟏𝟏𝟒𝟒 𝐒𝐒𝐜𝐜 × 𝟏𝟏 𝐒𝐒𝐜𝐜

= 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏

Area sides = (𝟏𝟏𝟑𝟑 𝐒𝐒𝐜𝐜 × 𝟏𝟏 𝐒𝐒𝐜𝐜) + (𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜 × 𝟏𝟏 𝐒𝐒𝐜𝐜)

= 𝟑𝟑𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏 Area front and back = 𝟏𝟏 �𝟏𝟏

𝟏𝟏(𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜 + 𝟏𝟏𝟑𝟑 𝐒𝐒𝐜𝐜)(𝟏𝟏𝟒𝟒 𝐒𝐒𝐜𝐜)�

= 𝟏𝟏(𝟒𝟒𝟗𝟗𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏)

= 𝟗𝟗𝟑𝟑𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Surface area = 𝟏𝟏𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟑𝟑𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟗𝟗𝟑𝟑𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

= 𝟏𝟏𝟗𝟗𝟒𝟒𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Top and bottom = 𝟏𝟏(𝟏𝟏𝟏𝟏 𝐜𝐜 × 𝟓𝟓 𝐜𝐜) = 𝟏𝟏𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏 Extra interior sides = 𝟏𝟏(𝟓𝟓 𝐜𝐜 × 𝟕𝟕 𝐜𝐜) = 𝟕𝟕𝟗𝟗 𝐜𝐜^𝟏𝟏 Left and right sides = 𝟏𝟏(𝟓𝟓 𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐜𝐜) = 𝟏𝟏𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏 Front and back sides = 𝟏𝟏�(𝟏𝟏𝟏𝟏 𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐜𝐜) − (𝟏𝟏 𝐜𝐜 × 𝟕𝟕 𝐜𝐜)�

= 𝟏𝟏(𝟏𝟏𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏− 𝟓𝟓𝟗𝟗 𝐜𝐜^𝟏𝟏)

= 𝟏𝟏(𝟏𝟏𝟗𝟗𝟗𝟗 𝐜𝐜^𝟏𝟏)

= 𝟑𝟑𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏

Surface area = 𝟏𝟏𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏+ 𝟕𝟕𝟗𝟗 𝐜𝐜^𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏+ 𝟑𝟑𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏

= 𝟗𝟗𝟗𝟗𝟗𝟗 𝐜𝐜^𝟏𝟏

269 A STORY OF RATIOS

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3.

There are two such rectangular prisms, so the surface area of both is 𝟕𝟕𝟑𝟑𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏.

4. In addition to your calculation, explain how the surface area was determined.

The 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.

5. A hexagonal prism has the following base and has a height of 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮. Determine the surface area of the prism.

Area of bases = 𝟏𝟏(𝟒𝟒𝟏𝟏 + 𝟗𝟗 + 𝟏𝟏𝟗𝟗 + 𝟏𝟏𝟗𝟗) 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

= 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏 Area of 𝟓𝟓 unit sides = 𝟒𝟒(𝟓𝟓 × 𝟏𝟏) 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

= 𝟏𝟏𝟗𝟗𝟗𝟗 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

Area of other sides = (𝟒𝟒 × 𝟏𝟏) 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏+ (𝟏𝟏𝟏𝟏 × 𝟏𝟏) 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

= 𝟏𝟏𝟏𝟏𝟏𝟏 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

Surface area = 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏+ 𝟏𝟏𝟗𝟗𝟗𝟗 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟏𝟏 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

= 𝟒𝟒𝟓𝟓𝟗𝟗 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

Surface area of prisms on the sides:

Area of front and back = 𝟏𝟏(𝟏𝟏 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢. )

= 𝟓𝟓𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏

Area of top and bottom = 𝟏𝟏(𝟏𝟏 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟗𝟗 𝐢𝐢𝐢𝐢. )

= 𝟒𝟒𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏

Area of side = 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟗𝟗 𝐢𝐢𝐢𝐢. = 𝟏𝟏𝟒𝟒𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏 Area of side with hole = 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟗𝟗 𝐢𝐢𝐢𝐢. −𝟑𝟑 𝐢𝐢𝐢𝐢.× 𝟑𝟑 𝐢𝐢𝐢𝐢.

