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Chapter Area, Pythagorean Theorem, and Volume 14 Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

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Presentation on theme: "Chapter Area, Pythagorean Theorem, and Volume 14 Copyright © 2013, 2010, and 2007, Pearson Education, Inc."— Presentation transcript:

1 Chapter Area, Pythagorean Theorem, and Volume 14 Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

2 14-4Surface Areas  Surface Area of Right Prisms  Surface Area of a Cylinder  Surface Area of a Pyramid  Surface Area of a Cone  Surface Area of a Sphere Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

3 Surface Area of Right Prisms Lateral area the sum of the areas of the lateral faces Surface area The sum of the lateral surface areas and the area of the bases. Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

4 Surface Area of aCube Cube Surface area = 6e 2 Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

5 Right pentagonal prism Surface area = ph + 2B, where p = the perimeter of the base, h is the height of the prism, and B is the area of the base. Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

6 Find the surface area of the prism. Example 14-14a Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

7 Surface Area of a Cylinder Right regular prismsRight circular cylinder Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

8 Surface Area of a Right Circular Cylinder Right circular cylinder Surface area = 2πr 2 + 2πrh Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

9 Surface Area of a Pyramid Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

10 Right regular pyramid Surface area = where n = the number of faces, is the slant height, and B is the area of the base. Surface Area of a Pyramid The formula for the surface area of a right regular pyramid can be simplified because nb is the perimeter, p, of the base. So, Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

11 Find the surface area of the pyramid. Example Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

12 Example The Great Pyramid of Cheops is a right square pyramid with a height of 148 m and a square base with perimeter of 940 m. The altitude of each triangular face is 189 m. The basic shape of the Transamerica Building in San Francisco is a right square pyramid that has a height of 260 m and a square base with a perimeter of 140 m. The altitude of each triangular face is 261 m. How do the lateral surface areas of the two structures compare? Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

13 Example (continued) The length of one side of the square base of the Great Pyramid is The lateral surface area of the Great Pyramid is Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

14 Example (continued) The length of one side of the square base of the Transamerica Building is The lateral surface area of the Transamerica Building is The lateral surface area of the Great Pyramid is approximately or 4.9 times as great as that of the Transamerica Building. Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

15 Surface Area of a Cone Cone Surface area = πr 2 + πr where r = radius of the cone and is the slant height. Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

16 Surface Area of a Cone Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

17 Find the surface area of the cone. Example Copyright © 2013, 2010, and 2007, Pearson Education, Inc.

18 Surface Area of a Sphere Sphere Surface area = 4πr 2 Copyright © 2013, 2010, and 2007, Pearson Education, Inc.


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