12 Surface Area and Volume 12.1 Three-Dimensional Figures 12.2 Surface Areas of Prisms and Cylinders 12.3 Surface Areas of Pyramids and Cones 12.4 Volumes of Prisms and Cylinders 12.5 Volumes of Pyramids and Cones 12.6 Surface Areas and Volumes of Spheres 12.7 Spherical Geometry
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TTrTrafficafffic CCoConene ((p(p.. 65656)6) Mathematical Thinking: Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. MMaintainingaintaining MathematicalMathematical ProficiencyProficiency
Finding the Area of a Circle (7.9.B)
Example 1 Find the area of the circle.
A = πr2 Formula for area of a circle 8 in. = π ⋅ 82 Substitute 8 for r. = 64π Simplify. ≈ 201.06 Use a calculator. The area is about 201.06 square inches.
Find the area of the circle.
1. 2. 3.
20 cm 6 m 9 ft
Finding the Area of a Composite Figure (7.9.C)
Example 2 Find the area of the composite figure. 17 ft The composite figure is made up of a rectangle, a triangle, and a semicircle. 16 ft Find the area of each figure. 32 ft 30 ft
Area of rectangle Area of triangle Area of semicircle 1 πr2 A =ℓw A = — bh A = — 2 2 1 π(17)2 = 32(16) = — (30)(16) = — 2 2 = 512 = 240 ≈ 453.96
So, the area is about 512 + 240 + 453.96 = 1205.96 square feet.
Find the area of the composite figure.
4. 7 m 5. 6 in. 6. 6 cm
7 m 3 in. 6 cm 10 in. 6 cm 7 m 5 in. 3 3 cm 20 m 9 cm
7. ABSTRACT REASONING A circle has a radius of x inches. Write a formula for the area of the circle when the radius is multiplied by a real number a.
637 MMathematicalathematical Mathematically profi cient students create and use representations to TThinkinghinking organize, record, and communicate mathematical ideas. (G.1.E) Creating a Coherent Representation CCoreore CConceptoncept Nets for Three-Dimensional Figures A net for a three-dimensional fi gure is a two-dimensional pattern that can be folded to form the three-dimensional fi gure.
base w w w h lateral lateral lateral lateral h face face face face w base
Drawing a Net for a Pyramid
Draw a net of the pyramid. 20 in.
19 in. 19 in. SOLUTION The pyramid has a square base. Its four lateral faces are congruent isosceles triangles. 19 in. 20 in. 19 in.
MMonitoringonitoring PProgressrogress Draw a net of the three-dimensional fi gure. Label the dimensions. 1. 2. 3. 5 m 4 ft 15 in.
12 m 4 ft 8 m 10 in. 2 ft 10 in.
638 Chapter 12 Surface Area and Volume 12.1 Three-Dimensional Figures
EEssentialssential QQuestionuestion What is the relationship between the numbers TEXAS ESSENTIAL of vertices V, edges E, and faces F of a polyhedron? KNOWLEDGE AND SKILLS G.10.A A polyhedron is a solid that is bounded edge by polygons, called faces.
• Each vertex is a point. face • Each edge is a segment of a line. • Each face is a portion of a plane. vertex
Analyzing a Property of Polyhedra Work with a partner. The fi ve Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid.
tetrahedron cube octahedron
dodecahedron icosahedron
Solid Vertices, V Edges, E Faces, F
tetrahedron cube octahedron MAKING dodecahedron MATHEMATICAL ARGUMENTS icosahedron To be profi cient in math, you need to reason inductively about data. CCommunicateommunicate YourYour AnswerAnswer 2. What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707–1783) discovered a formula that relates these quantities.) 3. Draw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron. Then verify that the relationship you found in Question 2 is valid for each polyhedron.
Section 12.1 Three-Dimensional Figures 639 12.1 Lesson WWhathat YYouou WWillill LLearnearn Classify solids. Describe cross sections. Core VocabularyVocabulary Sketch and describe solids of revolution. polyhedron, p. 640 face, p. 640 Classifying Solids edge, p. 640 A three-dimensional fi gure, or solid, is bounded by vertex, p. 640 face fl at or curved surfaces that enclose a single region cross section, p. 641 of space. A polyhedron is a solid that is bounded by solid of revolution, p. 642 polygons, called faces. An edge of a polyhedron is a axis of revolution, p. 642 line segment formed by the intersection of two faces. Previous A vertex of a polyhedron is a point where three or more vertex solid edges meet. The plural of polyhedron is polyhedra edge prism or polyhedrons. pyramid cylinder cone CCoreore CConceptoncept sphere Types of Solids base Polyhedra Not Polyhedra
prism cylinder cone
pyramid sphere
Pentagonal prism To name a prism or a pyramid, use the shape of the base. The two bases of a prism are congruent polygons in parallel planes. For example, the bases of a pentagonal Bases are prism are pentagons. The base of a pyramid is a polygon. For example, the base of a pentagons. triangular pyramid is a triangle.
Classifying Solids Triangular pyramid Tell whether each solid is a polyhedron. If it is, name the polyhedron. a. b. c. Base is a triangle.
SOLUTION a. The solid is formed by polygons, so it is a polyhedron. The two bases are congruent rectangles, so it is a rectangular prism. b. The solid is formed by polygons, so it is a polyhedron. The base is a hexagon, so it is a hexagonal pyramid. c. The cone has a curved surface, so it is not a polyhedron.
640 Chapter 12 Surface Area and Volume MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com Tell whether the solid is a polyhedron. If it is, name the polyhedron.
1. 2. 3.
Describing Cross Sections Imagine a plane slicing through a solid. The intersection of the plane and the solid is called a cross section. For example, three different cross sections of a cube are shown below.
square rectangle triangle
Describing Cross Sections
Describe the shape formed by the intersection of the plane and the solid. a. b. c.
d. e. f.
SOLUTION a. The cross section is a hexagon. b. The cross section is a triangle. c. The cross section is a rectangle. d. The cross section is a circle. e. The cross section is a circle. f. The cross section is a trapezoid.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com Describe the shape formed by the intersection of the plane and the solid.
4. 5. 6.
Section 12.1 Three-Dimensional Figures 641 Sketching and Describing Solids of Revolution A solid of revolution is a three-dimensional fi gure that is formed by rotating a two-dimensional shape around an axis. The line around which the shape is rotated is called the axis of revolution. For example, when you rotate a rectangle around a line that contains one of its sides, the solid of revolution that is produced is a cylinder.
Sketching and Describing Solids of Revolution
Sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid. a. 9 b. 4 4
9 5
2
SOLUTION a. 9 b.
4
5
2 The solid is a cylinder with a height of 9 and a base The solid is a cone with radius of 4. a height of 5 and a base radius of 2.
MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com Sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid.
7. 8. 6 9. 7 3 7 8 8 4
6
642 Chapter 12 Surface Area and Volume 12.1 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
VVocabularyocabulary andand CoreCore ConceptConcept CheckCheck 1. VOCABULARY A(n) ______is a solid that is bounded by polygons.
2. WHICH ONE DOESN’T BELONG? Which solid does not belong with the other three? Explain your reasoning.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics
In Exercises 3–6, match the polyhedron with its name. In Exercises 11−14, describe the cross section formed by the intersection of the plane and the solid. 3. 4. (See Example 2.) 11. 12.
5. 6.
13. 14.
A. triangular prism B. rectangular pyramid
C. hexagonal pyramid D. pentagonal prism
In Exercises 7–10, tell whether the solid is a polyhedron. If it is, name the polyhedron. (See Example 1.) In Exercises 15–18, sketch the solid produced by 7. 8. rotating the fi gure around the given axis. Then identify and describe the solid. (See Example 3.)
15. 8 16.
8 8 6 9. 10. 8 6
17. 18. 5 3 2 2
3 5
Section 12.1 Three-Dimensional Figures 643 19. ERROR ANALYSIS Describe and correct the error in 28. ATTENDING TO PRECISION The fi gure shows a plane identifying the solid. intersecting a cube through four of its vertices. The ✗ edge length of the cube is 6 inches. The solid is a rectangular pyramid.
a. Describe the shape formed by the cross section. 20. HOW DO YOU SEE IT? Is the swimming pool shown b. What is the perimeter of the cross section? a polyhedron? If it is, name the polyhedron. If not, explain why not. c. What is the area of the cross section?
REASONING In Exercises 29–34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube. 29. circle 30. pentagon
31. rhombus 32. isosceles triangle
33. hexagon 34. scalene triangle
35. REASONING Sketch the composite solid produced In Exercises 21–26, sketch the polyhedron. by rotating the fi gure around the given axis. Then 21. triangular prism 22. rectangular prism identify and describe the composite solid. a. 23. pentagonal prism 24. hexagonal prism 2 3 3 25. square pyramid 26. pentagonal pyramid b. 8 27. MAKING AN ARGUMENT Your friend says that the polyhedron shown is a triangular prism. Your cousin 5 44 says that it is a triangular pyramid. Who is correct? Explain your reasoning. 11
36. THOUGHT PROVOKING Describe how Plato might have argued that there are precisely fi ve Platonic Solids (see page 639). (Hint: Consider the angles that meet at a vertex.)
MMaintainingaintaining MathematicalMathematical ProficiencyProficiency Reviewing what you learned in previous grades and lessons Decide whether enough information is given to prove that the triangles are congruent. If so, state the theorem you would use. (Sections 5.3, 5.5, and 5.6) 37. △ABD, △CDB 38. △JLK, △JLM 39. △RQP, △RTS AB J S Q R
D C T KLM P
644 Chapter 12 Surface Area and Volume 12.2 Surface Areas of Prisms and Cylinders
EEssentialssential QQuestionuestion How can you fi nd the surface area of a prism or TEXAS ESSENTIAL a cylinder? KNOWLEDGE AND SKILLS G.10.B Recall that the surface area of a polyhedron is the sum of the areas of its faces. The G.11.C lateral area of a polyhedron is the sum of the areas of its lateral faces.
Finding a Formula for Surface Area Work with a partner. Consider the polyhedron shown. APPLYING MATHEMATICS a. Identify the polyhedron. Then sketch its net so that the lateral faces form a rectangle with the same To be profi cient in math, you height h as the polyhedron. What types of fi gures height, h need to analyze relationships make up the net? mathematically to draw a c conclusions. b. Write an expression that represents the perimeter P of the base of the polyhedron. Show how you can b use P to write an expression that represents the lateral area L of the polyhedron. c. Let B represent the area of a base of the polyhedron. Write a formula for the surface area S.
Finding a Formula for Surface Area Work with a partner. Consider the solid shown. radius, r a. Identify the solid. Then sketch its net. What types of fi gures make up the net?
b. Write an expression that represents the perimeter P height, h of the base of the solid. Show how you can use P to write an expression that represents the lateral area L of the solid. c. Write an expression that represents the area B of a base of the solid. d. Write a formula for the surface area S. CCommunicateommunicate YourYour AnswerAnswer 3. How can you fi nd the surface area of a prism or a cylinder? 4. Consider the rectangular prism shown. a. Find the surface area of the rectangular prism by drawing its net and fi nding the sum of the areas 7 of its faces. b. Find the surface area of the rectangular prism by using the formula you wrote in Exploration 1. 5 c. Compare your answers to parts (a) and (b). 3 What do you notice?
Section 12.2 Surface Areas of Prisms and Cylinders 645 12.2 Lesson WWhathat YYouou WWillill LLearnearn Find lateral areas and surface areas of right prisms. Find lateral areas and surface areas of right cylinders. Core VocabularyVocabulary Use surface areas of right prisms and right cylinders. lateral faces, p. 646 lateral edges, p. 646 Finding Lateral Areas and Surface Areas of Right Prisms surface area, p. 646 lateral area, p. 646 Recall that a prism is a polyhedron with two congruent faces, called bases, that lie in parallel planes. The other faces, called lateral faces, are parallelograms formed by net, p. 646 connecting the corresponding vertices of the bases. The segments connecting these right prism, p. 646 vertices are lateral edges. Prisms are classifi ed by the shapes of their bases. oblique prism, p. 646 right cylinder, p. 647 base oblique cylinder, p. 647 Previous lateral edges prism bases of a prism lateral cylinder base faces composite solid The surface area of a polyhedron is the sum of the areas of its faces. The lateral area of a polyhedron is the sum of the areas of its lateral faces. Imagine that you cut some edges of a polyhedron and unfold it. The two-dimensional representation of the faces is called a net. The surface area of a prism is equal to the area of its net. The height of a prism is the perpendicular distance between its bases. In a right prism, each lateral edge is perpendicular to both bases. A prism with lateral edges that are not perpendicular to the bases is an oblique prism.
height height
Right rectangular prism Oblique triangular prism
CCoreore CConceptoncept Lateral Area and Surface Area of a Right Prism For a right prism with base perimeter P, base apothem a, height h, and base area B, the lateral area L and surface area S are as follows. h Lateral area L = Ph B Surface area S = 2B + L P = aP + Ph
646 Chapter 12 Surface Area and Volume Finding Lateral Area and Surface Area
Find the lateral area and the surface area of the 7.05 ft 6 ft right pentagonal prism.
