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Virginia SOL Mathematics Textbook

Grade 7 Workbook — Chapter 12: Volume and Surface Area: Prisms and Cylinders

SOL 7.MG.1 · Companion to Textbook Chapter 12

Each page below is one Canva page. Headings are sized for direct paste: page title as H1, section labels as H2. Figures referenced by filename live in ../figures/. Item numbers match the textbook exactly, and every problem uses the same numbers as the textbook problem with the same number.


PAGE 1 — Chapter opener

Chapter 12 · Volume and Surface Area: Prisms and Cylinders

Standard 7.MG.1

In this chapter you will:

Words to know: volume · cubic unit · base area · right cylinder · radius · diameter · surface area · square unit · face · net · rectangular prism · cube · lateral area · circumference · factor

Two rules for every page of this chapter

Pi. Give every answer with π\pi twice: exactly in terms of π\pi, then approximately using π3.14\pi \approx 3.14.

Units. Volume gets cubic units (cm3^3, in3^3, ft3^3, m3^3). Surface area gets square units (cm2^2, in2^2, ft2^2, m2^2). An answer without units is not finished.

The four formulas you will build

Vprism=lwhVcylinder=πr2hV_{\text{prism}} = lwh \qquad V_{\text{cylinder}} = \pi r^2 h

SAprism=2lw+2lh+2whSAcylinder=2πr2+2πrhSA_{\text{prism}} = 2lw + 2lh + 2wh \qquad SA_{\text{cylinder}} = 2\pi r^2 + 2\pi r h


PAGE 2 — Cubic units and square units

Filling Is Not Covering

FIGURE: fig9-cubic-versus-square-units.png (full width)

Fill in the blanks.

A square unit is ____________ . It measures ____________ , written with a small ______ .

A cubic unit is ____________ . It measures ____________ , written with a small ______ .

Label each unit as a volume unit or an area unit.

Unit cm3^3 in2^2 ft3^3 m2^2 in3^3 ft2^2
Volume or area?

Explain. Why does volume need three measurements while area needs only two?



PAGE 3 — Building the cylinder volume formula

12.1 Volume of a Right Cylinder

FIGURE: fig2-cylinder-layers.png (full width)

Complete the derivation.

For any prism or cylinder, volume =(area of the )×()= (\text{area of the } \underline{\hspace{2cm}}) \times (\underline{\hspace{2cm}}), or V=B×hV = B \times h.

The base of a cylinder is a ____________ , and the area of a circle is A=A = \underline{\hspace{2cm}}.

Substituting: V=V = \underline{\hspace{3cm}}

V=πr2h\boxed{V = \pi r^2 h}

FIGURE: fig1-cylinder-labeled.png (right half of page)

Label the figure. Write r=r = \underline{\hspace{1.5cm}} and h=h = \underline{\hspace{1.5cm}} on the arrows, then find the volume.

V=π()2()=πV = \pi(\underline{\hspace{1cm}})^2(\underline{\hspace{1cm}}) = \underline{\hspace{1.5cm}}\pi cm3^3 \approx \underline{\hspace{2cm}} cm3^3

Two traps. Circle the correct choice.

In πr2\pi r^2, you ( square the radius first / multiply by π\pi first ).

If a problem gives the diameter, you must ( double it / halve it ) to get the radius.


PAGE 4 — Guided practice 12.1

Volume of a Cylinder · Guided Practice

Formula frame: V=πr2hV = \pi r^2 h · units are cubic

1. Radius 22 cm, height 77 cm.

V=π()2()=πV = \pi(\underline{\hspace{1cm}})^2(\underline{\hspace{1cm}}) = \underline{\hspace{1.5cm}}\pi cm3^3 \approx \underline{\hspace{2cm}} cm3^3

2. Radius 66 in, height 44 in.

V=V = \underline{\hspace{6cm}}

3. Diameter 1010 m, height 33 m. First: r=r = \underline{\hspace{1cm}} m.

V=V = \underline{\hspace{6cm}}

4. Radius 11 ft, height 1212 ft.

V=V = \underline{\hspace{6cm}}

5. Base area 20π20\pi cm2^2, height 66 cm. Use V=BhV = Bh.

V=V = \underline{\hspace{6cm}}


PAGE 5 — Independent practice 12.1

Volume of a Cylinder · Independent Practice

Complete the table. Give each volume exactly in terms of π\pi and then with π3.14\pi \approx 3.14.

