Slide 49 / 311 20 What is the locus of points (cross-section) of a cube and a plane perpendicular to the base and parallel to the non-intersecting sides? square A B rectangle trapezoid C hexagon D rhombus E parallelogram F G triangle circle H Slide 50 / 311 21 If the length of each edge of the cube is 12 inches, what would be the area of the cross-section of the cube and a plane perpendicular to the base and parallel to the non- intersecting sides? A 72 sq inches B 144 sq inches C 187.06 sq inches 12 in. D 203.65 sq inches Slide 51 / 311 22 What is the locus of points of a cube and a plane that contains the diagonal of the base and is perpendicular to the base? square A rectangle B C trapezoid hexagon D rhombus E parallelogram F triangle G H circle
Slide 52 / 311 23 If the length of each edge of the cube is 12 inches, what would be the area of the cross-section of the cube and a plane that contains the diagonal of the base and is perpendicular to the base? A 72 sq inches B 144 sq inches 12 in. C 187.06 sq inches D 203.65 sq inches Slide 53 / 311 24 What is the locus of points of a cube and a plane that contains the diagonal of the base but does not intersect the opposite base? square A rectangle B trapezoid C hexagon D E rhombus parallelogram F triangle G circle H Slide 54 / 311 25 What is the locus of points of a cube and a plane that intersects all of the faces? square A rectangle B trapezoid C hexagon D rhombus E parallelogram F triangle G circle H
Slide 55 / 311 Views & Drawings of 3-D Solids Return to Table of Contents Slide 56 / 311 Views & Drawings Isometric drawings are drawings that look 3-D & are created on a grid of dots using 3 axes that intersect to form 120° & 60° angles. Slide 57 / 311 Views & Drawings Example: Create an Isometric drawing of a cube.
Slide 58 / 311 Views & Drawings An Orthographic projection is a 2-D drawing that shows the different viewpoints of an object, usually from the front, top & side. Each drawing depends on your position relative to the figure. Top (from front) Front Side Slide 59 / 311 Views & Drawings Consider these three people viewing a pyramid: Slide 60 / 311 Consider these three people viewing a pyramid: The orange person is standing in front of a face, so their view is a triangle.
Slide 61 / 311 Consider these three people viewing a pyramid: The green person is standing in front of a lateral edge, so from their view they can see 2 faces. Slide 62 / 311 Consider these three people viewing a pyramid: The purple person is flying over and can see the four lateral faces. Slide 63 / 311 26 Given the surface shown, what would be the view from point A? a Rectangle A a Square B a Circle C a Pentagon D A (front) E a Triangle a Parallelogram F a Hexagon G a Trapezoid H
Slide 64 / 311 27 Given the surface shown, what would be the view from point A? a Rectangle A A (top) a Square B a Circle C D a Pentagon a Triangle E a Parallelogram F a Hexagon G a Trapezoid H Slide 65 / 311 28 Given the surface shown, what would be the view from point A? A (top) a Rectangle A B a Square a Circle C a Pentagon D a Triangle E a Parallelogram F G a Hexagon right square prism a Trapezoid H Slide 66 / 311 29 Given the surface shown, what would be the view from point A? right square prism a Rectangle A a Square B a Circle C a Pentagon D E a Triangle a Parallelogram F A (front) a Hexagon G a trapezoid H
Slide 67 / 311 30 Given the surface shown, what would be the view from point A? a Rectangle A a Square B a Circle C D a Pentagon A (front) a Triangle E a Parallelogram F a Hexagon G a Trapezoid H Slide 68 / 311 31 Given the surface shown, what would be the view from point A? A (above) a Rectangle A B a Square a Circle C a Pentagon D a Triangle E a Parallelogram F G a Hexagon a Trapezoid H Slide 69 / 311 32 Given the surface shown, what would be the view from point A? A (above) a Rectangle A a Square B a Circle C a Pentagon D E a Triangle a Parallelogram F a Hexagon G a Trapezoid H
Slide 70 / 311 33 Given the surface shown, what would be the view from point A? a Rectangle A a Square B a Circle C A D a Pentagon (front) a Triangle E a Parallelogram F a Hexagon G a Trapezoid H Slide 71 / 311 34 Given the surface shown, what would be the view from point A? sphere a Rectangle A B a Square a Circle C a Pentagon D A a Triangle E a Parallelogram F G a Hexagon a Trapezoid H Slide 72 / 311 Views & Drawings C (Looking down from above) What would the view be like from each position? A B
Slide 73 / 311 Views & Drawings What would the view be like from each position? From A, how many columns of blocks are visible? - 3 columns A Click to reveal How tall is each column? - first one is 4 high - second & third columns are each 2 blocks high Click to reveal Slide 74 / 311 Views & Drawings What would the view be like from each position? From B, how many columns of blocks are visible? B - 2 columns Click to reveal How tall is each column? - left one is 3 high - right one is 4 high Click to reveal Slide 75 / 311 C (Looking down Views & Drawings from above) What would the view be like from each position? From C, how many columns of blocks are visible? - 3 columns Click to reveal How tall is each column? - all of them are 2 blocks high Click to reveal
Slide 76 / 311 Views & Drawings Above Draw the 3 views. Front Side Move for Answer Front View Side View Top View Slide 77 / 311 Views & Drawings Draw the 3 views. Above Front Side Move for Answer Side Above Front Slide 78 / 311 Views & Drawings Here are 3 views of a solid, draw a 3-dimensional representation. L R F Side Top Front Move for Answer
Slide 79 / 311 Views & Drawings Here are 3 views of a solid, draw a 3-dimensional representation. R L Front Side F Top Move for Answer Slide 80 / 311 Surface Area of a Prism Return to Table of Contents Slide 81 / 311 Net A Net is a 2-dimensional shape that folds into a 3-dimensional figure. The Net shows all of the faces of the surface. Shown is the net of a right rectangular prism. 4 6 12 12 4 6 4 6 4 6
Slide 82 / 311 Net The net shown is a right triangular prism. The lateral faces are rectangles. The bases are on opposite sides of the rectangles, although they do not need to be on the same rectangle. Slide 83 / 311 Net The nets shown are for the same right triangular prism. Slide 84 / 311 Nets Nets of oblique prisms have parallelograms as lateral faces.
