12.1
Exploring Solids Goals p Use properties of polyhedra. p Use Euler’s Theorem. VOCABULARY Polyhedron A polyhedron is a solid that is bounded by polygons that enclose a single region of space. Face The faces of a polyhedron are polygons. Edge An edge of a polyhedron is a line segment formed by the intersection of two faces of the polyhedron. Vertex A vertex of a polyhedron is a point where three or more edges of the polyhedron meet. Regular polyhedron A regular polyhedron is a polyhedron whose faces are all congruent regular polygons. Convex polyhedron A convex polyhedron is a polyhedron such that any two points on its surface can be connected by a line segment that lies entirely inside or on the polyhedron. Cross section A cross section is the intersection of a plane and a solid. Platonic solids A Platonic solid is one of five regular polyhedra: a regular tetrahedron, a cube, a regular octahedron, a regular dodecahedron, and a regular icosahedron. These solids are named after Plato, a Greek mathematician and philosopher. Tetrahedron A tetrahedron is a polyhedron with four faces. Octahedron An octahedron is a polyhedron with eight faces. Dodecahedron A dodecahedron is a polyhedron with twelve faces. Icosahedron An icosahedron is a polyhedron with twenty faces.
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TYPES OF SOLIDS Of the five solids below, the prism and pyramid are polyhedra. The cylinder, sphere, and cone are not polyhedra.
Prism
Example 1
Pyramid
Cylinder
Sphere
Cone
Identifying Polyhedra
Decide whether the solid is a polyhedron. If so, count the number of faces, vertices, and edges of the polyhedron. a.
b.
c.
Solution a. This is a polyhedron. It has 6 faces, 8 vertices, and 12 edges. b. This is a polyhedron. It has 8 faces, 12 vertices, and 18 edges. c. This is not a polyhedron. Some of the faces are not polygons.
Example 2
Classifying Polyhedra
Is the polyhedron convex? Is it regular? a.
b.
convex, regular
c.
convex, nonregular
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nonconvex, nonregular
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THEOREM 12.1: EULER’S THEOREM The number of faces (F), vertices (V ), and edges (E) of a polyhedron are related by the formula F V E 2 .
Example 3
Using Euler’s Theorem
The solid has 10 faces: 8 trapezoids and 2 octagons. How many vertices does the solid have? On their own, 8 trapezoids and 2 octagons have 8( 4 ) 2( 8 ) 48 sides. In the solid, each side is shared by exactly two polygons. So the number of edges is 24 . Use Euler’s Theorem to find the number of vertices. FVE2
Write Euler’s Theorem.
10 V 24 2
Substitute.
V 16
Solve for V.
Answer The solid has 16 vertices.
Checkpoint Is the solid a polyhedron? If so, is it convex? Is it regular?
1.
2.
Yes, convex; nonregular
3.
No
Yes, nonconvex; nonregular
4. Critical Thinking Is it possible for a polyhedron to have 16 faces, 34 vertices, and 50 edges? Explain. No; From Euler’s Theorem, the number of faces (F), vertices (V), and edges (E) of a polyhedron are related by the formula F V E 2.
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12.2
Surface Area of Prisms and Cylinders Goal
p Find the surface area of a prism and of a cylinder.
VOCABULARY Prism A prism is a polyhedron with two congruent faces, called bases, that lie in parallel planes. The other faces, called lateral faces, are parallelograms formed by connecting the corresponding vertices of the bases. The segments connecting these vertices are lateral edges. Right prism In a right prism, each lateral edge is perpendicular to both bases. Oblique prisms Oblique prisms are prisms that have lateral edges that are not perpendicular to the bases. The length of the oblique lateral edges is the slant height of the prism. Surface area of a polyhedron The surface area of a polyhedron is the sum of the areas of its faces. Lateral area of a polyhedron The lateral area of a polyhedron is the sum of the areas of its lateral faces. Net A net is a two-dimensional representation of all of the faces of a polyhedron. Cylinder A cylinder is a solid with congruent circular bases that lie in parallel planes. The altitude, or height, of a cylinder is the perpendicular distance between its bases. The radius of the base is also called the radius of the cylinder. Right cylinder A cylinder such that the segment joining the centers of the bases is perpendicular to the bases Lateral area of a cylinder The lateral area of a cylinder is the area of its curved surface. The lateral area is equal to the product of the circumference and the height, which is 2πrh. Surface area of a cylinder The sum of the lateral area and the areas of the two bases
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THEOREM 12.2: SURFACE AREA OF A RIGHT PRISM The surface area S of a right prism can be found using the formula S 2B Ph , where B is the area of a base, P is the perimeter of a base, and h is the height.