SOLUTION 9 ft Find the apothem and perimeter of a base. — — a = √62 − 3.5252 = √23.574375 6 ft 6 ft a P = 5(7.05) = 35.25
3.525 ft 3.525 ft Find the lateral area and the surface area. ATTENDING TO L = Ph Formula for lateral area of a right prism PRECISION = (35.25)(9) Substitute. Throughout this chapter, = 317.25 Multiply. round lateral areas, surface = + areas, and volumes to S aP Ph Formula for surface area of a right prism — the nearest hundredth, = ( √23.574375 ) (35.25) + 317.25 Substitute. if necessary. ≈ 488.40 Use a calculator. The lateral area is 317.25 square feet and the surface area is about 488.40 square feet.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 1. Find the lateral area and the surface area of a right rectangular prism with a height of 7 inches, a length of 3 inches, and a width of 4 inches.
Finding Lateral Areas and Surface Areas of Right Cylinders height Recall that a cylinder is a solid with congruent circular bases that lie in parallel planes. The height of a cylinder is the perpendicular distance between its bases. The radius of a base is the radius of the cylinder. In a right cylinder, the segment joining the centers right cylinder of the bases is perpendicular to the bases. In an oblique cylinder, this segment is not perpendicular to the bases. The lateral area of a cylinder is the area of its curved surface. For a right cylinder, it is equal to the product of the circumference and the height, or 2πrh. The surface area of height a cylinder is equal to the sum of the lateral area and the areas of the two bases. CCoreore CConceptoncept oblique cylinder Lateral Area and Surface Area of a Right Cylinder For a right cylinder with radius r, r base area height h, and base area B, the r =π 2 π π A r lateral area L and surface area S 2 r 2 r are as follows. lateral area hh Lateral area L = 2πrh A = 2π rh Surface area S = 2B + L base area = π 2 + π 2 r 2 rh A = π r2
Section 12.2 Surface Areas of Prisms and Cylinders 647 Finding Lateral Area and Surface Area
Find the lateral area and the surface area of the right cylinder. 4 m
SOLUTION Find the lateral area and the surface area. 8 m L = 2πrh Formula for lateral area of a right cylinder = 2π(4)(8) Substitute. = 64π Simplify. ≈ 201.06 Use a calculator. S = 2πr2 + 2πrh Formula for surface area of a right cylinder = 2π(4)2 + 64π Substitute. = 96π Simplify. ≈ 301.59 Use a calculator.
The lateral area is 64π, or about 201.06 square meters. The surface area is 96π, or about 301.59 square meters.
Solving a Real-Life Problem
You are designing a label for the cylindrical soup can shown. 9 cm The label will cover the lateral area of the can. Find the minimum amount of material needed for the label.
SOLUTION 12 cm Find the radius of a base. 1 = — = r 2 (9) 4.5 Find the lateral area. L = 2πrh Formula for lateral area of a right cylinder = 2π(4.5)(12) Substitute. = 108π Simplify. ≈ 339.29 Use a calculator.
You need a minimum of about 339.29 square centimeters of material.
MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com 2. Find the lateral area and the surface area of the right cylinder.
10 in.
18 in.
3. WHAT IF? In Example 3, you change the design of the can so that the diameter is 12 centimeters. Find the minimum amount of material needed for the label.
648 Chapter 12 Surface Area and Volume Using Surface Areas of Right Prisms and Right Cylinders
Finding the Surface Area of a Composite Solid
4 m Find the lateral area and the surface area of the composite solid. 3 m SOLUTION
Lateral area Lateral area Lateral area = + 12 m of solid of cylinder of prism
= 2πrh + Ph
= 2π(6)(12) + 14(12)
6 m = 144π + 168
≈ 620.39
Surface area Lateral area Area of a base Area of a base = + 2 ⋅ − of solid of solid ( of the cylinder of the prism ) = 144π + 168 + 2(πr2 −ℓw)
= 144π + 168 + 2[π(6)2 − 4(3)]
= 216π + 144
≈ 822.58
The lateral area is about 620.39 square meters and the surface area is about 822.58 square meters.
Changing Dimensions in a Solid
Describe how doubling all the linear dimensions affects 2 ft the surface area of the right cylinder.
SOLUTION
Before change After change 8 ft Dimensions r = 2 ft, h = 8 ft r = 4 ft, h = 16 ft
S = 2πr2 + 2πrh S = 2πr2 + 2πrh Surface = π 2 + π = π 2 + π area 2 (2) 2 (2)(8) 2 (4) 2 (4)(16) = 40π ft2 = 160π ft2
Doubling all the linear dimensions results in a surface area that 2 mm 160π is — = 4 = 22 times the original surface area. 40π
10 mm MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 4. Find the lateral area and the surface area of the composite solid at the left. 1 — 5. In Example 5, describe how multiplying all the linear dimensions by 2 affects the 6 mm surface area of the right cylinder. 8 mm
Section 12.2 Surface Areas of Prisms and Cylinders 649 12.2 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
Vocabulary and Core Concept Check 1. VOCABULARY Sketch a right triangular prism. Identify the bases, lateral faces, and lateral edges.
2. WRITING Explain how the formula S = 2B + L applies to fi nding the surface area of both a right prism and a right cylinder.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics In Exercises 3 and 4, fi nd the surface area of the solid 13. MODELING WITH MATHEMATICS The inside of the formed by the net. cylindrical swimming pool shown must be covered with a vinyl liner. The liner must cover the side 3. 4. and bottom of the swimming pool. What is the
8 cm minimum amount of vinyl needed for the liner? 4 in. (See Example 3.)
24 ft
10 in. 20 cm 4 ft In Exercises 5–8, fi nd the lateral area and the surface area of the right prism. (See Example 1.) 5. 6. 14. MODELING WITH MATHEMATICS 2 ft The tent shown has fabric 5 ft 3 m 3 ft 9.1 m covering all four sides 8 ft 8 m and the fl oor. What is 4 ft the minimum amount 8 ft 7. A regular pentagonal prism has a height of 3.5 inches of fabric needed to 6 ft and a base edge length of 2 inches. construct the tent?
8. A regular hexagonal prism has a height of 80 feet and In Exercises 15–18, fi nd the lateral area and the surface a base edge length of 40 feet. area of the composite solid. (See Example 4.) 15. 1 cm 16. 4 ft In Exercises 9–12, fi nd the lateral area and the surface area of the right cylinder. (See Example 2.) 7 ft 9. 10. 0.8 in. 8 cm 16 cm 4 ft 2 cm 2 in. 8 cm 6 ft 4 cm 4 cm
17. 2 in. 18. 11. A right cylinder has a diameter of 24 millimeters and 5 m 7 m a height of 40 millimeters. 11 in. 12. A right cylinder has a radius of 2.5 feet and a height 15 m of 7.5 feet. 9 m 6 m
5 in.
650 Chapter 12 Surface Area and Volume 19. ERROR ANALYSIS Describe and correct the error in 27. MATHEMATICAL CONNECTIONS A cube has a fi nding the surface area of the right cylinder. surface area of 343 square inches. Write and solve an equation to fi nd the length of each edge of the cube. 6 cm ✗ S = 2훑(6)2 + 2훑(6)(8) 28. MATHEMATICAL CONNECTIONS A right cylinder has a surface area of 108π square meters. The radius of 8 cm = 168훑 the cylinder is twice its height. Write and solve an ≈ 527.79 cm2 equation to fi nd the height of the cylinder.
29. MODELING WITH MATHEMATICS A company makes 20. ERROR ANALYSIS Describe and correct the error in two types of recycling bins, as shown. Both types of fi nding the surface area of the composite solid. bins have an open top. Which recycling bin requires more material to make? Explain. ✗ 6 in. 16 ft 7 ft
18 ft 20 ft 36 in. S = 2(20)(7) + 2(18)(7) + 2훑(8)(7) 36 in. + 2[(18)(20) + 훑(8)2] ≈ 2005.98 ft2
12 in. In Exercises 21–24, describe how the change affects 10 in. the surface area of the right prism or right cylinder. (See Example 5.) 30. MODELING WITH MATHEMATICS You are painting a rectangular room that is 13 feet long, 9 feet wide, and 21. doubling all the 22. multiplying all the 8.5 feet high. There is a window that is 2.5 feet wide linear dimensions linear dimensions 1 and 5 feet high on one wall. On another wall, there by — 3 is a door that is 4 feet wide and 7 feet high. A gallon 5 in. 9 mm of paint covers 350 square feet. How many gallons
4 in. of paint do you need to cover the four walls with one 17 in. coat of paint, not including the window and door?
31. ANALYZING RELATIONSHIPS Which creates a greater surface area, doubling the radius of a cylinder 24 mm or doubling the height of a cylinder? Explain 23. tripling 24. multiplying the base your reasoning. 1 2 yd — the radius edge lengths by 4 and the height by 4 32. MAKING AN ARGUMENT You cut a cylindrical piece of lead, forming two congruent cylindrical pieces 8 m of lead. Your friend claims the surface area of each 7 yd smaller piece is exactly half the surface area of 2 m the original piece. Is your friend correct? Explain 16 m your reasoning.
In Exercises 25 and 26, fi nd the height of the right prism 33. USING STRUCTURE The right triangular prisms or right cylinder. shown have the same surface area. Find the height h of prism B. 25. S = 1097 m2 26. S = 480 in.2 Prism A Prism B
h 24 cm 3 cm h 6 cm 15 in. 20 cm 20 cm h 8 in. 8 cm 8.2 m
Section 12.2 Surface Areas of Prisms and Cylinders 651 34. USING STRUCTURE The lateral surface area of a 38. THOUGHT PROVOKING You have 24 cube-shaped regular pentagonal prism is 360 square feet. The building blocks with edge lengths of 1 unit. What height of the prism is twice the length of one of the arrangement of blocks gives you a rectangular prism edges of the base. Find the surface area of the prism. with the least surface area? Justify your answer.
35. ANALYZING RELATIONSHIPS Describe how multiplying all the linear dimensions of the right 39. USING STRUCTURE Sketch the net of the oblique rectangular prism by each given value affects the rectangular prism shown. Then fi nd the surface area. surface area of the prism. 4 ft
h 7 ft 8 ft
w 15 ft
1 — a. 2 b. 3 c. 2 d. n 40. WRITING Use the diagram to write a formula that can be used to fi nd the surface area S of any cylindrical ring where 0 < r < r . 36. HOW DO YOU SEE IT? 2 1 An open gift box is shown.wn. r1 r2 a. Why is the area of the net of the box h larger than the minimum amount of wrapping paper neededdd to cover the closed box? b. When wrapping the box, why would you want 41. USING STRUCTURE The diagonal of a cube is a to use more than the minimum amount of segment whose endpoints are vertices that are not on paper needed? the same face. Find the surface area of a cube with a diagonal length of 8 units.
37. REASONING Consider a cube that is 42. USING STRUCTURE A cuboctahedron has 6 square built using 27 unit cubes, as shown. faces and 8 equilateral triangular faces, as shown. A a. Find the surface area of the cuboctahedron can be made by slicing off the corners solid formed when the red unit of a cube. cubes are removed from the a. Sketch a net for the solid shown. cuboctahedron. b. Find the surface area of the solid formed when the b. Each edge of a blue unit cubes are removed from the solid shown. cuboctahedron has a length of 5 millimeters. c. Explain why your answers are different in parts (a) Find its surface area. and (b).
Maintaining Mathematical Proficiency Reviewing what you learned in previous grades and lessons Find the area of the regular polygon. (Section 11.3)
43. 44. 10.6 in. 45. 7 m 9 in.
8 cm
6 cm
652 Chapter 12 Surface Area and Volume 12.3 Surface Areas of Pyramids and Cones
EEssentialssential QQuestionuestion How can you fi nd the surface area of a pyramid TEXAS ESSENTIAL or a cone? KNOWLEDGE AND SKILLS G.10.B A regular pyramid has a regular polygon for a base and the segment joining the G.11.C vertex and the center of the base is perpendicular to the base. In a right cone, the segment joining the vertex and the center of the base is perpendicular to the base.
Finding a Formula for Surface Area APPLYING Work with a partner. Consider the polyhedron shown. MATHEMATICS a. Identify the polyhedron. Then sketch its net. slant height, To be profi cient in math, you height, h What types of fi gures make up the net? need to analyze relationships mathematically to draw b. Write an expression that represents the conclusions. perimeter P of the base of the polyhedron. Show how you can use P to write an expression that represents the lateral area L of the polyhedron. base edge length, b c. Let B represent the area of a base of the polyhedron. Write a formula for the surface area S.
Finding a Formula for Surface Area Work with a partner. Consider the solid shown. a. Identify the solid. Then sketch its net. What types of fi gures make up the net? slant height, b. Write an expression that represents the area B of the base of the solid. c. What is the arc measure of the lateral surface of the solid? What is the circumference and area radius, r of the entire circle that contains the lateral surface of the solid? Show how you can use these three measures to fi nd the lateral area L of the solid. d. Write a formula for the surface area S. CCommunicateommunicate YourYour AnswerAnswer 3. How can you fi nd the surface area of a pyramid or cone? 4. Consider the rectangular pyramid shown. a. Find the surface area of the rectangular pyramid 12 ft by drawing its net and fi nding the sum of the areas of its faces. b. Find the surface area of the rectangular pyramid by using the formula you wrote in Exploration 1. c. Compare your answers to parts (a) and (b). 8 ft What do you notice?
