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# Surface area and volume

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### Surface area and volume

1. 1. Prism Pyramid Cone Cylinder Sphere
2. 2. Prism Pyramid Cone Cylinder Sphere Link
3. 3. Check It Out: Example 1B All faces and bases are congruent squares. The figure is a cube. Identify the bases and faces of the figure. Then name the figure. Surface Area and Volume
4. 4. Check It Out: Example 2B Classify each figure as a polyhedron or not a polyhedron. Then name the figure. There are two triangular bases for the figure. The faces are all polygons, so the figure is a polyhedron. The figure is a triangular prism.
5. 5. Identify the bases and faces of the figure. Then name the figure. There are two octagonal bases. The figure is an octagonal prism. There are eight rectangular faces.
6. 6. Lesson Quiz: Part I Identify the bases and faces of each figure. Then name each figure. Two pentagon bases, 5 rectangular faces; pentagonal prism One square base, 4 triangular faces; square pyramid 1. 2.
7. 7. Lesson Quiz: Part II Classify each figure as a polyhedron or not a polyhedron. Then name the figure. polyhedron, rectangular prism polyhedron, triangular prism 3. 4.
8. 8. 2. Identify the bases and faces, and then the name of the given figure. A. pentagon; triangles; pentagonal pyramid B. heptagon; triangles; heptagonal pyramid C. pentagon; rectangles; pentagonal prism D. heptagon; rectangles; heptagonal prism Lesson Quiz for Student Response Systems
9. 9. 3. Classify the figure as a polyhedron or not a polyhedron, and then name the given figure. A. polyhedron; pentagonal prism B. polyhedron; pentagonal pyramid C. not a polyhedron; cylinder D. not a polyhedron; cone Lesson Quiz for Student Response Systems
10. 10. Vertices (points) Edges (lines) Faces (planes) 6 9 5 The base has 3 sides.
11. 11. Vertices (points) Edges (lines) Faces (planes) 8 12 6 The base has sides.4
12. 12. Vertices (points) Edges (lines) Faces (planes) 10 15 7 The base has sides.5
13. 13. Vertices (points) Edges (lines) Faces (planes) 12 18 8 The base has sides.6
14. 14. Vertices (points) Edges (lines) Faces (planes) 16 24 10 The base has sides.8
15. 15. Name Picture Base Vertices Edges Faces Triangular Prism Rectangular Prism Pentagonal Prism Hexagonal Prism Heptagonal Prism Octagonal Prism 3 6 9 5 4 8 12 6 5 10 15 7 6 12 18 8 7 14 21 9 8 16 24 10 Any Prism n 2n 3n n + 2 Draw it No picture
16. 16. Prism Surface area of wall = Length around the base.  high = (3+4+5)  6 = 72 Square centimeter Base = 43 2 1  = 6 Square centimeter Surface area = Surface area of wall + (2  Area of base ) = 72 + (2 6) = 72+12 = 84 Square centimeter = 84 Square centimeter
17. 17. Prism surface Area of wall = Length around the base.  high = ( 5+5+5+5 ) 10 = 20  10 = 200 cm2 Area of base = 5  5 = 25 cm2 Surface area = surface Area of wall + (2  Area of base) = 200 + (2  25) = 200 + 50 = 250 cm2 = 250 cm2
18. 18. Prism Surface Area of wall = perimeter of base.  high = ( 2+5+2+5 ) 10 = 14  10 = 140 cm2 Area of base = 5  2 = 10 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 140 + (2  10) = 140 + 20 = 160 cm2 = 160 cm2
19. 19. J.Stewart Highcliffe This is a Pentagonal based Prism • This shape has 7 faces • Five faces are rectangular • Two faces are pentagons • It has 10 vertices and 15 edges
20. 20. Prism Surface Area of wall = perimeter of base.  high = ( 6+6+6+6+6) 15 = 30  15 = 450 cm2 Area of base = 62 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 450+ (2  62) = 450 + 124 = 574 cm2 = 574 cm2 Find the surface area of a regular pentagonal prism if its base area is 62 cm2
21. 21. Pythagoras : C2 = A2 + B2 C2 = 32 + 42 C2 = 9 + 16 C2 = 25 C = 5 Surface Area of wall = perimeter of base.  High = (3+4+5) 10 = 12  10 = 120 cm2 Area of base = 6 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 120+ (2  6) = 120 + 12 = 132 cm2 = 132 cm2
22. 22. Surface Area of wall = perimeter of base.  High = (3+6+3+6) 12 = 18  12 = 216 cm2 Area of base = 18 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 216+ (2  18) = 216 + 36 = 252 cm2 = 252 cm2
23. 23. Surface Area of wall = perimeter of base.  High = (5+5+5+5) 11 = 20  11 = 220 cm2 Area of base = 25 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 220+ (2  25) = 220 + 50 = 270 cm2 = 270 cm2 Find surface area square prism if it has side of base 5 and high 11 cm.