Item Radius Height VV exact VV \approx Units
6 44 cm 99 cm
7 (diameter 1414 in) 1010 in
8 33 m 33 m
9 1010 ft 11 ft

10. A cylinder has radius 55 in and volume 100π100\pi in3^3. Find its height.

100π=π()2h  100π=πh  h=100\pi = \pi(\underline{\hspace{1cm}})^2 h \ \Rightarrow \ 100\pi = \underline{\hspace{1cm}}\pi h \ \Rightarrow \ h = \underline{\hspace{1.5cm}} in

11. Application. A soup can is a right cylinder with radius 44 cm and height 1111 cm. How much soup does it hold when full?


12. Reasoning. Explain why the volume of a cylinder is measured in cubic units. Use πr2\pi r^2 and hh in your explanation.




PAGE 6 — Exit ticket 12.1

Exit Ticket · Lesson 12.1

Name: ________________________ Date: ____________

13. Radius 22 in, height 1010 in. V=V = \underline{\hspace{5cm}}

14. Diameter 66 cm, height 55 cm. V=V = \underline{\hspace{5cm}}

15. Base area 30π30\pi m2^2, height 44 m. V=V = \underline{\hspace{5cm}}

16. A cylinder has diameter 1212 cm and height 22 cm. Dana wrote V=π(12)2(2)=288πV = \pi(12)^2(2) = 288\pi cm3^3. Find her error and give the correct volume, exactly and approximately.

Her error: _______________________________________________

Correct volume: _______________________________________________


PAGE 7 — Unfolding a box

12.2 Surface Area of a Rectangular Prism

FIGURE: fig3-prism-net.png (full width)

Net-folding work. The prism above is l=5l = 5 cm, w=3w = 3 cm, h=4h = 4 cm. Label each face of the net with its name and its area.

Face pair Dimensions Area of one face Area of the pair
top and bottom l×w=×l \times w = \underline{\hspace{1cm}} \times \underline{\hspace{1cm}}
front and back l×h=×l \times h = \underline{\hspace{1cm}} \times \underline{\hspace{1cm}}
the two sides w×h=×w \times h = \underline{\hspace{1cm}} \times \underline{\hspace{1cm}}
Total SASA

Write the formula from your table.

SA=2+2+2SA = 2\underline{\hspace{1cm}} + 2\underline{\hspace{1cm}} + 2\underline{\hspace{1cm}}

A cube is a special case. All six faces are s×ss \times s, so SA=SA = \underline{\hspace{1.5cm}} .

Fold check. How many rectangles are in the net of a rectangular prism? ______ How many matching pairs? ______


PAGE 8 — Guided practice 12.2

Surface Area of a Prism · Guided Practice

Formula frame: SA=2(lw+lh+wh)SA = 2(lw + lh + wh) · units are square

17. 44 cm by 33 cm by 22 cm.

SA=2(++)=SA = 2(\underline{\hspace{1cm}} + \underline{\hspace{1cm}} + \underline{\hspace{1cm}}) = \underline{\hspace{1.5cm}} cm2^2

18. Cube with edge length 55 in.

SA=6()2=SA = 6(\underline{\hspace{1cm}})^2 = \underline{\hspace{1.5cm}} in2^2

19. 77 ft by 55 ft by 22 ft. SA=SA = \underline{\hspace{5cm}}

20. 1010 m by 66 m by 33 m. SA=SA = \underline{\hspace{5cm}}

21. A net is made of two 8×38 \times 3 rectangles, two 8×28 \times 2 rectangles, and two 3×23 \times 2 rectangles, in inches.

SA=2()+2()+2()=SA = 2(\underline{\hspace{1cm}}) + 2(\underline{\hspace{1cm}}) + 2(\underline{\hspace{1cm}}) = \underline{\hspace{1.5cm}} in2^2


PAGE 9 — Independent practice 12.2

Surface Area of a Prism · Independent Practice

Computation table.

