geometry

Geometry3 Dimension

3D representation

Classify three-dimensional figures according to their properties.

Use nets and cross sections to analyze three-dimensional figures.

Objectives

faceedgevertexprismcylinderpyramidconecubenetcross section

Vocabulary

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.

A cube is a prism with six square faces. Other prisms and pyramids are named for the shape of their bases.

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:

Example 1B: Classifying Three-Dimensional Figures


pentagonal pyramid

vertices: A, B, C, D, E, F

base: ABCDE

edges:

Example 1c


vertex: N

cone

edges: none

base: •M

M

Example 1d


triangular prism

bases: ∆TUV, ∆WXY

vertices: T, U, V, W, X, Y

edges:

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.

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.

Example 2B: Identifying a Three-Dimensional Figure From a 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.

Example 2c


The net has four congruent triangular faces. So the net forms a triangular pyramid.

Example 2d


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.

A cross section is the intersection of a three-dimensional figure and a plane.

Example 3A: Describing Cross Sections of Three-Dimensional Figures

Describe the cross section.

The cross section is a point.

Example 3B: Describing Cross Sections of Three-Dimensional Figures


The cross section is a pentagon.

Example 3c


The cross section is a hexagon.

Example 3d


The cross section is a triangle.

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.

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.

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.

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:

2. Describe the three-dimensional figure that can be made from this net.

square pyramid

Question 2

3. Describe the cross section.

a rectangle

Question 3

Draw representations of three-dimensional figures.

Recognize a three dimensional figure from a given representation.

Objectives

orthographic drawingisometric drawingperspective drawingvanishing pointhorizon

Vocabulary

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.

Example 1: Drawing Orthographic Views of an Object

Draw all six orthographic views of the given object. Assume there are no hidden cubes.

Example 1 Continued


Bottom

Example 1 Continued


Example 1B


Example 1B Continued

Isometric drawing is a way to show three sides ofa 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 arepeating triangular pattern.

Example 2: Drawing an Isometric View of an Object

Draw an isometric view of the given object. Assume there are no hidden cubes.

Example 2 B

Draw an isometric view of the given object. Assume there are no hidden cubes.

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.

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

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 .

Example 3A Continued


From each corner of the , lightly draw dashed segments to the vanishing point.



Lightly draw a smaller with vertices on the greyed segments. This is the back of the .



Draw the edges of the , using dashed segments for hidden edges. Erase any segments that are not part of the .

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.

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 .


Lightly draw dashed segments from the endpoints of each new vertical segment to the vanishing points.


Draw the edges of the , using dashed segments for hidden edges. Erase any segments that are not part of the .


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

From each corner of the L, lightly draw dashed segments to the vanishing point.


Example 3C Continued

Lightly draw a smaller L with vertices on the dashed segments. This is the back of the L.



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 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

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


Lightly draw dashed segments from the endpoints of each new vertical segment to the vanishing points.



Draw the edges of the L, using dashed segments for hidden edges to the vanishing points.



Erase any segments that are not part of the L.



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.

Example 4B: Relating Different Representations of an Object


Yes; the drawing is a two-point perspective view of the object.

Example 4C: Relating Different Representations of an Object


Yes; the drawing is an isometric view of the object.

Example 4D: Relating Different Representations of an Object


Yes; the drawing shows the six orthographic views of the object.

Example 4E

Determine whether the drawing represents thegiven object. Assume there are no hidden cubes.

no

A

1. Draw all six orthographic views of the object. Assume there are no hidden cubes.

2. Draw an isometric view of the object.

A

3. Determine whether each drawing represents the given object. Assume there are no hidden cubes.

yes

yesno

A

Warm UpFind 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 ft2 and bases 3 ft and 9 ft

7.5 m

3 ft

Apply Euler’s 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

polyhedronspace

Vocabulary

polyhedron - formed by four or more polygons that intersect only at their edges.

Prisms and pyramids are polyhedrons, but cylinders and cones are not.