Section 12.3 Surface Areas of Pyramids and Cones 653 12.3 Lesson WWhathat YYouou WWillill LLearnearn Find lateral areas and surface areas of regular pyramids. Find lateral areas and surface areas of right cones. Core VocabularyVocabulary Use surface areas of regular pyramids and right cones. vertex of a pyramid, p. 654 regular pyramid, p. 654 Finding Lateral Areas and Surface Areas of slant hieght of a regular Regular Pyramids pyramid, p. 654 vertex A pyramid is a polyhedron in which the base vertex of a cone, p. 655 height lateral edge right cone, p. 655 is a polygon and the lateral faces are triangles oblique cone, p. 655 with a common vertex, called the vertex of the slant height of a right cone, pyramid. The intersection of two lateral faces p. 655 is a lateral edge. The intersection of the base lateral surface of a cone, and a lateral face is a base edge. The height base edge of the pyramid is the perpendicular distance p. 655 base lateral faces between the base and the vertex. Previous Pyramid A regular pyramid has a regular polygon for a base pyramid slant height cone height and the segment joining the vertex and the center of composite solid the base is perpendicular to the base. The lateral faces of a regular pyramid are congruent isosceles triangles. The slant height of a regular pyramid is the height of a lateral face of the regular pyramid. A nonregular Regular pyramid pyramid does not have a slant height. CCoreore CConceptoncept Lateral Area and Surface Area of a Regular Pyramid For a regular pyramid with base perimeter P, slant heightℓ, and base area B, the lateral area L and surface area S are as follows.
1 = — Lateral area L 2 Pℓ 1 = + = + — B P Surface area S B L B 2 Pℓ
Finding Lateral Area and Surface Area
Find the lateral area and the surface area of the regular hexagonal pyramid. 14 ft SOLUTION — The perimeter P of the base is 6 ⋅ 10 = 60, feet and the apothem a is 5 √3 feet. The slant heightℓof a face is 14 feet. Find the lateral area and the surface area. 1 L = — Pℓ Formula for lateral area of a regular pyramid 10 ft 5 3 ft 2 1 = — 2 (60)(14) Substitute. = 420 Simplify. 1 = + — S B 2 Pℓ Formula for surface area of a regular pyramid — 1 = — √ + 2 ( 5 3 ) (60) 420 Substitute. — = 150 √3 + 420 Simplify. ≈ 679.81 Use a calculator. The lateral area is 420 square feet and the surface area is about 679.81 square feet.
654 Chapter 12 Surface Area and Volume 4.8 m MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 5.5 m 1. Find the lateral area and the surface area of the regular pentagonal pyramid.
Finding Lateral Areas and Surface Areas of 8 m Right Cones A cone has a circular base and a vertex that is not in the same plane as the base. The radius of the base is the radius of the cone. The height is the perpendicular distance between the vertex and the base. vertex In a right cone, the segment joining the vertex and the center of the base is perpendicular to the base. In an oblique cone, this segment is not perpendicular to the slant height height base. The slant height of a right cone is the distance between the vertex and a point on the edge of the base. An oblique cone does not have a slant height. lateral The lateral surface of a cone consists of all segments that connect the vertex with surface points on the edge of the base. r base Right cone CCoreore CConceptoncept vertex Lateral Area and Surface Area of a Right Cone lateral For a right cone with radius r, slant heightℓ, and base area B, surface the lateral area L and surface area S are as follows. height Lateral area L = πrℓ Surface area S = B + L = πr2ℓ + πrℓ r r base Oblique cone
Finding Lateral Area and Surface Area
Find the lateral area and the surface area of the right cone. 6 m SOLUTION Use the Pythagorean Theorem (Theorem 9.1) to fi nd the 8 m slant heightℓ. — ℓ = √62 + 82 = 10 Find the lateral area and the surface area. L = πrℓ Formula for lateral area of a right cone = π(6)(10) Substitute. = 60π Simplify. ≈ 188.50 Use a calculator. S = πr2 + πrℓ Formula for surface area of a right cone = π(6)2 + 60π Substitute. = 96π Simplify. ≈ 301.59 Use a calculator. The lateral area is 60π, or about 188.50 square meters. The surface area is 96π, about 301.59 square meters.
Section 12.3 Surface Areas of Pyramids and Cones 655 Solving a Real-Life Problem
The traffi c cone can be approximated by a right cone with a radius of 5.7 inches and a height of 18 inches. Find the lateral area of the traffi c cone.
SOLUTION Use the Pythagorean Theorem (Theorem 9.1) to fi nd the slant heightℓ. 18 —— — ℓ= √182 + (5.7)2 = √356.49 5.7 Find the lateral area. L = πrℓ Formula for lateral area of a right cone — = π(5.7) ( √356.49 ) Substitute. ≈ 388.10 Use a calculator.
The lateral area of the traffi c cone is about 338.10 square inches.
MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com 2. Find the lateral area and the surface area of the right cone. 8 ft
3. WHAT IF? The radius of the cone in Example 3 15 ft is 6.3 inches. Find the lateral area.
Using Surface Areas of Regular Pyramids and Right Cones
Finding the Surface Area of a Composite Solid
Find the lateral area and the surface area of the composite solid. 5 cm SOLUTION
Lateral area Lateral area Lateral area = + of solid of cone of cylinder 3 cm
= πrℓ + 2πrh 6 cm = π(3)(5) + 2π(3)(6) = 51π ≈ 160.22
Surface area Lateral area Area of a base = + of solid of solid of the cylinder
= 51π + πr2 = 51π + π(3)2 = 60π ≈ 188.50 The lateral area is about 160.22 square centimeters and the surface area is about 188.50 square centimeters.
656 Chapter 12 Surface Area and Volume Changing Dimensions in a Solid
Describe how the change affects the surface area of the right cone. 3 — a. multiplying the radius by 2 26 m 3 — b. multiplying all the linear dimensions by 2
ANALYZING SOLUTION 10 m MATHEMATICAL a. RELATIONSHIPS Before change After change Notice that while the Dimensions r = 10 m, ℓ= 26 m r = 15 m, ℓ= 26 m surface area does not scale 3 = π 2 + π = π 2 + π by a factor of — , the lateral S r rℓ S r rℓ Surface 2 = π 2 + π = π 2 + π surface area does scale area (10) (10)(26) (15) (15)(26) π(15)(26) 3 = π 2 = π 2 by a factor or — = — . 360 m 615 m π(10)(26) 2 3 615π 41 Multiplying the radius by — results in a surface area that is — = — times 2 360π 24 the original surface area. b. Before change After change Dimensions r = 10 m, ℓ= 26 m r = 15 m, ℓ= 39 m
S = 360π m2 S = πr2 + πrℓ Surface = π 2 + π area (15) (15)(39) = 810π m2
3 Multiplying all the linear dimensions by — results in a surface area that 2 810π 9 3 2 is — = — = — times the original surface area. 360π 4 ( 2)
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 4. Find the lateral area and the surface area of the composite solid.
10 ft
6 ft
5 ft 5 m 5. Describe how (a) multiplying the base edge 1 — lengths by 2 and (b) multiplying all the linear 1 — dimensions by 2 affects the surface area of the square pyramid. 6 m
Section 12.3 Surface Areas of Pyramids and Cones 657 12.3 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
Vocabulary and Core Concept Check 1. WRITING Describe the differences between pyramids and cones. Describe their similarities.
2. DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Find the slant height of the regular pyramid. A 5 in.
Find AB.
B Find the height of the regular pyramid. 6 in.
Find the height of a lateral face of the regular pyramid.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics
In Exercises 3–6, fi nd the lateral area and the surface 11. ERROR ANALYSIS Describe and correct the error in area of the regular pyramid. (See Example 1.) fi nding the surface area of the regular pyramid. 3. 4. 5 ft
8 in. 15.4 mm ✗ 4 ft 1 S = B + — Pℓ 7.2 mm 2 1 = 2 + — 6 2 (24)(4) = 84 ft2 6 ft 5 in.
5. A square pyramid has a height of 21 feet and a base 20. ERROR ANALYSIS Describe and correct the error in edge length of 40 feet. fi nding the surface area of the right cone. 6. A regular hexagonal pyramid has a slant height of 15 centimeters and a base edge length of ✗ 10 cm 8 cm S = 훑r2 + 훑r2 8 centimeters. ℓ = 훑(6)2 + 훑(6)2(10) In Exercises 7–10, fi nd the lateral area and the surface = 396훑 cm2 area of the right cone. (See Example 2.) 6 cm 7. 8. 7.2 cm 13. MODELING WITH MATHEMATICS You are 16 in. making cardboard party hats like the one shown. About how much cardboard 11 cm 5.5 in. do you need for each hat? (See Example 3.) 8 in. 3.5 in. 9. A right cone has a radius of 9 inches and a height of 12 inches. 14. MODELING WITH MATHEMATICS A candle is in the shape of a regular square pyramid with a 10. A right cone has a diameter of 11.2 feet and a height base edge length of 16 centimeters and a height of of 9.2 feet. 15 centimeters. Find the surface area of the candle.
658 Chapter 12 Surface Area and Volume In Exercises 15–18, fi nd the lateral area and the surface 24. USING STRUCTURE The area of the composite solid. (See Example 4.) sector shown can be rolled 150° to form the lateral surface 15. 4 yd 16. 3 yd 3 in. area of a right cone. The lateral surface area of the 5 in cone is 20 square meters. 5 in. a. Use the formula for the area of a sector to fi nd the 8 yd slant height of the cone. Explain your reasoning. 5 in. 7.5 in. b. Find the radius and the height of the cone.
17. 18. 10 cm In Exercises 25 and 26, fi nd the missing dimensions of 5 mm the regular pyramid or right cone. 7 mm 25. S = 864 in.2 26. S = 628.3 cm2
12 cm h 15 in. h 4 mm
In Exercises 19–22, describe how the change affects x 8 in. the surface area of the regular pyramid or right cone. (See Example 5.) 27. WRITING Explain why a nonregular pyramid does not 19. doubling the radius 20. multiplying the base have a slant height. 4 — edge lengths by 5 and the slant height by 5 28. WRITING Explain why an oblique cone does not have a slant height. 7.6 in. 4 mm 29. ANALYZING RELATIONSHIPS In the fi gure, AC = 4, AB = 3, and DC = 2.
10 mm 3 in. AB 21. tripling all the linear 22. multiplying all the 4 — dimensions linear dimensions by 3 D E
4 m 3.6 ft C a. Prove △ABC ∼ △DEC.
2.4 ft b. Find BC, DE, and EC. 2 m c. Find the surface areas of the larger cone and the smaller cone in terms of π. Compare the surface 23. PROBLEM SOLVING Refer to the regular pyramid and areas using a percent. right cone. 30. REASONING To make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then 4 4 follow the given steps. How does the surface area of the cup compare to the original paper circle? Find ∠ 3 m ABC. 3 A C 4 3 in. a. Which solid has the base with the greater area? B b. Which solid has the greater slant height? fold fold open cup c. Which solid has the greater lateral area?
Section 12.3 Surface Areas of Pyramids and Cones 659 31. MAKING AN ARGUMENT Your friend claims that the 35. USING STRUCTURE A right cone with a radius of lateral area of a regular pyramid is always greater than 4 inches and a square pyramid both have a slant the area of the base. Is your friend correct? Explain height of 5 inches. Both solids have the same surface your reasoning. area. Find the length of a base edge of the pyramid.
32. HOW DO YOU SEE IT? Name the fi gure that is 36. THOUGHT PROVOKING The surface area of a regular represented by each net. Justify your answer. 1 = + — pyramid is given by S B 2 Pℓ. As the number a. of lateral faces approaches infi nity, what does the pyramid approach? What does B approach? What 1 — does 2 Pℓ approach? What can you conclude from your three answers? Explain your reasoning.
b.
37. DRAWING CONCLUSIONS The net of the lateral surface of a cone is a circular sector with radius y, as shown.
x° y 33. REASONING In the fi gure, a right cone is placed in the smallest right cylinder that can fi t the cone. Which solid has a greater surface area? Explain your reasoning. a. Let y = 2. Copy and complete the table.
Angle measure of 30 90 120 180 210 lateral surface, x Slant height of cone, ℓ Circumference of base of cone, C Height of cone, h 34. CRITICAL THINKING A regular hexagonal pyramid with a base edge of 9 feet and a height of 12 feet is b. What conjectures can you make about the inscribed in a right cone. Find the lateral area of dimensions of the cone as x increases? the cone.
Maintaining Mathematical Proficiency Reviewing what you learned in previous grades and lessons Find the volume of the prism. (Skills Review Handbook) 38. 39.