Slide 85 / 311 Rectangular Prisms cube H H H w w w ℓ ℓ ℓ Slide 86 / 311 Rectangular Prisms Base Base height height Base Base A prism has 2 bases. The base of a rectangular prism is a rectangle. The height of the prism is the length between the two bases. Slide 87 / 311 Rectangular Prisms The Surface Area of a figure is the total amount of area that is needed to cover the entire figure (e.g. the amount of wrapping paper required to wrap a gift). The Surface Area of a figure is the sum of the areas of each side of the figure. Area Top Area Area Back Side Side Area Area Area Area Area Area Front Area Bottom Area Area
Slide 88 / 311 Finding the Surface Area of a Rectangular Prism H w ℓ Area of the Top = ℓ x w Area of the Front = ℓ x H Area of the Bottom = ℓ x w Area of the Back = ℓ x H Area of Left Side = w x H Area of Right Side = w x H The Surface Area is the sum of all the areas S.A. = ℓw + ℓw + ℓH + ℓH + wH + wH S.A. = 2 ℓw + 2 ℓH + 2wH Slide 89 / 311 Finding the Surface Area of a Rectangular Prism Example: Find the surface area of the prism 3 4 7 Area of Top & Bottom Area of Right & Left A = 7(4) = 28u 2 A = 3(4) = 12 u 2 Click Click Area of Front & Back A = 3(7) = 21 u 2 Click Total Surface Area = 2(28) + 2(12) + 2(21) Click = 56 + 24 + 42 = 122 units 2 Click Slide 90 / 311 35 What is the total surface area, in square units? 5 9 4
Slide 91 / 311 36 What is the total surface area, in square units? 8 8 8 Slide 92 / 311 37 Troy wants to build a cube out of straws. The cube is to have a total surface area of 96 in 2 , what is the total length of the straws, in inches? Slide 93 / 311 Another Way of Looking at Surface Area S.A. = 2B + PH The Surface Area is the sum of the areas of the 2 Bases plus the Lateral Area (Perimeter of the base, P, times the height of the prism, H) The Lateral Area is the area of the Lateral Surface. The Lateral Surface is the part that wraps around the middle of the figure (in between the two bases). e s a B Base Lateral Surface Base e s a B
Slide 94 / 311 Rectangular Prisms Another formula for Surface Area of a right prism: S.A. = 2B + PH B = Area of the base B = ℓw P = Perimeter of the base P = 2 ℓ + 2w H = Height of the prism Base H Base w ℓ S.A. = 2B + PH S.A. = 2 ℓw + (2 ℓ +2w)H S.A. = 2 ℓw + 2 ℓH + 2wH Slide 95 / 311 Rectangular Prisms Another formula for Surface Area of a right prism: S.A. = 2B + PH B = Area of the base B = ℓw P = Perimeter of the base P = 2 ℓ + 2w H = Height of the prism Base H Base w ℓ In the surface area formula, 2B is the sum of the area of the 2 bases. The area of lateral faces or What does PH represent? Lateral Area Click Slide 96 / 311 38 If the base of the prism is 12 by 6, what is the lateral area, in sq ft? 4 ft 6 ft 12 ft
Slide 97 / 311 39 The surface area of the rectangular prism is : 24 sq ft A 4 ft 144 sq ft B 6 ft 288 sq ft C 12 ft 48 sq ft D 72 sq ft E Slide 98 / 311 40 If 7 by 6 is base of the prism, what is the lateral area, in sq units? 7 6 9 Slide 99 / 311 41 What is the total square units of the surface area? 7 6 9
Slide 100 / 311 42 Find the value of y, if the lateral area is 144 sq units, and y by 6 is the base. y 6 8 Slide 101 / 311 43 What is the value of the missing variable if the surface area is 350 sq. ft. 7 ft A 8.3 ft B 10 ft 12 ft C 15 ft D 5 ft x ft Slide 102 / 311 44 Sharon was invited to Maria's birthday party. For a present, she purchased an iHome (a clock radio for an iPod or iPhone) which is contained in a box that measures 7 inches in length, 5 inches in width, and 4 inches in height. How much wrapping paper does Sharon need to wrap Maria's present?
Slide 103 / 311 Other Prisms Slide 104 / 311 Other Prisms base base base height height base base base base height base height A Prism has 2 Bases The Base of a Prism matches the first word in the name of the prism. e.g. the Base of a Triangular Prism is a Triangle The Height of the Prism is the length between the two bases Slide 105 / 311 Other Prisms The Surface Area of a figure is the total amount of Area that is needed to cover the entire figure (e.g. the amount of wrapping paper required to wrap a gift). The Surface Area of a figure is the sum of the areas of each side of the figure Area Area Area Area Area Area Area Area Area Area Triangular Prism Net of the Triangular Prism
Slide 106 / 311 Finding the Surface Area of a Right Prism Surface Area: S.A. = 2B + PH B = Area of the triangular base = ½bh P = Perimeter of the triangular base = a + b + c H = Height of the prism Lateral Area = PH = (a + b + c)H The Lateral Area is the area of the Lateral Surface, the rectangular area that wraps around the prism between the triangular bases. a c base c a b h a c Prism's Lateral Surface base H H b height b B = ½ bh P = a + b + c Note: The formula above will work for any right prism. Slide 107 / 311 Other Prisms Example: Find the lateral area and surface area of the right triangular prism. 10 6 11 Since it has a base that is a right triangle, we need to find the base of the triangle using Pythagorean Theorem. 6 2 + b 2 = 10 2 36 + b 2 = 100 b 2 = 64 b = 8 units Next, calculate the perimeter of your base. P = 6 + 8 + 10 = 24 units Use this to find the Lateral Area LA = PH = 24(11) = 264 units 2 Slide 108 / 311 Other Prisms Example: Find the lateral area and surface area of the right triangular prism. 10 6 11 Then, calculate the area of your base, B B = (1/2)(8)(6) = 24 units 2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(24) + (24)(11) SA = 48 + 264 = 312 units 2
Slide 109 / 311 Other Prisms Example: Find the lateral area and surface area of the triangular prism. 9 9 12 9 Since it has a base that is an equilateral triangle, we need to find the height of the triangle using Pythagorean Theorem or the 30-60-90 Triangle Theorem. 4.5 2 + b 2 = 9 2 20.25 + b 2 = 81 b 2 = 60.75 b = 4.5√3 units = 7.79 units Next, calculate the perimeter of your base. P = 9 + 9 + 9 = 27 units Use this to find the Lateral Area LA = PH = 27(12) = 324 units 2 Slide 110 / 311 Other Prisms Example: Find the lateral area and surface area of the triangular prism. 9 9 12 9 Then, calculate the area of your base, B B = (1/2)(9)(4.5√3) = 20.25√3 units 2 = 35.07 units 2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(35.07) + (27)(12) SA = 70.14 + 324 = 394.14 units 2 Slide 111 / 311 45 The height of the triangular prism below is 11 ft, the height of the base is 3 ft, and the triangular base is an isosceles triangle. Find the surface area. A 88 sq ft B 132 sq ft 5 ft C 198 sq ft 3 ft 11 ft D 222 sq ft
Slide 112 / 311 46 The height of the triangular prism below is 3, and the triangular base is an equilateral triangle. Find the surface area. A 64 sq ft B 127.43 sq ft C 72 sq ft 3 ft D 55.43 sq ft 8 ft Slide 113 / 311 47 Find the lateral area of the right prism. 5 6 5 Slide 114 / 311 Finding the Surface Area of a Right Prism Surface Area : S.A. = 2B + PH B = Area of the regular hexagonal base = ½aP - a is the apothem of the regular base P = Perimeter of the base = b + c + d + e + f + g H = Height of the prism = H Lateral Area = PH = (b + c + d + e + f + g)H d c e base b f d c e b g f Prism's H height base a P = b + c + d + e + f + g B = ½ aP
