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Holt McDougal Geometry 10-5 Surface Area of Pyramids and Cones 10-5 Surface Area of Pyramids and...

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Holt McDougal Geometry 10-5 Surface Area of Pyramids and Cones 10-5 Surface Area of Pyramids and Cones Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Geometry
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Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones10-5Surface Area of Pyramids and Cones

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt McDougal Geometry

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Warm UpFind 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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Example 1B Continued

Step 2 Find the lateral area.

Lateral area of a regular pyramid

Substitute 60 for P and 16 for ℓ.

Find the lateral area and surface area of the regular pyramid.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Check It Out! Example 1

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Check It Out! Example 1 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 ℓ.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Step 3 Find the surface area.

Surface area of a regular pyramid

Check It Out! Example 1 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.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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 perpendicularsegment from the vertex of the cone to the plane of the base.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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.

S = rℓ + r2 Surface area of a cone

= 136 + (8)2

= 200 in2

Substitute 8 for r and 17 for ℓ.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Check It Out! Example 2

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

Substitute 8 for r and 10 for ℓ.

S = rℓ + r2 Surface area of a cone

= 80 + (8)2

= 144 cm2

Substitute 8 for r and 10 for ℓ.

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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?

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Check It Out! Example 5

What if…? If the radius of the large circle were 12 in., what would be the radius of the cone?

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Check It Out! Example 5 Continued

Substitute 12 for ℓ, the slant height of the cone and the radius of the large circle.

r = 9 in. Solve for r.

The radius of the cone is 9 in.

What if…? If the radius of the large circle were 12 in., what would be the radius of the cone?

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

Lesson Quiz: Part I

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

Holt McDougal Geometry

10-5 Surface Area of Pyramids and Cones

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 ft2

Lesson Quiz: Part II


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