3 ft
10 mm 2 ft 7 ft
B = 29 mm2
660 Chapter 12 Surface Area and Volume 12.1–12.3 What Did You Learn?
CCoreore VVocabularyocabulary polyhedron, p. 640 surface area, p. 646 slant height of a regular pyramid, face, p. 640 lateral area, p. 646 p. 654 edge, p. 640 net, p. 646 vertex of a cone, p. 655 vertex, p. 640 right prism, p. 646 right cone, p. 655 cross section, p. 641 oblique prism, p. 646 oblique cone, p. 655 solid of revolution, p. 642 right cylinder, p. 647 slant height of a right cone, p. 655 axis of revolution, p. 642 oblique cylinder, p. 647 lateral surface of a cone, p. 655 lateral faces, p. 646 vertex of a pyramid, p. 654 lateral edges, p. 646 regular pyramid, p. 654
CCoreore CConceptsoncepts Section 12.1 Types of Solids, p. 640 Cross Section of a Solid, p. 641 Solids of Revolution, p. 642
Section 12.2 Lateral Area and Surface Area of a Right Prism, p. 646 Lateral Area and Surface Area of a Right Cylinder, p. 647
Section 12.3 Lateral Area and Surface Area of a Regular Pyramid, p. 654 Lateral Area and Surface Area of a Right Cone, p. 655
MMathematicalathematical TThinkinghinking 1. In Exercises 21–26 on page 644, describe the steps you took to sketch each polyhedron. 2. Sketch and label a diagram to represent the situation described in Exercise 32 on page 651. 3. In Exercise 13 on page 658, you need to make a new party hat using 4 times as much cardboard as you previously used for one hat. How should you change the given dimensions to create the new party hat? Explain your reasoning.
Study Skills Form a Final Exam Study Group Form a study group several weeks before the fi nal exam. The intent of this group is to review what you have already learned while continuing to learn new material.
666161 12.1–12.3 Quiz
Tell whether the solid is a polyhedron. If it is, name the polyhedron. (Section 12.1) 1. 2. 3.
4. Sketch the composite solid produced by rotating the fi gure around the given 7 axis. Then identify and describe the composite solid. (Section 12.1) 6 3 10
Find the lateral area and the surface area of the right prism or right cylinder. (Section 12.2) 5. 6. 7. 7 ft
7 ft 5 in. 9 m 9 in. 10 m 7 in. 12 m 10 cm
12 cm 8. Find the lateral area and the surface area of the composite solid. (Section 12.2) 32 cm
10 cm Find the lateral area and the surface area of the regular pyramid or right cone. (Section 12.3) 9. 10 cm 10. 11. 12 ft
10 m 4 3 m
16 ft 8 cm
8 m
12 ft 12. You are replacing the siding and the roofi ng on the house shown. You have 900 square feet of siding, 500 square feet of roofi ng material, and 2000 square feet of tarp, in case it rains. (Section 12.3)
a. Do you have enough siding to replace the siding on all four sides 12 ft of the house? Explain. 18 ft b. Do you have enough roofi ng material to replace the entire roof? Explain. 18 ft c. Do you have enough tarp to cover the entire house? Explain.
662 Chapter 12 Surface Area and Volume 12.4 Volumes of Prisms and Cylinders
EEssentialssential QQuestionuestion How can you fi nd the volume of a prism or TEXAS ESSENTIAL cylinder that is not a right prism or right cylinder? KNOWLEDGE AND SKILLS G.10.B Recall that the volume V of a right prisms right cylinder G.11.D right prism or a right cylinder is equal to the product of the area of a base B and the height h. V = Bh
Finding Volume Work with a partner. Consider a stack of square papers that is in the USING PRECISE form of a right prism. MATHEMATICAL a. What is the volume of the prism? LANGUAGE b. When you twist the stack of papers, To be profi cient in math, as shown at the right, do you change 8 in. you need to communicate the volume? Explain your reasoning. precisely to others. c. Write a carefully worded conjecture that describes the conclusion you reached in part (b). d. Use your conjecture to fi nd the 2 in. 2 in. volume of the twisted stack of papers.
Finding Volume Work with a partner. Use the conjecture you wrote in Exploration 1 to fi nd the volume of the cylinder.
a. 2 in. b. 5 cm
3 in. 15 cm
CCommunicateommunicate YourYour AnswerAnswer 3. How can you fi nd the volume of a prism or cylinder that is not a right prism or right cylinder? 4. In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning.
Section 12.4 Volumes of Prisms and Cylinders 663 12.4 Lesson WWhathat YYouou WWillill LLearnearn Find volumes of prisms and cylinders. Core VocabularyVocabulary Use volumes of prisms and cylinders. volume, p. 664 Finding Volumes of Prisms and Cylinders Cavalieri’s Principle, p. 664 The volume of a solid is the number of cubic units contained in its interior. Volume Previous is measured in cubic units, such as cubic centimeters (cm3). Cavalieri’s Principle, prism named after Bonaventura Cavalieri (1598–1647), states that if two solids have the cylinder same height and the same cross-sectional area at every level, then they have the same composite solid volume. The prisms below have equal heights h and equal cross-sectional areas B at every level. By Cavalieri’s Principle, the prisms have the same volume.
BBh
CCoreore CConceptoncept Volume of a Prism The volume V of a prism is = V Bh h h where B is the area of a base and h is the height. B B
Finding Volumes of Prisms
Find the volume of each prism.
a. 3 cm 4 cm b. 3 cm 14 cm
5 cm 2 cm
6 cm SOLUTION 1 = — = 2 = a. The area of a base is B 2 (3)(4) 6 cm and the height is h 2 cm. V = Bh Formula for volume of a prism = 6(2) Substitute. = 12 Simplify. The volume is 12 cubic centimeters.
1 = — + = 2 = b. The area of a base is B 2 (3)(6 14) 30 cm and the height is h 5 cm. V = Bh Formula for volume of a prism = 30(5) Substitute. = 150 Simplify. The volume is 150 cubic centimeters.
664 Chapter 12 Surface Area and Volume Consider a cylinder with height h and base radius r and a rectangular prism with the — same height that has a square base with sides of length r √π.
B B h
r π r r π
The cylinder and the prism have the same cross-sectional area, πr2, at every level and the same height. By Cavalieri’s Principle, the prism and the cylinder have the same volume. The volume of the prism is V = Bh = πr2h, so the volume of the cylinder is also V = Bh = πr2h. CCoreore CConceptoncept Volume of a Cylinder rr The volume V of a cylinder is = = π 2 V Bh r h hh where B is the area of a base, h is the BB height, and r is the radius of a base.
Finding Volumes of Cylinders
Find the volume of each cylinder. a. b. 4 cm 9 ft 6 ft 7 cm
SOLUTION a. The dimensions of the cylinder are r = 9 ft and h = 6 ft. V = πr2h = π(9)2(6) = 486π ≈ 1526.81
The volume is 486π, or about 1526.81 cubic feet. b. The dimensions of the cylinder are r = 4 cm and h = 7 cm. V = πr2h = π(4)2(7) = 112π ≈ 351.86
The volume is 112π, or about 351.86 cubic centimeters.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com Find the volume of the solid. 1. 2. 8 ft 8 m 14 ft 9 m 5 m
Section 12.4 Volumes of Prisms and Cylinders 665 Using Volumes of Prisms and Cylinders
Modeling with Mathematics
You are building a rectangular chest. You want the length to be 6 feet, the V = 72 ft3 width to be 4 feet, and the volume to be 72 cubic feet. What should the h height be?
SOLUTION 6 ft 4 ft 1. Understand the Problem You know the dimensions of the base of a rectangular prism and the volume. You are asked to fi nd the height. 2. Make a Plan Write the formula for the volume of a rectangular prism, substitute known values, and solve for the height h. 3. Solve the Problem The area of a base is B = 6(4) = 24 ft2 and the volume is V = 72 ft3. V = Bh Formula for volume of a prism
72 = 24h Substitute. 3 = h Divide each side by 24. The height of the chest should be 3 feet.
4. Look Back Check your answer. V = Bh = 24(3) = 72 ✓
MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com 3. WHAT IF? In Example 3, you want the length to be 5 meters, the width to be 3 meters, and the volume to be 60 cubic meters. What should the height be?
Changing Dimensions in a Solid
Describe how doubling all the linear dimensions affects the volume of the rectangular prism. 6 ft ANALYZING MATHEMATICAL 3 ft RELATIONSHIPS SOLUTION 4 ft Notice that when all the linear dimensions are Before change After change multiplied by k, the volume ======is multiplied by k3. Dimensions ℓ 4 ft, w 3 ft, h 6 ft ℓ 8 ft, w 6 ft, h 12 ft = = Voriginal Bh ℓwh V = Bh V = Bh = Volume = = Vnew (kℓ)(kw)(kh) (4)(3)(6) (8)(6)(12) = (k3)ℓwh = 72 ft3 = 576 ft3 = (k3)V original 576 Doubling all the linear dimensions results in a volume that is — = 8 = 23 times 72 the original volume.
666 Chapter 12 Surface Area and Volume Changing a Dimension in a Solid 3 cm Describe how tripling the radius affects the volume of the cylinder.
6 cm SOLUTION
Before change After change Dimensions r = 3 cm, h = 6 cm r = 9 cm, h = 6 cm
V = πr2h V = πr2h Volume = π(3)2(6) = π(9)2(6) = 54π cm3 = 486π cm3
486π Tripling the radius results in a volume that is — = 9 = 32 times the 54π original volume.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 1 — 4. In Example 4, describe how multiplying all the linear dimensions by 2 affects the volume of the rectangular prism. 5. In Example 4, describe how doubling the length and width of the bases affects the volume of the rectangular prism. 2 — 6. In Example 5, describe how multiplying the height by 3 affects the volume of the cylinder. 7. In Example 5, describe how multiplying all the linear dimensions by 4 affects the volume of the cylinder.
Finding the Volume of a Composite Solid 0.39 ft 0.33 ft Find the volume of the concrete block. 0.33 ft
SOLUTION 0.66 ft To fi nd the area of the base, subtract two times the 0.66 ft area of the small rectangle from the large rectangle. 1.31 ft B = Area of large rectangle − 2 ⋅ Area of small rectangle = 1.31(0.66) − 2(0.33)(0.39) = 0.6072 Using the formula for the volume of a prism, the volume is V = Bh = 0.6072(0.66) ≈ 0.40. The volume is about 0.40 cubic foot. 3 ft MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 6 ft 8. Find the volume of the composite solid. 10 ft
Section 12.4 Volumes of Prisms and Cylinders 667 12.4 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
VVocabularyocabulary andand CoreCore ConceptConcept CheckCheck 1. VOCABULARY In what type of units is the volume of a solid measured?
2. COMPLETE THE SENTENCE Cavalieri’s Principle states that if two solids have the same ______and the same ______at every level, then they have the same ______.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics
In Exercises 3–6, fi nd the volume of the prism. 12. A pentagonal prism has a height of 9 feet and each (See Example 1.) base edge is 3 feet. 3. 4. 1.2 cm In Exercises 13–18, fi nd the missing dimension of the 1.5 m prism or cylinder. (See Example 3.) 1.8 cm 13. Volume = 560 ft3 14. Volume = 2700 yd3 2 cm 2 m 2.3 cm 4 m
u v 5. 7 in. 10 in. 6.
8 ft 15 yd 5 in. 14 m 7 ft 12 yd 15. Volume = 80 cm3 16. Volume = 72.66 in.3 11 m 6 m 8 cm x In Exercises 7–10, fi nd the volume of the cylinder. w (See Example 2.) 5 cm 2 in. 7. 3 ft 8. 26.8 cm 17. Volume = 3000 ft3 18. Volume = 1696.5 m3 z 9.3 ft 10.2 ft 9.8 cm y 15 m
9. 5 ft 10. 12 m
19. ERROR ANALYSIS Describe and correct the error in 8 ft 18 m fi nding the volume of the cylinder.
60° ✗ 4 ft V = 2πrh 3 ft In Exercises 11 and 12, make a sketch of the solid = 2π(4)(3) and fi nd its volume. = 24π 11. A prism has a height of 11.2 centimeters and an So, the volume of the cylinder is equilateral triangle for a base, where each base edge 24π cubic feet. is 8 centimeters.
668 Chapter 12 Surface Area and Volume 20. OPEN-ENDED Sketch two rectangular prisms that 31. MODELING WITH MATHEMATICS The Great Blue have volumes of 100 square inches but different Hole is a cylindrical trench located off the coast surface areas. Include dimensions in your sketches. of Belize. It is approximately 1000 feet wide and 400 feet deep. About how many gallons of water does In Exercises 21–26, describe how the change affects the the Great Blue Hole contain? (1 ft3 ≈ 7.48 gallons) volume of the prism or cylinder. (See Examples 4 and 5.) 21. tripling all the 22. multiplying all the linear dimensions linear dimensions 3 — by 4
12 m
16 m 3 in. 8 in.
4 in. 32. COMPARING METHODS The Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its nonoverlaping parts. Use this 23. multiplying the 24. tripling the base and 1 postulate to fi nd the volume of the block of concrete radius by — the height of the 2 in Example 6 by subtracting the volume of each triangular bases hole from the volume of the large rectangular prism. 8 cm Which method do you prefer? Explain your reasoning.
