CorrectionKey=TX-B LESSON 17 . 2 DO NOT EDIT--Changes must ... · 3. prism as part of their plan for finding the surface A gift box is a rectangular prism with length 9.8 cm, width
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G.11.C Apply the formulas for the total and lateral surface area of three-dimensional figures, including prisms, ... cylinders, ... to solve problems using appropriate units of measure. Also G.10.B
Explore Developing a Surface Area FormulaSurface area is the total area of all the faces and curved surfaces of a three-dimensional figure. The lateral area of a prism is the sum of the areas of the lateral faces.
Consider the right prism shown here and the net for the right prism. Complete the figure by labeling the dimensions of the net.
In the net, what type of figure is formed by the lateral faces of the prism?
Write an expression for the length of the base of the rectangle.
How is the base of the rectangle related to the perimeter of the base of the prism?
The lateral area L of the prism is the area of the rectangle. Write a formula for L in terms of h, a, b, and c.
rectangle
a + b + c
They are equal.
L = h (a + b + c)
Module 17 1021 Lesson 2
17 . 2 Surface Area of Prisms and Cylinders
Essential Question: How can you find the surface area of a prism or cylinder?
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Texas Math StandardsThe student is expected to:
G.11.C
Apply the formulas for the total and lateral surface area of three-dimensional figures, including prisms, . . . cylinders, . . . to solve problems using appropriate units of measure. Also G.10.B
Mathematical Processes
G.1.F
Analyze mathematical relationships to connect and communicate mathematical ideas.
Language Objective
1.B, 2.E.3, 3.E, 3.H.3, 4.D
Explain to a partner how to find the surface area of prisms and cylinders.
HARDCOVER PAGES 823832
Turn to these pages to find this lesson in the hardcover student edition.
Surface Area of Prisms and Cylinders
ENGAGE Essential Question: How can you find the surface area of a prism or a cylinder?You find the lateral area and then add twice the
area of a base.
PREVIEW: LESSON PERFORMANCE TASKView the Engage section online. Discuss the photograph. Ask students to identify the subject of the photo and to speculate on the significance of the surface area on determining how items are packaged. Then preview the Lesson Performance Task. 1021
HARDCOVER PAGES
Turn to these pages to find this lesson in the hardcover student edition.
G.11.C Apply the formulas for the total and lateral surface area of three-dimensional figures,
including prisms, ... cylinders, ... to solve problems using appropriate units of measure.
F Write the formula for L in terms of P, where P is the perimeter of the base of the prism.
G Let B be the area of the base of the prism. Write a formula for the surface area S of the prism in terms of B and L. Then write the formula in terms of B, P, and h.
Reflect
1. Explain why the net of the lateral surface of any right prism will always be a rectangle.
2. Suppose a rectangular prism has length ℓ, width w, and height h, as shown. Explain how you can write a formula for the surface area of the prism in terms of ℓ, w, and h.
Explain 1 Finding the Surface Area of a Prism
Lateral Area and Surface Area of Right Prisms
The lateral area of a right prism with height h and base perimeter P is L = Ph.
The surface area of a right prism with lateral area L and base area B is S = L + 2B, or S = Ph + 2B.
S = L + 2B; S = Ph + 2B
Sample answer: Each lateral face of any right prism is a rectangle. The net of the lateral
surface of any right prism is composed of rectangles joined end-to-end. Straight angles
are formed when the rectangles are joined in this manner resulting in one long
rectangular shape.
Sample answer: There are two faces with area ℓw, two faces with area wh, and two faces
with area ℓh, so the surface area can be written as S = 2ℓw + 2wh + 2ℓh.
L = Ph
Module 17 1022 Lesson 2
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Integrate Mathematical ProcessesThis lesson provides an opportunity to address TEKS G.1.F, which calls for students to “analyze relationships.” In this lesson, students analyze three-dimensional figures to determine how they “decompose” into two-dimensional faces, each with its own area, and to find that the sum of the areas of the faces is equal to the surface area of the figure. Since the faces of the figures are polygons or circles, the combined areas generate the lateral area and surface area formulas students will use in this lesson.
