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Activating Prior Knowledge

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Presentation on theme: "Activating Prior Knowledge"— Presentation transcript:

1 Activating Prior Knowledge
M7:LSN20 Lesson 20: Truncated Cones Activating Prior Knowledge Find the area of the following right triangle: Find the length of the chord Tie to LO

2 Learning Objective Today, we will know that a truncated cone or pyramid is the solid obtained by removing the top portion of a cone or a pyramid above a plane parallel to its base and be able to find their volumes. CFU

3 CFU Opening Exercise Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Opening Exercise Examine the bucket below. It has a height of 𝟗 inches and a radius at the top of the bucket of 𝟒 inches Describe the shape of the bucket. What is it similar to? The bucket is cone-shaped but not a cone or that it is cylinder-shaped but tapered Estimate the volume of the bucket. Any estimate between 48𝜋 in 3 (the volume of a cone with the given dimensions) and 144𝜋 in 3 (the volume of a cylinder with the given dimensions) is reasonable. Then, continue with the discussion below. CFU

4 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Since the triangles are similar, we will let 𝑥 inches represent the height of the cone that has been removed. 4 10 = 𝑥 𝑥+8 The 4 is the radius of the small cone. The 10 is the radius of the large cone. The 𝑥 represents the height of the small cone. The expression 𝑥+8 represents the height of the large cone. CFU

5 CFU Concept Development Lesson 20: Truncated Cones Solve for x
M7:LSN20 Lesson 20: Truncated Cones Concept Development Solve for x 4 𝑥+8 =10𝑥 4𝑥+32=10𝑥 32=6𝑥 32 6 =𝑥 5. 3 =𝑥 CFU

6 1 3 𝜋 10 2 13.3 − 1 3 𝜋 4 2 5.3 CFU Concept Development
M7:LSN20 Lesson 20: Truncated Cones Concept Development Now we can find the volume of the large cone, find the volume of the small cone that was removed, and then subtract the volumes. What will be left is the volume of the truncated cone. The volume of the truncated cone is given by the expression 1 3 𝜋 − 1 3 𝜋 1 3 𝜋 is the volume of the large cone, and 1 3 𝜋 is the volume of the smaller cone. The difference in the volumes will be the volume of the truncated cone. CFU

7 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development The volume of the small cone is 𝑉≈ 1 3 𝜋 ≈ 1 3 𝜋 ≈ 𝜋. The volume of the large cone is 𝑉≈ 1 3 𝜋 ≈ 𝜋. CFU

8 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development The volume of the truncated cone is 𝜋− 𝜋= − 𝜋 = 𝜋. The volume of the truncated cone is approximately 𝜋 in 3 . CFU

9 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated cone. Write a proportion that will allow you to determine the height of the cone that has been removed. Explain what all parts of the proportion represent. 𝟔 𝟏𝟐 = 𝒙 𝒙+𝟒 Let 𝒙 𝐜𝐦 represent the height of the small cone. Then, 𝒙+𝟒 is the height of the large cone (with the removed part included). The 𝟔 represents the base radius of the removed cone, and the 𝟏𝟐 represents the base radius of the large cone. CFU

10 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated cone. Solve your proportion to determine the height of the cone that has been removed. 𝟔 𝒙+𝟒 =𝟏𝟐𝒙 𝟔𝒙+𝟐𝟒=𝟏𝟐𝒙 𝟐𝟒=𝟔𝒙 𝟒=𝒙 CFU

11 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated cone. Write an expression that can be used to determine the volume of the truncated cone. Explain what each part of the expression represents. 𝟏 𝟑 𝝅 (𝟏𝟐) 𝟐 𝟖 − 𝟏 𝟑 𝝅 𝟔 𝟐 𝟒 The expression 𝟏 𝟑 𝝅 𝟏𝟐 𝟐 𝟖 is the volume of the large cone, and 𝟏 𝟑 𝝅 𝟔 𝟐 𝟒 is the volume of the small cone. The difference of the volumes gives the volume of the truncated cone. CFU

12 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated cone. Calculate the volume of the truncated cone The volume of the small cone is 𝑽= 𝟏 𝟑 𝝅 𝟔 𝟐 𝟒 = 𝟏𝟒𝟒 𝟑 𝝅. The volume of the large cone is 𝑽= 𝟏 𝟑 𝝅 𝟏𝟐 𝟐 𝟖 = 𝟏𝟏𝟓𝟐 𝟑 𝝅. CFU

13 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated cone. Calculate the volume of the truncated cone The volume of the truncated cone is 𝟏𝟏𝟓𝟐 𝟑 𝝅− 𝟏𝟒𝟒 𝟑 𝝅= 𝟏𝟏𝟓𝟐 𝟑 − 𝟏𝟒𝟒 𝟑 𝝅 = 𝟏𝟎𝟎𝟖 𝟑 𝝅 =𝟑𝟑𝟔𝝅. The volume of the truncated cone is 𝟑𝟑𝟔𝝅 𝐜𝐦 𝟑 . CFU

14 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated pyramid with a square base Write a proportion that will allow you to determine the height of the cone that has been removed. Explain what all parts of the proportion represent. 𝟏 𝟓 = 𝒙 𝒙+𝟐𝟐 Let 𝒙 𝐦 represent the height of the small pyramid. Then 𝒙+𝟐𝟐 is the height of the large pyramid. The 𝟏 represents half of the length of the base of the small pyramid, and the 𝟓 represents half of the length of the base of the large pyramid. CFU

15 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated pyramid with a square base Solve your proportion to determine the height of the pyramid that has been removed. 𝒙+𝟐𝟐=𝟓𝒙 𝟐𝟐=𝟒𝒙 𝟓.𝟓=𝒙 CFU

16 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated pyramid with a square base Write an expression that can be used to determine the volume of the truncated pyramid. Explain what each part of the expression represents. 𝟏 𝟑 𝟏𝟎𝟎 𝟐𝟕.𝟓 − 𝟏 𝟑 𝟒 𝟓.𝟓 The expression 𝟏 𝟑 𝟏𝟎𝟎 𝟐𝟕.𝟓 is the volume of the large pyramid, and 𝟏 𝟑 𝟒 𝟓.𝟓 is the volume of the small pyramid. The difference of the volumes gives the volume of the truncated pyramid. CFU

17 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated pyramid with a square base Calculate the volume of the truncated pyramid. The volume of the small pyramid is 𝑽= 𝟏 𝟑 𝟒 𝟓.𝟓 = 𝟐𝟐 𝟑 . The volume of the large pyramid is 𝑽= 𝟏 𝟑 𝟏𝟎𝟎 𝟐𝟕.𝟓 = 𝟐𝟕𝟓𝟎 𝟑 . CFU

18 CFU Concept Development Lesson 20: Truncated Cones
M7:LSN20 Lesson 20: Truncated Cones Concept Development Find the volume of the truncated pyramid with a square base The volume of the truncated pyramid is 𝟐𝟕𝟓𝟎 𝟑 − 𝟐𝟐 𝟑 = 𝟐𝟕𝟐𝟖 𝟑 . The volume of the truncated pyramid is 𝟐𝟕𝟐𝟖 𝟑 𝐦 𝟑 . CFU

19 Closure Homework – Module 7 page 111-112 Problem Set #1-3 CFU
What did we learn today? How do you find the volume of a truncated cone? Homework – Module 7 page Problem Set #1-3 CFU

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