= 𝟏𝟏𝟑𝟑𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

Area of base = 𝟗𝟗 𝐒𝐒𝐮𝐮.× 𝟗𝟗 𝐒𝐒𝐮𝐮.

= 𝟏𝟏𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏 Area of rectangular sides = 𝟒𝟒(𝟗𝟗 𝐒𝐒𝐮𝐮.× 𝟓𝟓 𝐒𝐒𝐮𝐮. )

= 𝟏𝟏𝟏𝟏𝟗𝟗 𝐒𝐒𝐮𝐮^𝟏𝟏

Area of triangular sides = 𝟒𝟒 �𝟏𝟏𝟏𝟏(𝟗𝟗 𝐒𝐒𝐮𝐮. )(𝟗𝟗 𝐒𝐒𝐮𝐮. )�

= 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏

Surface area = 𝟏𝟏𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏+ 𝟏𝟏𝟏𝟏𝟗𝟗 𝐒𝐒𝐮𝐮^𝟏𝟏+ 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏

= 𝟑𝟑𝟗𝟗𝟗𝟗 𝐒𝐒𝐮𝐮^𝟏𝟏 Surface area of middle prism:

Area of front and back = 𝟏𝟏(𝟑𝟑 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. ) = 𝟕𝟕𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏 Area of sides = 𝟏𝟏(𝟑𝟑 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟏𝟏 𝐢𝐢𝐢𝐢. ) = 𝟕𝟕𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏 Surface area = 𝟕𝟕𝟑𝟑𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏+ 𝟏𝟏𝟒𝟒𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏= 𝟏𝟏𝟕𝟕𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

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7•6

Lesson 24

6. Determine the surface area of each figure.

a.

b. A cube with a square hole with 𝟑𝟑 𝐜𝐜 side lengths has been drilled through the cube.

c. A second square hole with 𝟑𝟑 𝐜𝐜 side lengths has been drilled through the cube.

7. The figure below shows 𝟏𝟏𝟏𝟏 cubes with an edge length of 𝟏𝟏 unit. Determine the surface area.

𝑺𝑺𝑺𝑺 = 𝟗𝟗𝒔𝒔^𝟏𝟏

= 𝟗𝟗(𝟗𝟗 𝐜𝐜)^𝟏𝟏

= 𝟗𝟗(𝟏𝟏𝟏𝟏𝐜𝐜^𝟏𝟏)

= 𝟒𝟒𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏

Lateral sides of the hole = 𝟒𝟒(𝟗𝟗 𝐜𝐜 × 𝟑𝟑 𝐜𝐜) = 𝟏𝟏𝟗𝟗𝟏𝟏 𝐜𝐜^𝟏𝟏

Surface area of cube with holes = 𝟒𝟒𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏− 𝟏𝟏(𝟑𝟑 𝐜𝐜 × 𝟑𝟑 𝐜𝐜) + 𝟏𝟏𝟗𝟗𝟏𝟏 𝐜𝐜^𝟏𝟏

= 𝟓𝟓𝟕𝟕𝟗𝟗 𝐜𝐜^𝟏𝟏

𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒 𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒 = 𝟓𝟓𝟕𝟕𝟗𝟗 𝐜𝐜^𝟏𝟏− 𝟒𝟒(𝟑𝟑 𝐜𝐜 × 𝟑𝟑 𝐜𝐜) + 𝟏𝟏�𝟒𝟒(𝟑𝟑 𝐜𝐜 × 𝟑𝟑 𝐜𝐜)�

= 𝟗𝟗𝟏𝟏𝟏𝟏 𝐜𝐜^𝟏𝟏

Area top and bottom = 𝟏𝟏𝟒𝟒 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏 Area sides = 𝟏𝟏𝟏𝟏 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏 Area front and back = 𝟏𝟏𝟏𝟏 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

Surface area = 𝟏𝟏𝟒𝟒 + 𝟏𝟏𝟏𝟏 + 𝟏𝟏𝟏𝟏

= 𝟕𝟕𝟗𝟗 𝐒𝐒𝐢𝐢𝐢𝐢𝐮𝐮𝐮𝐮^𝟏𝟏

271 A STORY OF RATIOS

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8. The base rectangle of a right rectangular prism is 𝟒𝟒 𝐒𝐒𝐮𝐮.× 𝟗𝟗 𝐒𝐒𝐮𝐮. The surface area is 𝟏𝟏𝟏𝟏𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏. Find the height.