EXPLORE Developing a Surface Area Formula
INTEGRATE TECHNOLOGYStudents have the option of doing the Explore activity either in the book or online.
QUESTIONING STRATEGIESIn a prism, how is the lateral area formula related to the surface area formula? The
surface area formula consists of the lateral area plus
the area of the bases.
INTEGRATE MATHEMATICAL PROCESSESFocus on ReasoningHave students brainstorm how to determine what three-dimensional figure can be made from a given net and how the net can be used to find the surface area of the figure. Emphasize that prisms have parallelograms for sides, and cylinders have congruent circular bases.
EXPLAIN 1 Finding the Surface Area of a Prism
QUESTIONING STRATEGIESHow can you use the formula for the area of a parallelogram to find the lateral area of a
prism? Because the lateral faces of a prism are
parallelograms, you can use the parallelogram
formula to find the areas of the lateral faces and
then add them together.
PROFESSIONAL DEVELOPMENT
Surface Area of Prisms and Cylinders 1022
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Example 1 Each gift box is a right prism. Find the total amount of paper needed to wrap each box, not counting overlap.
Step 1 Find the lateral area.
Lateral area formula L = Ph
P = 2 (8) + 2 (6) = 28 cm = 28(12)
Multiply. = 336 c m 2
Step 2 Find the surface area.
Surface area formula S = L + 2B
Substitute the lateral area. = 336 + 2(6)(8)
Simplify. = 432 c m 2
Step 1 Find the length c of the hypotenuse of the base.
Pythagorean Theorem c 2 = a 2 + b 2
Substitute. = 2
+ 2
Simplify. =
Take the square root of each side. c =
Step 2 Find the lateral area.
Lateral area formula L = Ph
Substitute. = ( ) Multiply. = i n 2
10
60 20
26
24
676
1200
Module 17 1023 Lesson 2
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COLLABORATIVE LEARNING
Small Group ActivityHave students work in groups to find the surface areas of various prisms and cylinders. Have students each choose a prism or a cylinder and conjecture how to find the surface area. Then have them draw and label a model or a net and describe how to find the surface area. Ask them to verify or disprove their conjectures, and present their results to the group.
QUESTIONING STRATEGIESWhen can the Pythagorean Theorem be used to find the area of the bases of a triangular
prism? If the bases are right triangles, then the
Pythagorean Theorem can be used to find the
lengths of the legs of the triangles, which are
necessary to find the area of the triangles.
1023 Lesson 17 . 2
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3. A gift box is a rectangular prism with length 9.8 cm, width 10.2 cm, and height 9.7 cm. Explain how to estimate the amount of paper needed to wrap the box, not counting overlap.
Your Turn
Each gift box is a right prism. Find the total amount of paper needed to wrap each box, not counting overlap.
4. 5.
24 101200
1440
Sample answer: Round each dimension to 10 cm. Then each face has an area of
approximately 1 0 2 = 100 c m 2 , and the surface area is approximately 6 (100) = 600 c m 2 .
The lateral area is L = Ph.
P = 2 (18) + 2 (5) = 46 in.
So, L = 46 (5) = 230 i n 2 .
The surface area is S = L + 2B.
B = 18 (5) = 90 i n 2
So, S = 230 + 2 (90) = 410 i n 2 .
Let b be the unknown length of the leg of the base.
By the Pythagorean Theorem, c 2 = a 2 + b 2 ,
so 6 2 = 3. 6 2 + b 2 , 36 = 12.96 + b 2 , and b 2 = 23.04.
Taking the square root of each side shows that b = 4.8 in.
The lateral area is L = Ph.
P = 3.6 + 4.8 + 6 = 14.4 in.
So, L = 14.4 (8.5) = 122.4 i n 2 .
