Holt Geometry 11.4 Surface Area of Pyramids & Cones Learn and apply the formula for the surface area of a pyramid. Learn and apply the formula for the.

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Presentation transcript:

Holt Geometry 11.4 Surface Area of Pyramids & 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 Geometry 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 Geometry 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 Geometry

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

Holt Geometry 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 Geometry 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 Geometry 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 Geometry 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 Geometry 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.

Holt Geometry 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 = πrℓLateral area of a cone = π (9)(5) = 45π cm 2 Substitute 9 for r and 5 for ℓ. S = πrℓ + πr 2 Surface area of a cone = 45π + π (9) 2 = 126π cm 2 Substitute 5 for ℓ and 9 for r.

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

Holt Geometry Example 3: Finding Surface Area of Composite Three- Dimensional Figures Find the surface area of the composite figure. The lateral area of the cone is L = πr l = π (6)(12) = 72π in 2. Left-hand cone: Right-hand cone: Using the Pythagorean Theorem, l = 10 in. The lateral area of the cone is L = πr l = π (6)(10) = 60π in 2.

Holt Geometry Example 3 Continued Composite figure: S = (left cone lateral area) + (right cone lateral area) Find the surface area of the composite figure. = 60π in π in 2 = 132π in 2

Holt Geometry 11.4 Assignment Day 1 ex. 6 – 8, 10, 12, 14 – 20, 22 – 24, 28 – 30

Holt Geometry You try! 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 Geometry You try! 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 Geometry Step 3 Find the surface area. Surface area of a regular pyramid You try! 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 Geometry You try! Example 2 Find the lateral area and surface area of the right cone. Use the Pythagorean Theorem to find ℓ. L = πr ℓ = π (8)(10) = 80π cm 2 Lateral area of a right cone Substitute 8 for r and 10 for ℓ. S = πr ℓ + πr 2 Surface area of a cone = 80π + π (8) 2 = 144π cm 2 Substitute 8 for r and 10 for ℓ. ℓ

Holt Geometry You try! Example 3 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

Holt Geometry You try! Example 3 Continued Surface Area of Pyramid without base: Surface Area of Composite: Surface of Composite = SA of Cube + SA of Pyramid

Holt Geometry Example 4: 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.

Holt Geometry 3 in. 2 in. Example 4 Continued original dimensions: base edge length and slant height divided by 5: S = Pℓ + B

Holt Geometry Example 4 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.

Holt Geometry You try! Example 4 The base edge length and slant height of the regular square pyramid are both multiplied by. Describe the effect on the surface area.

Holt Geometry You try! Example 4 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

Holt Geometry 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 Geometry 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 Geometry You try! 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 Geometry You try! 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 Geometry 11.4 Assignment Day 2 ex. 9, 11, 21, 25 – 27, 34, 37, 39, 40, 42, 44

Holt Geometry Lesson Quiz: 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 3. 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.

Holt Geometry 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 L = 216 ft 2 ; S = 297 ft 2 L = 180 cm 2 ; S ≈ cm 2