24. 24. Find Surface area of rectangular prism long side is 2 and 5 cm high 8 cm . Surface Area of wall = perimeter of base.  High = (2+5+2+5) 8 = 14  8 = 112 cm2 Area of base = 10 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 112+ (2  10) = 112 + 20 = 132 cm2 = 132 cm2
25. 25. 6.Find surface area of right triangular prism if adjacent side of base is 6 and 8 and high 15 cm. Pythagoras : C2 = A2 + B2 C2 = 62 + 82 C2 = 36 + 64 C2 = 100 C = 10 Surface Area of wall = perimeter of base.  High = (6+8+10) 15 = 24  15 = 360 cm2 Area of base = 24 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 360+ (2  24) = 360 + 48 = 408 cm2 = 408 cm2
26. 26. Find surface area of octagonal prism if side of base is 4 cm. Area of base is 60 and high 20 cm Surface Area of wall = perimeter of base.  High = (4+4+4+4+4+4+4+4) 20 = 32  20 = 640 cm2 Area of base = 60 cm2 Surface Area = Surface Area of wall + (2  Area of base) = 640+ (2  60) = 640 + 120 = 760 cm2 = 760 cm2
27. 27. J.Stewart Highcliffe Hexagonal Prism • This shape has 8 faces • Six faces are rectangular • Two faces are hexagons • It has 12 vertices and 18 edges
28. 28. Find surface area square prism if it have side of base 5 and high 11 cm 1 2 3 4 5 Find area of rectangular prism long side is 2 and 5 cm high 8 cm 6 Home work
29. 29. Prism
30. 30. Solution Area of Base = highBase 2 1 = 34 2 1  = 6 = 6 cm2 Volume =Area of base x High = 6 x 10 = 60 cm3 = 60 cm3
31. 31. Prism
32. 32. Solution Area of Base = Width X length. = 6 x 2 = 12 = 12 cm2 Volume =Area of base x High = 12 x 11 = 132 cm3 = 132 cm3
33. 33. Volume = 1.72 × (base side) 2 x high Area of base = 2 4 3 5 SideBase Pentagonal Prism
34. 34. Solution Area of Base = 2 4 3 5 SideBase = 1.7 x Base side 2 = 1.7 x 52 = 41.5 cm2 Volume =Area of base x High = 41.5 x 12 = 498 cm3 = 498 cm3 Pentagonal Prism
35. 35. Solution Area of Base = 2 4 3 5 SideBase = 1.72 x Base side 2 = 1.72 x 52 = 43 cm2 Volume =Area of base x High = 43 x 12 = 516 cm3 = 516 cm3 Pentagonal Prism
36. 36. Solution Area of Base = 2 4 3 5 SideBase = 1.7 x Base side 2 = 1.7 x 62 = 61.2 cm2 Volume =Area of base x High = 61.2 x 15 = 918 cm3 = 918 cm3
37. 37. Solution Area of Base = 2 4 3 5 SideBase = 1.7 x Base side 2 = 1.7 x 92 = 137.7 cm2 Volume =Area of base x High = 137.7 x 10 = 1377 cm3 = 1377 cm3
38. 38. Solution Area of Base = 2 4 3 5 SideBase = 1.7 x Base side 2 = 1.7 x 112 = 205.7 cm2 Volume =Area of base x High = 205.7 x 8 = 1645.6 cm3
39. 39. Base side High 1 9 17 2 6 23 3 3 19 4 8 77 5 3 6 6 By your self By your self Volume(cm3 ) 2340.9 1407.6 290.7 8377.6 4596.8
40. 40. Volume = 2.5 × (base side) 2 x high Area of base = 2 4 3 6 SideBase Hexagonal Prism
41. 41. Volume = 3.6 × (base side) 2 x high Area of base = 2 4 3 7 SideBase Heptagonal Prism
42. 42. Volume = 4.8 × (base side) 2 x high Area of base = 2 4 3 8 SideBase octagonal Prism
