Geometry 3 Dimension. Classify three-dimensional figures according to their properties. Use nets and...
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Geometry 3 Dimension. Classify three-dimensional figures according to their properties. Use nets and cross sections to analyze three- dimensional figures
Classify three-dimensional figures according to their
properties. Use nets and cross sections to analyze three-
dimensional figures. Objectives
Slide 3
face edge vertex prism cylinder pyramid cone cube net cross
section Vocabulary
Slide 4
Three-dimensional figures, or solids, can be made up of flat or
curved surfaces. Each flat surface is called a face. An edge is the
segment that is the intersection of two faces. A vertex is the
point that is the intersection of three or more faces.
Slide 5
Slide 6
A cube is a prism with six square faces. Other prisms and
pyramids are named for the shape of their bases.
Slide 7
Example 1A: Classifying Three-Dimensional Figures Classify the
figure. Name the vertices, edges, and bases. cube vertices: A, B,
C, D, E, F, G, H bases: ABCD, EFGH, ABFE, DCGH, ADHE, BCGF
edges:
Slide 8
Example 1B: Classifying Three-Dimensional Figures Classify the
figure. Name the vertices, edges, and bases. pentagonal pyramid
vertices: A, B, C, D, E, F base: ABCDE edges:
Slide 9
Example 1c Classify the figure. Name the vertices, edges, and
bases. vertex: N cone edges: none base: M M
Slide 10
Example 1d Classify the figure. Name the vertices, edges, and
bases. triangular prism bases: TUV, WXY vertices: T, U, V, W, X, Y
edges:
Slide 11
A net is a diagram of the surfaces of a three-dimensional
figure that can be folded to form the three-dimensional figure. To
identify a three-dimensional figure from a net, look at the number
of faces and the shape of each face.
Slide 12
Example 2A: Identifying a Three-Dimensional Figure From a Net
Describe the three-dimensional figure that can be made from the
given net. The net has six congruent square faces. So the net forms
a cube.
Slide 13
Example 2B: Identifying a Three-Dimensional Figure From a Net
Describe the three-dimensional figure that can be made from the
given net. The net has one circular face and one semicircular face.
These are the base and sloping face of a cone. So the net forms a
cone.
Slide 14
Example 2c Describe the three-dimensional figure that can be
made from the given net. The net has four congruent triangular
faces. So the net forms a triangular pyramid.
Slide 15
Example 2d Describe the three-dimensional figure that can be
made from the given net. The net has two circular faces and one
rectangular face. These are the bases and curved surface of a
cylinder. So the net forms a cylinder.
Slide 16
A cross section is the intersection of a three-dimensional
figure and a plane.
Slide 17
Example 3A: Describing Cross Sections of Three-Dimensional
Figures Describe the cross section. The cross section is a
point.
Slide 18
Example 3B: Describing Cross Sections of Three-Dimensional
Figures Describe the cross section. The cross section is a
pentagon.
Slide 19
Example 3c Describe the cross section. The cross section is a
hexagon.
Slide 20
Example 3d Describe the cross section. The cross section is a
triangle.
Slide 21
Example 4A: Food Application A piece of cheese is a prism with
equilateral triangular bases. How can you slice the cheese to make
each shape? an equilateral triangle Cut parallel to the bases.
Slide 22
Example 4B: Food Application A piece of cheese is a prism with
equilateral triangular bases. How can you slice the cheese to make
each shape? a rectangle Cut perpendicular to the bases.
Slide 23
Example 4c How can a chef cut a cube-shaped watermelon to make
slices with triangular faces? Cut through the midpoints of 3 edges
that meet at 1 vertex.
Slide 24
Question 1 1. Classify the figure. Name the vertices, edges,
and bases. triangular prism; vertices: A, B, C, D, E, F; bases: ABC
and DEF edges:
Slide 25
2. Describe the three-dimensional figure that can be made from
this net. square pyramid Question 2
Slide 26
3. Describe the cross section. a rectangle Question 3
Slide 27
Draw representations of three-dimensional figures. Recognize a
three dimensional figure from a given representation.
Objectives
Slide 28
orthographic drawing isometric drawing perspective drawing
vanishing point horizon Vocabulary
Slide 29
There are many ways to represent a three dimensional object. An
orthographic drawing shows six different views of an object: top,
bottom, front, back, left side, and right side.
Slide 30
Example 1: Drawing Orthographic Views of an Object Draw all six
orthographic views of the given object. Assume there are no hidden
cubes.
Slide 31
Example 1 Continued Draw all six orthographic views of the
given object. Assume there are no hidden cubes. Bottom
Slide 32
Example 1 Continued Draw all six orthographic views of the
given object. Assume there are no hidden cubes.
Slide 33
Example 1 Continued Draw all six orthographic views of the
given object. Assume there are no hidden cubes.
Slide 34
Example 1B Draw all six orthographic views of the given object.
Assume there are no hidden cubes.
Slide 35
Example 1B Continued
Slide 36
Isometric drawing is a way to show three sides of a figure from
a corner view. You can use isometric dot paper to make an isometric
drawing. This paper has diagonal rows of dots that are equally
spaced in a repeating triangular pattern.
Slide 37
Example 2: Drawing an Isometric View of an Object Draw an
isometric view of the given object. Assume there are no hidden
cubes.
Slide 38
Example 2 B Draw an isometric view of the given object. Assume
there are no hidden cubes.
Slide 39
In a perspective drawing, nonvertical parallel lines are drawn
so that they meet at a point called a vanishing point. Vanishing
points are located on a horizontal line called the horizon. A
one-point perspective drawing contains one vanishing point. A
two-point perspective drawing contains two vanishing points.
Slide 40
Slide 41
In a one-point perspective drawing of a cube, you are looking
at a face. In a two- point perspective drawing, you are looking at
a corner. Helpful Hint
Slide 42
Example 3A: Drawing an Object in Perspective Draw the block
letter in one-point perspective. Draw a horizontal line to
represent the horizon. Mark a vanishing point on the horizon. Then
draw a shape below the horizon. This is the front of the.