Example 1A: Using Euler’s Formula

Find the number of vertices, edges, and faces of the polyhedron. Use your results to verify Euler’s formula.

V = 12, E = 18, F = 8

Use Euler’s Formula.

Simplify.

12 – 18 + 8 = 2 ?

2 = 2

Example 1B: Using Euler’s Formula


V = 5, E = 8, F = 5


Simplify.

5 – 8 + 5 = 2 ?

2 = 2

Example 1C


V = 6, E = 12, F = 8


Simplify. 2 = 2

6 – 12 + 8 = 2 ?

Example 1D


V = 7, E = 12, F = 7


Simplify. 2 = 2

7 – 12 + 7 = 2 ?

A diagonal of a three-dimensional figure connectstwo vertices of two different faces. Diagonal dof a rectangular prism is shown in the diagram.By the Pythagorean Theorem, 2 + w2 = x2, andx2 + h2 = d2. Using substitution, 2 + w2 + h2 = d2.

Box Problem

Box ProblemSolution

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.

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 + h2

h2 = 32

Substitute 15 for d, 12 for l, and 7 for w.

Square both sides of the equation.

Simplify.

Solve for h2.

Take the square root of both sides.

Example 2C

Find the length of the diagonal of a cube with edge length 5 cm.

d2 = 25 + 25 + 25

d2 = 75

Substitute 5 for each side.

Square both sides of the equation.

Simplify.

Solve for d2.

Take the square root of both sides.

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.

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)

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.

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.

You can find the distance between the two points (x1, y1, z1) and (x2, y2, z2) 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 (x1, y1, z1) and (x2, y2, z2) is related to the Midpoint Formula. (See Lesson 1-6.)

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:


midpoint:

M(1, 4, 2.5)


(0, 0, 0) and (2, 8, 5)

Example 4B: Finding Distances and Midpoints in Three Dimensions


(6, 11, 3) and (4, 6, 12)

distance:


midpoint:


(6, 11, 3) and (4, 6, 12)

M(5, 8.5, 7.5)

Example 4C


(0, 9, 5) and (6, 0, 12)

distance:

midpoint:


(0, 9, 5) and (6, 0, 12)


M(3, 4.5, 8.5)

Example 4D


(5, 8, 16) and (12, 16, 20)

distance:

midpoint:

Example 4D extended


(5, 8, 16) and (12, 16, 20)

M(8.5, 12, 18)

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).

Example 5 Continued

Use the Distance Formula to find the distance between the drivers.

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.

A

1. Find the number of vertices, edges, and faces of the polyhedron. Use your results to verify Euler’s formula.

V = 8; E = 12; F = 6;8 – 12 + 6 = 2

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)

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

lateral facelateral edgeright prismoblique prismaltitudesurface arealateral surfaceaxis of a cylinderright cylinderoblique cylinder

Vocabulary

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.

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 curvedsurfaces of a three-dimensional figure. The lateralarea of a prism is the sum of the areas of the lateral faces.

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 theperimeter of the base of the prism.

The surface area of a right rectangular prism with length ℓ, width w, and height h can be written asS = 2ℓw + 2wh + 2ℓh.

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!

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 ft2

S = Ph + 2B

= 448 + 2(7)(9) = 574 ft2

P = 2(9) + 2(7) = 32 ft

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 ft2

S = Ph + 2B

P = 3(10) = 30 cm

The base area is

Example 1C

Find the lateral area and surface area of a cube with edge length 8 cm.L = Ph

= 32(8) = 256 cm2

S = Ph + 2B

= 256 + 2(8)(8) = 384 cm2

P = 4(8) = 32 cm

•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.

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 = 2rh = 2(8)(10) = 160 in2

The radius is half the diameter, or 8 ft.

S = L + 2r2 = 160 + 2(8)2

= 288 in2

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 = 2r Circumference of a circle

24 = 2r Substitute 24 for C.

r = 12 Divide both sides by 2.


Step 2 Use the radius to find the lateral area and surface area. The height is half the radius, or 6 cm.

