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Warm Up Find 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

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Objectives Learn and apply the formula for the surface area of a pyramid. Learn and apply the formula for the surface area of a cone.

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**The base of a regular pyramid is a regular polygon.**

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.

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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.

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**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

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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

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**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.

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**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 3 Find the surface area. Surface area of a regular pyramid Substitute for B.

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**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.

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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.

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**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.

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**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 Lateral area of a right cone = (8)(17) = 136 in2 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.

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Check It Out! Example 2 Find the lateral area and surface area of the right cone. Use the Pythagorean Theorem to find ℓ. ℓ L = rℓ Lateral area of a right cone = (8)(10) = 80 cm2 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.

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**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.

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**base edge length and slant height divided by 5: original dimensions:**

Example 3 Continued base edge length and slant height divided by 5: original dimensions: S = Pℓ + B 12 S = Pℓ + B 12

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**base edge length and slant height divided by 5: original dimensions:**

Example 3 Continued base edge length and slant height divided by 5: original dimensions: Notice that If the base edge length and slant height are divided by 5, the surface area is divided by 52, or 25.

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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.

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**Check It Out! Example 3 Continued**

8 ft 10 ft 8 ft original dimensions: multiplied by two-thirds: S = Pℓ + B 12 S = Pℓ + B 12 = 585 cm2 = 260 cm2 By multiplying the dimensions by two-thirds, the surface area was multiplied by .

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**Example 4: Finding Surface Area of Composite Three-Dimensional Figures**

Find the surface area of the composite figure. Left-hand cone: The lateral area of the cone is L = rl = (6)(12) = 72 in2. Right-hand cone: Using the Pythagorean Theorem, l = 10 in. The lateral area of the cone is L = rl = (6)(10) = 60 in2.

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Example 4 Continued Find the surface area of the composite figure. Composite figure: S = (left cone lateral area) + (right cone lateral area) = 60 in2 + 72 in2 = 132 in2

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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 yd2

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**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

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**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

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Example 5 Continued If the pattern shown is used to make a paper cup, what is the diameter of the cup? 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.

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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

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**Check It Out! Example 5 Continued**

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

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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 ≈ cm2

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Lesson Quiz: Part II 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

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