The surface area is S = L + 2B.
B = 1 __ 2 (4.8) (3.6) = 8.64
So, S = 122.4 + 2 (8.64) = 139.68 i n 2 .
Module 17 1024 Lesson 2
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DIFFERENTIATE INSTRUCTION
Multiple RepresentationsHave students work in groups to cover boxes and cylinders with wrapping paper. Ask them to cut the wrap so that it does not overlap, and have them decompose the wraps into nets that they can use to find the surface area. Have groups discuss how the nets are related to the lateral area and the surface area formulas.
INTEGRATE MATHEMATICAL PROCESSESFocus on PatternsEncourage students to make an organized list of the dimensions of the lateral sides and the bases of a prism as part of their plan for finding the surface area. Then have them substitute the appropriate values into the formulas for lateral area and surface area of a prism.
Surface Area of Prisms and Cylinders 1024
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The lateral area of a cylinder is the area of the curved surface that connects the two bases.
The lateral area of a right cylinder with radius r and height h is L = 2πrh.
The surface area of a right cylinder with lateral area L and base area B is S = L + 2B, or S = 2πrh + 2π r 2 .
Example 2 Each aluminum can is a right cylinder. Find the amount of paper needed for the can’s label and the total amount of aluminum needed to make the can. Round to the nearest tenth.
Step 1 Find the lateral area.
Lateral area formula L = 2πrh
Substitute. L = 2π (3) (9)
Multiply. = 54π c m 2
Step 2 Find the surface area.
Surface area formula S = L + 2π r 2
Substitute the lateral area and radius. = 54π + 2r (3) 2
Simplify. = 72π c m 2
Step 3 Use a calculator and round to the nearest tenth.
The amount of paper needed for the label is the lateral area, 54π ≈ 169.6 c m 2 .
The amount of aluminum needed for the can is the surface area, 72π ≈ 226.2 c m 2 .
Module 17 1025 Lesson 2
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EXPLAIN 2 Finding the Surface Area of a Cylinder
QUESTIONING STRATEGIESHow is the height of a right cylinder used to find its surface area? The height is used to
find the lateral area. The lateral area is the
circumference of the base times the height. Adding
the lateral area to the area of the bases gives the
surface area.
1025 Lesson 17 . 2
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Substitute; the radius is half the diameter. = 2π ( ) ( ) Multiply. = π in 2
Step 2 Find the surface area.
Surface area formula S = L + 2π r 2
Substitute the lateral area and radius. = π + 2r ( ) 2 Simplify. = π in 2
Step 3 Use a calculator and round to the nearest tenth.
The amount of paper needed for the label is the lateral area, π ≈ i n 2 .
The amount of aluminum needed for the can is the surface area, π ≈ i n 2 .
Reflect
6. In these problems, why is it best to round only in the final step of the solution?
Your Turn
Each aluminum can is a right cylinder. Find the amount of paper needed for the can’s label and the total amount of aluminum needed to make the can. Round to the nearest tenth.
7. 8.
Sample answer: This results in a more accurate answer. If you round at an intermediate
step, the inaccuracies may be compounded as you perform subsequent operations.
2.5
10
10
10 31.4
22.5 70.7
2.5
22.5
2
The lateral area is L = 2πrh.So, L = 2π (6) (15) = 180π c m 2 .The surface area is S = L + 2π r 2 .So, S = 180π + 2π (6) 2 = 252π c m 2 .The amount of paper needed for the label is the lateral area, 180π ≈ 565.5 c m 2 .The amount of aluminum needed for the can is the surface area, 252π ≈ 791.7 c m 2 .
The radius of the cylinder is half the diameter, so r = 36 mm.The lateral area is L = 2πrh.So, L = 2π (36) (80) = 5760π m m 2 .The surface area is S = L + 2π r 2 .So, S = 5760π + 2π (36) 2 = 8352π m m 2 .The amount of paper needed for the label is the lateral area, 5760π ≈ 18,095.6 m m 2 .The amount of aluminum needed for the can is the surface area, 8352π ≈ 26,238.6 m m 2 .