Example 1
Using Theorem 12.2
Find the surface area of the right prism. 6 in.
Solution Each base is an equilateral triangle with a side length, s, of 6 inches. Using the formula for the area of an equilateral triangle, the area of each base is
6 in. 8 in. 6 in.
1 1 (s2) 3 ( 6 2) 9 3 in.2 B 3 4
4
The perimeter of each base is P 18 in. and the height is h 8 in.
6 in.
6 in. 6 in.
Answer So, the surface area is S 2B Ph 2( 9 3 ) 18 ( 8 ) ≈ 175 in.2
Checkpoint Find the surface area of the right prism.
1. 5m 4m 11 m
238 m2
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THEOREM 12.3: SURFACE AREA OF A RIGHT CYLINDER The surface area S of a right cylinder is
B r 2
S 2B Ch 2πr 2 2πrh ,
C 2r
where B is the area of a base, C is the circumference of a base, r is the radius of a base, and h is the height.
Example 2
h r
Finding the Surface Area of a Cylinder
Find the surface area of the right cylinder. Solution Each base has a radius of 4 meters, and the cylinder has a height of 5 meters. S 2πr2 2πrh
5m
4m
Formula for surface area of a cylinder
2π( 4 )2 2π( 4 )( 5 )
Substitute.
32 π 40 π
Simplify.
72 π
Add.
≈ 226.19
Use a calculator.
Answer The surface area is about 226 square meters.
Checkpoint Find the surface area of the right cylinder. Round your result to two decimal places.
2. 9 ft
6 ft
565.49 ft2
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12.3
Surface Area of Pyramids and Cones Goal
p Find the surface area of a pyramid and of a cone.
VOCABULARY Pyramid A pyramid is a polyhedron in which the base is a polygon and the lateral faces are triangles with a common vertex. The intersection of two lateral faces is a lateral edge. The intersection of the base and a lateral face is a base edge. The altitude, or height, of the pyramid is the perpendicular distance between the base and the vertex.
Regular pyramid A regular pyramid has a regular polygon for a base and its height meets the base at its center. The slant height of a regular pyramid is the altitude of any lateral face. A nonregular pyramid does not have a slant height.
Circular cone or cone A circular cone, or cone, has a circular base and a vertex that is not in the same plane as the base. The altitude, or height, is the perpendicular distance between the vertex and the base.
Right cone In a right cone, the height meets the base at its center and the slant height is the distance between the vertex and a point on the base edge.
Lateral surface of a cone The lateral surface of a cone consists of all segments that connect the vertex with points on the base edge.
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Example 1 A regular pyramid is considered a regular polyhedron only if all its faces, including the base, are congruent. So, the only pyramid that is a regular polyhedron is the regular triangular pyramid, or tetrahedron.
Finding the Area of a Lateral Face
Find the area of each lateral face of the regular pyramid shown at the right.
h 97 m
Solution To find the slant height of the pyramid, use the Pythagorean Theorem.
s 90 m
1 2 (Slant height)2 h2 s 2
Write formula.
(Slant height)2 97 2 45 2
Substitute.
(Slant height)2 11,434
Simplify.
slant height
1 s 2
Slant height 11,43 4
Take the positive square root.
Slant height ≈ 106.93
Use a calculator.
Answer So, the area of each lateral face is 1 1 (base of lateral face)(slant height), or about ( 90 )( 106.93 ), 2 2 which is about 4812 square meters.
Checkpoint Complete the following exercise.
1. Find the area of a lateral face of the regular pyramid. Round the result to one decimal place.
h 11 in.
slant height
40.4 in.2 s 7 in.
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1 s 2
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THEOREM 12.4: SURFACE AREA OF A REGULAR PYRAMID 1 The surface area S of a regular pyramid is S B Pl , where 2 B is the area of the base, P is the perimeter of the base, and l is the slant height.