43. 43. Prism Base side High Triangular 3,4,5 16 Rectangular 10,12 20 Pentagonal 12 30 hexagonal 14 50 heptagonal 21 24 octagonal 33 15 Volume 96 2400 7344 24500 38102.4 78408
44. 44. Prism Pyramid Cone Cylinder Sphere
45. 45. Cones and Pyramids Pyramid Slant Heigh t Heigh t h Slant Heigh t  Pyramid
46. 46. Theorem 10-3 • Lateral Area and Surface Area of a Regular Pyramid PL 2 1  Lateral Area Base Perimeter Slant Height BLS  Surfac e Area Lateral Area Base Area B 
47. 47. #1 Finding Surface Area of a Pyramid • Find the surface area of a square pyramid with base edges 5 m and slant height 3 m. m3 m5 BLS  PL 2 1  )5(4P 2530S 20 )3(20 2 1 L 2 m30 2 sB  2 5B 2 m25B 2 m55S
48. 48. Base edges Slant height 6 9 5 13 7 10 11 12 3 19 Surface Area 144 155 189 385 123
49. 49. #2 Finding Surface Area of a Pyramid • Find the surface area of a regular hexagonal pyramid with a slant height 20 in and sides 8 in. BLS  pL 2 1  )8(6P 160480 S 2 in48 )20(48 2 1 L 2 in480 2 5.2 BasesideB  2 85.2 B 2 in160B 2 in640S in8 in02
50. 50. BLS  pL 2 1  )9(5P cm45 )10(45 2 1 L 2 cm225 2 97.1 B 2 7.1 basesideB  2 7.137 cmB  7.137225S 2 7.362 cmS  Find the surface area of pentagonal pyramid with base Edges 9 cm and Slant height 10 cm
51. 51. Find the surface area of Hexagonal pyramid with base Edges 11 cm and Slant height 20 cm BLS  pL 2 1  )11(6P cm66 )20(66 2 1 L 2 cm660 2 115.2 B 2 5.2 basesideB  2 5.302 cmB  5.302660 S 2 5.962 cmS 
52. 52. Pyramid Base edges Slant Height Pentagonal 12 16 Hexagonal 21 24 Heptagonal 8 32 Octagonal 31 12 Triangular 13 42 Pentagonal 15 16 Octagonal 6 22 Find the surface area of ………. pyramid with base Edges ……. cm and Slant height ……..cm Surface area 724.8 2614.5 1126.4 6101.8 892 982.5 700.8
53. 53. A Pyramid is a three dimensional figure with a regular polygon as its base and lateral faces are identical isosceles triangles meeting at a point. Pyramids base = quadrilateral base = pentagon base = heptagon Identical isosceles triangles
54. 54. Volume of a Pyramid: V = (1/3) Area of the base x height V = (1/3) Ah Volume of a Pyramid = 1/3 x Volume of a Prism Volume of Pyramids + + =
55. 55. Theorem 10-8 • Volume of a Pyramid BhV 3 1  Volum e HeightBase Area B
56. 56. Finding Volume of a Pyramid • Find the volume of a square pyramid with base edges 5 m and height 3 m. m3 m5 BhV 3 1  )3)(25( 3 1 V 2 sB  2 5B 2 m25B 3 m25V
57. 57. • Pyramid – Is a polyhedron in which one face can be any polygon & the other faces are triangles. hVp = ⅓Bh Area of the Base A = l•w A = ½bh Height of the pyramid, not to be confused with the slant height (l) Volume of a Pyramid
58. 58. • Find the volume of a square pyramid with base edges of 15cm & a height of 22cm. 22cm 15cm 15cm V = (⅓)Bh = (⅓)l•w•h = (⅓)15•15•22 = (⅓)4950 = 1650cm3 Square Volume of a right Pyramid
59. 59. • Find the area of a square pyramid w/ base edges 16ft long & a slant height 17ft. h 17ft 16ft 8ft V = (⅓)Bh = (⅓)l•w•h = (⅓)16•16•___ = (⅓)3840 = 1280ft3 a2 + b2 = c2 h2 + 82 = 172 h2 = 225 h = 15 15 Another square pyramid