Slide 115 / 311 Finding the Surface Area of a Right Prism The Lateral Area is the area of the Lateral Surface, the rectangular area that wraps around the prism between the triangular bases. d c e base b f d c e b g f Prism's height H base P = b + c + d + e + f + g a B = ½ aP Slide 116 / 311 Other Prisms Example: Find the lateral area and surface area of the regular hexagonal prism. Because the base is a regular polygon, we need to calculate the apothem. To begin, figure out the central angle & top angle in the triangle. 360 7 in = 60° = central angle 6 Click 60 = 30° = top angle of the triangle. 2 8 in Click 30° a 4 in. Click Slide 117 / 311 Other Prisms Example: Find the lateral area and surface area of the regular hexagonal prism. Next find the apothem using trigonometry, or special right triangles (if it applies). 4 7 in tan 30 = a Click atan30 = 4 8 in tan30 tan30 Click a = 4√3 = 6.93 in. Click
Slide 118 / 311 Other Prisms Example: Find the lateral area and surface area of the regular hexagonal prism. Next, calculate the perimeter of your base. P = 8(6) = 48 in Click Use this to find the Lateral Area LA = PH = 48(7) = 336 in 2 7 in Click Then, calculate the area of your base, B B = (1/2)aP = (1/2)(4√3)(48) = 8 in Click Click 96√3 in 2 = 166.28in 2 Click Click Finally, calculate your Surface Area. SA = 2B + PH Click SA = 2(166.28) + (48)(7) Click SA = 332.56 + 336 = 668.56 in 2 Click Slide 119 / 311 Other Prisms Example: Find the lateral area and surface area of the right prism. Because the base is a regular polygon, 6 ft we need to calculate the apothem. To begin, figure out the central angle & top angle in the triangle. 360 = 72° = central angle 10 ft 5 Click 72 = 36° = top angle of the triangle. 2 Click The base is a regular pentagon. 36° a Click 3 in. Slide 120 / 311 Other Prisms Example: Find the lateral area and surface area of the right prism. Next find the apothem using 6 ft trigonometry, or special right triangles (if it applies). 10 ft 3 tan 36 = a Click atan36 = 3 tan36 tan36 Click The base is a regular a = 4.13 in. pentagon. Click
Slide 121 / 311 Other Prisms Example: Find the lateral area and surface area of the right prism. Next, calculate the perimeter of your base. 6 ft P = 5(6) = 30 in Click Use this to find the Lateral Area LA = PH = 30(10) = 300 in 2 10 ft Click Then, calculate the area of your base, B B = (1/2)aP = (1/2)(4.13)(30) = 61.95 in 2 Click Click Click The base is a regular pentagon. Finally, calculate your Surface Area. SA = 2B + PH Click SA = 2(61.95) + (30)(10) Click SA = 123.9 + 300 = 423.9 in 2 Click Slide 122 / 311 Other Prisms Example: Find the lateral area and surface area of the right prism. 3 8 6 7 5 Angles are right angles. Then, calculate the area of your base, B First, calculate the perimeter of B = 7(5)+3(3) = 44 units 2 your base. P = 8 + 7 + 5 + 4 + 3 + 3 Finally, calculate your Surface Area. P = 30 units SA = 2B + PH Use this to find the Lateral Area SA = 2(44) + (30)(6) LA = PH = 30(6) = 180 units 2 SA = 88 + 180 = 268 units 2 Slide 123 / 311 48 Find the lateral area of the right prism. 8 11 The base is a regular hexagon.
Slide 124 / 311 49 Find the total surface area of the right prism. 8 11 The base is a regular hexagon. Slide 125 / 311 50 Find the total surface area of the right prism. 10 4 2 3 4 9 All angles are right angles. Slide 126 / 311 51 The right triangular prism has a surface area of 150 sq ft. Find the height of the prism. 6 A 5 ft 5 B 6 ft C 7.81 ft D 6.38 ft y
Slide 127 / 311 Surface Area of a Cylinder Return to Table of Contents Slide 128 / 311 Cylinders height base radius height base base base radius A Cylinder is a solid w/ 2 circular bases that lie in || planes. Because each base is a circle, it contains a radius. The remaining measurement that connects the 2 bases is the height of the cylinder. Slide 129 / 311 Cylinders The net of a right cylinder is two circles and a rectangle that forms the lateral surface. radius 8 8 x radius What is the length of x? - The circumference of the circle (base) Click to reveal
Slide 130 / 311 Finding the Surface Area of a Right Cylinder Surface Area : S.A. = 2B + PH B = Area of the circular base = πr 2 C = Perimeter of the Circular base (Circumference) = 2πr H = Height of the prism Lateral Area = CH = 2πrH Base Base height height Lateral Surface Base Base Slide 131 / 311 Finding the Surface Area of a Right Cylinder The Lateral Area is the area of the Lateral Surface, the rectangular area that wraps around the cylinder between the circular bases. Base Base height height Lateral Surface Base Base Therefore, the Surface Area of a Cylinder can be simplified to the equation below. SA = 2πr 2 + 2πrH Slide 132 / 311 Finding the Surface Area of a Right Cylinder Example: Find the lateral area and surface area of the right cylinder. LA = 2πrh LA = 2π(4)(8) LA = 64π units 2 8 LA = 201.06 units 2 r = 4 SA = 2πr 2 + 2πrh SA = 2π(4) 2 + 2π(4)(8) SA = 32π + 64π SA = 96π units 2 SA = 301.59 units 2
Slide 133 / 311 Cylinders Example: Find the lateral area and surface area of the right cylinder. 16 2 + h 2 = 34 2 click 256 + h 2 = 1156 click h 2 = 900 34 click h = 30 click Note: 16-30-34 = 2(8-15-17) Pyth. Triple click d = 16 SA = 2πr 2 + 2πrh LA = 2πrh click click SA = 2π(8) 2 + 2π(8)(30) LA = 2π(8)(30) click click SA = 128π + 480π LA = 480π units 2 click click SA = 608π units 2 LA = 1507.96 units 2 click click SA = 1,910.09 units 2 click Slide 134 / 311 Cylinders Example: Find the lateral area and surface area of the right cylinder when the base circumference is 16π ft & the height is 10 ft. C = 2πr click 16π = 2πr 2π 2π click click 8 ft = r click LA = 2πrh SA = 2πr 2 + 2πrh click click LA = 2π(8)(10) SA = 2π(8) 2 + 2π(8)(10) click click LA = 160π ft 2 SA = 128π + 160π click click LA = 502.64 ft 2 SA = 288π ft 2 click click SA = 904.78 ft 2 click Slide 135 / 311 52 Find the lateral area of the right cylinder. h = 12 r = 7
Slide 136 / 311 53 Find the surface area of the right cylinder. Use 3.14 as your value of π & round to two decimal places. A 1200 sq in. B 307.72 sq in. C 835.24 sq in. h = 12 D 1670.48 sq in. r = 7 Slide 137 / 311 54 Find the lateral area of the right cylinder. 13 r = 5 Slide 138 / 311 55 Find the lateral area of the right cylinder. h = 12 Base area is 36π units 2
Slide 139 / 311 56 Find the surface area of the right cylinder. h = 12 Base area is 36π units 2 Slide 140 / 311 57 The surface area of the right cylinder is 653.12 sq in. Find the height of the cylinder. Use 3.14 as your value of π. h A 7 in. B 8 in. r = 8 in. C 5 in. D 6 in. Slide 141 / 311 58 A food company packages soup in aluminum cans that have a diameter of 2 1 / 2 inches and a height of 4 inches. Before shipping the cans off to the stores, they add their company label to the can which does not cover the top and bottom. If the company is shipping 200 cans of soup to one store, how much paper material is required to make the labels?