Module 17 1026 Lesson 2
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AVOID COMMON ERRORSCommon errors students make when applying the surface area formula include multiplying the height of the cylinder by the area of the base; using a diameter in the formula for cylinders instead of a radius; and forgetting to include the area of both bases. Caution students to look for these errors.
Surface Area of Prisms and Cylinders 1026
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Step 1 Find the surface area of the right rectangular prism.
Surface area formula S = Ph + 2B
Substitute. = ( ) + 2 ( ) ( ) Simplify. = c m 2
Step 2 Find the surface area of the cylinder.
Lateral area formula L = 2πrh
Substitute. = 2π ( ) ( ) Simplify. = π c m 2
Surface area formula S = L + 2π r 2
Substitute. = π + 2π ( ) 2
Simplify. = π c m 2
Step 3 Find the surface area of the composite figure. The surface area is the sum of the areas of all surfaces on the exterior of the figure.
S = (prism surface area) + (cylinder surface area) - 2(area of one cylinder base)
= + π - 2π ( ) 2 = + π ≈ c m 2
Reflect
9. Discussion A student said the answer in Part A must be incorrect since a part of the rectangular prism is removed, yet the surface area of the composite figure is greater than the surface area of the rectangular prism. Do you agree with the student? Explain.
202
2
12
12
20
2
202 20
12 239.7
2
202
3
26 5 9 4
No; removing part of the rectangular prism produces a hole through the prism and this
creates additional exposed area on the interior surface of the hole.
Module 17 1028 Lesson 2
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INTEGRATE MATHEMATICAL PROCESSESFocus on PatternsEncourage students to carefully decompose a figure as part of their plan to find its surface area. Have them make an organized list of the dimensions of the lateral sides and of the bases for each figure, along with a list of those areas that are overlapping in the composite figure. Then have them write an equation for the total surface area of the parts, including subtractions for overlapping parts, and substitute the appropriate values into the formulas.
Surface Area of Prisms and Cylinders 1028
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Find the surface area of each composite figure. Round to the nearest tenth.
10. 11.
Elaborate
12. Can the surface area of a cylinder ever be less than the lateral area of the cylinder? Explain.
13. Is it possible to find the surface area of a cylinder if you know the height and the circumference of the base? Explain.
14. Essential Question Check-In How is finding the surface area of a right prism similar to finding the surface area of a right cylinder?
The surface area of the large prism is S large = Ph + 2B.
So, S large = (32) (5) + 2 (9) (7) = 286 i n 2 .
The surface area of the small prism is S small = Ph + 2B.
So, S small = (16) (3) + 2 (5) (3) = 78 i n 2 .
The surface area of the composite figure is the surface area of the large prism plus the surface area of the small prism minus 2 times the area of the base of the small prism.
S = 286 + 78 - 2 (5) (3) = 344 i n 2
The surface area of the large cylinder is S large = 2πrh + 2π r 2 .
So, S large = 2π (7) (6) + 2π (7) 2 = 182π m m 2 .
The lateral area of the small prism is L small = 2πrh.
So, L small = 2π (3) (6) = 36π m m 2 .
The area of each base of the small cylinder is B = π r 2 = π 3 2 = 9π m m 2 .
The surface area of the composite figure is the surface area of the large cylinder plus the lateral area of the small cylinder minus 2 times the area of the base of the small cylinder.
S = 182π + 36π - 2 (9π) = 200π ≈ 628.3 m m 2
No. The surface area is the lateral area plus the area of the two bases. Since the area of the
two bases is greater than 0, the surface area must be greater than the lateral area.
Yes. You can use the circumference of the base to find the radius of the base. Then you can
use the height, circumference, and radius in the surface area formula.