60. 60. Finding the Volume of a Pyramid The base of a pyramid is a square. The side length of the square is 24 feet. The height of the pyramid is 9 feet. Find the volume of the pyramid. Write formula for volume of a pyramid. Substitute 242 for B and 9 for h. Simplify. ANSWER The volume of the pyramid is 1728 cubic feet. V = Bh 1 3 = 1728 = (242)(9) 1 3 Volumes of Pyramids and Cones
61. 61. Find the volume of a ………… pyramid with base edges ………. m and height …………. m. Pyramid Base edges Height Pentagonal 12 16 Hexagonal 21 24 Heptagonal 8 32 Octagonal 31 12 Triangular 13 42 Pentagonal 15 16 Octagonal 6 22 1305.60 8820.00 2457.60 18451.20 1017.38 2040.00 1267.20
62. 62. Surface Area of a Cylinder h A cylinder is a prism with a circular cross-section. 2r
63. 63. Surface Area of a Cylinder A cylinder is a prism with a circular cross- section. r2 r2  Removing top and bottom Open out2r h Surface Area = 2r2 + 2rh = 2r(r + h)
64. 64. Surface Area of a Cylinder 8cm Surface area = 2r(r + h) 3cm Find the surface area of the cylinder. SA = 2 x  x 3(3 + 8) = 6 x 11 = 66 = 207 cm2 (3 sf)
65. 65. 6 cm Calculate the surface area of the following cylinders. 15 cm 2.1mm 9.2 mm 3 cm 2 cm 1 2 3 SA = 2 x  x 3(3 + 6) = 6 x 9 = 54 = 169.6 cm2 SA = 2 x  x 2(2 + 15) = 4 x 17 = 68 = 213.6 m2 SA = 2 x  x 4.6(4.6 + 2.1) = 9.2 x 6.7 = 193.6 mm2 2r(r + h)
66. 66. 2 ( )SA r r h  The Surface Area of a Cylinder Find the surface area of the cylinder. 2 m 6m 2 3(3 2)x x  2 94 26 5 . mx 
67. 67. The Surface Area of a Cylinder Find the surface area of the cylinder. 0.7 m 1.5m 2 ( )SA r r h  2 0.7(0.7 1.5)x x  2 1.4 2. 9.2 7mx 
68. 68. The Surface Area of a Cylinder Find the surface area of the cylinder. 1.9 m 8m 2 ( )SA r r h  2 4(4 1.9)x x  2 8 5.9 148.3x m 
69. 69. The Surface Area of a Cylinder Find the surface area of the cylinder. 0.76 m 1.3m 2 ( )SA r r h  2 0.76(0.76 1.3)x x  2 1.52 2. 9.06 8x m 
70. 70. Radius • Volume of a Cylinder BhV  Volume Height 2 rB  Base Area Radius Base Areah Base Area
71. 71. #3 Finding the Volume of a Cylinder 10 cm 5 cm BhV  • Find the volume of a cylinder with height 10 cm and radius 5cm. 2 rB  2 )5(B 2 cm25B )10(25V 3 cm250V
72. 72. #4 Finding the Volume of a Cylinder 9 ft 6 ft BhV  • Find the volume of the cylinder. 2 rB  2 )6(B 2 ft36B )9(36V 3 ft324V
73. 73. Prism Pyramid Cone Cylinder Sphere
74. 74. SA = LSA + B or Lateral Surface Area: Total Surface Area: l Cones SA = rℓ + 2 r LSA = rℓ
75. 75. • Find the slant height then find the surface area of the cone. 222 cba  222 25.1  2 425.2  2 25.6  5.2
76. 76. Find the lateral area and surface area of the cone. LSA = rℓ = (8)(17) = 136 in2  427.04 in2 SA = rℓ + r2 = 136 + (8)2 = 200 in2  628 in2 Example