Slide 142 / 311 59 Maria's mom baked a cake for her daughter's birthday party. The diameter of the cake is 9 inches and the height is 2 inches. How much base frosting (pink in the picture below) was required to cover the cake? Slide 143 / 311 Surface Area of a Pyramid Return to Table of Contents Slide 144 / 311 Pyramids A Pyramid is a polyhedron in which the base is a polygon & the lateral faces are triangles with a common vertex. Lateral Edges are the intersection of 2 lateral faces Vertex Lateral Edge Lateral Face Base
Slide 145 / 311 Net This is a right square pyramid. Another name for it is pentahedron. Hedron is a suffix that means face. Why is this a pentahedron? Slide 146 / 311 Surface Area = Sum of the Areas of all the sides Height of the Triangle Slant Height ℓ The Pyramid has a square base and 4 triangular faces The triangular faces are all isosceles triangles if its a right pyramid. The Height of each triangular face is the Slant Height of the pyramid if it ℓ is a regular pyramid (labeled as , or a cursive lower case L). Slide 147 / 311 Segment Lengths in a Pyramid Pyramid's Height (h) ℓ Slant Height ( ) Square Base (B)
Slide 148 / 311 Segment Lengths in a Pyramid Example: Find the value of x. a 2 + 12 2 = 13 2 a 2 + 144 = 169 a 2 = 25 13 a = 5 12 Note: 5-12-13 Right Triangle Therefore x = 2(5) = 10 x Slide 149 / 311 Segment Lengths in a Pyramid Example: Find the value of x. Square Base has an area of 64, so 64 = y 2 y = 8, so a = 4 of the right x triangle. 8 4 2 + 8 2 = x 2 16 + 64 = x 2 x 2 = 80 x = 8.94 units Base Area of the right square pyramid is 64 u 2 . Slide 150 / 311 Segment Lengths in a Pyramid Example: Find the length of the slant height. This is a regular hexagonal pyramid. r = 6 lateral edge = 12 r
Slide 151 / 311 Segment Lengths in a Pyramid First, find the height of the pyramid using Pythagorean Theorem. 6 2 + h 2 = 122 12 h click 36 + h 2 = 144 click h 2 = 108 6 click h = 6√3 = 10.39 click r Note: 30-60-90 triangle click Slide 152 / 311 Segment Lengths in a Pyramid Second, find the apothem of the hexagonal base. 360 = 60° = central 6 click Note: equilateral click 60 = 30° = top of the . r 2 click 3 2 + a 2 = 6 2 click 9 + a 2 = 36 6 6 click a 2 = 27 a click a = 3√3 = 5.20 3 3 click Note: 30-60-90 triangle click Slide 153 / 311 Segment Lengths in a Pyramid Last, find the slant height of your pyramid w/ the apothem & height. ℓ (3√3) 2 + (6√3) 2 = 2 click ℓ 27 + 108 = 2 click ℓ 2 = 135 r click ℓ = 3√15 = 11.62 click ℓ h = 6√3 Click a = 3√3 Click
Slide 154 / 311 60 Find the value of the variable. x 6 16 Slide 155 / 311 61 Find the value of the variable. x 11 12 Slide 156 / 311 62 Find the value of the variable. 6 x area of the base is 36 u 2
Slide 157 / 311 63 Find the value of the slant height. Regular Hexagonal Pyramid r r = 8 lateral edge = 17 Slide 158 / 311 64 Find the value of the slant height. Regular Hexagonal Pyramid a a = 9 lateral edge = 12 Slide 159 / 311 Surface Area of a Regular Pyramid Pyramid's Height (h) ℓ Slant Height ( ) Square Base (B) ℓ ℓ Surface Area = B + ½P and Lateral Area = ½P ℓ = Slant Height P = Perimeter of Base B = Area of Base
Slide 160 / 311 Surface Area of a Regular Pyramid 1 ℓ Why is the Surface Area SA = B + P ? 2 Pyramid's Height (h) Slant Height ( ) ℓ Square Base (B) Surface Area is the sum of all of the areas that make up the solid. In our diagram, these are 4 triangles & 1 square. A square = s s = s 2 = B 1 ℓ A ∆ = s 2 Slide 161 / 311 Surface Area of a Regular Pyramid 1 ℓ Why is the Surface Area SA = B + P ? 2 s ℓ Net of Pyramid Since there are 4 ∆s, we can multiply the area of each ∆ by 4. Therefore, our Surface Area for the Pyramid above is ℓℓ SA = s 2 + 4( 1 / 2 )s SA = s 2 + ( 1 / 2 )(4s) ℓ SA = B + 1 / 2 P Slide 162 / 311 Surface Area of a Regular Pyramid Example: Find the lateral area and the surface area of the pyramid. LA = 1 / 2 P ℓ LA = 1 / 2 (24)(7) ℓ = 7 LA = 12(7) LA = 84 units 2 SA = B + 1 / 2 P ℓ SA = 6 2 + 1 / 2 (24)(7) SA = 36 + 84 SA = 120 units 2 s = 6
Slide 163 / 311 Surface Area of a Regular Pyramid Example: Find the lateral area and the surface area of the pyramid. First, calculate the slant height. 3 2 + 8 2 = ℓ 2 9 + 64 = ℓ 2 73 = ℓ 2 ℓ = 8.54 Next, calculate the LA & SA h = 8 LA = 1 / 2 P ℓ LA = 1 / 2 (24)(8.54) LA = 12(8.54) s = 6 LA = 102.48 units 2 SA = B + 1 / 2 P ℓ SA = 6 2 + 1 / 2 (24)(8.54) SA = 36 + 102.48 SA = 138.48 units 2 Slide 164 / 311 Surface Area of a Regular Pyramid Example: Find the lateral area and the surface area of the pyramid. 10 ℓ 8 e = 10 First, calculate the slant height. 8 2 + ℓ 2 = 10 2 click 64 + ℓ 2 = 100 s = 16 click ℓ 2 = 36 ℓ = 6 click Slide 165 / 311 Surface Area of a Regular Pyramid Example: Find the lateral area and the surface area of the pyramid. Next, calculate the LA & SA LA = 1 / 2 P ℓ click LA = 1 / 2 (64)(6) click e = 10 LA = 32(6) click LA = 192 units 2 click SA = B + 1 / 2 P ℓ click SA = 16 2 + 1 / 2 (64)(6) s = 16 click SA = 256 + 192 click SA = 448 units 2 click
Slide 166 / 311 Surface Area of a Regular Pyramid Example: Find the lateral area and the surface area of the pyramid. Regular Pentagonal Pyramid a a = 4 lateral edge = 8 Slide 167 / 311 Surface Area of a Regular Pyramid Example: Find the lateral area and the surface area of the pyramid. First, find the radius & side length of the regular pentagon using the apothem & trigonometric ratios 360 = 72° = central 5 Click 72° Click 72 r 36°36° = 36° = top 2 Click Click of the . 4 Click x x tan36 = 4 Click 4 cos36 = r Click x = 4tan36 = 2.91 rcos36 = 4 Click cos36 cos36 Click Therefore, s = 2(2.91) = 5.82 Click r = 4.94 Click Slide 168 / 311 Surface Area of a Regular Pyramid Next, find the slant height of the pyramid using the lateral edge, the value of x from the previous slide & Pythagorean Theorem. 2.912 + ℓ 2 = 8 2 click 8.4681 + ℓ 2 = 64 8 ℓ click ℓ 2 = 55.5319 click ℓ = 7.45 click 2.91