7 cm 33. WRITING Both of the fi gures shown are made up of 5 ft 12 ft the same number of congruent rectangles. Explain 12 ft how Cavalieri’s Principle can be adapted to compare the areas of these fi gures. 25. multiplying the 26. multiplying the height 1 — by 4 height by 3 6 in.
1 in.
3 m 5 m 5 m 34. HOW DO YOU SEE IT? Each stack of memo papers contains 500 equally-sized sheets of paper. Compare In Exercises 27–30, fi nd the volume of the composite their volumes. Explain your reasoning. solid. (See Example 6.)
27. 5 ft 28. 2 ft 3 ft 2 ft 4 in. 6 ft 10 ft 4 in. 4 in. 35. PROBLEM SOLVING An aquarium shaped like a 29. 3 in. 30. 1 ft 8 in. rectangular prism has a length of 30 inches, a width of 10 inches, and a height of 20 inches. You fi ll the 3 — aquarium 4 full with water. When you submerge a 5 ft 11 in. rock in the aquarium, the water level rises 0.25 inch. a. Find the volume of the rock. 2 ft b. How many rocks of this size can you place in the 4 ft aquarium before water spills out?
Section 12.4 Volumes of Prisms and Cylinders 669 36. MODELING WITH MATHEMATICS Which box gives 41. MATHEMATICAL CONNECTIONS You drill a circular you more cereal for your money? Explain. hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.
16 in. 10 in. 42. THOUGHT PROVOKING Cavalieri’s Principle states that the two solids shown below have the same 2 in. volume. Do they also have the same surface area? Explain your reasoning. 4 in. 10 in. 8 in.
37. CRITICAL THINKING A 3-inch by 5-inch index card BBh is rotated around a horizontal line and a vertical line to produce two different solids. Which solid has a greater volume? Explain your reasoning.
5 in. 43. PROBLEM SOLVING A barn is in the shape of a 3 in. 3 in. pentagonal prism with the dimensions shown. The volume of the barn is 9072 cubic feet. Find the 5 in. dimensions of each half of the roof.
Not drawn to scale 38. CRITICAL THINKING The height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is x ft half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer. 8 ft 8 ft 18 ft 36 ft 39. USING STRUCTURE Find the volume of the solid shown. The bases of the solid are sectors of circles. 44. PROBLEM SOLVING A wooden box is in the shape of ° 2π in. 60 3 a regular pentagonal prism. The sides, top, and bottom of the box are 1 centimeter thick. Approximate the volume of wood used to construct the box. Round 3.5 in. your answer to the nearest tenth.
4 cm
40. ANALYZING RELATIONSHIPS How can you change 6 cm the height of a cylinder so that the volume is increased by 25% but the radius remains the same?
MMaintainingaintaining MathematicalMathematical ProficiencyProficiency Reviewing what you learned in previous grades and lessons Find the surface area of the regular pyramid. (Section 12.3) 45. 46. 47. 10 cm 20 in. 3 m
8 cm 18 in. 15.6 in. 2 m
670 Chapter 12 Surface Area and Volume 12.5 Volumes of Pyramids and Cones
EEssentialssential QQuestionuestion How can you fi nd the volume of a pyramid TEXAS ESSENTIAL or a cone? KNOWLEDGE AND SKILLS G.10.B Finding the Volume of a Pyramid G.11.D Work with a partner. The pyramid and the prism have the same height and the same square base. h
When the pyramid is fi lled with sand and poured into the prism, it takes three pyramids to fi ll the prism.
ANALYZING MATHEMATICAL RELATIONSHIPS To be profi cient in math, you need to look closely to discern a pattern or structure. Use this information to write a formula for the volume V of a pyramid.
Finding the Volume of a Cone Work with a partner. The cone and the cylinder have the same height and the same circular base. h
When the cone is fi lled with sand and poured into the cylinder, it takes three cones to fi ll the cylinder.
Use this information to write a formula for the volume V of a cone. CCommunicateommunicate YourYour AnswerAnswer 3. How can you fi nd the volume of a pyramid or a cone? Section 12.5 Volumes of Pyramids and Cones 671 12.5 Lesson WWhathat YYouou WWillill LLearnearn Find volumes of pyramids. Find volumes of cones. Core VocabularyVocabulary Use volumes of pyramids and cones. Previous pyramid Finding Volumes of Pyramids cone Consider a triangular prism with parallel, congruent bases △JKL and △MNP. You can composite solid divide this triangular prism into three triangular pyramids. P L P N M N M M L L N M KJK KJL Triangular Triangular Triangular Triangular prism pyramid 1 pyramid 2 pyramid 3 L You can combine triangular pyramids 1 and 2 to form a pyramid with a base that is a parallelogram, as shown at the left. Name this pyramid Q. Similarly, you can combine triangular pyramids 1 and 3 to form pyramid R with a base that is a parallelogram. — In pyramid Q, diagonal KM divides ▱JKNM into two congruent triangles, so the K J bases of triangular pyramids 1 and 2 are congruent. Similarly, you can divide any cross ▱ N M section parallel to JKNM into two congruent triangles that are the cross sections of triangular pyramids 1 and 2. Pyramid Q By Cavalieri’s Principle, triangular pyramids 1 and 2 have the same volume. Similarly, M using pyramid R, you can show that triangular pyramids 1 and 3 have the same volume. By the Transitive Property of Equality, triangular pyramids 2 and 3 have the same volume. P L 1 = — The volume of each pyramid must be one-third the volume of the prism, or V 3 Bh. N K You can generalize this formula to say that the volume of any pyramid with any base is 1 — Pyramid R equal to 3 the volume of a prism with the same base and height because you can divide any polygon into triangles and any pyramid into triangular pyramids. CCoreore CConceptoncept Volume of a Pyramid The volume V of a pyramid is h 1 h = — V 3 Bh where B is the area of a base B B and h is the height.
Finding the Volume of a Pyramid
Find the volume of the pyramid.
SOLUTION 9 m 1 = — V 3 Bh Formula for volume of a pyramid 1 1 = — — ⋅ 4 ⋅ 6 (9) Substitute. 3( 2 ) 6 m = 36 Simplify. 4 m The volume is 36 cubic meters.
672 Chapter 12 Surface Area and Volume Finding Volumes of Cones Consider a cone with a regular polygon inscribed in the base. The pyramid with the 1 = — same vertex as the cone has volume V 3 Bh. As you increase the number of sides of the polygon, it approaches the base of the cone and the pyramid approaches the cone. 1 — π 2 π 2 The volume approaches 3 r h as the base area B approaches r .
CCoreore CConceptoncept Volume of a Cone The volume V of a cone is 1 1 = — = — π 2 V 3 Bh 3 r h h h where B is the area of a base, BBrr h is the height, and r is the radius of the base.
Finding the Volume of a Cone
Find the volume of the cone.
4.5 cm
SOLUTION 2.2 cm 1 = — π 2 V 3 r h Formula for volume of a cone 1 = — π 2 3 ⋅(2.2) ⋅4.5 Substitute. = 7.26π Simplify. ≈ 22.81 Use a calculator.
The volume is 7.26π, or about 22.81 cubic centimeters.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com Find the volume of the solid.
1. 2. 5 m 20 cm
8 m
12 cm
Section 12.5 Volumes of Pyramids and Cones 673 Using Volumes of Pyramids and Cones
Using the Volume of a Pyramid
Originally,O Khafre’s Pyramid had a height of about 144 meters and a volume of about 2,218,8002 cubic meters. Find the side length of the square base.
SOLUTION 1 = — V 3 Bh Formula for volume of a pyramid 1 ≈ — 2 2,218,800 3 x (144) Substitute. 6,656,400 ≈ 144x2 Multiply each side by 3. 46,225 ≈ x2 Divide each side by 144. 215 ≈ x Find the positive square root. Khafre’s Pyramid, Egypt Originally, the side length of the square base was about 215 meters.
MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com 3. The volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base. 4. Find the height of the 5. Find the radius of the cone. triangular pyramid.
V = 24 m3 13 in.
h r
3 m V = 351 π in.3 6 m
Changing Dimensions in a Solid
1 — Describe how multiplying all the linear dimensions by 3 affects the volume of the rectangular pyramid.
SOLUTION 18 m Before change After change Dimensions ℓ= 9 m, w = 6 m, h = 18 m ℓ= 3 m, w = 2 m, h = 6 m 1 1 9 m V = — Bh V = — Bh 3 3 6 m 1 1 Volume = — (9)(6)(18) = — (3)(2)(6) 3 3 = 324 m3 = 12 m3
1 Multiplying all the linear dimensions by — results in a volume that 3 12 1 1 3 is — = — = — times the original volume. 324 27 ( 3)
674 Chapter 12 Surface Area and Volume Changing a Dimension in a Solid
Describe how doubling the height affects the volume of the cone. 5 ft ANALYZING MATHEMATICAL RELATIONSHIPS 3 ft SOLUTION Notice that when the height is multiplied by Before change After change k, the volume is also multiplied by k. Dimensions r = 3 ft, h = 5 ft r = 3 ft, h = 10 ft 1 1 1 V = — πr2h V = — πr2h V = — πr2h original 3 3 3 1 2 V = — πr (kh) Volume 1 2 1 2 new 3 = — π(3) (5) = — π(3) (10) 1 3 3 = (k) — πr2h = π 3 = π 3 3 15 ft 30 ft = (k)V original 30π Doubling the height results in a volume that is — = 2 times the 15π original volume.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 6. In Example 4, describe how multiplying all the linear dimensions by 4 affects the volume of the rectangular pyramid. 1 — 7. In Example 4, describe how multiplying the height by 2 affects the volume of the rectangular prism. 8. In Example 5, describe how doubling the radius affects the volume of the cone. 9. In Example 5, describe how tripling all the linear dimensions affects the volume of the cone.
Finding the Volume of a Composite Solid
Find the volume of the composite solid. 6 m SOLUTION
Volume of Volume of Volume of = + 6 m solid cube pyramid
1 = 3 + — 6 m s 3 Bh Write formulas. 6 m 1 = 3 + — 2 6 3 (6) ⋅ 6 Substitute. = 216 + 72 Simplify. 5 cm = 288 Add. The volume is 288 cubic meters.
10 cm MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 3 cm 10. Find the volume of the composite solid.
Section 12.5 Volumes of Pyramids and Cones 675 12.5 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
VVocabularyocabulary andand CoreCore ConceptConcept CheckCheck 1. REASONING A square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.
1 — 2. COMPLETE THE SENTENCE The volume of a cone with radius r and height h is 3 the volume of a(n) ______with radius r and height h.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics
In Exercises 3 and 4, fi nd the volume of the pyramid. 11. Volume = 224 in.3 12. Volume = 198 yd3 (See Example 1.) 3. 4. 7 m 3 in. 16 m 12 in. 9 yd 12 m 4 in. 8 in. 11 yd 3 in. 13. ERROR ANALYSIS Describe and correct the error in In Exercises 5 and 6, fi nd the volume of the cone. fi nding the volume of the pyramid. (See Example 2.)
5. 6. 1 ✗ V = — (6)(5) 13 mm 1 m 5 ft 3 2 m 1 = — 10 mm 3 (30) = 10 ft3 6 ft
In Exercises 7–12, fi nd the missing dimension of the 14. ERROR ANALYSIS Describe and correct the error in pyramid or cone. (See Example 3.) fi nding the volume of the cone. 7. Volume = 912 ft3 8. Volume = 105 cm3 ✗ 1 10 m = — 훑 2 V 3 (6) (10) 19 ft 15 cm = 120훑 m3 6 m
w s 7 cm 15. OPEN-ENDED Give an example of a pyramid and a 9. Volume = 24π m3 10. Volume = 216π in.3 prism that have the same base and the same volume. Explain your reasoning.
16. OPEN-ENDED Give an example of a cone and a 18 in. cylinder that have the same base and the same h r volume. Explain your reasoning.
3 m
676 Chapter 12 Surface Area and Volume In Exercises 17–22, describe how the change affects the 27. 28. volume of the pyramid or cone. (See Examples 4 and 5.) 17. doubling all the 18. multiplying all the 12 in. 5.1 m 1 — linear dimensions linear dimensions by 4 4 in. 12 in. 5.1 m 20 cm 12 in. 5.1 m
7 in. 29. REASONING A snack stand serves a small order of popcorn in a cone-shaped container and a large order of popcorn in a cylindrical container. Do not perform 16 cm any calculations. 12 cm 3 in. 3 in.
19. multiplying the base 20. tripling the radius 1 — 8 in. 8 in. edge lengths by 3
$1.25 $2.50 9 m 11 ft a. How many small containers of popcorn do you have to buy to equal the amount of popcorn in a 5 m large container? Explain. 9 ft b. Which container gives you more popcorn for your money? Explain. 21. multiplying the 22. multiplying the height 3 height by 4 by — 2 30. HOW DO YOU SEE IT? 6 in. 14 cm The cube shown is formed by three pyramids, each with 10 in. the same square base 12 cm and the same height. 18 cm How could you use this to verify the formula In Exercises 23–28, fi nd the volume of the composite for the volume of solid. (See Example 6.) a pyramid? 23. 24. 7 cm In Exercises 31 and 32, fi nd the volume of the right cone. 5 cm 31. 32. 32° 10 cm 22 ft 5 cm 60° 14 yd 8 cm 9 cm 12 cm
33. MODELING WITH MATHEMATICS 2.5 in. 25. 3 cm 26. A cat eats half a cup 3 cm 8 ft of food, twice per day. Will the automatic 10 cm 9 ft pet feeder hold enough 7.5 in. food for 10 days? 12 ft Explain your reasoning. (1 cup ≈ 14.4 in.3)
4 in.