77. 77. Ex: Find the lateral & surface areas of the right cone. LA = rl LA = (6)(10) LA = 60 in2 S = r2 + rl S = (62) + 60 S = 36 + 60 S = 96 in2 6 in 8in l How do you find the slant height? Use Pythag. Thm! 82 + 62 = l 2 10 = l l = 10 S  301.44 in2
78. 78. • Find the surface area of the cone. Surface Area of Pyramids and Cones 2 rrSA    2 5135  SA  2565 SA 2 in90SA 2 in6.282SA
79. 79. • Find the slant height then find the surface area of the cone. Surface Area of Pyramids and Cones 2 rrSA     25.275.3 SA 2 m6SA 2 5.15.25.1  SA 2 m84.18SA
80. 80. No. high Radius 1 3 4 2 6 8 3 5 12 4 10 24 5 9 12 6 25 60 7 15 20 SA 113.0 452.2 942.0 3768.0 1017.4 23550.0 2826.0 Slant high 5 10 13 26 15 65 25
81. 81. V = or Volume: Cones Bh 3 1 V = hr2 3 1  h
82. 82. • Volume of a Cone BhV 3 1  Volume HeightBase Arear B
83. 83. Find the volume of the cone. V = 1/3 Bh or 1/3 (TTr2h) V  1/3 (3.14  42  16) V  1/3 (3.14  16  16) V  1/3 (50.24  16) V  1/3 (803.84) V  267.95 cm3
84. 84. Volume of a Cone • Formula • Find the volume of a cone with a diameter of 13.5 m and a height of 10m hrV 2 3 1 
85. 85. 91 EXAMPLE 2 Finding the Volume of a Cone Find the volume of the cone shown. Round to the nearest cubic millimeter. Write formula for volume of a cone. Substitute 6.75 for r and 10 for h. Evaluate. Use a calculator. The radius is one half of the diameter, so r = 6.75. V = πr2h 1 3 = π(6.75)2(10) 1 3 ≈ 477.1 ANSWER The volume of the cone is about 477 cubic millimeters. Volumes of Pyramids and Cones
86. 86. Finding the Volume of a Cone • The radius of the base of a cone is 6 m. Its height is 13 m. Find the volume. m6 BhV 3 1  )13)(36( 3 1 V 2 rB   2 6B 2 m36B 3 m156V m13 3 m84.489V
87. 87. Find the volume of a cone. Step 1 Use the Pythagorean Theorem to find the height. 162 + h2 = 342 h2 = 900 h = 30 Example Step 2 Use the radius and height to find the volume. = 2560 cm3  8050.96 cm3
88. 88. 11in V = ⅓Bh = (⅓)r2h = (⅓)(3)2(11) = (⅓)99 = 33  103.7in3 Circle 3in Ex.3: Find the volume of the following right cone w/ a diameter of 6 in.
89. 89. • The following cone has a volume of 110. What is its radius. 10cm r V = ⅓Bh V = ⅓(r2)h 110 = (⅓)r2(10) 110 = (⅓)r2(10) 11 = (⅓)r2 33 = r2 r = √(33) = 5.7cm Ex.5: Solve for the missing variable.
90. 90. Find the volume of a cone with base circumference 25 in. and a height 2 in. more than twice the radius. Step 1 Use the circumference to find the radius. Step 2 Use the radius to find the height. h = 2(12.5) + 2 = 27 in. 2r = 25 r = 12.5 Step 3 Use the radius and height to find the volume. = 1406.25 in3 = 4415.63 in3
91. 91. Finding Volumes of Cones Step 2 Use the radius and height to find the volume. Volume of a cone Substitute 16 for r and 30 for h.  2560 cm3  8042.5 cm3 Simplify. Find the volume of a cone.
92. 92. Finding Volumes of Cones Find the volume of a cone. Step 1 Use the Pythagorean Theorem to find the height. 162 + h2 = 342 Pythagorean Theorem h2 = 900 Subtract 162 from both sides. h = 30 Take the square root of both sides.