Slide 43
Example 3A Continued Draw the block letter in one-point
perspective. From each corner of the, lightly draw dashed segments
to the vanishing point.
Slide 44
Example 3A Continued Draw the block letter in one-point
perspective. Lightly draw a smaller with vertices on the greyed
segments. This is the back of the.
Slide 45
Example 3A Continued Draw the block letter in one-point
perspective. Draw the edges of the, using dashed segments for
hidden edges. Erase any segments that are not part of the.
Slide 46
Draw the block letter in two-point perspective. Example 3B:
Drawing an Object in Perspective Draw a horizontal line to
represent the horizon. Mark two vanishing points on the horizon.
Then draw a vertical segment below the horizon and between the
vanishing points. This is the front edge of the. Lightly mark a
point of the way down the segment, for the lower part of the
shape.
Slide 47
From the marked point and the endpoints of the segment, lightly
draw dashed segments to each vanishing point. Draw vertical
segments connecting the dashed lines. These are other vertical
edges of the. Example 3B Continued
Slide 48
Lightly draw dashed segments from the endpoints of each new
vertical segment to the vanishing points. Example 3B Continued
Slide 49
Draw the edges of the, using dashed segments for hidden edges.
Erase any segments that are not part of the. Example 3B
Continued
Slide 50
Draw the block letter L in one-point perspective. Draw a
horizontal line to represent the horizon. Mark a vanishing point on
the horizon. Then draw a L shape below the horizon. This is the
front of the L. Example 3C
Slide 51
From each corner of the L, lightly draw dashed segments to the
vanishing point. Draw the block letter L in one-point perspective.
Example 3C Continued
Slide 52
Lightly draw a smaller L with vertices on the dashed segments.
This is the back of the L. Draw the block letter L in one-point
perspective. Example 3C Continued
Slide 53
Draw the edges of the L, using dashed segments for hidden
edges. Erase any segments that are not part of the L. Draw the
block letter L in one-point perspective. Example 3C Continued
Slide 54
Draw the block letter L in two-point perspective. Draw a
horizontal line to represent the horizon. Mark two vanishing points
on the horizon. Then draw a vertical segment below the horizon and
between the vanishing points. This is the front edge of the L.
Lightly mark a point of the way down the segment, for the lower
part of the L shape. Example 3D
Slide 55
From the marked point and the endpoints of the segment, lightly
draw dashed segments to each vanishing point. Draw vertical
segments connecting the dashed lines. These are other vertical
edges of the L. Example 3D Continued Draw the block letter L in
two-point perspective.
Slide 56
Lightly draw dashed segments from the endpoints of each new
vertical segment to the vanishing points. Example 3D Continued Draw
the block letter L in two-point perspective.
Slide 57
Draw the edges of the L, using dashed segments for hidden edges
to the vanishing points. Example 3D Continued Draw the block letter
L in two-point perspective.
Slide 58
Erase any segments that are not part of the L. Example 3D
Continued Draw the block letter L in two-point perspective.
Slide 59
Example 4A: Relating Different Representations of an Object
Determine whether the drawing represents the given object. Assume
there are no hidden cubes. No; the base has one cube too many.
Slide 60
Example 4B: Relating Different Representations of an Object
Determine whether the drawing represents the given object. Assume
there are no hidden cubes. Yes; the drawing is a two-point
perspective view of the object.
Slide 61
Example 4C: Relating Different Representations of an Object
Determine whether the drawing represents the given object. Assume
there are no hidden cubes. Yes; the drawing is an isometric view of
the object.
Slide 62
Example 4D: Relating Different Representations of an Object
Determine whether the drawing represents the given object. Assume
there are no hidden cubes. Yes; the drawing shows the six
orthographic views of the object.
Slide 63
Example 4E Determine whether the drawing represents the given
object. Assume there are no hidden cubes. no
Slide 64
A 1. Draw all six orthographic views of the object. Assume
there are no hidden cubes.
Slide 65
2. Draw an isometric view of the object. A
Slide 66
3. Determine whether each drawing represents the given object.
Assume there are no hidden cubes. yes no A
Slide 67
Warm Up Find the unknown lengths. 1. the diagonal of a square
with side length 5 cm 2. the base of a rectangle with diagonal 15 m
and height 13 m 3. the height of a trapezoid with area 18 ft 2 and
bases 3 ft and 9 ft 7.5 m 3 ft
Slide 68
Apply Eulers formula to find the number of vertices, edges, and
faces of a polyhedron. Develop and apply the distance and midpoint
formulas in three dimensions. Objectives
Slide 69
polyhedron space Vocabulary
Slide 70
polyhedron - formed by four or more polygons that intersect
only at their edges. Prisms and pyramids are polyhedrons, but
cylinders and cones are not.
Slide 71
Slide 72
Example 1A: Using Eulers Formula Find the number of vertices,
edges, and faces of the polyhedron. Use your results to verify
Eulers formula. V = 12, E = 18, F = 8 Use Eulers Formula. Simplify.
12 18 + 8 = 2 ? 2 = 2
Slide 73
Example 1B: Using Eulers Formula Find the number of vertices,
edges, and faces of the polyhedron. Use your results to verify
Eulers formula. V = 5, E = 8, F = 5 Use Eulers Formula. Simplify. 5
8 + 5 = 2 ? 2 = 2
Slide 74
Example 1C Find the number of vertices, edges, and faces of the
polyhedron. Use your results to verify Eulers formula. V = 6, E =
12, F = 8 Use Eulers Formula. Simplify. 2 = 2 6 12 + 8 = 2 ?
Slide 75
Example 1D Find the number of vertices, edges, and faces of the
polyhedron. Use your results to verify Eulers formula. V = 7, E =
12, F = 7 Use Eulers Formula. Simplify. 2 = 2 7 12 + 7 = 2 ?
