1. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Warm Up Warm Up California Standards California Standards Lesson Presentation Lesson PresentationPreview

2. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Warm Up 1. A rectangular prism is 0.6 m by 0.4 m by 1.0 m. What is the surface area? 2. A cylindrical can has a diameter of 14 cm and a height of 20 cm. What is the surface area to the nearest tenth? Use 3.14 for . 2.48 m 2 1186.9 cm 2

3. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Extension of MG2.1 Use formulas routinely for finding the perimeter and area of basic two-dimensional figures and the surface area and volume of basic three- dimensional figures, including rectangles, parallelograms, trapezoids, squares, triangles, circles, prisms, and cylinders. California Standards

4. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Vocabulary slant height regular pyramid right cone

5. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones The slant height of a pyramid or cone is measured along its lateral surface. In a right cone, a line perpendicular to the base through the vertex passes through the center of the base. The base of a regular pyramid is a regular polygon, and the lateral faces are all congruent. Right cone Regular Pyramid

6. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones

7. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Additional Example 1: Finding Surface Area Find the surface area of the figure to the nearest tenth. Use 3.14 for . = 20.16 ft 2 S = B + Pl 1212 = (2.4 2.4) + (9.6)(3) 1212

8. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Check It Out! Example 1 = (3 3) + (12)(5) 1212 B. S = r 2 + rl = 39 m 2 = (7 2 ) + (7)(18) = 175  549.5 ft 2 5 m 3 m 7 ft 18 ft A. S = B + Pl 1212 Find the surface area of each figure to the nearest tenth. Use 3.14 for .

9. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Additional Example 2: Exploring the Effects of Changing Dimensions A cone has diameter 8 in. and slant height 3 in. Explain whether tripling only the slant height would have the same effect on the surface area as tripling only the radius. Use 3.14 for . They would not have the same effect. Tripling the radius would increase the surface area more than tripling the slant height.

10. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Check It Out! Example 2 Original Dimensions Triple the Slant HeightTriple the Radius S = r 2 + rl = (4.5) 2 + (4.5)(2) = 29.25in 2  91.8 in 2 S = r 2 + r(3l) = (4.5) 2 + (4.5)(6) = 47.25in 2  148.4 in 2 S = r) 2 + r)l = (13.5) 2 + (13.5)(2) = 209.25in 2  657.0 in 2 A cone has diameter 9 in. and a slant height 2 in. Explain whether tripling only the slant height would have the same effect on the surface area as tripling only the radius. Use the 3.14 for . They would not have the same effect. Tripling the radius would increase the surface area more than tripling the height.

11. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Additional Example 3: Application The upper portion of an hourglass is approximately an inverted cone with the given dimensions. What is the lateral surface area of the upper portion of the hourglass? = (10)(26)  816.8 mm 2 Pythagorean Theorem Lateral surface area L = rl a 2 + b 2 = l 2 10 2 + 24 2 = l 2 l = 26

12. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Check It Out! Example 3 A large road construction cone is almost a full cone. With the given dimensions, what is the lateral surface area of the cone? = (9)(15)  424.1 in 2 12 in. 9 in. Pythagorean Theorem a 2 + b 2 = l 2 9 2 + 12 2 = l 2 l = 15 Lateral surface area L = rl

13. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones Lesson Quiz: Part I Find the surface area of each figure to the nearest tenth. Use 3.14 for . 1. the triangular pyramid 2. the cone 175.8 in 2 6.2 m 2

14. Holt CA Course 1 10-5 Surface Area of Pyramids and Cones 3. Tell whether doubling the dimensions of a cone will double the surface area. Lesson Quiz: Part II It will more than double the surface area because you square the radius to find the area of the base.