150 likes | 352 Views
Surface and Area of Pyramids and Cones. 6-9. Warm Up. Problem of the Day. Lesson Presentation. Course 3. Surface and Area of Pyramids and Cones. 6-9. Course 3. Warm Up 1. A rectangular prism is 0.6 m by 0.4 m by 1.0 m. What is the surface area?
E N D
Surface and Area of Pyramids and Cones 6-9 Warm Up Problem of the Day Lesson Presentation Course 3
Surface and Area of Pyramids and Cones 6-9 Course 3 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 m2 1186.9 cm2
Surface and Area of Pyramids and Cones 6-9 Course 3 Problem of the Day Sandy is building a model of a pyramid with a hexagonal base. If she uses a toothpick for each edge, how many toothpicks will she need? 12
Surface and Area of Pyramids and Cones 6-9 Course 3 Learn to find the surface area of pyramids and cones.
Surface and Area of Pyramids and Cones 6-9 Course 3 Insert Lesson Title Here Vocabulary slant height regular pyramid right cone
Surface and Area of Pyramids and Cones 6-9 Course 3 Regular Pyramid The slant height of a pyramid or cone is measured along its lateral surface. Right cone The base of a regular pyramid is a regular polygon, and the lateral faces are all congruent. In a right cone, a line perpendicular to the base through the tip of the cone passes through the center of the base.
Surface and Area of Pyramids and Cones 6-9 Course 3
Surface and Area of Pyramids and Cones 6-9 1 2 = (2.4 • 2.4) + (9.6)(3) Course 3 Additional Example 1: Finding Surface Area Find the surface area of each figure 1 2 A. S = B + Pl = 20.16 ft2 B. S = pr2 + prl = p(32) + p(3)(6) = 27p 84.8 cm2
Surface and Area of Pyramids and Cones 6-9 1 2 = (3 • 3) + (12)(5) Course 3 Try This: Example 1 Find the surface area of each figure. 1 2 5 m A. S = B + Pl = 39 m2 3 m 3 m B. S = pr2 + prl 18 ft = p(72) + p(7)(18) 7 ft = 175p 549.5 ft2
Surface and Area of Pyramids and Cones 6-9 Course 3 Additional Example 2: Exploring the Effects of Changing Dimensions A cone has diameter 8 in. and slant height 3 in. Explain whether tripling the slant height would have the same effect on the surface area as tripling the radius. They would not have the same effect. Tripling the radius would increase the surface area more than tripling the slant height.
Surface and Area of Pyramids and Cones 6-9 Course 3 Try This: Example 2 A cone has diameter 9 in. and a slant height 2 in. Explain whether tripling the slant height would have the same effect on the surface area as tripling the radius. S = pr2 + pr(3l) S = pr2 + prl S = p(3r)2 + p(3r)l = p(4.5)2 + p(4.5)(2) = p(4.5)2 + p(4.5)(6) = p(13.5)2 + p(13.5)(2) = 29.25p in2 91.8 in2 = 47.25p in2 148.4 in2 = 209.25p in2 657.0 in2 They would not have the same effect. Tripling the radius would increase the surface area more than tripling the height.
Surface and Area of Pyramids and Cones 6-9 Course 3 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? a2 + b2 = l2 Pythagorean Theorem 102 + 262 = l2 l 27.9 Lateral surface area L = prl =p(10)(27.9)876.1 mm2
Surface and Area of Pyramids and Cones 6-9 Course 3 Try This: Example 3 A road construction cone is almost a full cone. With the given dimensions, what is the lateral surface area of the cone? a2 + b2 = l2 Pythagorean Theorem 12 in. 42 + 122 = l2 4 in. l 12.65 Lateral surface area L = prl =p(4)(12.65)158.9 in2
Surface and Area of Pyramids and Cones 6-9 Course 3 Insert Lesson Title Here Lesson Quiz: Part 1 Find the surface area of each figure to the nearest tenth. Use 3.14 for p. 1. the triangular pyramid 2. the cone 6.2 m2 175.8 in2
Surface and Area of Pyramids and Cones 6-9 Course 3 Insert Lesson Title Here Lesson Quiz: Part 2 3. Tell whether doubling the dimensions of a cone will double the surface area. It will more than double the surface area because you square the radius to find the area of the base.