Item ll ww hh lwlw lhlh whwh SASA Units
22 99 cm 44 cm 55 cm
23 1010 ft 1010 ft 1010 ft
24 1212 in 55 in 22 in
25 66 m 66 m 22 m

26. A prism is 77 cm by 44 cm by 33 cm. List the three different face shapes in its net, how many of each, and the surface area.

Face shape How many Area of one Total
SASA

27. Application. An aquarium with no lid is 88 ft long, 44 ft wide, and 33 ft tall. Count the bottom and the four sides only.

bottom == \underline{\hspace{1.5cm}} · front and back == \underline{\hspace{1.5cm}} · two sides == \underline{\hspace{1.5cm}} · total == \underline{\hspace{1.5cm}} ft2^2

28. Reasoning. Explain why surface area is measured in square units and not cubic units, and how a net makes that obvious.




PAGE 10 — Exit ticket 12.2

Exit Ticket · Lesson 12.2

Name: ________________________ Date: ____________

29. 55 in by 44 in by 33 in. SA=SA = \underline{\hspace{4cm}}

30. Cube with edge length 44 cm. SA=SA = \underline{\hspace{4cm}}

31. A net of two 6×26 \times 2, two 6×56 \times 5, and two 2×52 \times 5 rectangles, in feet. SA=SA = \underline{\hspace{4cm}}

32. For a prism that is 55 in by 44 in by 33 in, Owen wrote SA=2(20)+15+12=67SA = 2(20) + 15 + 12 = 67 in2^2.

His error: _______________________________________________

Correct surface area: _______________________________________________


PAGE 11 — Unrolling a can

12.3 Surface Area of a Right Cylinder

FIGURE: fig4-cylinder-net.png (full width)

Net-folding work. Cut apart a cylinder and you get three pieces. Name them and give each area.

Piece of the net How many Area of one piece
circle
rectangle width ×\times height =×= \underline{\hspace{2cm}} \times \underline{\hspace{1cm}}

Why is the rectangle that wide? The label travels once around the circle, and the distance once around a circle is its ____________ , which equals ______________ .

Add the pieces.

two circles == \underline{\hspace{2cm}} ++ rectangle == \underline{\hspace{2cm}}

SA=2πr2+2πrh\boxed{SA = 2\pi r^2 + 2\pi r h}

Name the parts. 2πr22\pi r^2 is the area of the ____________ . 2πrh2\pi r h is the ____________ area, which is what a label covers.


PAGE 12 — Guided practice 12.3

Surface Area of a Cylinder · Guided Practice

Formula frame: SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi r h · units are square

33. Radius 11 m, height 44 m.

SA=2π()2+2π()()=π+π=πSA = 2\pi(\underline{\hspace{1cm}})^2 + 2\pi(\underline{\hspace{1cm}})(\underline{\hspace{1cm}}) = \underline{\hspace{1cm}}\pi + \underline{\hspace{1cm}}\pi = \underline{\hspace{1.5cm}}\pi m2^2 \approx \underline{\hspace{2cm}} m2^2

34. Radius 55 cm, height 22 cm. SA=SA = \underline{\hspace{5cm}}

35. Diameter 88 in, height 33 in. First r=r = \underline{\hspace{1cm}} in. SA=SA = \underline{\hspace{5cm}}

36. Lateral area only: radius 66 ft, height 55 ft. 2πrh=2\pi r h = \underline{\hspace{4cm}}

37. Radius 22 cm, height 1010 cm. SA=SA = \underline{\hspace{5cm}}


PAGE 13 — Independent practice 12.3

Surface Area of a Cylinder · Independent Practice

Computation table.

Item rr hh 2πr22\pi r^2 2πrh2\pi r h SASA exact SASA \approx
38 33 m 44 m
39 1010 in 11 in
40 (diameter 66 cm) 99 cm

41. Lateral area only: radius 22 ft, height 77 ft. \underline{\hspace{5cm}}

42. A bucket with no lid has radius 55 in and height 66 in. Count one circle and the side.

one circle == \underline{\hspace{1.5cm}} ++ side == \underline{\hspace{1.5cm}} == \underline{\hspace{2cm}} in2^2 \approx \underline{\hspace{2cm}} in2^2

43. Application. A label wraps once around a can of radius 44 cm and height 1010 cm, covering the side only.

Area of label == \underline{\hspace{4cm}}

The flat rectangle it is cut from is ____________ cm wide by ____________ cm tall.