Let 𝒃𝒃 be the height in feet.

Area of one base: 𝟒𝟒 𝐒𝐒𝐮𝐮.× 𝟗𝟗 𝐒𝐒𝐮𝐮. = 𝟏𝟏𝟒𝟒 𝐒𝐒𝐮𝐮^𝟏𝟏 Area of two bases: 𝟏𝟏(𝟏𝟏𝟒𝟒 𝐒𝐒𝐮𝐮^𝟏𝟏) = 𝟒𝟒𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏 Numeric area of four lateral faces: 𝟏𝟏𝟏𝟏𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏− 𝟒𝟒𝟏𝟏 𝐒𝐒𝐮𝐮^𝟏𝟏= 𝟏𝟏𝟒𝟒𝟗𝟗 𝐒𝐒𝐮𝐮^𝟏𝟏 Algebraic area of four lateral faces: 𝟏𝟏(𝟗𝟗𝒃𝒃 + 𝟒𝟒𝒃𝒃)

Solve for 𝒃𝒃.

𝟏𝟏(𝟗𝟗𝒃𝒃 + 𝟒𝟒𝒃𝒃) = 𝟏𝟏𝟒𝟒𝟗𝟗 𝟏𝟏𝟗𝟗𝒃𝒃 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝒃𝒃 = 𝟏𝟏𝟏𝟏 The height is 𝟏𝟏𝟏𝟏 feet.

Closing (2 minutes)

 Write down three tips that you would give a friend that is trying to calculate surface area.

Exit Ticket (5 minutes)

Lesson Summary

 To calculate the surface area of a composite figure, determine the surface area of each prism separately, and add them together. From the sum, subtract the area of the sections that were covered by another prism.

 To calculate the surface area with a missing section, find the total surface area of the whole figure. From the total surface area, subtract the area of the missing parts. Then, add the area of the lateral faces of the cutout prism.

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7•6

Lesson 24

Name Date

Lesson 24: Surface Area

Exit Ticket

Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area.

273 A STORY OF RATIOS

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Exit Ticket Sample Solutions

Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area.

Area of top and

bottom = 𝟏𝟏(𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟗𝟗 𝐒𝐒𝐜𝐜)

= 𝟏𝟏𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏 Area of sides = 𝟏𝟏(𝟗𝟗 𝐒𝐒𝐜𝐜 × 𝟏𝟏 𝐒𝐒𝐜𝐜)

= 𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Area of front and back = 𝟏𝟏(𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟏𝟏 𝐒𝐒𝐜𝐜) − 𝟒𝟒(𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟓𝟓 𝐒𝐒𝐜𝐜)

= 𝟏𝟏𝟒𝟒𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏 Area inside = 𝟏𝟏(𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟗𝟗 𝐒𝐒𝐜𝐜)

= 𝟏𝟏𝟒𝟒𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Surface area = 𝟏𝟏𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟏𝟏𝟒𝟒𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟏𝟏𝟒𝟒𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

= 𝟗𝟗𝟓𝟓𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Take 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 bases of the main rectangular prism. Finally, add the lateral faces of the prisms that were cut out of the main prism.

Problem Set Sample Solutions

Determine the surface area of each figure.

1. In addition to the calculation of the surface area, describe how you found the surface area.

Area of top = 𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟑𝟑 𝐒𝐒𝐜𝐜

= 𝟑𝟑𝟗𝟗𝟒𝟒 𝐒𝐒𝐜𝐜^𝟏𝟏 Area of bottom = 𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜

= 𝟑𝟑𝟑𝟑𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Area of left and right sides = 𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜 + 𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜

= 𝟗𝟗𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Area of front and back sides = 𝟏𝟏 �(𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜) + 𝟏𝟏𝟏𝟏(𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜)�

= 𝟏𝟏(𝟏𝟏𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟑𝟑𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏)

= 𝟏𝟏(𝟏𝟏𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏)

= 𝟒𝟒𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Surface area = 𝟑𝟑𝟗𝟗𝟒𝟒 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟑𝟑𝟑𝟑𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟗𝟗𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟒𝟒𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

= 𝟏𝟏𝟏𝟏𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

Split 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.