Example 2
Finding the Surface Area of a Pyramid
To find the surface area of the regular pyramid shown, start by finding the area of the base. Use the formula for the area of a regular 1 polygon, (apothem)(perimeter). 2
18 in.
8 in.
A diagram of the base is shown at the right. After substituting, the area of the base is 1 (43 )(6 p 8 ), or 96 3 square inches. 2
4 3 in. 4 3 in.
8 in.
Now you can find the surface area, using for the area of the base, B. 96 3 1 S B Pl 2
Write formula.
1 ( 48 )( 18 ) 96 3 2
Substitute.
432 96 3
Simplify.
≈ 598.3
Use a calculator.
Answer So, the surface area is about 598.3 square inches.
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THEOREM 12.5: SURFACE AREA OF A RIGHT CONE The surface area S of a right cone is S πr 2 πrl ,
l
where r is the radius of the base and l is the slant height.
r
Finding the Surface Area of a Right Cone
Example 3
To find the surface area of the right cone shown, use the formula for the surface area. S πr2 πrl
5m
Write formula.
π( 3 )2 π( 3 )( 5 )
Substitute.
9 π 15 π
Simplify.
24 π
Add.
3m
Answer The surface area is 24 π square meters, or about 75.4 square meters.
Checkpoint Find the surface area of the solid. Round your result to two decimal places.
2. Regular pyramid
3. Right cone
30 mm 10 ft 6 ft 20 mm
1600 mm2
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301.60 ft2
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12.4
Volume of Prisms and Cylinders Goals p Use volume postulates. p Find the volumes of prisms and cylinders. VOCABULARY Volume of a solid The volume of a solid is the number of cubic units contained in the solid’s interior.
POSTULATE 27: VOLUME OF A CUBE The volume of a cube is the cube of the length of its side, or V s3 . POSTULATE 28: VOLUME CONGRUENCE POSTULATE If two polyhedra are congruent, then they have the same volume . POSTULATE 29: VOLUME ADDITION POSTULATE The volume of a solid is the sum of the volumes of all its nonoverlapping parts. THEOREM 12.6: CAVALIERI’S PRINCIPLE If two solids have the same height and the same cross-sectional area at every level, then they have the same volume . THEOREM 12.7: VOLUME OF A PRISM The volume V of a prism is V Bh , where B is the area of a base and h is the height. THEOREM 12.8: VOLUME OF A CYLINDER The volume V of a cylinder is V Bh πr 2h , where B is the area of a base, h is the height, and r is the radius of a base. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
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Example 1
Finding Volumes
Find the volume of the right prism and the right cylinder. a.
3 ft
b.
1 ft
5m
6m
2 ft
Solution 1 3 a. The area B of the base is ( 1 )( 3 ), or ft 2. Use h 2 to 2 2 find the volume. 3 V Bh ( 2 ) 3 ft 3 2 b. The area B of the base is π p 5 2, or 25 π m 2. Use h 6 to find the volume. V Bh 25 π( 6 ) 150 π ≈ 471.24 m 3
Checkpoint Find the volume of the solid. Round your result to two decimal places.
1. Right prism
2. Right cylinder 6 ft 5m
9 ft 4m 11 m
220 m3
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1017.88 ft3
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Using Volumes
Example 2
Use the measurements given to solve for x. a. Cube, V 90 ft3
b. Right cylinder, V 1253 m3 xm x ft 10 m
x ft x ft
Solution a. A side length of the cube is x feet. V s3 90 x 3 4.48 ≈ x
Formula for volume of cube Substitute. Take the cube root.
Answer So, the height, width, and length of the cube are about 4.48 feet. b. The area of the base is πx 2 square meters. V Bh
Formula for volume of cylinder
1253 πx 2( 10 )
Substitute.
1253 10 πx 2
Rewrite.
1253 10 π
x2
39.88 ≈ x 2 6.32 ≈ x
Divide each side by 10 π. Simplify. Find the positive square root.
Answer So, the radius of the cylinder is about 6.32 meters.
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12.5
Volume of Pyramids and Cones Goal
p Find the volume of pyramids and cones.
