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Holt Geometry
10-3 Formulas in Three Dimensions
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.
CN#3 Objectives
Holt Geometry
10-3 Formulas in Three Dimensions
polyhedron space
Vocabulary
Holt Geometry
10-3 Formulas in Three Dimensions
A polyhedron is formed by four or more polygons that intersect only at their edges. Prisms and pyramids are polyhedrons, but cylinders and cones are not.
Holt Geometry
10-3 Formulas in Three Dimensions 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
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Holt Geometry
10-3 Formulas in Three Dimensions Example 1B: Using Euler’s Formula
Find the number of vertices, edges, and faces of the polyhedron. Use your results to verify Euler’s formula.
V = 5, E = 8, F = 5
Use Euler’s Formula.
Simplify.
5 – 8 + 5 = 2 ?
2 = 2
Holt Geometry
10-3 Formulas in Three Dimensions 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 + w2 = x2, and x2 + h2 = d2. Using substitution, 2 + w2 + h2 = d2.
Holt Geometry
10-3 Formulas in Three Dimensions
Holt Geometry
10-3 Formulas in Three Dimensions 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.
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Holt Geometry
10-3 Formulas in Three Dimensions 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.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
I will learn and apply the formula for the surface area of a prism and a cylinder. Modeling with Geometry
Modeling with Geometry G.MG.1 Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).*
G.MG.2 Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).
G.MG.3 Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios).*
CN#4 Objectives
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
lateral face lateral edge right prism oblique prism altitude surface area lateral surface axis of a cylinder right cylinder oblique cylinder
Vocabulary
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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.
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Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
The surface area of a right rectangular prism with length ℓ, width w, and height h can be written as S = 2ℓw + 2wh + 2ℓh.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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!
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Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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 π.
L = 2πrh = 2π(8)(10) = 160π in2
The radius is half the diameter, or 8 ft.
S = L + 2πr2 = 160π + 2π(8)2
= 288π in2
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 2B: Finding Lateral Areas and Surface Areas of Right Cylinders
Find the lateral area and surface area of a right cylinder with circumference 24π cm and a height equal to half the radius. Give your answers in terms of π.
Step 1 Use the circumference to find the radius.
C = 2πr Circumference of a circle
24π = 2πr Substitute 24π for C. r = 12 Divide both sides by 2π.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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 = 2πrh = 2π(12)(6) = 144π cm2
S = L + 2πr2 = 144π + 2π(12)2
= 432π in2
Lateral area
Surface area
Find the lateral area and surface area of a right cylinder with circumference 24π cm and a height equal to half the radius. Give your answers in terms of π.
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 3: Finding Surface Areas of Composite Three-Dimensional Figures
Find the surface area of the composite figure.
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Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Always round at the last step of the problem. Use the value of π given by the π key on your calculator.
Remember!
Holt Geometry
10-4 Surface Area of Prisms and Cylinders
Example 4: Exploring Effects of Changing Dimensions
The edge length of the cube is tripled. Describe the effect on the surface area.
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Holt Geometry
10-4 Surface Area of Prisms and Cylinders
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