Slide 76
A diagonal of a three-dimensional figure connects two vertices
of two different faces. Diagonal d of a rectangular prism is shown
in the diagram. By the Pythagorean Theorem, 2 + w 2 = x 2, and x 2
+ h 2 = d 2. Using substitution, 2 + w 2 + h 2 = d 2. Box
Problem
Slide 77
Solution
Slide 78
Example 2A: Using the Pythagorean Theorem in Three Dimensions
Find the unknown dimension in the figure. the length of the
diagonal of a 6 cm by 8 cm by 10 cm rectangular prism Substitute 6
for l, 8 for w, and 10 for h. Simplify.
Slide 79
Example 2B: Using the Pythagorean Theorem in Three Dimensions
Find the unknown dimension in the figure. the height of a
rectangular prism with a 12 in. by 7 in. base and a 15 in. diagonal
225 = 144 + 49 + h 2 h 2 = 32 Substitute 15 for d, 12 for l, and 7
for w. Square both sides of the equation. Simplify. Solve for h 2.
Take the square root of both sides.
Slide 80
Example 2C Find the length of the diagonal of a cube with edge
length 5 cm. d 2 = 25 + 25 + 25 d 2 = 75 Substitute 5 for each
side. Square both sides of the equation. Simplify. Solve for d 2.
Take the square root of both sides.
Slide 81
Space is the set of all points in three dimensions. Three
coordinates are needed to locate a point in space. A
three-dimensional coordinate system has 3 perpendicular axes: the
x-axis, the y-axis, and the z-axis. An ordered triple (x, y, z) is
used to locate a point. To locate the point (3, 2, 4), start at (0,
0, 0). From there move 3 units forward, 2 units right, and then 4
units up.
Slide 82
Example 3A: Graphing Figures in Three Dimensions Graph a
rectangular prism with length 5 units, width 3 units, height 4
units, and one vertex at (0, 0, 0). The prism has 8 vertices: (0,
0, 0), (5, 0, 0), (0, 3, 0), (0, 0, 4), (5, 3, 0), (5, 0, 4), (0,
3, 4), (5, 3, 4)
Slide 83
Example 3B: Graphing Figures in Three Dimensions Graph a cone
with radius 3 units, height 5 units, and the base centered at (0,
0, 0) Graph the center of the base at (0, 0, 0). Since the height
is 5, graph the vertex at (0, 0, 5). The radius is 3, so the base
will cross the x-axis at (3, 0, 0) and the y-axis at (0, 3, 0).
Draw the bottom base and connect it to the vertex.
Slide 84
Example 3C Graph a cone with radius 5 units, height 7 units,
and the base centered at (0, 0, 0). Graph the center of the base at
(0, 0, 0). Since the height is 7, graph the vertex at (0, 0, 7).
The radius is 5, so the base will cross the x-axis at (5, 0, 0) and
the y-axis at (0, 5, 0). Draw the bottom base and connect it to the
vertex.
Slide 85
You can find the distance between the two points (x 1, y 1, z 1
) and (x 2, y 2, z 2 ) by drawing a rectangular prism with the
given points as endpoints of a diagonal. Then use the formula for
the length of the diagonal. You can also use a formula related to
the Distance Formula. (See Lesson 1-6.) The formula for the
midpoint between (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ) is related
to the Midpoint Formula. (See Lesson 1-6.)
Slide 86
Slide 87
Example 4A: Finding Distances and Midpoints in Three Dimensions
Find the distance between the given points. Find the midpoint of
the segment with the given endpoints. Round to the nearest tenth,
if necessary. (0, 0, 0) and (2, 8, 5) distance:
Slide 88
Example 4A Continued midpoint: M(1, 4, 2.5) Find the distance
between the given points. Find the midpoint of the segment with the
given endpoints. Round to the nearest tenth, if necessary. (0, 0,
0) and (2, 8, 5)
Slide 89
Example 4B: Finding Distances and Midpoints in Three Dimensions
Find the distance between the given points. Find the midpoint of
the segment with the given endpoints. Round to the nearest tenth,
if necessary. (6, 11, 3) and (4, 6, 12) distance:
Slide 90
Example 4B Continued midpoint: Find the distance between the
given points. Find the midpoint of the segment with the given
endpoints. Round to the nearest tenth, if necessary. (6, 11, 3) and
(4, 6, 12) M(5, 8.5, 7.5)
Slide 91
Example 4C Find the distance between the given points. Find the
midpoint of the segment with the given endpoints. Round to the
nearest tenth, if necessary. (0, 9, 5) and (6, 0, 12)
distance:
Slide 92
midpoint: Find the distance between the given points. Find the
midpoint of the segment with the given endpoints. Round to the
nearest tenth, if necessary. (0, 9, 5) and (6, 0, 12) Example 4C
Continued M(3, 4.5, 8.5)
Slide 93
Example 4D Find the distance between the given points. Find the
midpoint of the segment with the given endpoints. Round to the
nearest tenth, if necessary. (5, 8, 16) and (12, 16, 20)
distance:
Slide 94
midpoint: Example 4D extended Find the distance between the
given points. Find the midpoint of the segment with the given
endpoints. Round to the nearest tenth, if necessary. (5, 8, 16) and
(12, 16, 20) M(8.5, 12, 18)
Slide 95
Example 5: Recreation Application Trevor drove 12 miles east
and 25 miles south from a cabin while gaining 0.1 mile in
elevation. Samira drove 8 miles west and 17 miles north from the
cabin while gaining 0.15 mile in elevation. How far apart were the
drivers? The location of the cabin can be represented by the
ordered triple (0, 0, 0), and the locations of the drivers can be
represented by the ordered triples (12, 25, 0.1) and ( 8, 17,
0.15).
Slide 96
Example 5 Continued Use the Distance Formula to find the
distance between the drivers.
Slide 97
Example 6 If both divers swam straight up to the surface, how
far apart would they be? Use the Distance Formula to find the
distance between the divers.