Section 12.5 Volumes of Pyramids and Cones 677 34. MODELING WITH MATHEMATICS During a chemistry 39. MAKING AN ARGUMENT In the fi gure, the two lab, you use a funnel to pour a solvent into a fl ask. cylinders are congruent. The combined height of The radius of the funnel is 5 centimeters and its height the two smaller cones equals the height of the larger is 10 centimeters. You pour the solvent into the funnel cone. Your friend claims that this means the total at a rate of 80 milliliters per second and the solvent volume of the two smaller cones is equal to the fl ows out of the funnel at a rate of 65 milliliters volume of the larger cone. Is your friend correct? per second. How long will it be before the funnel Justify your answer. overfl ows? (1 mL = 1 cm3)
35. ANALYZING RELATIONSHIPS A cone has height h and a base with radius r. You want to change the cone so its volume is doubled. What is the new height if you change only the height? What is the new radius if you change only the radius? Explain. 40. MODELING WITH MATHEMATICS Nautical deck prisms were used as a safe way to illuminate decks 36. REASONING The fi gure shown is a cone that has been on ships. The deck prism shown here is composed of warped but whose cross sections still have the same the following three solids: a regular hexagonal prism area as a right cone with equal radius and height. Find with an edge length of 3.5 inches and a height of the volume of this solid. Explain your reasoning. 1.5 inches, a regular hexagonal prism with an edge length of 3.25 inches and a height of 0.25 inch, and a regular 3 cm hexagonal pyramid with an edge length of 3 inches and a height of 3 inches. 2 cm Find the volume of the deck prism. 37. CRITICAL THINKING Find the volume of the regular pentagonal pyramid. Round your answer to the nearest hundredth. In the diagram, m∠ABC = 35°.
A
41. CRITICAL THINKING When the given triangle is C 3 ft rotated around each of its sides, solids of revolution B are formed. Describe the three solids and fi nd their volumes. Give your answers in terms of π.
38. THOUGHT PROVOKING A frustum of a cone is the part of the cone that lies between the base and a plane 15 20 parallel to the base, as shown. Write a formula for the volume of the frustum of a cone in terms of a, b, and h. (Hint: Consider the “missing” top of the cone 25 and use similar triangles.) 42. CRITICAL THINKING A square pyramid is inscribed in a a right cylinder so that the base of the pyramid is on h a base of the cylinder, and the vertex of the pyramid is on the other base of the cylinder. The cylinder has b a radius of 6 feet and a height of 12 feet. Find the volume of the pyramid.
MMaintainingaintaining MathematicalMathematical ProficiencyProficiency Reviewing what you learned in previous grades and lessons Find the indicated measure. (Section 11.2) 43. area of a circle with a radius of 7 feet 44. area of a circle with a diameter of 22 centimeters 45. diameter of a circle with an area of 256π 46. radius of a circle with an area of 529π
678 Chapter 12 Surface Area and Volume 12.6 Surface Areas and Volumes of Spheres
EEssentialssential QQuestionuestion How can you fi nd the surface area and the TEXAS ESSENTIAL volume of a sphere? KNOWLEDGE AND SKILLS G.10.B G.11.C Finding the Surface Area of a Sphere G.11.D Work with a partner. Remove the covering from a baseball or softball.
r SELECTING TOOLS To be profi cient in math, you need to identify You will end up with two “fi gure 8” pieces of material, as shown above. From the relevant external amount of material it takes to cover the ball, what would you estimate the surface area mathematical resources, S of the ball to be? Express your answer in terms of the radius r of the ball. such as content located S = Surface area of a sphere on a website. Use the Internet or some other resource to confi rm that the formula you wrote for the surface area of a sphere is correct.
Finding the Volume of a Sphere Work with a partner. A cylinder is circumscribed about a r sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r. r V = Volume of cylinder 2r
When half of the sphere (a hemisphere) is fi lled with sand and poured into the cylinder, it takes three hemispheres to fi ll the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.
V = Volume of a sphere
CCommunicateommunicate YourYour AnswerAnswer 3. How can you fi nd the surface area and the volume of a sphere? 4. Use the results of Explorations 1 and 2 to fi nd the surface area and the volume of a sphere with a radius of (a) 3 inches and (b) 2 centimeters.
Section 12.6 Surface Areas and Volumes of Spheres 679 12.6 Lesson WWhathat YYouou WWillill LLearnearn Find surface areas of spheres. Core VocabularyVocabulary Find volumes of spheres. chord of a sphere, p. 680 Finding Surface Areas of Spheres great circle, p. 680 A sphere is the set of all points in space equidistant from a given point. This point is Previous called the center of the sphere. A radius of a sphere is a segment from the center to sphere a point on the sphere. A chord of a sphere is a segment whose endpoints are on the center of a sphere sphere. A diameter of a sphere is a chord that contains the center. radius of a sphere diameter of a sphere chord hemisphere C C radius diameter center
As with circles, the terms radius and diameter also represent distances, and the diameter is twice the radius. If a plane intersects a sphere, then the intersection is either a single point or a circle. If the plane contains the center of the sphere, hemispheres then the intersection is a great circle of the great sphere. The circumference of a great circle circle is the circumference of the sphere. Every great circle of a sphere separates the sphere into two congruent halves called hemispheres.
CCoreore CConceptoncept Surface Area of a Sphere The surface area S of a sphere is r S = 4πr2 where r is the radius of the sphere. S = 4π r2
To understand the formula for the surface area of a sphere, think of a baseball. The surface area of a baseball is sewn from two congruent shapes, each of which resembles two joined circles. r So, the entire covering of the baseball consists of four circles, each with radius r. The area A of a circle with radius r is A = πr2. So, the area of the leather covering covering can be approximated by 4πr2. This is the formula for the surface area of a sphere.
680 Chapter 12 Surface Area and Volume Finding the Surface Areas of Spheres
Find the surface area of each sphere. a. b. 8 in. C = 12π ft
SOLUTION a. S= 4πr2 Formula for surface area of a sphere = 4π(8)2 Substitute 8 for r. = 256π Simplify. ≈ 804.25 Use a calculator. The surface area is 256π, or about 804.25 square inches. 12π b. The circumference of the sphere is 12π, so the radius of the sphere is — = 6 feet. 2π S = 4πr2 Formula for surface area of a sphere = 4π(6)2 Substitute 6 for r. = 144π Simplify. ≈ 452.39 Use a calculator. The surface area is 144π, or about 452.39 square feet.
Finding the Diameter of a Sphere
Find the diameter of the sphere.
SOLUTION S = 4πr2 Formula for surface area of a sphere π = π 2 π COMMON ERROR 20.25 4 r Substitute 20.25 for S. S = 20.25π cm2 2 Be sure to multiply the 5.0625 = r Divide each side by 4π. value of r by 2 to fi nd 2.25 = r Find the positive square root. the diameter. The diameter is 2r = 2 • 2.25 = 4.5 centimeters.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com Find the surface area of the sphere. 1. 40 ft 2. C = 6π ft
3. Find the radius of the sphere.
S = 30π m2
Section 12.6 Surface Areas and Volumes of Spheres 681 Finding Volumes of Spheres The fi gure shows a hemisphere and a cylinder with a cone removed. A plane parallel to their bases intersects the solids z units above their bases.
r2 − z2
r z r r
Using the AA Similarity Theorem (Theorem 8.3), you can show that the radius of the cross section of the cone at height z is z. The area of the cross section formed by the plane is π(r2 − z2) for both solids. Because the solids have the same height and the same cross-sectional area at every level, they have the same volume by Cavalieri’s Principle. = − Vhemisphere Vcylinder Vcone
1 = π 2 − — π 2 r (r) 3 r (r)
2 = — π 3 3 r So, the volume of a sphere of radius r is
2 4 = — π 3 = — π 3 2 ⋅ Vhemisphere 2 ⋅ 3 r 3 r .
CCoreore CConceptoncept Volume of a Sphere The volume V of a sphere is r 4 V = — πr3 3 V = 4 π r3 where r is the radius of the sphere. 3
Finding the Volume of a Sphere
Find the volume of the soccer ball. 4.5 in.
SOLUTION 4 = — π 3 V 3 r Formula for volume of a sphere 4 = — π 3 3 (4.5) Substitute 4.5 for r. = 121.5π Simplify. ≈ 381.70 Use a calculator.
The volume of the soccer ball is 121.5π, or about 381.70 cubic inches.
682 Chapter 12 Surface Area and Volume Changing Dimensions in a Solid
1 — Describe how multiplying the radius by 4 affects the volume of the sphere. 12 in.
SOLUTION
Before change After change Dimensions r = 12 in. r = 3 in. 4 4 V = — πr3 V = — πr3 3 3 4 4 Volume = — π(12)3 = — π(3)3 3 3 = 2304π in.3 = 36π in.3
1 36π 1 1 3 Multiplying the radius by — results in a volume that is — = — = — 4 2304π 64 ( 4) times the original volume.
Finding the Volume of a Composite Solid
Find the volume of the composite solid.
SOLUTION 2 in. 2 in. Volume = Volume of − Volume of of solid cylinder hemisphere
1 4 = π 2 − — — π 3 r h 2 ( 3 r ) Write formulas. 2 = π 2 − — π 3 (2) (2) 3 (2) Substitute. 16 = π − — π 8 3 Multiply. 24 16 Rewrite fractions using least = — π − — π 3 3 common denominator. 8 = — π 3 Subtract. ≈ 8.38 Use a calculator.
8 — π The volume is 3 , or about 8.38 cubic inches.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 4. The radius of a sphere is 5 yards. Find the volume of the sphere. 1 m 5. The diameter of a sphere is 36 inches. Find the volume of the sphere. 6. In Example 4, describe how doubling the radius affects the volume of the sphere. 5 m 7. A sphere has a radius of 18 centimeters. Describe how multiplying the 1 — radius by 3 affects the volume of the sphere. 8. Find the volume of the composite solid at the left.
Section 12.6 Surface Areas and Volumes of Spheres 683 12.6 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
VVocabularyocabulary andand CoreCore ConceptConcept CheckCheck 1. VOCABULARY When a plane intersects a sphere, what must be true for the intersection to be a great circle?
2. WRITING Explain the difference between a sphere and a hemisphere.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics
In Exercises 3–6, fi nd the surface area of the sphere. In Exercises 13–18, fi nd the volume of the sphere. (See Example 1.) (See Example 3.) 3. 4. 13. 14. 7.5 cm 4 ft 8 m 4 ft
5. 6. 15. 16.
22 yd 14 ft 18.3 m C = 4π ft
17. C = 20π cm 18. C = 7π in. In Exercises 7–10, fi nd the indicated measure. (See Example 2.) 7. Find the radius of a sphere with a surface area of 4π square feet.
8. Find the radius of a sphere with a surface area of 1024π square inches. In Exercises 19 and 20, fi nd the volume of the sphere with the given surface area. 9. Find the diameter of a sphere with a surface area of 19. Surface area = 16π ft2 900π square meters. 20. Surface area = 484π cm2 10. Find the diameter of a sphere with a surface area of 196π square centimeters. 21. ERROR ANALYSIS Describe and correct the error in fi nding the volume of the sphere. In Exercises 11 and 12, fi nd the surface area of the hemisphere.
11. 12. ✗ = —4 π 2
V 3 (6) 6 ft = π 5 m 12 in. 48 ≈ 150.80 ft3
684 Chapter 12 Surface Area and Volume 22. ERROR ANALYSIS Describe and correct the error in 33. MAKING AN ARGUMENT Your friend claims that if fi nding the volume of the sphere. the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is your friend correct? Explain your reasoning. 4 ✗ V = — π(3)3 3 34. REASONING A semicircle with a diameter of 3 in. = 36π 18 inches is rotated about its diameter. Find the ≈ 3 113.10 in. surface area and the volume of the solid formed.
35. MODELING WITH MATHEMATICS A silo has the dimensions shown. The top of the silo is a In Exercises 23 and 24, describe how the change affects hemispherical shape. Find the volume of the silo. the volume of the sphere. (See Example 4.) 23. tripling the radius 24. multiplying the radius 2 — by 3
1 m 60 ft 36 cm
20 ft
In Exercises 25–28, fi nd the volume of the composite solid. (See Example 5.) 36. MODELING WITH MATHEMATICS Three tennis balls are stored in a cylindrical container 25. 26. 6 ft with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches. 9 in. 5 in. 12 ft a. Find the volume of a tennis ball. b. Find the amount of space within the cylinder not taken up by the tennis balls. 27. 18 cm 28. 14 m 10 cm 6 m 37. ANALYZING RELATIONSHIPS Use the table shown for a sphere.