L = 2rh = 2(12)(6) = 144 cm2

S = L + 2r2 = 144 + 2(12)2

= 432 in2

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 .

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.


A = r2

49 = r2

r = 7

Area of a circle

Substitute 49 for A.

Divide both sides by and take the square root.

Step 2 Use the radius to find the lateral area and surface area. The height is twice the radius, or 14 cm.

L = 2rh = 2(7)(14)=196 in2

S = L + 2r2 = 196 + 2(7)2 =294 in2

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 3: Finding Surface Areas of Composite Three-Dimensional Figures

Find the surface area of the composite figure.

Example 3 Continued

Two copies of the rectangular prism base are removed. The area of the base is B = 2(4) = 8 cm2.

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

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 cm2

Example 3B

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 cm2.

The surface area of the cylinder is

S =Ph + 2B = 2(2)(3) + 2(2)2 = 20 ≈ 62.8 cm2.

The surface area of the composite figure is the sum of the areas of all surfaces on the exterior of the figure.

S = (rectangular surface area) +

(cylinder surface area) – 2(cylinder base area)

S = 202 + 62.8 — 2()(22) = 239.7 cm2



Always round at the last step of the problem.

Use the value of given by the key on your calculator.

Remember!

Example 4: Exploring Effects of Changing Dimensions

The edge length of the cube is tripled. Describe the effect on the surface area.

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 32, or 9.

S = 6ℓ2

= 6(8)2 = 384 cm2

S = 6ℓ2

= 6(24)2 = 3456 cm2

24 cm

Example 4B

The height and diameter of the cylinder are

multiplied by .

Describe the effect on the

surface area.

original dimensions: height and diameter halved:

S = 2(112) + 2(11)(14)

= 550 cm2

S = 2(5.52) + 2(5.5)(7)

= 137.5 cm2

11 cm

7 cm


Notice than 550 = 4(137.5).

If the dimensions are halved, the surface area is multiplied by

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?

Example 5 Continued

Pup tent:

Tunnel tent:

The tunnel tent requires less nylon.

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 cm2, 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.

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 cm2 and a height that is the radius

L = 400 cm2 ; S = 600 cm2

L = 720 in2; S 1052.6 in2

L 301.6 cm2; S = 1206.4 cm2

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 m2

PracticeFind 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

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

vertex of a pyramidregular pyramidslant height of a regular pyramidaltitude of a pyramidvertex of a coneaxis of a coneright coneoblique coneslant height of a right conealtitude of a cone

Vocabulary

•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.

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.

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 = 142 = 196 cm2

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


Step 2 Find the lateral area.


Substitute 60 for P and 16 for ℓ.



Step 3 Find the surface area.


Substitute for B.


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



Step 2 Find the lateral area.


Substitute 18 for P and 10 for ℓ.

Step 3 Find the surface area.




Substitute for B.

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.

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.

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 = rℓ Lateral area of a cone

= (9)(5) = 45 cm2 Substitute 9 for r and 5 for ℓ.

S = rℓ + r2 Surface area of a cone

= 45 + (9)2 = 126 cm2 Substitute 5 for ℓ and 9 for r.

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.

L = rℓ

= (8)(17)

= 136 in2

Lateral area of a right cone

Substitute 8 for r and 17 for ℓ.


= 136 + (8)2

= 200 in2


Example 2C

Find the lateral area and surface area of the right cone.

Use the Pythagorean Theorem to find ℓ.

L = rℓ

= (8)(10)

= 80 cm2

Lateral area of a right cone



= 80 + (8)2

= 144 cm2


ℓ


The base edge length and slant height of the regular hexagonal pyramid are both divided by 5.

Describe the effect on the surface area.

3 in.

2 in.

Example 3 Continued

original dimensions: base edge length and slant height divided by 5:

S = Pℓ + B 12

S = Pℓ + B 12

in

in

in

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 52, or 25.

in2

in2

in2

in2

Example 3B

The base edge length and slant height of the regular square pyramid are both multiplied by .



original dimensions: multiplied by two-thirds:

By multiplying the dimensions by two-thirds, the surface area was multiplied by .