Slide 169 / 311 Surface Area of a Regular Pyramid Last, find the lateral area & surface area of the pyramid. SA = B + 1 / 2 P ℓ LA = 1 / 2 P ℓ click click SA = 1 / 2 (4)(29.1) + 1 / 2 (29.1)(7.45) LA = 1 / 2 (29.1)(7.45) click click SA = 58.2 + 108.40 LA = 108.40 units 2 click click SA = 166.6 units 2 click Slide 170 / 311 65 Find the lateral area of the right pyramid. ℓ = 9 s = 10 Slide 171 / 311 66 Find the surface area of the right pyramid. ℓ = 9 s = 10
Slide 172 / 311 67 Find the lateral area of the right pyramid. e = 10 base area = 16 Slide 173 / 311 68 Find the surface area of the right pyramid. e = 10 base area = 16 Slide 174 / 311 69 Find the lateral area of the right pyramid. a = 5 h = 12 a Regular Octagonal Pyramid
Slide 175 / 311 70 Find the surface area of the right pyramid. a = 5 h = 12 a Regular Octagonal Pyramid Slide 176 / 311 71 Find the surface area of the right pyramid. 8 12 30 Hint: The pyramid is NOT regular. Hint So, B + 1/2 P ℓ doesn't work. Instead, draw a net of the pyramid & find each area. Slide 177 / 311 Surface Area of a Cone Return to Table of Contents
Slide 178 / 311 Cones S Base l a n t H e i g height h t ℓ r Lateral Surface Slant Height The Base of the cone is a circle ℓ The length of the circular portion of the Lateral Surface is the same as the Circumference of the Circlular Base. The Slant Height is the length of the diagonal slant of the cone from the top to the edge of the base. The Height of the cone is the length from the top to the center of the circular base. Slide 179 / 311 Finding the Surface Area of a Right Cone Surface Area = Area of the Base + Lateral Area Base Lateral Area= ½P ℓ S.A. = B + ½P ℓ ℓ = Slant Height P = Perimeter of Circular Base B = Area of Circular Base Lateral Surface Because the base is a circle. P = Circumference = 2πr Slant Height L.A. = ½(2πr) ℓ = πr ℓ ℓ S.A. = πr 2 + πr ℓ Slide 180 / 311 Cones Example: Find the lateral area and surface area of the right cone. r = 6 LA = πr ℓ click = π(6)(8) click LA = 48π units 2 ℓ = 8 click LA = 150.80 units 2 click SA = πr 2 + πr ℓ click = π(6) 2 + π(6)(8) click = 36π + 48π click SA = 84π units 2 click SA = 263.89 units 2 click
Slide 181 / 311 Cones Example: Find the lateral area and surface area of the right cone. C = 12π units C = 2πr click 12π = 2πr click 2π 2π h = 8 click 6 units = r click 6 2 + 8 2 = ℓ 2 click 36 + 64 = ℓ 2 click 100 = ℓ 2 click 10 units = ℓ click Slide 182 / 311 Cones Example: Find the lateral area and surface area of the right cone. C = 12π units LA = πr ℓ click = π(6)(10) click LA = 60π units 2 h = 8 click LA = 188.50 units 2 click SA = πr 2 + πr ℓ click = π(6) 2 + π(6)(10) click = 36π + 60π click SA = 96π units 2 click SA = 301.59 units 2 click Slide 183 / 311 72 Find the lateral area of the right cone, in square units. r = 4 ℓ = 9
Slide 184 / 311 73 Find the surface area of the right cone, in square units. r = 4 ℓ = 9 Slide 185 / 311 74 Find the lateral area of the right cone, in square units. Base Area = 16π units 2 h = 9 Slide 186 / 311 75 Find the surface area of the right cone, in square units. Base Area = 16π units 2 h = 9
Slide 187 / 311 76 Find the length of the radius of the right cone if the lateral area is 50π units 2 ? ℓ = 10 Slide 188 / 311 77 Find the height of the right cone if the lateral area is 50π units 2 ? ℓ = 10 Slide 189 / 311 78 Find the slant height of the right cone if the surface area is 45π units 2 and the diameter of the base is 6 units?
Slide 190 / 311 79 Find the height of the right cone if the surface area is 45π units 2 and the diameter of the base is 6 units? Slide 191 / 311 80 The Department of Transportation keeps 4 piles of road salt for snowy days. Each conical shaped pile is 20 feet high and 30 feet across at the base. During the summer the piles are covered with tarps to prevent erosion. How much tarp is needed to cover the conical shaped piles so that no part of them are exposed? Slide 192 / 311 Volume of a Prism Return to Table of Contents
Slide 193 / 311 Prisms The volume of a solid is the amount of cubic units that a solid can hold. Where area used square units, volume will use cubic units. Slide 194 / 311 Prisms Finding the Volume of a Prism Base V = BH H height Specific Prisms Rectangular Prism: V = ℓwH Cube: V = s 3 Base w ℓ Slide 195 / 311 Prisms Does a prism need to be a right prism for the volume formula to work? Think of a ream of paper If the stack is fanned, it Stacked nicely it still has 500 sheets. has 500 sheets. So the volume doesn't change if the prism, stack of paper, is right or oblique. The formula V = BH works for all prisms.
Slide 196 / 311 Prisms Example: Find the volume of the rectangular prism with a length of 2, a width of 6, and a height of 5. V = ℓ w H V = 2(6)(5) V = 60 units 3 Slide 197 / 311 Prisms Example: The volume of a box is 48 ft 3 . If the height is 4 ft and width is 6 ft, what is the length? V = ℓ w H 48 = ℓ(6)(4) 48 = 24 ℓ 24 24 2 ft = ℓ Slide 198 / 311 Prisms Example: Find the volume of the prism shown below. 10 6 11 Since it has a base that is a right triangle, we need to find the base of the triangle using Pythagorean Theorem. 6 2 + b 2 = 10 2 36 + b 2 = 100 b 2 = 64 b = 8 units Next, calculate the area of your base, B B = (1/2)(8)(6) = 24 units 2 Finally, calculate your Volume. V = BH V = 24(11) V = 264 units 3
Slide 199 / 311 Prisms Example: The volume of a cube is 64 m 3 , what is area of one face? V = s 3 64 = s 3 4 m = s Area of one face A = 4(4) A = 16 m 2 Slide 200 / 311 Prisms Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Because the base is a regular polygon, we need to calculate the side length. To begin, figure out the central angle & top angle in the triangle. 7 in 360 = 60° = central angle 6 Click 60 4 in = 30° = top angle of the triangle. 2 Click 30° 4 in. x in. Click Slide 201 / 311 Prisms Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Next, find the value of x using trigonometry, or special right triangles (if it applies). x tan 30 = 4 7 in Click 4tan30 = x x = 4√3 = 2.31 in. 3 Click Click 4 in Then, calculate the side length of your base. s = 2(2.31) = 4.62 in Click 30° 4 in. x in.