Slide 98
A 1. Find the number of vertices, edges, and faces of the
polyhedron. Use your results to verify Eulers formula. V = 8; E =
12; F = 6; 8 12 + 6 = 2
Slide 99
B Find the unknown dimension in each figure. Round to the
nearest tenth, if necessary. 2. the length of the diagonal of a
cube with edge length 25 cm 3. the height of a rectangular prism
with a 20 cm by 12 cm base and a 30 cm diagonal 4. Find the
distance between the points (4, 5, 8) and (0, 14, 15). Find the
midpoint of the segment with the given endpoints. Round to the
nearest tenth, if necessary. 43.3 cm 18.9 cm d 12.1 units; M (2,
9.5, 11.5)
Slide 100
Learn and apply the formula for the surface area of a prism.
Learn and apply the formula for the surface area of a cylinder.
Objectives
Slide 101
lateral face lateral edge right prism oblique prism altitude
surface area lateral surface axis of a cylinder right cylinder
oblique cylinder Vocabulary
Slide 102
Prisms and cylinders have 2 congruent parallel bases. A lateral
face is not a base. The edges of the base are called base edges. A
lateral edge is not an edge of a base. The lateral faces of a right
prism are all rectangles. An oblique prism has at least one
nonrectangular lateral face.
Slide 103
An altitude of a prism or cylinder is a perpendicular segment
joining the planes of the bases. The height of a three-dimensional
figure is the length of an altitude. Surface area is the total area
of all faces and curved surfaces of a three-dimensional figure. The
lateral area of a prism is the sum of the areas of the lateral
faces.
Slide 104
The net of a right prism can be drawn so that the lateral faces
form a rectangle with the same height as the prism. The base of the
rectangle is equal to the perimeter of the base of the prism.
Slide 105
The surface area of a right rectangular prism with length ,
width w, and height h can be written as S = 2w + 2wh + 2h.
Slide 106
The surface area formula is only true for right prisms. To find
the surface area of an oblique prism, add the areas of the faces.
Caution!
Slide 107
Example 1A: Finding Lateral Areas and Surface Areas of Prisms
Find the lateral area and surface area of the right rectangular
prism. Round to the nearest tenth, if necessary. L = Ph = 32(14) =
448 ft 2 S = Ph + 2B = 448 + 2(7)(9) = 574 ft 2 P = 2(9) + 2(7) =
32 ft
Slide 108
Example 1B: Finding Lateral Areas and Surface Areas of Prisms
Find the lateral area and surface area of a right regular
triangular prism with height 20 cm and base edges of length 10 cm.
Round to the nearest tenth, if necessary. L = Ph = 30(20) = 600 ft
2 S = Ph + 2B P = 3(10) = 30 cm The base area is
Slide 109
Example 1C Find the lateral area and surface area of a cube
with edge length 8 cm. L = Ph = 32(8) = 256 cm 2 S = Ph + 2B = 256
+ 2(8)(8) = 384 cm 2 P = 4(8) = 32 cm
Slide 110
The lateral surface of a cylinder is the curved surface that
connects the two bases. The axis of a cylinder is the segment with
endpoints at the centers of the bases. The axis of a right cylinder
is perpendicular to its bases. The axis of an oblique cylinder is
not perpendicular to its bases. The altitude of a right cylinder is
the same length as the axis.
Slide 111
Slide 112
Example 2A: Finding Lateral Areas and Surface Areas of Right
Cylinders Find the lateral area and surface area of the right
cylinder. Give your answers in terms of . L = 2 rh = 2 (8)(10) =
160 in 2 The radius is half the diameter, or 8 ft. S = L + 2 r 2 =
160 + 2 (8) 2 = 288 in 2
Slide 113
Example 2B: Finding Lateral Areas and Surface Areas of Right
Cylinders Find the lateral area and surface area of a right
cylinder with circumference 24 cm and a height equal to half the
radius. Give your answers in terms of . Step 1 Use the
circumference to find the radius. C = 2 r Circumference of a circle
24 = 2 r Substitute 24 for C. r = 12 Divide both sides by 2 .
Slide 114
Example 2B Continued Step 2 Use the radius to find the lateral
area and surface area. The height is half the radius, or 6 cm. L =
2 rh = 2 (12)(6) = 144 cm 2 S = L + 2 r 2 = 144 + 2 (12) 2 = 432 in
2 Lateral area Surface area Find the lateral area and surface area
of a right cylinder with circumference 24 cm and a height equal to
half the radius. Give your answers in terms of .
Slide 115
Example 2C Find the lateral area and surface area of a cylinder
with a base area of 49 and a height that is 2 times the radius.
Step 1 Use the circumference to find the radius. A = r 2 49 = r 2 r
= 7 Area of a circle Substitute 49 for A. Divide both sides by and
take the square root.
Slide 116
Step 2 Use the radius to find the lateral area and surface
area. The height is twice the radius, or 14 cm. L = 2 rh = 2
(7)(14)=196 in 2 S = L + 2 r 2 = 196 + 2 (7) 2 =294 in 2 Lateral
area Surface area Find the lateral area and surface area of a
cylinder with a base area of 49 and a height that is 2 times the
radius. Example 2C Continued
Slide 117
Example 3: Finding Surface Areas of Composite Three-Dimensional
Figures Find the surface area of the composite figure.
Slide 118
Example 3 Continued Two copies of the rectangular prism base
are removed. The area of the base is B = 2(4) = 8 cm 2. The surface
area of the rectangular prism is.. A right triangular prism is
added to the rectangular prism. The surface area of the triangular
prism is
Slide 119
The surface area of the composite figure is the sum of the
areas of all surfaces on the exterior of the figure. Example 3
Continued S = (rectangular prism surface area) + (triangular prism
surface area) 2(rectangular prism base area) S = 52 + 36 2(8) = 72
cm 2
Slide 120
Example 3B Find the surface area of the composite figure. Round
to the nearest tenth.