44. Reasoning. Why is the width of the rectangle 2πr2\pi r? What would go wrong if you used rr instead?




PAGE 14 — Exit ticket 12.3

Exit Ticket · Lesson 12.3

Name: ________________________ Date: ____________

45. Radius 22 cm, height 33 cm. SA=SA = \underline{\hspace{5cm}}

46. Diameter 1212 in, height 55 in. SA=SA = \underline{\hspace{5cm}}

47. Lateral area only: radius 11 m, height 99 m. \underline{\hspace{5cm}}

48. A cylinder has radius 33 cm and height 88 cm. Malik answered 48π48\pi cm2^2.

Part of the net he forgot: _______________________________________________

Correct surface area: _______________________________________________


PAGE 15 — Inside or outside?

12.4 Volume or Surface Area?

FIGURE: fig5-volume-or-surface-area.png (full width)

One question decides it. Is the problem about the ____________ or the ____________ ?

Complete the sorting chart.

The problem asks about... Volume or surface area? Units
how much water it holds
how much wrapping paper
how much sand fills it
how much paint for the outside
the label around a can
how much sheet metal to build a can
how many cubic feet of concrete
how much cereal is in the box

Units are the double check. If a question asks how much paint and your answer says ft3^3, you found the ____________ instead of the ____________ .

The four formulas. Write each one from memory.

Vprism=V_{\text{prism}} = \underline{\hspace{3cm}} Vcylinder=V_{\text{cylinder}} = \underline{\hspace{3cm}}

SAprism=SA_{\text{prism}} = \underline{\hspace{4cm}} SAcylinder=SA_{\text{cylinder}} = \underline{\hspace{4cm}}


PAGE 16 — Guided practice 12.4

Volume or Surface Area · Guided Practice

For each item, first circle V or SA, then compute.

49. Wrapping paper to cover a closed box 88 in by 55 in by 22 in. V / SA

Answer: _______________________________________________

50. Water held by a cylindrical tank, radius 33 m and height 1010 m. V / SA

Answer: _______________________________________________

51. Cereal that fills a box 1010 in by 66 in by 33 in. V / SA

Answer: _______________________________________________

52. Foil to cover a closed cylinder completely, radius 11 in and height 66 in. V / SA

Answer: _______________________________________________

53. Concrete that fills a cylindrical post hole, radius 11 ft and depth 44 ft. V / SA

Answer: _______________________________________________


PAGE 17 — Independent practice 12.4

Volume or Surface Area · Independent Practice

54. Write volume or surface area and name the units.

Situation Volume or surface area? Units
a) Painting the outside of a shipping crate (feet)
b) Filling a swimming pool with water (meters)
c) Sheet metal to build a can (centimeters)
d) Grain a silo can hold (feet)

Same solid, two questions. A box is 1212 in by 33 in by 22 in.

55. Gift wrap to cover it exactly: _______________________________________________

56. Rice that fills it: _______________________________________________

Same solid, two questions. A cylinder has radius 22 in and height 99 in.

57. Juice it holds: _______________________________________________

58. Metal to make it, top and bottom included: _______________________________________________

59. Application. An open-top cylindrical trough has radius 22 ft and height 55 ft.

a) Water it holds: _______________________________________________

b) Metal for the bottom and curved side: _______________________________________________

60. Reasoning. A classmate hands you the answer 150150 cm2^2 but forgot the question. What kind of question was it, and how do you know?