Radius Surface area Volume In Exercises 29–32, fi nd the surface area and volume of 3 in. 36π in.2 36π in.3 the ball. 6 in. 29. bowling ball 30. basketball 9 in. 12 in.
a. Copy and complete the table. Leave your answers in terms of π. b. What happens to the surface area of the sphere = d 8.5 in. C = 29.5 in. when the radius is doubled? tripled? quadrupled? c. What happens to the volume of the sphere when 31. softball 32. golf ball the radius is doubled? tripled? quadrupled?
38. MATHEMATICAL CONNECTIONS A sphere has a diameter of 4(x + 3) centimeters and a surface area of 784π square centimeters. Find the value of x. = = C 12 in. d 1.7 in.
Section 12.6 Surface Areas and Volumes of Spheres 685 39. MODELING WITH MATHEMATICS The radius of Earth 43. CRITICAL THINKING Let V be the volume of a sphere, is about 3960 miles. The radius of the moon is about S be the surface area of the sphere, and r be the radius 1080 miles. of the sphere. Write an equation for V in terms of r V a. Find the surface area of Earth and the moon. and S. Hint: Start with the ratio — . ( S ) b. Compare the surface areas of Earth and the moon. 44. THOUGHT PROVOKING A spherical lune is the c. About 70% of the surface of Earth is water. How region between two great circles of a sphere. Find many square miles of water are on Earth’s surface? the formula for the area of a lune. 40. MODELING WITH MATHEMATICS The Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the r distance as the height of a cylindrical belt around the θ Earth at the equator. Tropic of Cancer
Torrid 3250 mi equator Zone 45. CRITICAL THINKING The volume of a right cylinder is the same as the volume of a sphere. The radius of Tropic of the sphere is 1 inch. Give three possibilities for the Capricorn dimensions of the cylinder.
a. Estimate the surface area of the Torrid Zone. 46. PROBLEM SOLVING A spherical cap is a portion of a (The radius of Earth is about 3960 miles.) sphere cut off by a plane. The formula for the volume πh b. A meteorite is equally likely to hit anywhere on of a spherical cap is V = — (3a2 + h2), where a is 6 Earth. Estimate the probability that a meteorite the radius of the base of the cap and h is the height will land in the Torrid Zone. of the cap. Use the diagram and given information to fi nd the volume of each spherical cap. 41. ABSTRACT REASONING A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the h surface area of the sphere? Explain your reasoning. a r 42. HOW DO YOU SEE IT? The formula for the volume of a hemisphere and a cone are shown. If each solid has the same radius and r = h, which solid will have a greater volume? Explain your reasoning. a. r = 5 ft, a = 4 ft b. r = 34 cm, a = 30 cm rr c. r = 13 m, h = 8 m d. r = 75 in., h = 54 in.
h 47. CRITICAL THINKING A sphere with a radius of 2 inches is inscribed in a right cone with a height = 2π 3 = 1π 2 V 3 r V 3 r h of 6 inches. Find the surface area and the volume of the cone.
MMaintainingaintaining MathematicalMathematical ProficiencyProficiency Reviewing what you learned in previous grades and lessons
Use the diagram. (Section 1.1) U 48. Name four points. 49. Name two line segments. S P R 50. Name two lines. 51. Name a plane. T
Q
686 Chapter 12 Surface Area and Volume 12.7 Spherical Geometry
EEssentialssential QQuestionuestion How can you represent a line on a sphere? TEXAS ESSENTIAL KNOWLEDGE AND SKILLS The endpoints of a diameter of a sphere G.4.D are called antipodal points. AB great circle
Points A and B are antipodal.
Finding the Shortest Distance Between Two Points Work with a partner. Use a washable marker, a piece of paper, a ball, and a piece of string. a. Draw two points on the piece of paper. Label the points A and B. Use the string to create the shortest path between the points. What geometric term describes the path of the string? When you extend the path infi nitely in either direction, what ANALYZING geometric term describes the path? MATHEMATICAL b. Draw two points that are not antipodal on the ball. Label RELATIONSHIPS the points C and D. Use the string to create the shortest C To be profi cient in math path between the points. What geometric term describes D you need to look closely the path of the string? Can you extend the path infi nitely to discern a pattern or in either direction? What geometric term describes the structure. path when you extend it all the way around the ball? c. Compare and contrast the paths on the piece of paper in part (a) and the paths on the ball in part (b).
Exploring Lines on a Sphere Work with a partner. Use a ball and three rubber bands. a. Construct a great circle on the ball using a rubber band. b. Construct a second distinct great circle. At how many points do the two great circles intersect? Compare this to the number of points at which two distinct lines can intersect in the plane. c. How many great circles can you construct through two points that are not antipodal? Explain. d. How many great circles can you construct through two points that are antipodal? Explain. e. Is the Two Point Postulate (Postulate 2.1) true for great circles on a sphere? Explain. Two Point Postulate: Through any two points, there exists exactly one line. f. A polygon on the surface of a sphere is a shape enclosed by arcs of great circles. Describe how you can construct a polygon with two sides on the surface of a sphere. Then construct a two-sided polygon on the ball using rubber bands. g. Describe how you can construct a triangle on the surface of a sphere. Then construct a triangle on the ball using rubber bands. CCommunicateommunicate YourYour AnswerAnswer 3. How can you represent a line on a sphere?
Section 12.7 Spherical Geometry 687 12.7 Lesson WWhathat YYouou WWillill LLearnearn Compare Euclidean and spherical geometry. Find distances on a sphere. Core VocabularyVocabulary Find areas of spherical triangles. antipodal points, p. 688 Previous Comparing Euclidean and Spherical Geometry great circle In Euclidean geometry, a plane is a fl at surface that extends without end in all directions, and a line in the plane is a set of points that extends without end in two directions. Geometry on a sphere is different. In spherical geometry, a plane is the surface center of a sphere, a line is a great circle, and the great circles angle between two lines is the angle between the planes containing the two corresponding great circles.
CCoreore CConceptoncept Euclidean Geometry and Spherical Geometry Euclidean Geometry Spherical Geometry A center
P A S m
Plane P contains lineℓand point A Sphere S contains great circle m and not on the lineℓ. point A not on m. Great circle m is a line.
A
A P B C B C S
The vertices of △ABC are points in The vertices of △ABC are points on plane P and the sides are segments. sphere S and the sides are arcs of The sum of the interior angles of a great circles. The sum of the interior triangle is 180°. angles of a spherical triangle is greater than 180°. m∠A + m∠B + m∠C = 180° m∠A + m∠B + m∠C > 180°
Some properties and postulates in Euclidean geometry are true in spherical geometry. Others are not, or are true only under certain circumstances. For example, in Euclidean geometry, the Two Point Postulate (Postulate 2.1) states that through any two points, there exists exactly one line. In spherical geometry, this postulate is true only for points that are not the endpoints of a diameter of the sphere. The endpoints of a diameter of a sphere are called antipodal points.
688 Chapter 12 Surface Area and Volume Comparing Euclidean and Spherical Geometry
Tell whether the following postulate in Euclidean geometry is also true in spherical geometry. Draw a diagram to support your answer. Parallel Postulate (Postulate 3.1): If there is a line and a point not on the line, then there is exactly one line through the point parallel to the given line.
SOLUTION Parallel lines do not intersect. The sphere shows a line ℓ (a great circle) and a point A not onℓ. Several lines are drawn through A. Each great A circle containing A intersectsℓ. So, there can be no line parallel toℓ. The Parallel Postulate is not true in spherical geometry.
MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com 1. Draw sketches to show that the Two Point Postulate (Postulate 2.1), discussed on the previous page, is not true for antipodal points and is true for two points that are not antipodal points.
Finding Distances on a Sphere
In Euclidean geometry, there is exactly one distance that can be measured between any two points. On a sphere, there are two distances that can be measured between any two points. These distances are the lengths of the major and minor arcs of the great circle drawn through the points.
Finding Distances on a Sphere
The diameter of the sphere is 15 inches, and ២ 15 in. m AB = 60°. Find the distances between AC points A and B. P B SOLUTION ២ ២ Find the lengths of the minor២ arc AB and the major arc ACB ២ of the great circle shown. Let x be the arc length of AB and let y be the arc length of ACB . ២ ២ ២ ២ Arc length of AB mAB Arc length of ACB mACB —— = — —— = — 2πr 360° 2πr 360° x 60° y 360° − 60° — = — — = — 15π 360° 15π 360° x = 2.5π y = 12.5π x ≈ 7.85 y ≈ 39.27
The distances are about 7.85 inches and about 39.27 inches.
20 cm MMonitoringonitoring ProgressProgress Help in English and Spanish at BigIdeasMath.com AC P ២ 2. The diameter of the sphere is 20 centimeters, and mAB = 120°. Find the distances B between points A and B.
Section 12.7 Spherical Geometry 689 Finding Areas of Spherical Triangles CCoreore CConceptoncept Area of a Spherical Triangle A The area of △ABC on a sphere is r πr2 A = — ( m∠A + m∠B + m∠C − 180°) B C 180° where r is the radius of the sphere.
Finding Areas of Spherical Triangles
Find the area of each spherical triangle.
a. △DEF D 95° b. △JKL J 12 m 10 in. 102°
90° 78° 115° 105° K L E F
SOLUTION πr2 a. A = — (m∠D + m∠E + m∠F − 180°) Formula for area of a spherical triangle 180° π (10)2 = — (95 ° + 115° + 105° − 180°) Substitute. 180° 100π = — (135°) Simplify. 180° = 75π Simplify. ≈ 235.62 Use a calculator. The area of △DEF is 75π, or about 235.62 square inches. πr2 b. A = — ( m∠J + m∠K + m∠L − 180°) Formula for area of a spherical triangle 180° π(12)2 = — (102 ° + 90° + 78° − 180°) Substitute. 180° 144π = — (90°) Simplify. 180° = 72π Simplify. ≈ 226.19 Use a calculator. The area of △JKL is 72π, or about 226.19 square meters.
MMonitoringonitoring PProgressrogress Help in English and Spanish at BigIdeasMath.com Find the area of the spherical triangle.
3. △MNP 4. △QRS 5. △TUV M Q 9 mm 12 m 8 cm 110° 90° T 120° ° ° 70° ° ° ° 60 60 30 90 90 U V N P R S
690 Chapter 12 Surface Area and Volume 12.7 Exercises Tutorial Help in English and Spanish at BigIdeasMath.com
VVocabularyocabulary andand CoreCore ConceptConcept CheckCheck 1. WRITING How is a line in Euclidean geometry different from a line in spherical geometry?
2. WHICH ONE DOESN’T BELONG? Which statement does not belong with the other three? Explain your reasoning. There are no parallel lines. Two distances can be measured between any two points.
The sum of the interior angles of a triangle is greater than 180°. A line extends without end.
MMonitoringonitoring ProgressProgress andand ModelingModeling withwith MathematicsMathematics ២ ២ In Exercises 3–8, the statement is true in Euclidean 13. m AB = 150° 14. m AB = 135° geometry. Rewrite the statement to be true for spherical geometry. Explain your reasoning. (See Example 1.) 3. 14 yd 40 in. Given a line and a point not on the line, there is exactly ACAC one line through the point parallel to the given line. P P B B 4. If two lines intersect, then their intersection is exactly one point. In Exercises 15–20, fi nd the area of the spherical 5. The length of a line is infi nite. triangle. (See Example 3.) △ △ 6. A line divides a plane into two infi nite regions. 15. ABC 16. DEF D ° 2 m 7. A triangle can have no more than one right angle. 127 A 6 ft 45° B 8. Two distinct lines in a plane are either parallel or 45° 53° 90° C 90° intersect at exactly one point. E F In Exercises 9–14, use the diagram and the given arc measure to fi nd the distances between points A and B. 17. △JKL 18. △MNP (See Example 2.) M ២ ២ J 9. m AB = 90° 10. m AB = 140° ° 5 yd 108 ° 4 in. 100 ° K 75 132° L 16 cm 30 mm 120° 110° ACAC P P N P B B 19. △QRS 20. △TUV ២ ២ 11. m AB = 30° 12. m AB = 45° T 140° Q 85° 16 mm 30 cm 60° 60° 12 ft 18 m RS ACAC 111° 119° P P V B B U
Section 12.7 Spherical Geometry 691 21. ERROR ANALYSIS Describe and correct the error in 24. REASONING Determine whether each of the theorems fi nding the distances between points A and B. from Euclidean geometry is true in spherical ២ ២ geometry. Explain your reasoning. Arc length of AB m AB —— = — ✗ 2훑r 360° a. Linear Pair Perpendicular Theorem (Theorem 3.10): If two lines intersect to form a linear pair of x 90° — = — congruent angles, then the lines are perpendicular. 2훑(24) 360° ° —x = 90 b. Lines Perpendicular to a Transversal Theorem 24 cm — AC48훑 360° (Theorem 3.12): In a plane, if two lines are P = 훑 perpendicular to the same line, then they are B x 12 parallel to each other. ≈ 37.70 ២ ២ 25. MODELING WITH MATHEMATICS The radius of ——Arc length of ACB = —mACB Earth is about 3960 miles. Find the distance between 훑 ° 2 r 360 Pontianak, Indonesia, and the North Pole. y 360° − 90° — = —— 2훑(24) 360° North Pole y 270° — = — 48훑 360° y = 36훑 The distances are about ≈ 37.70 centimeters and 113.10 Equator about 113.10 centimeters. Pontianak, Indonesia
22. ERROR ANALYSIS Describe and correct the error in fi nding the area of △ABC. 26. HOW DO YOU SEE IT? In Euclidean geometry, when three points are collinear, exactly one of the points 훑r2 lies between the other two. Is this true in spherical A = — ( m∠A + m∠B + m∠C) ✗ 180° geometry? Explain. 훑(5)2 = — (114° + 106° + 104°) 180° C 25훑 B = — 324° C ( 180° ) A A A m = 45훑 B 114° ≈ 141.37 5 in. 106° 104° The area of △ABC B C is 45훑, or about 27. REASONING In Euclidean geometry, two 141.37 square inches. perpendicular lines form four right angles. How many right angles do two perpendicular lines form in spherical geometry? 23. MAKING AN ARGUMENT A polygon on a sphere has arcs of great circles for sides. Your friend claims that a 28. THOUGHT PROVOKING What are the possible sums polygon on a sphere can have two sides. Is your friend of the interior angles of a spherical triangle? Explain. correct? Explain your reasoning.