Slide 202 / 311 Prisms Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Next, use your value of s to find the Perimeter of your base P = 6(4.62) = 27.72 in Click 7 in Then, calculate the area of your base, B B = ( 1 / 2 )aP = ( 1 / 2 )(2.31)(27.72) = 32.02 in 2 Click Click Click 4 in Finally, calculate your Volume. V = Bh V = 32.02(7) Click Click V = 224.14 in 3 Click Slide 203 / 311 81 What is the volume of a rectangular prism with edges of 4, 5, and 7? Slide 204 / 311 82 What is the volume of a cube with edges of 5 units?
Slide 205 / 311 83 If the volume of a rectangular prism is 64 u 3 and has height 8 and width 4, what is the length? Slide 206 / 311 84 If a cube has volume 27 u 3 , what is the cubes surface area? Slide 207 / 311 85 Find the volume of the prism. 15 20 12
Slide 208 / 311 86 Find the volume of the prism. 6 6 7 6 2 Slide 209 / 311 87 Find the volume of the prism. 8 11 The base is a regular hexagon. Slide 210 / 311 88 A high school has a pool that is 25 yards in length, 60 feet in width, and contains the depth dimensions shown in the figure below. 2 yds 19 yds 4 yds 3 ft 9 ft Shallow end Deep end If one cubic yard is about 201.974 gallons, how much water is required to fill the pool?
Slide 211 / 311 Volume of a Cylinder Return to Table of Contents Slide 212 / 311 Finding the Volume of a Cylinder r base V = Bh V = πr 2 h height r base Slide 213 / 311 Cylinders Example: Find the volume of the cylinder with a radius of 4 and a height of 11. V = π(4) 2 (11) V = 176π units 3 V = 552.92 units 3
Slide 214 / 311 Cylinders Example: The surface area of a cylinder is 96π units 2 , and its radius is 4 units. What is the volume? SA = 2πr 2 + 2πrh 96π = 2π(4) 2 + 2π(4)h 96π = 32π + 8πh -32π -32π 64π = 8πh 8π 8π h = 8 units V = π(4) 2 (8) V = 128π units 3 V = 402.12 units 3 Slide 215 / 311 89 Find the volume of the cylinder with radius 6 and height 8. Slide 216 / 311 90 Find the volume of the cylinder with a circumference of 18π units and a height of 6.
Slide 217 / 311 91 Find the volume of the cylinder with a surface area of 653.12 u 2 & a radius of 8 units. Use 3.14 as your value of π. h r = 8 Slide 218 / 311 92 The volume of a cylinder is 108π u 3 , and the height is 12 units. What is the surface area? Slide 219 / 311 93 The height of a cylinder doubles, what happens to the volume? Doubles A Quadruples B Depends on the cylinder C Cannot be determined D
Slide 220 / 311 94 The radius of a cylinder doubles, what happens to the volume? Doubles A Quadruples B Depends on the cylinder C D Cannot be determined Slide 221 / 311 95 A 3" hole is drilled through a solid cylinder with a diameter of 4" forming a tube. What is the volume of the tube? 4" 3" 24" Slide 222 / 311 Volume of a Pyramid Return to Table of Contents
Slide 223 / 311 Finding the Volume of a Pyramid Pyramid's Height (h) ℓ Slant Height ( ) V = 1 / 3 Bh Square Base (B) Slide 224 / 311 Volume of Pyramids Example: Find the volume of the pyramid. 6 V = 1 / 3 Bh 4 B = 5(4) = 20 h = 6 units 5 V = 1 / 3 (20)(6) V = 40 units 3 Slide 225 / 311 Volume of Pyramids Example: Find the volume of the pyramid. 5 8 8 h 5 4 8 8 click for extra diagram
Slide 226 / 311 96 Find the volume of the pyramid. 5 6 7 Slide 227 / 311 97 Find the volume of the pyramid. 8 6 6 Slide 228 / 311 98 Find the volume of the pyramid. 10 12 12
Slide 229 / 311 Volume of Pyramids Example: Find the volume of the pyramid. Regular Pentagonal Pyramid First, find the side length of the regular pentagon using the apothem & trigonometric ratios. 360 = 72° = central 5 Click a 72 = 36° = top angle of the . 2 a = 4 Click lateral edge = 8 x tan36 = 4 Click x = 4tan36 = 2.91 Click Therefore, s = 2(2.91) = 5.82 Click Slide 230 / 311 Volume of Pyramids Example: Find the volume of the pyramid. Next, find the slant height of Regular Pentagonal Pyramid the pyramid using the lateral edge, the value of x from the previous slide & Pyth. Theorem. 2.912 + ℓ 2 = 8 2 Click a 8.4681 + ℓ 2 = 64 Click ℓ 2 = 55.5319 8 ℓ a = 4 Click ℓ = 7.45 lateral edge = 8 Click Then, use the slant 2.91 4 2 + h 2 = 7.45 2 height & apothem w/ Click Click Pyth. Theorem to 16 + h 2 = 55.5319 5 find the height. k 4 c Click i . l h C h 2 = 39.5319 7 Click h = 6.29 4 Click Slide 231 / 311 Volume of Pyramids Example: Find the volume of the pyramid. Regular Pentagonal Pyramid Last, find the Area of your Base & Volume. B = 1 / 2 aP Click B = 1 / 2 (4)(29.1) Click B = 58.2 units 2 a Click a = 4 V = 1 / 3 Bh lateral edge = 8 Click V = 1 / 3 (58.2)(6.29) Click V = 122.03 units 3 Click
Slide 232 / 311 99 Find the volume of the right pyramid. a = 5 h = 12 a Regular Octagonal Pyramid Slide 233 / 311 100 Find the volume of the right pyramid. 11 8 The base is a regular hexagon. Slide 234 / 311 Volume of Pyramids A truncated pyramid is a pyramid with its top cutoff parallel to its base. Find the volume of the truncated pyramid shown. V truncated = V big - V small B big = 6(6) = 36 3 2 h big = 3 + 9 = 12 2 V big = 1 / 3 (36)(12) V big = 144 units 3 9 B small = 2(2) = 4 6 h small = 3 V small = 1 / 3 (4)(3) 6 V small = 4 units 3 V truncated = 144 - 4 V truncated = 140 units 3
Slide 235 / 311 101 Find the volume of the truncated pyramid. 3 2 2 12 8 8 Slide 236 / 311 102 The table shows the approximate measurements of the Red Pyramid in Egypt and the Great Pyramid of Cholula in Mexico. Length Width Height Red Pyramid 220 m 220m 104 m Great Pyramid of 450 m 450 m 66 m Cholula Approximately, what is the difference between the Answer volume of the Red Pyramid and the volume of the Great Pyramid of Cholula? A 6,132,867 cubic meters B 4,455,000 cubic meters C 2,777,133 cubic meters D 1,677,867 cubic meters Slide 237 / 311 The Geometryville Aquarium is building a new tank space for coral reef fish shown in the figure below. The laws say that the dimensions of the tank must have a maximum length of 14 feet, a maximum width of 10 feet and a maximum height of 16 feet. h 2 h 1 w ℓ 103 Salt water comes in cylindrical containers that measure 10 feet high and have a diameter of 8 feet. Determine the height of the aquarium that should be used in the design. Show that your design will be able to store at least 3 cylindrical containers of water. When you finish, enter your value for h 1 into your SMART Responder.