PAGE 18 — Exit ticket 12.4

Exit Ticket · Lesson 12.4

Name: ________________________ Date: ____________

61. Volume or surface area: how much wrapping paper covers a shoebox? ______________

62. A cereal box is 88 in by 22 in by 1212 in. How much cereal fills it? \underline{\hspace{4cm}}

63. A closed cardboard tube is a cylinder with radius 33 in and height 44 in. How much cardboard? \underline{\hspace{4cm}}

64. A student says a room needs 9696 cm3^3 of paint. Explain what is wrong without doing arithmetic.



PAGE 19 — One dimension, one factor

12.5 Changing One Dimension: Volume

FIGURE: fig6-double-height-volume.png (full width)

Compute before and after.

ll ww hh VV Units
before 44 33 22
after (height ×2\times 2) 44 33

Vafter÷Vbefore=V_{\text{after}} \div V_{\text{before}} = \underline{\hspace{1cm}} , so the volume was multiplied by ______ .

FIGURE: fig8-half-length-volume.png (full width)

ll ww hh VV Units
before 1212 66 22
after (length ×12\times \tfrac12) 66 22

Vafter÷Vbefore=V_{\text{after}} \div V_{\text{before}} = \underline{\hspace{1cm}} , so the volume was multiplied by ______ .

State the pattern.

Multiplying one measured attribute of a rectangular prism by a factor multiplies the volume by ______________ .

Why? V=l×w×hV = l \times w \times h. If the height becomes 2h2h, then Vnew=l×w×2h=×(lwh)V_{\text{new}} = l \times w \times 2h = \underline{\hspace{1cm}} \times (lwh).

Caution. This rule is for changing ______ attribute. Allowed factors for volume: 14\tfrac14, 13\tfrac13, 12\tfrac12, 2, 3, 4.


PAGE 20 — Guided practice 12.5

Changing One Dimension: Volume · Guided Practice

Items 65–68 all start from the same prism: 88 cm by 33 cm by 44 cm, volume == \underline{\hspace{1.5cm}} cm3^3.

Item What changes New dimensions New VV Volume factor
65 height ×2\times 2 8×3×8 \times 3 \times \underline{\hspace{0.7cm}}
66 length ×12\times \tfrac12 ×3×4\underline{\hspace{0.7cm}} \times 3 \times 4
67 width ×3\times 3 8××48 \times \underline{\hspace{0.7cm}} \times 4
68 height ×14\times \tfrac14 8×3×8 \times 3 \times \underline{\hspace{0.7cm}}

Use the shortcut. New volume == old volume ×\times factor.

69. V=200V = 200 in3^3; width ×4\times 4. New V=×=V = \underline{\hspace{1cm}} \times \underline{\hspace{1cm}} = \underline{\hspace{2cm}} in3^3

70. V=72V = 72 m3^3; length ×13\times \tfrac13. New V=V = \underline{\hspace{3cm}} m3^3


PAGE 21 — Independent practice 12.5

Changing One Dimension: Volume · Independent Practice

Item Prism Change VV before VV after Factor
71 5×4×35 \times 4 \times 3 cm height ×2\times 2
72 12×6×212 \times 6 \times 2 in length ×13\times \tfrac13
73 10×4×510 \times 4 \times 5 ft width ×14\times \tfrac14

74. A prism has volume 9090 m3^3. One attribute is multiplied by 33. New V=V = \underline{\hspace{3cm}}

75. A prism has volume 4848 in3^3. One attribute is multiplied by 12\tfrac12. New V=V = \underline{\hspace{3cm}}

76. Application. A shipping box is 99 in by 44 in by 55 in. The height is doubled.

VV before == \underline{\hspace{2cm}} VV after == \underline{\hspace{2cm}} How many more cubic inches? \underline{\hspace{2cm}}

77. Reasoning. Use V=lwhV = lwh to explain why multiplying just the width by 33 multiplies the volume by 33. Then explain why multiplying all three dimensions by 33 is a different situation.