In both cases, you can find the surface area by finding the lateral area and then adding
twice the area of a base.
Module 17 1029 Lesson 2
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LANGUAGE SUPPORT
Connect VocabularyTo help students remember the vocabulary in the lesson, including lateral area and surface area, have students make note cards of several different solid figures and their lateral and surface areas. Then have them use colored pencils to mark the dimensions of each in one color, and the formulas they will use in another color. Have them label the figures with the units and show the substitutions for the formulas. Ask them to share their note cards with other students
ELABORATE QUESTIONING STRATEGIES
How do you find the surface area of a prism? You add the perimeter of the base
times the height to twice the area of the base.
How do you find the surface area of a cylinder? You add the circumference of the
base times the height to twice the area of the base.
SUMMARIZE THE LESSONWhat is the same about finding the surface area of a prism and a cylinder? What is
different? For both a prism and a cylinder, you find
the surface area by finding the lateral area and then
adding twice the area of the base; the bases of
prisms and cylinders are different, so finding the
lateral areas and base areas will require
different processes.
1029 Lesson 17 . 2
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Example 3Finding the Surface Area of a Composite Figure
Exercises 7–10
INTEGRATE MATHEMATICAL PROCESSESFocus on ModelingSome students may benefit from a hands-on approach for finding the surface area of solids. Have students draw simple figures like prisms and cylinders and then discuss in groups how they can find the lateral areas and the surface areas. Have them include a discussion of the properties of the faces of the figures that will help them find the lateral areas or the surface areas.
INTEGRATE MATHEMATICAL PROCESSESFocus on TechnologySome students may benefit from using the programming features of a graphing calculator to find the surface areas of right rectangular prisms and right cylinders. Have students enter the formulas for the surface areas of these simple solids as output from a program, with the dimensions of the solids as inputs.
Surface Area of Prisms and Cylinders 1030
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Find the lateral area and surface area of the cylinder. Leave your answer in terms of π.
5. 6.
Find the total surface area of the composite figure. Round to the nearest tenth.
7. 8.
L = 2πrh
= 2π (3) (4)
= 24π f t 2
S = L + 2π r 2
= 24π + 2π (3) 2
= 24π + 18π = 42π f t 2
L = 24π f t 2
S = 42π f t 2
L = 2πrh
= 2π (5.5) (7)
= 77π i n 2
S = L + 2π r 2
= 77π + 2π (5.5) 2
= 77π + 60.5π = 137.5π i n 2
L = 77π i n 2
S = 137.5π i n 2
Surface Area of Cylinder
L = 2πrh S = L + 2π r 2
= 2π (4) (8) = 64π + 2π (4) 2
= 64π f t 2 = 96π f t 2
Surface Area of Prism
L = Ph S = L + 2B
= (44) 12 = 528 + 2 (14) (8)
= 528 f t 2 = 752 f t 2
96π - π (4) 2 + 752 - π (4) 2 ≈ 953.1 f t 2
S ≈ 953.1 f t 2
Surface Area of Cylinder
L = 2πrh S = L + 2π r 2
= 2π (14) (14) = 392π + 2π 14 2
= 392π f t 2 = 784π f t 2
Lateral Surface Area of Prism
L = Ph
= (40) 14
= 560 f t 2
784π + 560 - 2 (14 ⋅ 6) ≈ 2855.0 f t 2
S ≈ 2855.0 f t 2
Module 17 1031 Lesson 2
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AVOID COMMON ERRORSAs students find the surface area of cylinders, caution them to avoid the common errors of forgetting to include the areas of both bases, or using the diameter of the base instead of the radius in the formula.
1031 Lesson 17 . 2
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Find the total surface area of the composite figure. Round to the nearest tenth.
9. 10.
11. The greater the lateral area of a florescent light bulb, the more light the bulb produces. One cylindrical light bulb is 16 inches long with a 1-inch radius. Another cylindrical bulb is 23 inches long with a 3 __ 4 -inch radius. Which bulb will produce more light?