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32 m

7•6

Lesson 24

2.

Area of front and back = 𝟏𝟏(𝟏𝟏𝟏𝟏. 𝟒𝟒 𝐢𝐢𝐢𝐢.× 𝟏𝟏. 𝟗𝟗 𝐢𝐢𝐢𝐢. ) = 𝟑𝟑𝟏𝟏𝟗𝟗. 𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏 Area of sides = 𝟏𝟏(𝟏𝟏. 𝟗𝟗 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢. ) = 𝟒𝟒𝟏𝟏𝟏𝟏. 𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

Area of top and bottom = 𝟏𝟏�(𝟏𝟏𝟏𝟏. 𝟒𝟒 𝐢𝐢𝐢𝐢.× 𝟏𝟏𝟒𝟒 𝐢𝐢𝐢𝐢. ) − (𝟒𝟒. 𝟏𝟏 𝐢𝐢𝐢𝐢.× 𝟏𝟏. 𝟏𝟏 𝐢𝐢𝐢𝐢. )�

= 𝟏𝟏(𝟒𝟒𝟒𝟒𝟏𝟏. 𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏− 𝟏𝟏𝟗𝟗. 𝟓𝟓𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏)

= 𝟏𝟏(𝟒𝟒𝟑𝟑𝟏𝟏. 𝟗𝟗𝟒𝟒 𝐢𝐢𝐢𝐢^𝟏𝟏)

= 𝟏𝟏𝟗𝟗𝟏𝟏. 𝟗𝟗𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

Surface area = 𝟑𝟑𝟏𝟏𝟗𝟗. 𝟒𝟒𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏+ 𝟒𝟒𝟏𝟏𝟏𝟏. 𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏+ 𝟏𝟏𝟗𝟗𝟏𝟏. 𝟗𝟗𝟏𝟏 𝐢𝐢𝐢𝐢^𝟏𝟏

= 𝟏𝟏𝟓𝟓𝟗𝟗𝟏𝟏. 𝟑𝟑𝟗𝟗 𝐢𝐢𝐢𝐢^𝟏𝟏

3.

Area of front and back = 𝟏𝟏 �𝟏𝟏

𝟏𝟏(𝟑𝟑𝟏𝟏 𝐜𝐜 + 𝟏𝟏𝟗𝟗 𝐜𝐜)𝟏𝟏𝟓𝟓 𝐜𝐜�

= 𝟕𝟕𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏 Area of top = 𝟏𝟏𝟗𝟗 𝐜𝐜 × 𝟑𝟑𝟗𝟗 𝐜𝐜

= 𝟓𝟓𝟕𝟕𝟗𝟗 𝐜𝐜^𝟏𝟏 Area of left and right sides = 𝟏𝟏(𝟏𝟏𝟕𝟕 𝐜𝐜 × 𝟑𝟑𝟗𝟗 𝐜𝐜)

= 𝟏𝟏(𝟗𝟗𝟏𝟏𝟏𝟏 𝐜𝐜^𝟏𝟏)

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜^𝟏𝟏 Area of bottom = 𝟑𝟑𝟏𝟏 𝐜𝐜 × 𝟑𝟑𝟗𝟗 𝐜𝐜

= 𝟏𝟏𝟏𝟏𝟓𝟓𝟏𝟏 𝐜𝐜^𝟏𝟏

Surface area = 𝟕𝟕𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏+ 𝟏𝟏, 𝟏𝟏𝟓𝟓𝟏𝟏 𝐜𝐜^𝟏𝟏+ 𝟏𝟏, 𝟏𝟏𝟏𝟏𝟒𝟒 𝐜𝐜^𝟏𝟏+ 𝟓𝟓𝟕𝟕𝟗𝟗 𝐜𝐜^𝟏𝟏

= 𝟑𝟑𝟗𝟗𝟕𝟕𝟏𝟏 𝐜𝐜^𝟏𝟏

4. Determine the surface area after two square holes with a side length of 𝟏𝟏 𝐜𝐜 are drilled through the solid figure composed of two rectangular prisms.