PAGE 22 — Exit ticket 12.5

Exit Ticket · Lesson 12.5

Name: ________________________ Date: ____________

78. 66 cm by 55 cm by 22 cm; height ×3\times 3. New V=V = \underline{\hspace{3cm}}

79. 66 cm by 55 cm by 22 cm; length ×12\times \tfrac12. New V=V = \underline{\hspace{3cm}}

80. V=100V = 100 ft3^3; one attribute ×4\times 4. New V=V = \underline{\hspace{3cm}}

81. Rosa says doubling the length makes the volume 88 times as large.

Her error: _______________________________________________

Correct factor: ______


PAGE 23 — A rule that does not carry over

12.6 Changing One Dimension: Surface Area

FIGURE: fig7-double-height-surface-area.png (full width)

Compute before and after for the prism 44 cm by 33 cm by 22 cm with the height doubled.

lwlw lhlh whwh SA=2(lw+lh+wh)SA = 2(lw + lh + wh)
before (4×3×24 \times 3 \times 2)
after (4×3×44 \times 3 \times 4)

Twice the original surface area would be 2×=2 \times \underline{\hspace{1cm}} = \underline{\hspace{1cm}} cm2^2. The actual new surface area is \underline{\hspace{1cm}} cm2^2, so the surface area ( did / did not ) double.

Sort the six faces.

Faces Do they contain the height? Area before Area after
top and bottom (l×wl \times w)
four side faces (l×hl \times h and w×hw \times h)

Say why. Part of the surface ______________ and part of it ______________ , so the total grows by ( more / less ) than a factor of 2.

Allowed factors for surface area: 12\tfrac12 and 22 only.


PAGE 24 — Guided practice 12.6

Changing One Dimension: Surface Area · Guided Practice

Items 82–85 all start from the same prism: 55 cm by 44 cm by 33 cm, SA=94SA = 94 cm2^2.

Item Change New dimensions New SASA Did it change by the factor?
82 height ×2\times 2 5×4×5 \times 4 \times \underline{\hspace{0.7cm}}
83 length ×2\times 2 ×4×3\underline{\hspace{0.7cm}} \times 4 \times 3
84 width ×12\times \tfrac12 5××35 \times \underline{\hspace{0.7cm}} \times 3

85. In item 82, which two faces did not change, and what is the area of each?


86. A cube has edge length 44 in and its height is multiplied by 12\tfrac12.

SASA before == \underline{\hspace{2cm}} SASA after == \underline{\hspace{2cm}}


PAGE 25 — Independent practice 12.6

Changing One Dimension: Surface Area · Independent Practice

Item Prism Change SASA before SASA after
87 6×3×26 \times 3 \times 2 cm height ×2\times 2
88 8×5×48 \times 5 \times 4 in length ×12\times \tfrac12
89 10×2×310 \times 2 \times 3 ft width ×2\times 2
90 cube, edge 66 m one edge ×12\times \tfrac12

91. A cube has edge length 44 in and its height is multiplied by 22.

SASA before == \underline{\hspace{2cm}} SASA after == \underline{\hspace{2cm}} Did it double? ______ Actual change: \underline{\hspace{2cm}}

92. Application. A shipping box is 1212 in by 66 in by 44 in. The maker multiplies the height by 12\tfrac12.

Cardboard before == \underline{\hspace{2cm}} after == \underline{\hspace{2cm}} saved per box == \underline{\hspace{2cm}}

93. Reasoning. Explain why doubling one attribute doubles the volume but not the surface area. Name the faces that change and the faces that do not.




PAGE 26 — Exit ticket 12.6

Exit Ticket · Lesson 12.6

Name: ________________________ Date: ____________

Items 94–96 use the prism 66 cm by 44 cm by 22 cm.

94. Height ×2\times 2. SASA before == \underline{\hspace{2cm}} SASA after == \underline{\hspace{2cm}}

95. Starting again from 6×4×26 \times 4 \times 2, length ×12\times \tfrac12. SASA before == \underline{\hspace{2cm}} SASA after == \underline{\hspace{2cm}}

96. VV of 6×4×2=6 \times 4 \times 2 = \underline{\hspace{1.5cm}} ; after height ×2\times 2, V=V = \underline{\hspace{1.5cm}} . Volume factor == \underline{\hspace{1cm}} . Did the surface area change by that same factor? ______

97. In one or two sentences, explain why volume and surface area respond differently when one attribute is doubled.



PAGE 27 — Review Parts A and B

Chapter 12 Review · Volume and Surface Area Formulas

Vocabulary check. volume · cubic unit · base area · right cylinder · radius · diameter · surface area · square unit · face · net · rectangular prism · cube · lateral area · circumference · factor

Part A — Volume of right cylinders (7.MG.1a). Exact, then π3.14\pi \approx 3.14.