12. Find the lateral and surface area of a cube with edge length 9 inches.
13. Find the lateral and surface area of a cylinder with base area 64π m 2 and a height 3 meters less than the radius.
Surface Area of PrismL = Ph = (24) 9 = 216 c m 2 The base is a 6–8–10, (3–4–5) , right triangle, so in the area formula b = 8 and h = 6.S = L + 2B = 216 + 2 ( 1 _ 2 (8) (6) ) = 264 c m 2 Lateral Surface Area of CylinderL = 2πrh = 2π (2) (9) = 36π c m 2 264 + 36π - 2 (π 2 2 ) ≈ 352.0 c m 2 S ≈ 352.0 c m 2
Surface Area of PrismL = Ph = (8) 0.5 = 4 f t 2 S = L + 2B = 4 + 2 (2) (2) = 12 f t 2 Total Area of CylinderL = 2πrh = 2π (0.5) (2) = 2π f t 2 S = L + 2π r 2 = 2π + 2π0. 5 2 = 2.5π f t 2 12 + 2.5π - 2 (π0. 5 2 ) ≈ 18.3 f t 2 S ≈ 18.3 f t 2
Lateral Area of 16 inch bulb L = 2πrh = 2π (1) (16) = 32π i n 2
Lateral Area of 23 inch bulb L = 2πrh = 2π (0.75) (23) = 34.5π i n 2 The 23 inch bulb will produce more light.
L = Ph = (36) 9 = 324 i n 2 S = L + 2B = 324 + 2 (9) (9) = 324 + 162 = 486 i n 2 L = 324 i n 2 S = 486 i n 2
Find the Radius A = π r 2 64π = π r 2 64π _ π = π r 2 _ π 64 = r 2 8 = r h = r - 3 h = 8 - 3 h = 5
L = 2πrh = 2π (8) (5) = 80π m 2 S = L + 2π r 2 = 80π + 2π (8) 2 = 208π m 2 L = 80π m 2 S = 208π m 2
Module 17 1032 Lesson 2
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INTEGRATE MATHEMATICAL PROCESSESFocus on Critical ThinkingBecause a cylinder has circular bases, the circumference of the bases is the perimeter of the bases. Therefore, the lateral area of the right cylinder depends on the circumference of the base. If students think about the net for a cylinder, the net includes a rectangle and two circles. That means that the rectangle must have length equal to the circumference of the base.
Surface Area of Prisms and Cylinders 1032
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14. Biology Plant cells are shaped approximately like a right rectangular prism. Each cell absorbs oxygen and nutrients through its surface. Which cell can be expected to absorb at a greater rate? (Hint: 1 μm = 1 micrometer = 0.000001 meter)
15. Find the height of a right cylinder with surface area 160π f t 2 and radius 5 ft.
16. Find the height of a right rectangular prism with surface area 286 m 2 , length 10 m, and width 8 m.
17. Represent Real-World Problems If one gallon of paint covers 250 square feet, how many gallons of paint will be needed to cover the shed, not including the roof? If a gallon of paint costs $25, about how much will it cost to paint the walls of the shed?
Surface Area of Cell 1 Surface Area of Cell 2L = Ph S = L + 2B L = Ph S = L + 2B = (90) 7 = 630 + 2 (35) (10) = (52) 15 = 780 + 2 (15) (11) = 630 μ m 2 = 630 + 700 = 780 μ m 2 = 780 + 330 = 1330 μ m 2 = 1110 μ m 2 The cell that measures 35 μm by 7 μm by 10 μm will absorb at a greater rate.