8 ft8 ft

10 ft

S = Pℓ + B 12

= 260 cm2

S = Pℓ + B 12

= 585 cm2ft2 ft2

Example 4: Finding Surface Area of Composite Three-Dimensional Figures


The lateral area of the cone isL = rl = (6)(12) = 72 in2.

Left-hand cone:

Right-hand cone:

Using the Pythagorean Theorem, l = 10 in. The lateral area of the cone isL = rl = (6)(10) = 60 in2.

Example 4 Continued

Composite figure:

S = (left cone lateral area) + (right cone lateral area)


= 60 in2 + 72 in2 = 132 in2

Example 4B


Surface Area of Cube without the top side:

S = 4wh + B

S = 4(2)(2) + (2)(2) = 20 yd2

Example 4b Continued

Surface Area of Pyramid without base:

Surface Area of Composite:

Surface of Composite = SA of Cube + SA of Pyramid

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

Example 5 Continued

Substitute 4 for ℓ, 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?

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 ft2; S = 297 ft2

L = 180 cm2; S ≈ 242.4 cm2

4. A right cone has radius 3 and slant height 5. The radius and slant height are both multiplied by .


5. Find the surface area of the composite figure. Give your answer in terms of .

The surface area is multiplied by .

S = 24 ft2

5C

PracticeFind the volume of each figure. Round to the nearest tenth, if necessary.

1. a square prism with base area 189 ft2 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 ft3

14,964.9 m3

4423.4 in3

Learn and apply the formula for the volume of a pyramid.

Learn and apply the formula for the volume of a cone.

Objectives

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.

The square pyramids are congruent, so they have the same volume. The volume of each pyramid is one third the volume of the cube.

Example 1A: Finding Volumes of Pyramids

Find the volume a rectangular pyramid with length 11 m, width 18 m, and height 23 m.

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.

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.


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 ft3

Find the volume of the regular hexagonal pyramid with height equal to the apothem of the base

Simplify.

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.


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 cm3 Simplify.

An art gallery is a 6-story square pyramid with base area of acre (1 acre = 4840 yd2, 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 yd2. The base edge length is . The height is about 6(10) = 60 ft or about 20 yd.

Example 2 Continued

Substitute 2420 for B and 20 for h.

16,133 yd3 16,100 yd3

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 ft3. Use the conversion factor to find the volume in cubic feet.

Example 2B

What if…? What would be the volume of the Rainforest Pyramid if the height were doubled?


Substitute 70 for B and 66 for h.

= 107,800 yd3

or 107,800(27) = 2,910,600 ft3

= 245 cm3 ≈ 769.7 cm3

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.

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 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.

2r = 25 Substitute 25 for the circumference.

r = 12.5 Solve for r.


Step 3 Use the radius and height to find the volume.


Substitute 12.5 for r and 27 for h.

= 1406.25 in3 ≈ 4417.9 in3 Simplify.

Find the volume of a cone with base circumference 25 in. and a height 2 in. more than twice the radius.

Example 3C: Finding Volumes of Cones

Find the volume of a cone.

Step 1 Use the Pythagorean Theorem to find the height.

162 + h2 = 342 Pythagorean Theorem

h2 = 900 Subtract 162 from both sides.

h = 30 Take the square root of both sides.


Step 2 Use the radius and height to find the volume.

Volume of a cone


2560 cm3 8042.5 cm3 Simplify.

Find the volume of a cone.

Example 3D

Find the volume of the cone.

Volume of a cone


≈ 216 m3 ≈ 678.6 m3 Simplify.


original dimensions:

radius and height divided by 3:

Notice that .

If the radius and height are divided by 3, the volume is divided by 33, or 27.

The diameter and height of the cone are divided by 3. Describe the effect on the volume.

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.

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

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.

Vcylinder = r2h = (21)2(35)=15,435 cm3.

V = 5145 + 15,435 + 5,880 = 26,460 83,126.5 cm3

Example 5B

Find the volume of the composite figure.

