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9-8 Surface Area Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day Lesson Quizzes Lesson Quizzes.

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Presentation on theme: "9-8 Surface Area Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day Lesson Quizzes Lesson Quizzes."— Presentation transcript:

1 9-8 Surface Area Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day Lesson Quizzes Lesson Quizzes

2 9-8 Surface Area Warm Up Identify the figure described. 1. two parallel congruent faces, with the other faces being parallelograms 2. a polyhedron that has a vertex and a face at opposite ends, with the other faces being triangles prism pyramid

3 9-8 Surface Area Problem of the Day Which figure has the longer side and by how much: a square with an area of 81 ft 2 or a square with perimeter of 84 ft? a square with a perimeter of 84 ft; by 12 ft

4 9-8 Surface Area Preview of MA.7.G.2.1 Justify and apply formulas for…surface area of…prisms, pyramids, cylinders… Sunshine State Standards

5 9-8 Surface Area Vocabulary surface area net

6 9-8 Surface Area The surface area of a three- dimensional figure is the sum of the areas of its surfaces. To help you see all the surfaces of a three-dimensional figure, you can use a net. A net is the pattern made when the surface of a three-dimensional figure is layed out flat showing each face of the figure.

7 9-8 Surface Area Additional Example 1A: Finding the Surface Area of a Prism Find the surface area S of the prism. Method 1: Use a net. Draw a net to help you see each face of the prism. Use the formula A = lw to find the area of each face.

8 9-8 Surface Area Additional Example 1A Continued A: A = 5  2 = 10 B: A = 12  5 = 60 C: A = 12  2 = 24 D: A = 12  5 = 60 E: A = 12  2 = 24 F: A = 5  2 = 10 S = 10 + 60 + 24 + 60 + 24 + 10 = 188 Add the areas of each face. The surface area is 188 in 2.

9 9-8 Surface Area Additional Example 1B: Finding the Surface Area of a Prism Find the surface area S of each prism. Method 2: Use a three-dimensional drawing. Find the area of the front, top, and side, and multiply each by 2 to include the opposite faces.

10 9-8 Surface Area Additional Example 1B Continued Front: 9  7 = 63 Top: 9  5 = 45 Side: 7  5 = 35 63  2 = 126 45  2 = 90 35  2 = 70 S = 126 + 90 + 70 = 286Add the areas of each face. The surface area is 286 cm 2.

11 9-8 Surface Area Check It Out: Example 1A Find the surface area S of the prism. Method 1: Use a net. Draw a net to help you see each face of the prism. Use the formula A = lw to find the area of each face. 3 in. 11 in. 6 in. 11 in. 6 in. 3 in. A B C DE F

12 9-8 Surface Area Check It Out: Example 1A A: A = 6  3 = 18 B: A = 11  6 = 66 C: A = 11  3 = 33 D: A = 11  6 = 66 E: A = 11  3 = 33 F: A = 6  3 = 18 S = 18 + 66 + 33 + 66 + 33 + 18 = 234 Add the areas of each face. The surface area is 234 in 2. 11 in. 6 in. 3 in. A B C DE F

13 9-8 Surface Area Check It Out: Example 1B Find the surface area S of each prism. Method 2: Use a three-dimensional drawing. Find the area of the front, top, and side, and multiply each by 2 to include the opposite faces. 6 cm 10 cm 8 cm top front side

14 9-8 Surface Area Check It Out: Example 1B Continued Side: 10  8 = 80 Top: 10  6 = 60 Front: 8  6 = 48 80  2 = 160 60  2 = 120 48  2 = 96 S = 160 + 120 + 96 = 376Add the areas of each face. The surface area is 376 cm 2. 6 cm 10 cm 8 cm top front side

15 9-8 Surface Area The surface area of a pyramid equals the sum of the area of the base and the areas of the triangular faces. To find the surface area of a pyramid, think of its net.

16 9-8 Surface Area Additional Example 2: Finding the Surface Area of a Pyramid Find the surface area S of the pyramid. S = area of square + 4  (area of triangular face) S = 49 + 4  28 S = 49 + 112 Substitute. S = s 2 + 4  ( bh) 1 2 __ S = 7 2 + 4  (  7  8) 1 2 __ S = 161 The surface area is 161 ft 2.

17 9-8 Surface Area Check It Out: Example 2 Find the surface area S of the pyramid. S = area of square + 4  (area of triangular face) S = 25 + 4  25 S = 25 + 100 Substitute. S = s 2 + 4  ( bh) 1 2 __ S = 5 2 + 4  (  5  10) 1 2 __ S = 125 The surface area is 125 ft 2. 5 ft 10 ft 5 ft


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