S = 2πrh + 2π r 2 160π = 2π (5) h + 2π (5) 2 160π = 10πh + 50π 110π = 10πh
110π _ 10π = 10πh _
10π
11 = h h = 11 ft
S = Ph + 2B286 = 36h + 2 (10) (8) 286 = 36h + 160126 = 36h 3.5 = h h = 3.5 m
Front/Back Rectangles + Left/Right Rectangles +
Top Front/Back Triangles
S = 2 (18 ⋅ 12) + 2 (12 ⋅ 12) + 2 ( 1 _ 2
⋅ 18 ⋅ 6) = 432 + 288 + 108 = 828 f t 2
828 f t 2 ⋅ 1 gal
_ 250 f t 2
≈ 3.3 gal
Since you can’t get half a gallon, 4 total gallons will be needed.
4 ⋅ $25 = 100
4 gallons; $100
Module 17 1033 Lesson 2
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1033 Lesson 17 . 2
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18. Match the Surface Area with the appropriate coin in the table.
Coin Diameter (mm) Thickness (mm) Surface Area (m m 2 )
Penny 19.05 1.55
Nickel 21.21 1.95
Dime 17.91 1.35
Quarter 24.26 1.75
A. 836.58
B. 579.82
C. 662.81
D. 1057.86
19. Algebra The lateral area of a right rectangular prism is 144 c m 2 . Its length is three times its width, and its height is twice its width. Find its surface area.
20. A cylinder has a radius of 8 cm and a height of 3 cm. Find the height of another cylinder that has a radius of 4 cm and the same surface area as the first cylinder.
C
A
B
D
Penny L = 2πrh = 2π (9.525) (1.55) = 29.5275π mm 2 S = L + 2π r 2 = 29.5275π + 2π 9.525 2 ≈ 662.81 m m 2 Nickel L = 2πrh = 2π (10.605) (1.95) = 41.3595π mm 2 S = L + 2π r 2 = 41.3595π + 2π 10.605 2 ≈ 836.58 m m 2 Dime L = 2πrh = 2π (8.955) (1.35) = 24.1785π mm 2 S = L + 2π r 2 = 24.1785π + 2π 8.955 2 ≈ 579.82 m m 2 Quarter L = 2πrh = 2π (12.13) (1.75) = 42.455π mm 2 S = L + 2π r 2 = 42.455π + 2π 12.13 2 ≈ 1057.86 m m 2
ℓ = 3w, h = 2w L = Ph 144 = 2 (w + ℓ) h 144 = 2 (w + 3w) 2w 144 = 1 6w 2 3 = w
w = 3 cm, ℓ = 9 cm, h = 6 cm S = L + 2B = 144 + 2 (9) (3) = 144 + 54 = 198 cm 2
L = 2πrh = 2π (8) (3) = 48π cm 2 S = L + 2π r 2 = 48π + 2π (8) 2 = 48π + 128π = 176π cm 2
S = 2πrh + 2π r 2 176π = 2π (4) h + 2π (4) 2 176π = 8πh + 32π 144π = 8πh 144π ____ 8π = 8πh ___ 8π 18 = h
Module 17 1034 Lesson 2
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INTEGRATE MATHEMATICAL PROCESSESFocus on ReasoningHave students brainstorm how they would find the surface area of a prism whose dimensions have all been doubled. Does the surface area double? no If not, what is the relationship? The area is 4 times as
great. Have students also consider how the surface area changes if only the height of the prism changes. Ask students to use examples to justify their reasoning.
Surface Area of Prisms and Cylinders 1034
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21. Analyze Relationships Ingrid is building a shelter to protect her plants from freezing. She is planning to stretch plastic sheeting over the top and the ends of the frame. Assume that the triangles in the frame on the left are equilateral. Which of the frames shown will require more plastic? Explain how finding the surface area of these figures is different from finding the lateral surface area of a figure.
22. Communicate Mathematical Ideas Explain how to use the net of a three-dimensional figure to find its surface area.
23. Draw Conclusions Explain how the edge lengths of a rectangular prism can be changed so that the surface area is multiplied by 9.