Slide 121
Example 3B Continued Find the surface area of the composite
figure. Round to the nearest tenth. The surface area of the
rectangular prism is S =Ph + 2B = 26(5) + 2(36) = 202 cm 2. The
surface area of the cylinder is S =Ph + 2B = 2 (2)(3) + 2 (2) 2 =
20 62.8 cm 2. The surface area of the composite figure is the sum
of the areas of all surfaces on the exterior of the figure.
Slide 122
S = (rectangular surface area) + (cylinder surface area)
2(cylinder base area) S = 202 + 62.8 2( )(2 2 ) = 239.7 cm 2
Example 3B Continued Find the surface area of the composite figure.
Round to the nearest tenth.
Slide 123
Always round at the last step of the problem. Use the value of
given by the key on your calculator. Remember!
Slide 124
Example 4: Exploring Effects of Changing Dimensions The edge
length of the cube is tripled. Describe the effect on the surface
area.
Slide 125
Example 4 Continued original dimensions:edge length tripled:
Notice than 3456 = 9(384). If the length, width, and height are
tripled, the surface area is multiplied by 3 2, or 9. S = 6 2 =
6(8) 2 = 384 cm 2 S = 6 2 = 6(24) 2 = 3456 cm 2 24 cm
Slide 126
Example 4B The height and diameter of the cylinder are
multiplied by. Describe the effect on the surface area.
Slide 127
original dimensions:height and diameter halved: S = 2 (11 2 ) +
2 (11)(14) = 550 cm 2 S = 2 (5.5 2 ) + 2 (5.5)(7) = 137.5 cm 2 11
cm 7 cm Example 4B Continued Notice than 550 = 4(137.5). If the
dimensions are halved, the surface area is multiplied by
Slide 128
Example 5: Recreation Application A sporting goods company
sells tents in two styles, shown below. The sides and floor of each
tent are made of nylon. Which tent requires less nylon to
manufacture?
Slide 129
Example 5 Continued Pup tent: Tunnel tent: The tunnel tent
requires less nylon.
Slide 130
Example 5B A piece of ice shaped like a 5 cm by 5 cm by 1 cm
rectangular prism has approximately the same volume as the pieces
below. Compare the surface areas. Which will melt faster? The 5 cm
by 5 cm by 1 cm prism has a surface area of 70 cm 2, which is
greater than the 2 cm by 3 cm by 4 cm prism and about the same as
the half cylinder. It will melt at about the same rate as the half
cylinder.
Slide 131
C Find the lateral area and the surface area of each figure.
Round to the nearest tenth, if necessary. 1.a cube with edge length
10 cm 2. a regular hexagonal prism with height 15 in. and base edge
length 8 in. 3. a right cylinder with base area 144 cm 2 and a
height that is the radius L = 400 cm 2 ; S = 600 cm 2 L = 720 in 2
; S 1052.6 in 2 L 301.6 cm 2 ; S = 1206.4 cm 2
Slide 132
C 4.A cube has edge length 12 cm. If the edge length of the
cube is doubled, what happens to the surface area? 5. Find the
surface area of the composite figure. The surface area is
multiplied by 4. S = 3752 m 2
Slide 133
Practice Find the missing side length of each right triangle
with legs a and b and hypotenuse c. 1. a = 7, b = 24 2. c = 15, a =
9 3. b = 40, c = 41 4. a = 5, b = 5 5. a = 4, c = 8 c = 25 b = 12 a
= 9
Slide 134
Learn and apply the formula for the surface area of a pyramid.
Learn and apply the formula for the surface area of a cone.
Objectives
Slide 135
vertex of a pyramid regular pyramid slant height of a regular
pyramid altitude of a pyramid vertex of a cone axis of a cone right
cone oblique cone slant height of a right cone altitude of a cone
Vocabulary
Slide 136
The vertex of a pyramid is the point opposite the base of the
pyramid. The base of a regular pyramid is a regular polygon, and
the lateral faces are congruent isosceles triangles. The slant
height of a regular pyramid is the distance from the vertex to the
midpoint of an edge of the base. The altitude of a pyramid is the
perpendicular segment from the vertex to the plane of the
base.
Slide 137
The lateral faces of a regular pyramid can be arranged to cover
half of a rectangle with a height equal to the slant height of the
pyramid. The width of the rectangle is equal to the base perimeter
of the pyramid.
Slide 138
Slide 139
Example 1A: Finding Lateral Area and Surface Area of Pyramids
Find the lateral area and surface area of a regular square pyramid
with base edge length 14 cm and slant height 25 cm. Round to the
nearest tenth, if necessary. Lateral area of a regular pyramid P =
4(14) = 56 cm Surface area of a regular pyramid B = 14 2 = 196 cm
2
Slide 140
Example 1B: Finding Lateral Area and Surface Area of Pyramids
Step 1 Find the base perimeter and apothem. Find the lateral area
and surface area of the regular pyramid. The base perimeter is
6(10) = 60 in. The apothem is, so the base area is
Slide 141
Example 1B Continued Step 2 Find the lateral area. Lateral area
of a regular pyramid Substitute 60 for P and 16 for . Find the
lateral area and surface area of the regular pyramid.
Slide 142
Example 1B Continued Step 3 Find the surface area. Surface area
of a regular pyramid Substitute for B. Find the lateral area and
surface area of the regular pyramid.
Slide 143
Example 1C Find the lateral area and surface area of a regular
triangular pyramid with base edge length 6 ft and slant height 10
ft. Step 1 Find the base perimeter and apothem. The base perimeter
is 3(6) = 18 ft. The apothem is so the base area is
Slide 144
Example 1C Continued Find the lateral area and surface area of
a regular triangular pyramid with base edge length 6 ft and slant
height 10 ft. Step 2 Find the lateral area. Lateral area of a
regular pyramid Substitute 18 for P and 10 for .
Slide 145
Step 3 Find the surface area. Surface area of a regular pyramid
Example 1C Continued Find the lateral area and surface area of a
regular triangular pyramid with base edge length 6 ft and slant
height 10 ft. Substitute for B.