93. 93. No. Radius high 1 9 13 2 6 16 3 8 10 4 2 21 5 5 12 6 8 15 Volume 1102.1 602.9 669.9 87.9 314.0 1004.8
94. 94. 8cm 10cm 4cm Volume of Cone first! Vc = ⅓Bh = (⅓)r2h = (⅓)(8)2(10) = (⅓)(640) = 213.3 = 670.2cm3 Volume of Cylinder NEXT! Vc = Bh = r2h = (8)2(4) = 256  804.2cm3 VT = Vc + Vc VT = 670.2cm3 + 804.2cm3 VT = 1474.4cm3 Volume of a Composite Figure
95. 95. Check It Out! Find the volume of the cone. Volume of a cone Substitute 9 for r and 8 for h. ≈ 216 m3 ≈ 678.6 m3 Simplify.
96. 96. Example 4: Exploring Effects of Changing Dimensions original dimensions: radius and height divided by 3: Notice that . If the radius and height are divided by 3, the volume is divided by 33, or 27. The diameter and height of the cone are divided by 3. Describe the effect on the volume.
97. 97. Check It Out! Example 4 original dimensions: radius and height doubled: The volume is multiplied by 8. The radius and height of the cone are doubled. Describe the effect on the volume.
98. 98. Example 5: Finding Volumes of Composite Three-Dimensional Figures Find the volume of the composite figure. Round to the nearest tenth. The volume of the upper cone is
99. 99. Example 5: Finding Volumes of Composite Three-Dimensional Figures The volume of the cylinder is The volume of the lower cone is The volume of the figure is the sum of the volumes. Find the volume of the composite figure. Round to the nearest tenth. Vcylinder = r2h = (21)2(35)=15,435 cm3. V = 5145 + 15,435 + 5,880 = 26,460  83,126.5 cm3
100. 100. The volume of a sphere is two thirds the volume of its corresponding cylinder. Spheres and Cylinders r Sphere r Cylinder h=2r CylinderVolume 3 2 =SphereVolume cylindersphere V 3 2 =V h)r( 3 2 =V 2 sphere  2r)r( 3 2 =V 2 sphere  3 sphere r 3 4 =V  © Treap
101. 101. The Sphere © Treap
102. 102. A sphere is a three dimensional figure in which every point on the surface is equally distant from the centre O. Volume of a Sphere: The cross- section of a sphere is a circle. Surface Area of a Sphere: SA = 4r2 r O 3 r 3 4 =V  Sphere © Treap
103. 103. The volume of a sphere is twice the volume of its corresponding cone. Spheres and Cones r Sphere r Cone h=2r ConeVolume2=SphereVolume conesphere V2=V ) 3 hr (2=V 2 sphere  3 sphere r 3 4 =V  ) 3 2rr (2=V 2 sphere  © Treap
104. 104. 2r 2r  r The Surface Area of a SphereThe formula for the surface area of a sphere was discovered by Archimedes. In the diagram below a cylinder just encloses a sphere of radius r. Archimedes was able to determine the formula by showing that a pair of parallel planes perpendicular to the vertical axis of the cylinder, would enclose equal areas on both shapes. 2r Surface area = 2r x 2r Surface area = 4r2
105. 105. Surface Area 4r2 Archimedes did not have the advantage of a sophisticated algebra like we use today. He had to express relationships in terms of simpler geometric shapes. For him the surface area of a sphere was equal to the area of 4 of the greatest circles that it could contain. r2 r2 r2 r2 Archimedes was intrigued by this amazing discovery. Why is the answer exactly 4 and not 4.342? Painting the surface of a sphere uses the same amount of paint as painting four of its greatest circles!