Slide 238 / 311 Volume of a Cone Return to Table of Contents Slide 239 / 311 Finding the Volume of a Cone S l a n t H V = 1 / 3 Bh e i g height h t V = 1 / 3 πr 2 h ℓ r Slide 240 / 311 Volume of a Cone Example: Find the volume of the cone. 7 9 V = 1 / 3 πr 2 h V = 1 / 3 π(7) 2 (9) V = 147π units 3 V = 461.81 units 3
Slide 241 / 311 Volume of a Cone Example: Find the volume of the cone. 12 r = 4, so d = 8 With the right triangle, use Pythagorean Theorem to find the 4 height of the pyramid. h 2 + 8 2 = 12 2 h 2 + 64 = 144 h 2 = 80, h = √80 = 8.94 V = 1 / 3 πr 2 h V = 1 / 3 π(4) 2 (8.94) V = 47.68π units 3 V = 149.79 units 3 Slide 242 / 311 Volume of a Cone Example: Find the volume of the cone, with lateral area of 15π units 2 and a slant height 5 units. LA = πr ℓ 1) You know the Lateral area & slant height, so use the Lateral Click 15π = πr(5) Area formula to calculate the Click 15π = 5πr radius. Click 5π 5π Click 3 units = r Click Slide 243 / 311 Volume of a Cone 2) Next, use the slant height & radius to calculate the height of the cone using Pythagorean Theorem. h 2 + 3 2 = 5 2 Click h 2 + 9 = 25 Click h 2 = 16 Click h = 4 Click Note: 3-4-5 Pyth. Triple Click
Slide 244 / 311 Volume of a Cone 3) Last, calculate the volume of the cone. V = 1 / 3 πr 2 h Click V = 1 / 3 π(3) 2 (4) Click V = 12π units 3 Click V = 37.70 units 3 Click Slide 245 / 311 104 What is the volume of the cone? 8 d = 10 Slide 246 / 311 105 What is the volume of the cone? r = 4 ℓ = 9
Slide 247 / 311 106 What is the volume of the cone? 40° 10 Slide 248 / 311 107 What is the volume of the truncated cone? 6 r = 4 6 r = 8 Slide 249 / 311 Surface Area & Volume of Spheres Return to Table of Contents
Slide 250 / 311 Spheres Recall the Definition of a Circle The locus of points in a plane that are the same distance from a point called the center of the circle. Y X X Y Every point on the above circle is the same distance from the origin in the x, y plane. Slide 251 / 311 Spheres The locus of points in space that are the same distance from a point. Z Y X X X Y Y Every point on the sphere above on the left side, is the same distance from the origin in space, the x, y, z plane. Slide 252 / 311 Spheres The Great Circle of a sphere is found at the intersection of a plane and a sphere when the plane contains the center of the sphere. Z X Y
Slide 253 / 311 Great Circles Z X Y Each of these planes intersects the sphere, and the plane contains the center of the sphere Slide 254 / 311 Great Circles The Earth has 2 Great Circles: Can you name them? The Equator click The Prime Meridian w/ the International Date Line click International Date Line Click to reveal picture Slide 255 / 311 Great Circle The Great Circle separates the Sphere into two equal halves at the center of the sphere.
Slide 256 / 311 Each half is called a Hemisphere Slide 257 / 311 Cross Sections A Cross Section is found by the intersection of a plane and a solid. Cross - Section (Click the top hemisphere to see the cross section.) Slide 258 / 311 Cross Sections The farther the cross section of the sphere is taken from its center the smaller the circle. . small great circles circle
Slide 259 / 311 Cross Sections Example: Find the radius of the cross section of the sphere that has a radius of 8 if the cross section is 2 from the center. r 2 8 8 2 2 + r 2 = 8 2 4 + r 2 = 64 r 2 = 60 r = √60 = 2√15 = 7.75 Slide 260 / 311 Cross Sections Example: A cross section of a sphere is 4 units from the center of the sphere and has an area of 16π units 2 . What is area of the great circle? Leave your answer in terms of π. 16π = πr 2 4 r = 4 units in the cross section 4 2 + 4 2 = r 2 32 = r 2 r =√32 = 4√2 = 2.83 = radius of sphere A = π(√32) 2 A = 32π units 2 Slide 261 / 311 108 What is the area of the cross section of a sphere that is 6 units from the center of the sphere if the sphere has radius 8 units?
Slide 262 / 311 109 What is the area of the great circle if a cross section that is 3 from the center has a circumference of 10π? Slide 263 / 311 110 The circumference of the great circle of a sphere is 12π units and a cross section has a circumference of 8π units. How far is the cross section from the center? Slide 264 / 311 Finding the Surface Area of the Sphere S.A. = 4πr 2 r Why is there no formula for lateral area? A sphere doesn't have any bases, so the lateral area is the same as the surface area. Click to reveal
Slide 265 / 311 Finding the Volume of the Sphere 4 V = πr 3 r 3 Slide 266 / 311 Finding the Volume of the Sphere Example: Find the surface area & volume of a sphere with radius of 6 ft. SA = 4π(6) 2 SA = 144π units 2 SA = 452.39 units 2 4 V = π(6) 3 3 V = 288π units 3 V = 904.78 units 3 Slide 267 / 311 Finding the Volume of the Sphere Example: Find the surface area & volume of a sphere that a great circle with area 24π units 2 ? 24π = πr 2 π π r 2 = 24 r = 4.90 units SA = 4π(4.9) 2 SA = 96.04π units 2 SA = 301.72 units 2 4 V = π(4.9) 3 3 V = 156.87π units 3 V = 492.81 units 3
Slide 268 / 311 Finding the Volume of the Sphere Example: A cross section of a sphere has area 36π units 2 and is 10 units from the center, what is the surface area & volume of the sphere? Radius of Cross Section 36π = πr 2 click π π click r 2 = 36 click r = 6 units click Radius of Sphere 10 2 + 6 2 = R 2 click 136 = R 2 click R = √136 = 11.66 units click Slide 269 / 311 Finding the Volume of the Sphere Example: A cross section of a sphere has area 36π units 2 and is 10 units from the center, what is the surface area & volume of the sphere? SA = 4π(√136) 2 click SA = 544π units 2 click SA = 1,709.03 units 2 click 4 V = π(√136) 3 3 V = 2,114.69π units 3 Click click V = 6,643.50 units 3 click Slide 270 / 311 111 Find the surface area of a sphere with radius 10.