THEOREM 12.9: VOLUME OF A PYRAMID 1 The volume V of a pyramid is V Bh , where 3 B is the area of the base and h is the height.
h
B
THEOREM 12.10: VOLUME OF A CONE 1 1 The volume V of a cone is V Bh πr 2h , 3 3 where B is the area of the base, h is the height, and r is the radius of the base.
Example 1
h B r
Finding the Volume of a Pyramid
Find the volume of the pyramid with the regular base. 10 in.
Solution The base can be divided into six equilateral triangles. Using the formula for the area of an 1 p s 2, the area of the equilateral triangle, 3 4 base B can be found as follows: 1 1 6 p 3 p s2 6 p 3 p 4 2 24 3 in.2 4 4 Use Theorem 12.9 to find the volume of the pyramid. 1 1 V Bh ( 243 )( 10 ) 80 3 3 3 Answer The volume of the pyramid is 80 3 , or about 138.6 cubic inches.
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4 in.
4 in.
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Example 2
Finding the Volume of a Cone
Find the volume of each cone. a. Right circular cone
b. Oblique circular cone
16.8 mm 12 ft 9.5 mm
4.5 ft
Solution a. Use the formula for the volume of a cone. 1 V Bh Formula for volume of cone 3 1 (πr2)h 3
Base area equals πr 2.
1 (π 9.5 2) 16.8 3
Substitute.
505.4 π
Simplify.
Answer The volume of the cone is 505.4 π, or about 1588 cubic millimeters. b. Use the formula for the volume of a cone. 1 V Bh Formula for volume of cone 3 1 (πr 2)h 3
Base area equals πr 2.
1 (π 4.5 2) 12 3
Substitute.
81 π
Simplify.
Answer The volume of the cone is 81 π, or about 254 cubic feet.
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Checkpoint Find the volume of the solid. Round your result to two decimal places.
1. Pyramid with regular base
2. Right circular cone
15 cm
10 ft 6 ft 9 cm
1052.22 cm3
Example 3
301.60 ft3
Using the Volume of a Cone
Use the given measurements to solve for x. 8m
Solution 1 V (πr 2)h 3
Formula for volume
1 135 (πx 2)( 8 ) 3
Substitute.
405 8 πx 2
Multiply each side by 3 .
16.11 ≈ x 2
Divide each side by 8π .
4.01 ≈ x
Find positive square root.
x Volume 135 m3
Answer The radius of the cone is about 4.01 meters. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
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12.6
Surface Area and Volume of Spheres Goals p Find the surface area of a sphere. p Find the volume of a sphere. VOCABULARY Sphere A sphere is the locus of points in space that are a given distance from a point. The point is called the center of the sphere.
Radius of a sphere A radius of a sphere is a segment from the center to a point on the sphere.
Chord of a sphere A chord of a sphere is a segment whose endpoints are on the sphere.
Diameter of a sphere A diameter is a chord that contains the center of the sphere.
Great circle A great circle is the intersection of a sphere and a plane that contains the center of the sphere.
Hemisphere Half of a sphere, formed when a great circle separates a sphere into two congruent halves
THEOREM 12.11: SURFACE AREA OF A SPHERE The surface area S of a sphere with radius r is S 4πr 2 . Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
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Example 1
Finding the Surface Area of a Sphere
Find the surface area. When the radius doubles, does the surface area double? a.
b.
3 cm
6 cm
Solution a. S 4πr 2 4π( 3 )2 36 π cm2 b. S 4πr 2 4π( 6 )2 144 π cm2 The surface area of the sphere in part (b) is four times greater than the surface area of the sphere in part (a) because 36 π p 4 144 π. Answer When the radius of a sphere doubles, the surface area does not double .
Example 2
Using a Great Circle
The circumference of a great circle of a sphere is 7.4π feet. What is the surface area of the sphere? Solution Begin by finding the radius of the sphere. C 2πr 7.4π 2πr 3.7 r
Formula for circumference of circle Substitute for C. Divide each side by 2π.
Using a radius of 3.7 feet, the surface area is S 4πr 2 4π( 3.7 )2 54.76 π ft2. Answer The surface area of the sphere is 54.76 π ft2, or about 172 ft2.
THEOREM 12.12: VOLUME OF A SPHERE 4 The volume V of a sphere with radius r is V πr 3 . 3
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Example 3
Finding the Volume of a Sphere 5 cm
What is the radius of a sphere made from the cylinder of modeling clay shown? Assume the sphere has the same volume as the cylinder.