106. 106. Example 2 A spherical balloon is blown up from a radius of 7 cm to one of 21 cm. What is the change in its surface area? Step 1 Find the surface area of the small balloon A = 4  r2 = 4(3.14)(72) = 615.44 cm2 Step 2 Find the surface area of the big balloon A = 4  r2 = 4(3.14)(212) = 5538.96 cm2 The change in the surface area 5538.96 – 615.44 = 4923.52 cm2 Step 3 © Treap
107. 107. 12 cm 7.3 cm SA = 4r2 SA = 4 x  x 7.32 = 669.7cm2 SA = 4r2 SA = 4 x  x 122 = 1809.6 cm2 Example Questions: Calculate the surface area of the spheres below. (to 1 dp) 1 2 SA = 4r2
108. 108. radius 3.14 22/7 7 21 28 35 42 radius 3.14 22/7 7 615.44 616 21 5538.96 5544 28 9847.04 9856 35 15386 15400 42 22155.84 22176
109. 109. Questions: Calculate the surface area of the spheres below. (to 1 dp) SA = 4r2 SA = 4 x  x 3.22 = 128.7 m2 SA = 4r2 SA = 4 x  x 2.42 = 72.4 m2 3.2 m 2.4 m 1 2 SA = 4r2
110. 110. Example Questions: Calculate the radius of the spheres shown below. (to 1 dp) SA = 1500 cm2 1 2 4r2 = 1500 SA = 3500 cm2 4 15002  r cmr 910 4 1500 .  4r2 = 3500 4 35002  r cmr 716 4 3500 .  SA = 4r2
111. 111. Questions: Calculate the radius of the spheres shown below. (to 1 dp) SA = 8.4 m2 4r2 = 8.4 SA = 1200 cm2 4 482 .  r mr 820 4 48 . .   4r2 = 1200 4 12002  r cmr 89 4 1200 .  1 2 SA = 4r2
112. 112. = 324 Simplify. Find the surface area of the sphere below. Leave your answer in terms of . S.A. = 4 r 2 Use the formula for the surface area of a sphere. = 4 • 92 Substitute r = = 9.18 2 The surface area of the sphere is 324 ft2 = 1073.88 ft2
113. 113. S.A. = 4 r 2 Use the formula for the surface area of a sphere. 13 = 2 r Substitute 13 for C. C = 2 r Use the formula for circumference. Step 1: Use the circumference to find the radius in terms of . The circumference of a rubber ball is 13 cm. Calculate its surface area to the nearest whole number. = r Solve for r. 13 2 Step 2: Use the radius to find the surface area. = 4 • ( )2 Substitute for r.13 2 13 2 To the nearest whole number, the surface area of the rubber ball is 54 cm2 . = Simplify. 169 53.794371 Use a calculator.
114. 114. Find the volume of the sphere. Leave your answer in terms of . V = r 3 Use the formula for the volume of a sphere. 4 3 30 2 = • 153 Substitute r = = 15. 4 3 = 4500 Simplify. The volume of the sphere is 4500 cm3.
115. 115. The volume of a sphere is 1 in.3. Find its surface area to the nearest tenth. Step 1: Use the volume to find the radius r. V = r 3 Use the formula for the volume of a sphere. 4 3 1 = r 3 Substitute. 4 3 = r 3 Solve for r 3. 3 4 = r Find the cube root of each side. 3 4 3 r 0.62035049 Use a calculator.
116. 116. Step 2: Use the radius to find the surface area. S.A. = 4 r 2 Use the formula for the surface area of a sphere. 4 (0.62035049)2 Substitute. 4.8359744 Use a calculator. To the nearest tenth, the surface area of the sphere is 4.8 in.2. (continued)
117. 117. A cylindrical container used to store chocolate is 12 cm high with a diameter of 6 cm. Each chocolate is a sphere with a diameter of 3 cm. Each chocolate sells for \$1.50 What is the cost of one contai- ner full of chocolates? Example 1 Step 1 Find the radius of the container R = 6  2 = 3 cm 12 cm 6 cm Container Step 2 Find the radius of the chocolate r = 3  2 = 1.5 cm © Treap 3 cm
118. 118. Example 1 Cont’d Step 3 Find the volume of the container V = r2h = (3.14)(32)(12) = 339.12 cm3 12 cm 6 cm 3 cm Container Chocolate Step 4 Find the volume of one chocolate 3 r 3 4 =V  )5.1)(14.3( 3 4 = 3 3 cm13.14= © Treap
119. 119. Example 1 Cont’d Step 5 Find how many chocolates fit into the container V container  V chocolate = 339.12  14.13 = 2412 cm 6 cm 3 cm Container Chocolate Step 6 Find the cost of one container Cost = 24 (\$1.50) = \$36 © Treap