98. r=4r = 4 cm, h=5h = 5 cm \underline{\hspace{5cm}}

99. d=10d = 10 in, h=6h = 6 in \underline{\hspace{5cm}}

100. r=2r = 2 m, h=11h = 11 m \underline{\hspace{5cm}}

101. r=3r = 3 ft, V=63πV = 63\pi ft3^3. Find hh. h=h = \underline{\hspace{2cm}}

102. Application. A cylindrical drum has r=3r = 3 ft and h=4h = 4 ft. Water held: \underline{\hspace{4cm}}

Part B — Surface area (7.MG.1b).

103. Prism 77 cm by 33 cm by 22 cm \underline{\hspace{4cm}}

104. Cube, edge 88 in \underline{\hspace{4cm}}

105. Cylinder r=5r = 5 m, h=3h = 3 m \underline{\hspace{5cm}}

106. Cylinder d=4d = 4 in, h=10h = 10 in \underline{\hspace{5cm}}

107. Net of two 9×59 \times 5, two 9×29 \times 2, two 5×25 \times 2 rectangles (feet) \underline{\hspace{4cm}}

108. Lateral area only, cylinder r=6r = 6 cm, h=2h = 2 cm \underline{\hspace{5cm}}


PAGE 28 — Review Parts C, D, and E

Chapter 12 Review · Deciding and Scaling

Part C — Volume or surface area? (7.MG.1c)

109. Write volume or surface area and the units.

Situation Which measure? Units
a) Paint for the outside of a closed crate (feet)
b) Water that fills a cylindrical tank (meters)
c) Paper for a label around a can (centimeters)
d) Soil that fills a rectangular planter box (inches)

A box is 1010 in by 44 in by 33 in.

110. Paint to cover its outside: \underline{\hspace{4cm}}

111. Sand that fills it: \underline{\hspace{4cm}}

112. Application. An open-top cylindrical barrel has r=2r = 2 ft and h=6h = 6 ft.

a) Water held: \underline{\hspace{4cm}} b) Metal for bottom and curved side: \underline{\hspace{4cm}}

Part D — Changing one attribute: volume (7.MG.1d)

Item Prism Change VV before VV after
113 8×5×28 \times 5 \times 2 cm height ×3\times 3
114 12×3×412 \times 3 \times 4 in length ×14\times \tfrac14

115. V=60V = 60 m3^3; one attribute ×12\times \tfrac12. New V=V = \underline{\hspace{3cm}}

116. Application. A grain bin is 66 ft by 55 ft by 44 ft; the width is doubled.

before == \underline{\hspace{2cm}} after == \underline{\hspace{2cm}} more == \underline{\hspace{2cm}}

Part E — Changing one attribute: surface area (7.MG.1e)

Item Prism Change SASA before SASA after
117 5×4×25 \times 4 \times 2 cm height ×2\times 2
118 8×6×38 \times 6 \times 3 in length ×12\times \tfrac12
119 cube, edge 1010 ft one edge ×12\times \tfrac12

PAGE 29 — Review Part F

Chapter 12 Review · Mixed Application and Reasoning

120. A closed cylindrical can has r=3r = 3 in and h=10h = 10 in.

a) Volume: \underline{\hspace{5cm}}

b) Surface area: \underline{\hspace{5cm}}

c) Which answer tells how much juice it holds? ______________

121. Explain why volume is reported in cubic units and surface area in square units. Use V=lwhV = lwh and SA=2lw+2lh+2whSA = 2lw + 2lh + 2wh.



122. A student claims doubling the height of a prism doubles both the volume and the surface area. Use 44 cm by 33 cm by 22 cm to test both claims.

VV SASA
before
after (height ×2\times 2)

Which claim is right? ______________ Which is wrong? ______________

123. Explain how the net of a cylinder shows where each term of SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi r h comes from.


124. Application. A cylindrical can has r=4r = 4 cm and h=10h = 10 cm.

a) How much it holds: \underline{\hspace{5cm}}

b) Label covering the side only: \underline{\hspace{5cm}}

c) Which answer is in cm3^3, and why? _______________________________________________