Surface Area of Triangular Prism (minus bottom side)
L = Ph
= (30) 10
= 300 cm 2
a 2 + b 2 = c 2
5 2 + b 2 = 10 2
25 + b 2 = 100
b 2 = 75
b = √ ― 75
S = Ph + 2B - Square
= 300 + 2 ( 1 _ 2 (5) ( √ ― 75 ) ) - 10 ⋅ 10
= 300 + 5 √ ― 75 - 100
≈ 243.3 f t 2 The triangular-prism-shaped frame will take more plastic; In lateral surface area, the area of the bases are not used. In this case, it is not the area of the bases that need to be removed.
Surface Area of Half Cylinder
1 _ 2 L = 1 _ 2 (2πrh)
= 1 _ 2 (2π (5) (10) )
= 50π f t 2
1 _ 2 S = 1 _ 2 L + 1 _ 2 2π r 2
= 50π + 1 _ 2 (2π 5 2 )
= 50π + 25π
= 75π
≈ 235.6 ft 2
Find the area of each part of the net, then add the areas.
Triple all the edge lengths.
Module 17 1035 Lesson 2
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JOURNALHave students illustrate and describe how to use formula S = L + 2B to find the surface area of a prism and of a right cylinder. Ask them to include all of the steps as well as the substitutions they will use in the formula.
1035 Lesson 17 . 2
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A manufacturer of number cubes has the bright idea of packaging them individually in cylindrical boxes. Each number cube measures 2 inches on a side.
1. What is the surface area of each cube?
2. What is the surface area of the cylindrical box? Assume the cube fits snugly in the box and that the box includes a top. Use 3.14 for π.
Lesson Performance Task
1. The cube has 6 faces each with an area of 2 × 2 = 4 in 2 . Total surface area of the cube: 6 × 4 i n 2 = 24 i n 2
2. The top and bottom of the cylinder are circles, each with a diameter equal to a diagonal of one side of the cube, or 2 √ ― 2 inches.
The radius of the top and bottom is half the diameter, or √ ― 2 inches.
Area of cylinder top = π r 2 = 3.14 ( √ ― 2 ) 2 = 6.28. Total
area of top and bottom: 2 × 6.28 = 12.56 i n 2 Lateral area of cylinder: 2πrh = 2 (3.14) √ ― 2 (2) = 12.56 √ ― 2 i n 2 Total surface area of cylindrical box: (12.56 + 12.56 √ ― 2 ) i n 2
Module 17 1036 Lesson 2
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EXTENSION ACTIVITY
A packaging engineer is designing a rectangular-prism-shaped container with a surface area of 64 square inches. Find the possible dimensions for at least three containers that have surface areas of 64 square inches.
Find the volumes of your containers. Then propose a hypothesis about the shape of a rectangular prism with the greatest volume for a given surface area. Sample answer: The rectangular prism with the greatest volume for a given
surface area is a cube.
AVOID COMMON ERRORSTo find the length of a diagonal of one side of the cube, students must use the Pythagorean Theorem to find h, the hypotenuse of a right triangle with 2-inch sides, and then must simplify the resulting square root. Here are the steps:
h 2 = 2 2 + 2 2
= 4 + 4
= 8
h = √ ― 8
= √ ―― 2 2 · 2
= √ ― 2 2 √ ― 2
= 2 √ ― 2
INTEGRATE MATHEMATICAL PROCESSESFocus on Math ConnectionsDescribe how you could find the volume of an empty cylindrical number-cube container. Then find that volume. Use 3.14 for π. Subtract the volume of a
number cube from the volume of a cylindrical
container; about 4.56 cubic inches.
V (cylinder) - V (cube) = π r 2 h - s 3
≈ 3.14 ( √ ― 2 ) 2 (2) - (2) 3
= 3.14 (2) (2) - 8
= 4.56 i n 3
Scoring Rubric2 points: Student correctly solves the problem and explains his/her reasoning.1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning.0 points: Student does not demonstrate understanding of the problem.
Surface Area of Prisms and Cylinders 1036
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