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Holt 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.

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Presentation on theme: "Holt 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 transcript:

1 Holt 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

2 Holt Geometry 10-5 Surface Area of Pyramids and Cones Warm Up Find the missing side length of each right triangle with legs a and b and hypotenuse c. 1. a = 7, b = c = 15, a = 9 3. b = 40, c = a = 5, b = 5 5. a = 4, c = 8 c = 25 b = 12 a = 9

3 Holt 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

4 Holt Geometry 10-5 Surface Area of Pyramids and Cones 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

5 Holt 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.

6 Holt 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.

7 Holt Geometry 10-5 Surface Area of Pyramids and Cones

8 Holt 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 = 14 2 = 196 cm 2

9 Holt 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

10 Holt 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.

11 Holt 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.

12 Holt 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

13 Holt 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.

14 Holt 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.

15 Holt 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.

16 Holt 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 perpendicular segment from the vertex of the cone to the plane of the base.

17 Holt 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 cm 2 Substitute 9 for r and 5 for ℓ. S = rℓ + r 2 Surface area of a cone = 45 + (9) 2 = 126 cm 2 Substitute 5 for ℓ and 9 for r.

18 Holt 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 = r ℓ = (8)(17) = 136 in 2 Lateral area of a right cone Substitute 8 for r and 17 for ℓ. S = r ℓ + r 2 Surface area of a cone = 136 + (8) 2 = 200 in 2 Substitute 8 for r and 17 for ℓ.

19 Holt 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 = r ℓ = (8)(10) = 80 cm 2 Lateral area of a right cone Substitute 8 for r and 10 for ℓ. S = r ℓ + r 2 Surface area of a cone = 80 + (8) 2 = 144 cm 2 Substitute 8 for r and 10 for ℓ. ℓ

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

21 Holt Geometry 10-5 Surface Area of Pyramids and Cones 3 in. 2 in. Example 3 Continued original dimensions: base edge length and slant height divided by 5: S = Pℓ + B

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

23 Holt Geometry 10-5 Surface Area of Pyramids and Cones Check It Out! Example 3 The base edge length and slant height of the regular square pyramid are both multiplied by. Describe the effect on the surface area.

24 Holt Geometry 10-5 Surface Area of Pyramids and Cones Check It Out! Example 3 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 = Pℓ + B 1212 = 260 cm 2 S = Pℓ + B 1212 = 585 cm 2

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

26 Holt Geometry 10-5 Surface Area of Pyramids and Cones Example 4 Continued Composite figure: S = (left cone lateral area) + (right cone lateral area) Find the surface area of the composite figure. = 60 in  in 2 = 132 in 2

27 Holt Geometry 10-5 Surface Area of Pyramids and Cones Check It Out! Example 4 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

28 Holt Geometry 10-5 Surface Area of Pyramids and Cones Check It Out! Example 4 Continued Surface Area of Pyramid without base: Surface Area of Composite: Surface of Composite = SA of Cube + SA of Pyramid

29 Holt 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

30 Holt 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?

31 Holt 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

32 Holt 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?

33 Holt 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 ft 2 ; S = 297 ft 2 L = 180 cm 2 ; S ≈ cm 2

34 Holt 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 ft 2 Lesson Quiz: Part II


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