Surface area of top prism before the hole is drilled:

Area of top = 𝟒𝟒 𝐜𝐜 × 𝟓𝟓 𝐜𝐜

= 𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏 Area of front and back = 𝟏𝟏(𝟒𝟒 𝐜𝐜 × 𝟓𝟓 𝐜𝐜)

= 𝟒𝟒𝟗𝟗 𝐜𝐜^𝟏𝟏 Area of sides = 𝟏𝟏(𝟓𝟓 𝐜𝐜 × 𝟓𝟓 𝐜𝐜)

= 𝟓𝟓𝟗𝟗 𝐜𝐜^𝟏𝟏

Surface area of bottom prism before the hole is drilled:

Area of top = 𝟏𝟏𝟗𝟗 𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐜𝐜 − 𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏

= 𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏 Area of bottom = 𝟏𝟏𝟗𝟗 𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐜𝐜

= 𝟏𝟏𝟗𝟗𝟗𝟗 𝐜𝐜^𝟏𝟏 Surface area of interiors:

Area of front and back = 𝟏𝟏(𝟏𝟏𝟗𝟗 𝐜𝐜 × 𝟑𝟑 𝐜𝐜)

= 𝟗𝟗𝟗𝟗 𝐜𝐜^𝟏𝟏

Area of interiors = 𝟒𝟒(𝟏𝟏 𝐜𝐜 × 𝟒𝟒 𝐜𝐜) + 𝟒𝟒(𝟏𝟏 𝐜𝐜 × 𝟑𝟑 𝐜𝐜)

= 𝟓𝟓𝟗𝟗 𝐜𝐜^𝟏𝟏 Area of sides = 𝟏𝟏(𝟏𝟏𝟗𝟗 𝐜𝐜 × 𝟑𝟑 𝐜𝐜)

= 𝟗𝟗𝟗𝟗 𝐜𝐜^𝟏𝟏

𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒 𝐒𝐒𝐒𝐒𝐒𝐒𝐒𝐒 = 𝟏𝟏𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏+ 𝟑𝟑𝟗𝟗𝟗𝟗 𝐜𝐜^𝟏𝟏+ 𝟓𝟓𝟗𝟗 𝐜𝐜^𝟏𝟏− 𝟏𝟏𝟗𝟗 𝐜𝐜^𝟏𝟏

= 𝟒𝟒𝟓𝟓𝟗𝟗 𝐜𝐜^𝟏𝟏

275 A STORY OF RATIOS

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5. The base of a right prism is shown below. Determine the surface area if the height of the prism is 𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜. Explain how you determined the surface area.

Take 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.

Area of sides = (𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜) + (𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜) + (𝟏𝟏𝟒𝟒 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜) + (𝟕𝟕 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜)

= 𝟏𝟏𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟏𝟏𝟓𝟓𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟏𝟏𝟒𝟒𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟕𝟕𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

= 𝟗𝟗𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏 Area of bases = 𝟏𝟏 �𝟏𝟏

𝟏𝟏(𝟕𝟕 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟒𝟒 𝐒𝐒𝐜𝐜) +𝟏𝟏

𝟏𝟏(𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜)�

= (𝟕𝟕 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟒𝟒 𝐒𝐒𝐜𝐜) + (𝟏𝟏𝟗𝟗 𝐒𝐒𝐜𝐜 × 𝟏𝟏𝟓𝟓 𝐒𝐒𝐜𝐜)

= 𝟏𝟏𝟗𝟗𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟑𝟑𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏

= 𝟒𝟒𝟗𝟗𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏

Surface area = 𝟗𝟗𝟗𝟗𝟗𝟗 𝐒𝐒𝐜𝐜^𝟏𝟏+ 𝟒𝟒𝟗𝟗𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏

= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐒𝐒𝐜𝐜^𝟏𝟏