Slide 146
The vertex of a cone is the point opposite the base. The axis
of a cone is the segment with endpoints at the vertex and the
center of the base. The axis of a right cone is perpendicular to
the base. The axis of an oblique cone is not perpendicular to the
base.
Slide 147
The slant height of a right cone is the distance from the
vertex of a right cone to a point on the edge of the base. The
altitude of a cone is a perpendicular segment from the vertex of
the cone to the plane of the base.
Slide 148
Example 2A: Finding Lateral Area and Surface Area of Right
Cones Find the lateral area and surface area of a right cone with
radius 9 cm and slant height 5 cm. L = r Lateral area of a cone =
(9)(5) = 45 cm 2 Substitute 9 for r and 5 for . S = r + r 2 Surface
area of a cone = 45 + (9) 2 = 126 cm 2 Substitute 5 for and 9 for
r.
Slide 149
Example 2B: Finding Lateral Area and Surface Area of Right
Cones Find the lateral area and surface area of the cone. Use the
Pythagorean Theorem to find . L = r = (8)(17) = 136 in 2 Lateral
area of a right cone Substitute 8 for r and 17 for . S = r + r 2
Surface area of a cone = 136 + (8) 2 = 200 in 2 Substitute 8 for r
and 17 for .
Slide 150
Example 2C Find the lateral area and surface area of the right
cone. Use the Pythagorean Theorem to find . L = r = (8)(10) = 80 cm
2 Lateral area of a right cone Substitute 8 for r and 10 for . S =
r + r 2 Surface area of a cone = 80 + (8) 2 = 144 cm 2 Substitute 8
for r and 10 for .
Slide 151
Example 3: Exploring Effects of Changing Dimensions The base
edge length and slant height of the regular hexagonal pyramid are
both divided by 5. Describe the effect on the surface area.
Slide 152
3 in. 2 in. Example 3 Continued original dimensions: base edge
length and slant height divided by 5: S = P + B 1212 1212 in
Slide 153
Example 3 Continued original dimensions: base edge length and
slant height divided by 5: 3 in. 2 in. Notice that. If the base
edge length and slant height are divided by 5, the surface area is
divided by 5 2, or 25. in 2
Slide 154
Example 3B The base edge length and slant height of the regular
square pyramid are both multiplied by. Describe the effect on the
surface area.
Slide 155
Example 3B Continued original dimensions:multiplied by
two-thirds: By multiplying the dimensions by two-thirds, the
surface area was multiplied by. 8 ft 10 ft S = P + B 1212 = 260 cm
2 S = P + B 1212 = 585 cm 2 ft 2
Slide 156
Example 4: Finding Surface Area of Composite Three-Dimensional
Figures Find the surface area of the composite figure. The lateral
area of the cone is L = r l = (6)(12) = 72 in 2. Left-hand cone:
Right-hand cone: Using the Pythagorean Theorem, l = 10 in. The
lateral area of the cone is L = r l = (6)(10) = 60 in 2.
Slide 157
Example 4 Continued Composite figure: S = (left cone lateral
area) + (right cone lateral area) Find the surface area of the
composite figure. = 60 in 2 + 72 in 2 = 132 in 2
Slide 158
Example 4B Find the surface area of the composite figure.
Surface Area of Cube without the top side: S = 4wh + B S = 4(2)(2)
+ (2)(2) = 20 yd 2
Slide 159
Example 4b Continued Surface Area of Pyramid without base:
Surface Area of Composite: Surface of Composite = SA of Cube + SA
of Pyramid
Slide 160
Example 5: Manufacturing Application If the pattern shown is
used to make a paper cup, what is the diameter of the cup? The
radius of the large circle used to create the pattern is the slant
height of the cone. The area of the pattern is the lateral area of
the cone. The area of the pattern is also of the area of the large
circle, so
Slide 161
Example 5 Continued Substitute 4 for , the slant height of the
cone and the radius of the large circle. r = 2 in. Solve for r. The
diameter of the cone is 2(2) = 4 in. If the pattern shown is used
to make a paper cup, what is the diameter of the cup?
Slide 162
5B Find the lateral area and surface area of each figure. Round
to the nearest tenth, if necessary. 1. a regular square pyramid
with base edge length 9 ft and slant height 12 ft 2. a regular
triangular pyramid with base edge length 12 cm and slant height 10
cm L = 216 ft 2 ; S = 297 ft 2 L = 180 cm 2 ; S 242.4 cm 2
Slide 163
4. A right cone has radius 3 and slant height 5. The radius and
slant height are both multiplied by. Describe the effect on the
surface area. 5. Find the surface area of the composite figure.
Give your answer in terms of . The surface area is multiplied by. S
= 24 ft 2 5C
Slide 164
Practice Find the volume of each figure. Round to the nearest
tenth, if necessary. 1. a square prism with base area 189 ft 2 and
height 21 ft 2. a regular hexagonal prism with base edge length 24
m and height 10 m 3. a cylinder with diameter 16 in. and height 22
in. 3969 ft 3 14,964.9 m 3 4423.4 in 3
Slide 165
Learn and apply the formula for the volume of a pyramid. Learn
and apply the formula for the volume of a cone. Objectives
Slide 166
The volume of a pyramid is related to the volume of a prism
with the same base and height. The relationship can be verified by
dividing a cube into three congruent square pyramids, as
shown.
Slide 167
The square pyramids are congruent, so they have the same
volume. The volume of each pyramid is one third the volume of the
cube.
Slide 168
Example 1A: Finding Volumes of Pyramids Find the volume a
rectangular pyramid with length 11 m, width 18 m, and height 23
m.
Slide 169
Example 1B: Finding Volumes of Pyramids Find the volume of the
square pyramid with base edge length 9 cm and height 14 cm. The
base is a square with a side length of 9 cm, and the height is 14
cm.