The volume of the rectangular prism is

V = ℓwh = 25(12)(15) = 4500 ft3.

The volume of the pyramid is

The volume of the composite is the rectangular prism subtract the pyramid.

4500 — 1500 = 3000 ft3

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 cm3

207.8 in3

V 3801.3 cm3

V 117.3 m2

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 yd3

PracticeFind 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 cm2

12.5 cm

85 in.

169 ft2

50 cm

Learn and apply the formula for the volume of a sphere.Learn and apply the formula for the surface area of a sphere.

Objectives

spherecenter of a sphereradius of a spherehemispheregreat circle

Vocabulary

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

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 Cavalieri’s Principle. Y

The height of the hemisphere is equal to the radius.

V(hemisphere) = V(cylinder) – V(cone)

The volume of a sphere with radius r is twice the volume of the hemisphere, or .

Example 1A: Finding Volumes of Spheres

Find the volume of the sphere. Give your answer in terms of .

= 2304 in3 Simplify.

Volume of a sphere.

Example 1B: Finding Volumes of Spheres

Find the diameter of a sphere with volume 36,000 cm3.

Substitute 36,000 for V.

27,000 = r3

r = 30

d = 60 cm d = 2r

Take the cube root of both sides.

Volume of a sphere.

Example 1C: Finding Volumes of Spheres

Find the volume of the hemisphere.

Volume of a hemisphere

Substitute 15 for r.

= 2250 m3 Simplify.

Example 1D

Find the radius of a sphere with volume 2304 ft3.

Volume of a sphere

Substitute for V.

r = 12 ft Simplify.

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.

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.

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.

4r2 ≈ 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 4r2 ≈ S. As the number of pyramids increases, the approximation gets closer to the actual surface area.

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 = 4r2

S = 4(38)2 = 5776 cm2

Surface area of a sphere

Example 3B: Finding Surface Area of Spheres

Find the volume of a sphere with surface area 324 in2. Give your answers in terms of .

Substitute 324 for S.324 = 4r2

r = 9 Solve for r.

Substitute 9 for r.

The volume of the sphere is 972 in2.

S = 4r2 Surface area of a sphere

Example 3C: Finding Surface Area of Spheres

Find the surface area of a sphere with a great circle that has an area of 49 mi2.

Substitute 49 for A.49 = r2

r = 7 Solve for r.

S = 4r2

= 4(7)2 = 196 mi2 Substitute 7 for r.

A = r2 Area of a circle

Example 3D

Find the surface area of the sphere.

Substitute 25 for r.

S = 2500 cm2

S = 4r2

S = 4(25)2

Surface area of a sphere


The radius of the sphere is multiplied by . Describe the effect on the volume.


radius multiplied by :

Notice that . If the radius is multiplied by ,

the volume is multiplied by , or .

Example 4

The radius of the sphere is divided by 3. Describe the effect on the surface area.


dimensions divided by 3:

The surface area is divided by 9.

S = 4r2

= 4(3)2 = 36 m3

S = 4r2

= 4(1)2 = 4 m3

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.

Example 5 Continued

The surface area of the composite figure is

L(cylinder) = 2rh = 2(6)(9) = 108 in2

B(cylinder) = r2 = (6)2 = 36 in2

72 + 108 + 36 = 216 in2.


Step 2 Find the volume of the composite figure.

Example 5 Continued


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 in3.

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.

Example 5 Continued

The surface area of the composite figure is


L(cylinder) = 2rh = 2(3)(5) = 30 ft2

B(cylinder) = r2 = (3)2 = 9 ft2

18 + 30 + 9 = 57 ft2.

Step 2 Find the 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 ft3

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 in2

3. the volume and surface areaof the hemisphere

V = 36 cm3; S = 36 cm2

V = 288 in3;S = 144 in2

V = 23,958 ft3;S = 3267 ft2

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 cm3; S = 267 cm2

geometry

Documents

net forms

given net

equilateral triangular

congruent triangular

circular faces

number of faces

congruent square faces

d representation