Slide 271 / 311 112 Find the volume of a sphere with radius 10. Slide 272 / 311 113 What is the surface area of a sphere if a cross section 7 units from the center has an area of 50.26 units 2 ? Slide 273 / 311 114 What is the volume of a sphere if a cross section 7 units from the center has an area of 50.26 units 2 ?
Slide 274 / 311 115 The volume of a sphere is 24π units 3 . What is the area of a great circle of the sphere? Slide 275 / 311 116 A recipe calls for half of an orange. Shelly use an orange that has a diameter of 3 inches. She wraps the remaining half of orange in plastic wrap. What is the amount of area that Shelly has to cover? Slide 276 / 311 Cavalieri's Principle Return to Table of Contents
Slide 277 / 311 Cavalieri's Principle If two solids are the same height, and the area of their cross sections are equal, then the two solids will have the same volume. Slide 278 / 311 Cavalieri's Principle Which solid has the greatest volume? 4π 4 8 4 2π 14 14 14 224π 224π 224π 703.72 703.72 703.72 Click Click Click None: All of the solids have the same volume. Click Slide 279 / 311 Cavalieri's Principle Example: A sphere is submerged in a cylinder. Both solids have a radius of 4. What is the volume of the cylinder not occupied by the sphere? volume of cylinder - volume of sphere Click π(4) 2 (8) - 4 / 3 π(4) 3 Click 128π - 256 / 3 π Click 128 / 3 π units 3 Click
Slide 280 / 311 Cavalieri's Principle The result shows that the left over volume is equal to what other solid? cone Click According to Cavalieri, what can be said about the cross section? The cross section of the great circle of the sphere is equal to the circle cross section of the cylinder. Click Slide 281 / 311 Cavalieri's Principle Example: What is the radius of a sphere made from the 5 cylinder of modeling clay shown? 15 If you are using clay to model both solids, what measurement is the same? Volume Click Slide 282 / 311 Cavalieri's Principle Therefore, calculate the volume of the cylinder 5 first. V = π(5) 2 (15) Click V = 375π units 3 Click 15 Then create an equation to represent the problem and solve for r. 375π = 4 / 3 πr 3 Click 375 = 4 / 3 r 3 Click 281.25 = r 3 Click r = 6.55 units Click
Slide 283 / 311 117 These 2 solids have the same volume, find the value of x. r = 6 9 x 11 11 Slide 284 / 311 118 These 2 solids have the same volume, find the value of x. 10 x 8 12 12 Slide 285 / 311 Two prisms each with a base area of 20 square inches are shown. Which statements about prisms C and D are true. Select all that apply. (Statements are on the next slide.) x y B = 20 in 2 B = 20 in 2 Prism C Prism D
Slide 286 / 311 119 Two prisms each with a base area of 20 square inches are shown. Which statements about prisms C and D are true. Select all that apply. A If x > y, the area of a vertical cross section of prism C is greater than the area of a vertical cross section of prism D. B If x > y, the area of a vertical cross section of prism C is equal to the area of a vertical cross section of prism D. C If x > y, the area of a vertical cross section of prism C is less than the area of a vertical cross section of prism D. D If x = y, the volume of prism C is greater than the volume of prism D, because prism C is a right prism. E If x = y, the volume of prism C is equal to the volume of prism D because the prisms have the same base area. F If x = y, the volume of prism C is less than the volume of prism D because prism D is an oblique prism. Slide 287 / 311 Similar Solids Return to Table of Contents Slide 288 / 311 Similar Solids Corresponding sides of similar figures are similar. The prisms shown are similar. Find the values of x and y. 4 6 2 y x 9 4 x 4 2 6 = 6 = 9 y Click Click 36 = 6x 4y = 12 Click 6 6 Click 4 4 Click Click 6 = x y = 3 Click Click
Slide 289 / 311 Similar Solids 4 6 2 y x 9 The ratio of similarity, k, is the common value that is multiplied to preimage to get to the image. - Hint: it's the ratio of image : preimage click for the hint If the smaller prism is the preimage, then the 3 / 2 value of k is Click If the larger prism is the preimage, then the value 2 / 3 of k is Click Slide 290 / 311 120 The pyramid on the left is the preimage and is similar to the image on the right. Find the value of x. y 16 h 3 2 8 x 8 Slide 291 / 311 121 The pyramid on the left is the preimage and is similar to the image on the right. Find the value of y. y 16 3 h 2 x 8 8
Slide 292 / 311 122 The pyramid on the left is the preimage and is similar to the image on the right. Find the value of h. y 16 3 h 2 x 8 8 Slide 293 / 311 Similar Solids Consider the example of the prisms from earlier. The ratio of similarity from the smaller solid to the larger is 2:3. 6 4 3 2 9 6 Calculate the surface area of both solids. How do they compare? SA small = 2(6)(2) + 16(4) = 88 units 2 Click Click SA big = 2(3)(9) + 24(6) = 198 units 2 Click Click SA Similarity ratio = 88:198 = 4:9 = 2 2 :3 2 Click Click Click How do their volumes compare? V small = 2(4)(6) = 48 units 3 V big = 6(3)(9) = 162 units 3 Click V Similarity ratio = 48:162 = 8:27 = 2 3 :3 3 Click Click Click Click Slide 294 / 311 Comparing Similar Figures length in image = k length in preimage area in image = k 2 area in preimage volume in image = k 3 volume in preimage
Slide 295 / 311 Comparing Similar Figures Example: r = 3 r = 9 How many times bigger is the radius of the sphere to the right? 3 times bigger Click How many times bigger is the surface area of the sphere to the right? SA small = 4π(3) 2 = 36π units 2 Click Click SA big = 4π(9) 2 = 324π units 2 9 times bigger Click Click Click How many times bigger is the volume of the sphere to the right? V small = 4 / 3 π(3) 3 = 36π units 3 Click 27 times bigger Click V big = 4 / 3 π(9) 3 = 972π units 3 Click Click Click Slide 296 / 311 123 The scale factor of 2 similar pyramids is 4. If the surface area of the larger one is 64 units 2 , what is surface area of the smaller one? Slide 297 / 311 124 The scale factor of 2 similar right square pyramids is 3. If the area of the base of the larger one is 36 u 2 and its height is 12, what is the volume of the smaller one?
Slide 298 / 311 125 An architect builds a scale model of a home using a scale of 2 in to 5 ft. Given the view of the roof of the model, how much roofing material is needed for the house? 12 in 6 in 8 in 5 in 4 in 3 in Slide 299 / 311 PARCC Sample Questions The remaining slides in this presentation contain questions from the PARCC Sample Test. After finishing this unit, you should be able to answer these questions. Good Luck! Return to Table of Contents Slide 300 / 311 Topic: Intro to 3-D Solids Question 6/11 Daniel buys a block of clay for an art project. The block is shaped like a cube with edge lengths of 10 inches. Daniel decides to cut the block of clay into two pieces. He places a wire across the diagonal of one face of the cube, as shown in the figure. Then he pulls the wire straight back to create two congruent chunks of clay. PARCC Released Question - PBA - Calculator Section
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