Slide 170
Example 1C: Finding Volumes of Pyramids Find the volume of the
regular hexagonal pyramid with height equal to the apothem of the
base Step 1 Find the area of the base. Area of a regular polygon
Simplify.
Slide 171
Example 1C Continued Step 2 Use the base area and the height to
find the volume. The height is equal to the apothem,. Volume of a
pyramid. = 1296 ft 3 Find the volume of the regular hexagonal
pyramid with height equal to the apothem of the base Simplify.
Slide 172
Example 1D Find the volume of a regular hexagonal pyramid with
a base edge length of 2 cm and a height equal to the area of the
base. Step 1 Find the area of the base. Area of a regular polygon
Simplify.
Slide 173
Example 1D Continued Step 2 Use the base area and the height to
find the volume. Volume of a pyramid Find the volume of a regular
hexagonal pyramid with a base edge length of 2 cm and a height
equal to the area of the base. = 36 cm 3 Simplify.
Slide 174
An art gallery is a 6-story square pyramid with base area of
acre (1 acre = 4840 yd 2, 1 story 10 ft). Estimate the volume in
cubic yards and cubic feet. Example 2: Architecture Application
First find the volume in cubic yards. Volume of a pyramid The base
is a square with an area of about 2420 yd 2. The base edge length
is. The height is about 6(10) = 60 ft or about 20 yd.
Slide 175
Example 2 Continued Substitute 2420 for B and 20 for h. 16,133
yd 3 16,100 yd 3 Volume of a pyramid Then convert your answer to
find the volume in cubic feet. The volume of one cubic yard is (3
ft)(3 ft)(3 ft) = 27 ft 3. Use the conversion factor to find the
volume in cubic feet.
Slide 176
Example 2B What if? What would be the volume of the Rainforest
Pyramid if the height were doubled? Volume of a pyramid. Substitute
70 for B and 66 for h. = 107,800 yd 3 or 107,800(27) = 2,910,600 ft
3
Slide 177
Slide 178
= 245 cm 3 769.7 cm 3 Example 3A: Finding Volumes of Cones Find
the volume of a cone with radius 7 cm and height 15 cm. Give your
answers both in terms of and rounded to the nearest tenth. Volume
of a pyramid Substitute 7 for r and 15 for h. Simplify.
Slide 179
Example 3B: Finding Volumes of Cones 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. The
height is 2 in. more than twice the radius. 2 r = 25 Substitute 25
for the circumference. r = 12.5 Solve for r.
Slide 180
Example 3B Continued Step 3 Use the radius and height to find
the volume. Volume of a pyramid. Substitute 12.5 for r and 27 for
h. = 1406.25 in 3 4417.9 in 3 Simplify. Find the volume of a cone
with base circumference 25 in. and a height 2 in. more than twice
the radius.
Slide 181
Example 3C: Finding Volumes of Cones Find the volume of a cone.
Step 1 Use the Pythagorean Theorem to find the height. 16 2 + h 2 =
34 2 Pythagorean Theorem h 2 = 900 Subtract 16 2 from both sides. h
= 30 Take the square root of both sides.
Slide 182
Example 3C Continued Step 2 Use the radius and height to find
the volume. Volume of a cone Substitute 16 for r and 30 for h. 2560
cm 3 8042.5 cm 3 Simplify. Find the volume of a cone.
Slide 183
Example 3D Find the volume of the cone. Volume of a cone
Substitute 9 for r and 8 for h. 216 m 3 678.6 m 3 Simplify.
Slide 184
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 3 3,
or 27. The diameter and height of the cone are divided by 3.
Describe the effect on the volume.
Slide 185
Example 4B 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.
Slide 186
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
Slide 187
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. ound to the nearest tenth. V
cylinder = r 2 h = (21) 2 (35)=15,435 cm 3. V = 5145 + 15,435 +
5,880 = 26,460 83,126.5 cm 3
Slide 188
Example 5B Find the volume of the composite figure. The volume
of the rectangular prism is V = wh = 25(12)(15) = 4500 ft 3. The
volume of the pyramid is The volume of the composite is the
rectangular prism subtract the pyramid. 4500 1500 = 3000 ft 3
Slide 189
5C Find the volume of each figure. Round to the nearest tenth,
if necessary. 1. a rectangular pyramid with length 25 cm, width 17
cm, and height 21 cm 2. a regular triangular pyramid with base edge
length 12 in. and height 10 in. 3. a cone with diameter 22 cm and
height 30 cm 4. a cone with base circumference 8 m and a height 5 m
more than the radius 2975 cm 3 207.8 in 3 V 3801.3 cm 3 V 117.3 m
2
Slide 190
5D 5. A cone has radius 2 in. and height 7 in. If the radius
and height are multiplied by, describe the effect on the volume. 6.
Find the volume of the composite figure. Give your answer in terms
of . The volume is multiplied by. 10,800 yd 3
Slide 191
Practice Find each measurement. 1. the radius of circle M if
the diameter is 25 cm 2. the circumference of circle X if the
radius is 42.5 in. 3. the area of circle T if the diameter is 26 ft
4. the circumference of circle N if the area is 625 cm 2 12.5 cm 85
in. 169 ft 2 50 cm
Slide 192
Learn and apply the formula for the volume of a sphere. Learn
and apply the formula for the surface area of a sphere.
Objectives
Slide 193
sphere center of a sphere radius of a sphere hemisphere great
circle Vocabulary
Slide 194
A sphere is the locus of points in space that are a fixed
distance from a given point called the center of a sphere. A radius
of a sphere connects the center of the sphere to any point on the
sphere. A hemisphere is half of a sphere. A great circle divides a
sphere into two hemispheres
Slide 195
The figure shows a hemisphere and a cylinder with a cone
removed from its interior. The cross sections have the same area at
every level, so the volumes are equal by Cavalieris Principle. Y
The height of the hemisphere is equal to the radius.
Slide 196
V(hemisphere) = V(cylinder) V(cone) The volume of a sphere with
radius r is twice the volume of the hemisphere, or.