15 cm
Solution To find the volume of the cylinder of modeling clay, use the formula for the volume of a cylinder.
Cylinder of modeling clay
r
V πr 2h π( 5 )2( 15 ) 375 π cm3 To find the radius of the sphere, use the formula for the volume of a sphere and solve for r. 4 V πr 3 3 4 375 π πr 3 3
Formula for volume of sphere Substitute for V.
1125 π 4πr 3
Multiply each side by 3 .
281.25 r 3
Divide each side by 4π .
6.55 ≈ r
Sphere made from cylinder of modeling clay
Use a calculator to take the cube root.
Answer The radius of the sphere is about 6.55 centimeters.
Checkpoint Find the surface area and volume of the sphere. Round your results to two decimal places.
1.
5 ft
314.16 ft2; 523.60 ft3
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2.
6.5 m
530.93 m2; 1150.35 m3
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12.7
Similar Solids Goals p Find and use the scale factor of similar solids. p Use similar solids to solve real-life problems. VOCABULARY Similar solids Two solids with equal ratios of corresponding linear measures, such as heights or radii, are called similar solids.
Identifying Similar Solids
Example 1
Decide whether the two solids are similar. If so, compare the volumes of the solids. a.
b. 7
6
8
14 3 5
3
5
3
5 10
6
Solution a. The solids are not similar because the ratios of corresponding linear measures are not equal, as shown. 5 lengths: 3
5 widths: 3
8 4 heights: 6 3
b. The solids are similar because the ratios of corresponding linear measures are equal, as shown. The solids have a scale factor of 2 : 1 . 10 2 6 2 lengths: widths: 5 1 3 1
14 2 heights: 7 1
The volume of the larger prism is V Bh 60 ( 14 ) 840 . The volume of the smaller prism is V Bh 15 ( 7 ) 105 . The ratio of side lengths is 2 : 1 and the ratio of volumes is 840 : 105 , or 8 : 1 . Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
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Checkpoint Decide whether the two solids are similar.
1.
2.
5m
10 ft
15 m
5m 4 ft 5 ft
7.5 m
3 ft
not similar
similar
THEOREM 12.13: SIMILAR SOLIDS THEOREM If two similar solids have a scale factor of a : b, then corresponding areas have a ratio of a2 : b2 , and corresponding volumes have a ratio of a3 : b3 .
Example 2
Using the Scale Factor of Similar Solids
Cylinders A and B are similar with a scale factor of 2 : 5. Find the surface area and volume of cylinder B given that the surface area of cylinder A is 96π square feet and the volume of cylinder A is 128π cubic feet.
B A
Solution Begin by using Theorem 12.13 to set up two proportions. Surface area of A a2 Surface area of B b2
Volume of A a3 Vo lume of B b3
96π 4 Surface area of B 25
128π 8 Volume of B 125
Surface area of B 600π
Volume of B 2000π
Answer The surface area of cylinder B is 600π square feet and the volume of cylinder B is 2000π cubic feet. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved.
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Example 3
Finding the Scale Factor of Similar Solids
The two cones are similar. Find the scale factor. Solution Find the ratio of the two volumes.
V 108 cm3 V 256 cm3
a3 108π b3 256π
Write ratio of volumes.
a3 27 3 b 64
Simplify.
a 3 b 4
Find the cube root.
Answer The two cones have a scale factor of 3 : 4 .
Example 4
Comparing Similar Solids
Two punch bowls are similar with a scale factor of 2 : 3. The amount of concentrate to be added is proportional to the volume. How much concentrate does the smaller bowl require if the larger bowl requires 48 ounces? Solution Using the scale factor, the ratio of the volume of the smaller punch bowl to the larger punch bowl is a3 b3
2 3
8 ≈ 27 3 3
1
.
3.4
The ratio of the volumes of the concentrates is about 1 : 3.4 . The amount of concentrate for the smaller punch bowl can be found by multiplying the amount of concentrate for the larger punch bowl by 1 3.4
as follows: 48
1
3.4
≈ 14.1 ounces.
Answer The smaller bowl requires about 14.1 ounces of concentrate.
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