Maintainingg Mathematical Proficiency Reviewing what you learned in previous grades and lessons Find the areas of the sectors formed by ∠DFE. (Section 11.2) 29. 30. 31. G G 6 m E G 5 ft F F E F 120° 8 yd 150° 45° D DE D
692 Chapter 12 Surface Area and Volume 12.4–12.7 What Did You Learn?
CCoreore VVocabularyocabulary volume, p. 664 chord of a sphere, p. 680 antipodal points, p. 688 Cavalieri’s Principle, p. 664 great circle, p. 680
CCoreore CConceptsoncepts Section 12.4 Cavalieri’s Principle, p. 664 Volume of a Cylinder, p. 665 Volume of a Prism, p. 664 Section 12.5 Volume of a Pyramid, p. 672 Volume of a Cone, p. 673 Section 12.6 Surface Area of a Sphere, p. 680 Volume of a Sphere, p. 682 Section 12.7 Euclidean Geometry and Spherical Geometry, p. 688 Area of a Spherical Triangle, p. 690 Finding Distances on a Sphere, p. 689
MMathematicalathematical TThinkinghinking 1. Search online for advertisements for products that come in different sizes. Then compare the unit prices, as done in Exercise 36 on page 670. Do you get results similar to Exercise 36? Explain. 2. In Exercise 30 on page 677, the cube is formed by three oblique pyramids. If these pyramids were right pyramids, it would not be possible to form a cube. Explain why the formula for the volume of a right pyramid is the same as the formula for the volume of an oblique pyramid. 3. In Exercise 38 on page 685, explain the steps you used to fi nd the value ofof x.
Performancee TasTaskk Water Park Renovation The city council will consider reopening the closed water pparkark if your team can come up with a cost analysis for painting some of the structures, fi lling the pool water reservoirs, and resurfacingurfacing some of the surfaces. What is your plan to convince the cityty council to open the water park?
To explore the answer to this question and more, go to BigIdeasMath.com.
693693 1122 Chapter Review
12.1 Three-Dimensional Figures (pp. 639–644)
Sketch the solid produced by rotating the fi gure around 8 the given axis. Then identify and describe the solid. 3 3 8 8 3 The solid is a cylinder with a height of 8 and a radius of 3.
Sketch the solid produced by rotating the fi gure around the given axis. Then identify and describe the solid. 1. 2. 3.
7 8
9 7 6
5
Describe the cross section formed by the intersection of the plane and the solid. 4. 5. 6.
12.2 Surface Areas of Prisms and Cylinders (pp. 645–652)
Find the lateral area and the surface area of the right rectangular prism. L = Ph Formula for lateral area of a right prism
= [2(8) + 2(14)](9) Substitute.
= 396 Simplify. 9 in. 14 in. S = 2B + L Formula for surface area of a right prism 8 in. = 2(8)(14) + 396 Substitute.
= 620 Simplify.
The lateral area is 396 square inches and the surface area is 620 square inches.
694 Chapter 12 Surface Area and Volume Find the lateral area and the surface area of the right prism or right cylinder. 7. 8. 9. 5 m
17 ft 11 m 20 ft 18 cm 4 cm 16 ft
12.3 Surface Areas of Pyramids and Cones (pp. 653–660)
Find the lateral area and the surface area of the right cone. 15 m Use the Pythagorean Theorem (Theorem 9.1) to fi nd the slant heightℓ. — ℓ= √152 + 362 = 39 Find the lateral area and the surface area. 36 m L = πrℓ Formula for lateral area of a right cone = π(15)(39) Substitute. = 585π Simplify. ≈ 1837.83 Use a calculator.
S = πr 2 + πrℓ Formula for surface area of a right cone = π(15)2 + 585π Substitute. = 810π Simplify. ≈ 2544.69 Use a calculator.
The lateral area is 585π, or about 1837.83 square meters. The surface area is 810π, or about 2544.69 square meters.
Find the lateral area and the surface area of the regular pyramid or right cone. 10. 11. 12. 8 in. 10 ft 24 cm
6 ft 9 in. 10 cm 3 ft
13. Find the lateral area and the surface area of the 4 cm composite solid. 3 cm
12 cm
Chapter 12 Chapter Review 695 12.4 Volumes of Prisms and Cylinders (pp. 663–670)
Find the volume of the cylinder. 7 ft The dimensions of the cylinder are r = 7 ft and h = 10 ft. = π 2 V r h Formula for volume of a cylinder 10 ft = π(7)2(10) Substitute. = 490π Simplify. ≈ 1539.38 Use a calculator.
The volume is 490π, or about 1539.38 cubic feet.
Find the volume of the solid. 14. 15. 16.
3.6 m 8 mm
4 yd 2.1 m 2 mm 1.5 m 2 yd
17. Describe how the change affects the volume of the triangular prism. 1 — a. multiplying the height of the prism by 3 b. multiplying all the linear dimensions by 2
7 cm 30 cm 6 in. 6 in. 24 cm 10 in.
6 in. 18. Find the volume of the composite solid.
12.5 Volumes of Pyramids and Cones (pp. 671–678)
Find the volume of the pyramid. 1 = — V 3 Bh Formula for volume of a pyramid
1 1 = — — 5 8 (12) Substitute. 3( 2 ⋅ ⋅ ) 12 m = 80 Simplify.
The volume is 80 cubic meters. 8 m 5 m
696 Chapter 12 Surface Area and Volume Find the volume of the solid. 19. 20. 21. 7 m 5 m 34 cm 30 cm 18 m 13 m 16 cm 10 m
22. The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid.
23. A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the height of the cone.
12.6 Surface Areas and Volumes of Spheres (pp. 679–686)
Find the (a) surface area and (b) volume of the sphere. a. S = 4πr2 Formula for surface area of a sphere = 4π(18)2 Substitute 18 for r. 18 in. = 1296π Simplify. ≈ 4071.50 Use a calculator.
The surface area is 1296π, or about 4071.50 square inches.
4 = — π 3 b. V 3 r Formula for volume of a sphere 4 = — π 3 3 (18) Substitute 18 for r. = 7776π Simplify. ≈ 24,429.02 Use a calculator.
The volume is 7776π, or about 24,429.02 cubic inches.
Find the surface area and the volume of the sphere. 24. 25. 26.
= π 7 in. 17 ft C 30 ft
27. The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury. 28. A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.
Chapter 12 Chapter Review 697 12.7 Spherical Geometry (pp. 687–692) ២ a. The diameter of the sphere is 24 inches, and m AB = 30°. Find the distances between points A and B. ២ ២ 24 in. Find the lengths of the minor arc AB and the major arc ACB of the AC ២ P great circle shown.២ Let x be the arc length of AB and let y be the arc length of ACB . B ២ ២ ២ ២ Arc length of AB m AB Arc length of ACB m ACB —— = — —— = — 2πr 360° 2πr 360° x 30° y 360° − 30° — = — — = — 24π 360° 24π 360° x = 2π y = 22π x ≈ 6.28 y ≈ 69.12
The distances are about 6.28 inches and about 69.12 inches.
b. Find the area of △DEF. D πr2 Formula for area of A = — (m∠D + m∠E + m∠F − 180°) 116° 6 ft 180° a spherical triangle 82° 82° π (6)2 = — (116° + 82° + 82° − 180°) Substitute. E F 180° 36π = — (100°) Simplify. 180° = 20π Simplify.
≈ 62.83 Use a calculator.
The area of △DEF is 20π, or about 62.83 square feet. ២ 29. The diameter of the sphere is 18 feet, and mXY = 90°. 18 ft Find the distances between points X and Y. XZ P Y
Find the area of the spherical triangle. 30. △JKL 31. △RST 32. △XYZ J R X 88° 12 m ° 125° 110 3 in. 18 cm ° ° ° ° ° 85 90 124 88 102° 108 Z K L S T Y
698 Chapter 12 Surface Area and Volume 1122 Chapter Test Find the volume of the solid.
1. 2. 3. 3 m 4 m 4. 4 ft
3.2 ft 15.5 m 6 m 8 ft
4 m 8 m 3 m 5 ft 2 ft Find the lateral area and the surface area of the solid. 5. 6. 7. 4 in. 8. 8 m 9 ft 16 cm 8.3 m 12 cm
20 ft 30 cm 6 in. 12 m 7 in. 7 in.
9. Sketch the composite solid produced by rotating the fi gure around 6 the given axis. Then identify and describe the composite solid. 33
10. The shape of Mars can be approximated by a sphere with a diameter 9 of 6779 kilometers. Find the surface area and the volume of Mars.
11. You have a funnel with the dimensions shown. 6 cm a. Find the approximate volume of the funnel. b. You use the funnel to put oil in a car. Oil fl ows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full 10 cm of oil? (1 mL = 1 cm3) c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil fl ows out of the funnel at a rate of 45 milliliters per second? d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations.
12. Describe how the change affects the surface area and the volume of the right cone. 20 yd 3 — a. multiplying the height by 4 6 — b. multiplying all the linear dimensions by 5
13. A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. 15 yd The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.
14. Is it possible for a right triangular pyramid to have a cross section that is a quadrilateral? If so, describe or sketch how the plane could intersect the pyramid.
15. The soccer ball shown has a radius of 4.5 inches. Each interior angle of the spherical triangles has a measure of 70°. Find the area of one spherical triangle.
Chapter 12 Chapter Test 699 1122 Standards Assessment
1. Which cross section formed by the intersection of the plane and the solid is a rectangle? (TEKS G.10.A)
○A ○B
○C ○D
2. The coordinates of the vertices of △DEF are D(−8, 5), E(−5, 8), and F(−1, 4). Which set of coordinates describes a triangle that is similar to △DEF? (TEKS G.7.A)
○F A(−24, 15), B(−15, 24), and C(−3, 8) ○G J(16, −10), K(10, −16), and L(2, −8) ○H P(−8, 10), Q(−5, 16), and R(−1, 8) ○J U(−10, 5), V(−7, −2), and W(−2, −7)
3. The top of the Washington Monument in Washington, D.C., is a square pyramid, called a pyramidion. What is the volume of the pyramidion? (TEKS G.11.D) ○A 22,019.63 ft3 ○B 172,006.91 ft3 C 66,058.88 ft3 ○ 55.5 ft ○D 207,530.08 ft3
34.5 ft
4. Use the diagram. Which proportion is false? (TEKS G.8.B) C D DB DA CA AB F — = — G — = — ○ DC DB ○ AB AD CA BA DC BC H — = — J — = — A B ○ BA CA ○ BC CA
700 Chapter 12 Surface Area and Volume 5. The surface area of the right cone is 200π square feet. What is the slant height of the cone? (TEKS G.11.C)
○A 10.5 ft ○B 17 ft
○C 23 ft ○D 24 ft 16 ft
6. In the diagram, ABCD is a parallelogram. Which statement is not true? (TEKS G.6.B)
A B
E
D C
○F △AED ≅ △CEB ○G △DEC ≅ △BEA ○H △ABD ≅ △CDB ○J △AEB ≅ △DEC
7. Points X, Y, and Z lie on the surface of a sphere. What is a possible value of the sum of the interior angles of △XYZ? (TEKS G.4.D)
○A 90° ○B 167° ○C 180° ○D 204°
8. GRIDDED ANSWER The diagram shows a square pyramid and a cone. Both solids have a height of 23 inches, and the base of the cone has a radius of 8 inches. According to Cavalieri’s Principle, the solids will have the same volume if the square base has sides of length ___ inches. Round your answer to the nearest tenth of an inch. (TEKS G.11.D)
B B h
r
9. Which statement about the circle is not true? (TEKS G.12.A) — ○F AD is a diameter. B — ○G BH is a chord. A H — ○H AD is a chord. C D — G ○J CD is a radius. E F
Chapter 12 Standards Assessment 701