Slide 197
Slide 198
Example 1A: Finding Volumes of Spheres Find the volume of the
sphere. Give your answer in terms of . = 2304 in 3 Simplify. Volume
of a sphere.
Slide 199
Example 1B: Finding Volumes of Spheres Find the diameter of a
sphere with volume 36,000 cm 3. Substitute 36,000 for V. 27,000 = r
3 r = 30 d = 60 cm d = 2r Take the cube root of both sides. Volume
of a sphere.
Slide 200
Example 1C: Finding Volumes of Spheres Find the volume of the
hemisphere. Volume of a hemisphere Substitute 15 for r. = 2250 m 3
Simplify.
Slide 201
Example 1D Find the radius of a sphere with volume 2304 ft 3.
Volume of a sphere Substitute for V. r = 12 ftSimplify.
Slide 202
Example 2: Sports Application A sporting goods store sells
exercise balls in two sizes, standard (22- in. diameter) and jumbo
(34-in. diameter). How many times as great is the volume of a jumbo
ball as the volume of a standard ball? standard ball: jumbo ball: A
jumbo ball is about 3.7 times as great in volume as a standard
ball.
Slide 203
Example 2B A hummingbird eyeball has a diameter of
approximately 0.6 cm. How many times as great is the volume of a
human eyeball as the volume of a hummingbird eyeball? hummingbird:
human: The human eyeball is about 72.3 times as great in volume as
a hummingbird eyeball.
Slide 204
In the figure, the vertex of the pyramid is at the center of
the sphere. The height of the pyramid is approximately the radius r
of the sphere. Suppose the entire sphere is filled with n pyramids
that each have base area B and height r.
Slide 205
4 r 2 nB If the pyramids fill the sphere, the total area of the
bases is approximately equal to the surface area of the sphere S,
so 4 r 2 S. As the number of pyramids increases, the approximation
gets closer to the actual surface area.
Slide 206
Slide 207
Example 3A: Finding Surface Area of Spheres Find the surface
area of a sphere with diameter 76 cm. Give your answers in terms of
. S = 4 r 2 S = 4 (38) 2 = 5776 cm 2 Surface area of a sphere
Slide 208
Example 3B: Finding Surface Area of Spheres Find the volume of
a sphere with surface area 324 in 2. Give your answers in terms of
. Substitute 324 for S. 324 = 4 r 2 r = 9Solve for r. Substitute 9
for r. The volume of the sphere is 972 in 2. S = 4 r 2 Surface area
of a sphere
Slide 209
Example 3C: Finding Surface Area of Spheres Find the surface
area of a sphere with a great circle that has an area of 49 mi 2.
Substitute 49 for A. 49 = r 2 r = 7 Solve for r. S = 4 r 2 = 4 (7)
2 = 196 mi 2 Substitute 7 for r. A = r 2 Area of a circle
Slide 210
Example 3D Find the surface area of the sphere. Substitute 25
for r. S = 2500 cm 2 S = 4 r 2 S = 4 (25) 2 Surface area of a
sphere
Slide 211
Example 4: Exploring Effects of Changing Dimensions The radius
of the sphere is multiplied by. Describe the effect on the volume.
original dimensions: radius multiplied by : Notice that. If the
radius is multiplied by, the volume is multiplied by, or.
Slide 212
Example 4 The radius of the sphere is divided by 3. Describe
the effect on the surface area. original dimensions: dimensions
divided by 3: The surface area is divided by 9. S = 4 r 2 = 4 (3) 2
= 36 m 3 S = 4 r 2 = 4 (1) 2 = 4 m 3
Slide 213
Example 5: Finding Surface Areas and Volumes of Composite
Figures Find the surface area and volume of the composite figure.
Give your answer in terms of . Step 1 Find the surface area of the
composite figure. The surface area of the composite figure is the
sum of the curved surface area of the hemisphere, the lateral area
of the cylinder, and the base area of the cylinder.
Slide 214
Example 5 Continued The surface area of the composite figure is
L(cylinder) = 2 rh = 2 (6)(9) = 108 in 2 B(cylinder) = r 2 = (6) 2
= 36 in 2 72 + 108 + 36 = 216 in 2. Find the surface area and
volume of the composite figure. Give your answer in terms of .
Slide 215
Step 2 Find the volume of the composite figure. Example 5
Continued Find the surface area and volume of the composite figure.
Give your answer in terms of . The volume of the composite figure
is the sum of the volume of the hemisphere and the volume of the
cylinder. The volume of the composite figure is 144 + 324 = 468 in
3.
Slide 216
Example 5B Find the surface area and volume of the composite
figure. Step 1 Find the surface area of the composite figure. The
surface area of the composite figure is the sum of the curved
surface area of the hemisphere, the lateral area of the cylinder,
and the base area of the cylinder.
Slide 217
Example 5 Continued The surface area of the composite figure is
Find the surface area and volume of the composite figure.
L(cylinder) = 2 rh = 2 (3)(5) = 30 ft 2 B(cylinder) = r 2 = (3) 2 =
9 ft 2 18 + 30 + 9 = 57 ft 2.
Slide 218
Step 2 Find the volume of the composite figure. Find the
surface area and volume of the composite figure. Example 5
Continued The volume of the composite figure is the volume of the
cylinder minus the volume of the hemisphere. V = 45 18 = 27 ft
3
Slide 219
B Find each measurement. Give your answers in terms of . 1. the
volume and surface area of the sphere 2. the volume and surface
area of a sphere with great circle area 36 in 2 3. the volume and
surface area of the hemisphere V = 36 cm 3 ; S = 36 cm 2 V = 288 in
3 ; S = 144 in 2 V = 23,958 ft 3 ; S = 3267 ft 2
Slide 220
C 4. A sphere has radius 4. If the radius is multiplied by 5,
describe what happens to the surface area. 5. Find the volume and
surface area of the composite figure. Give your answer in terms of
. The surface area is multiplied by 25. V = 522 cm 3 ; S = 267 cm
2