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Surface Area 10-7 Warm Up Problem of the Day Lesson Presentation

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Presentation on theme: "Surface Area 10-7 Warm Up Problem of the Day Lesson Presentation"— Presentation transcript:

1 Surface Area 10-7 Warm Up Problem of the Day Lesson Presentation
Course 1 Warm Up Problem of the Day Lesson Presentation

2 Surface Area 10-7 Warm Up Identify the figure described.
Course 1 10-7 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 Surface Area 10-7 Problem of the Day
Course 1 10-7 Surface Area Problem of the Day Which figure has the longer side and by how much, a square with an area of 81 ft2 or a square with perimeter of 84 ft? A square with a perimeter of 84 ft; by 12 ft

4 Course 1 10-7 Surface Area Learn to find the surface areas of prisms, pyramids, and cylinders.

5 Insert Lesson Title Here
Course 1 10-7 Surface Area Insert Lesson Title Here Vocabulary surface area net

6 Course 1 10-7 Surface Area The surface area of a solid figure is the sum of the areas of its surfaces. To help you see all the surfaces of a solid figure, you can use a net. A net is the pattern made when the surface of a solid figure is layed out flat showing each face of the figure.

7 Additional Example 1A: Finding the Surface Area of a Prism
Course 1 10-7 Surface Area Additional Example 1A: Finding the Surface Area of a Prism Find the surface area S of the prism. A. 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 Additional Example 1A Continued
Course 1 10-7 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 Add the areas of each face. S = = 188 The surface area is 188 in2.

9 Additional Example 1B: Finding the Surface Area of a Prism
Course 1 10-7 Surface Area Additional Example 1B: Finding the Surface Area of a Prism Find the surface area S of each prism. B. 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 Additional Example 1B Continued
Course 1 10-7 Surface Area Additional Example 1B Continued Front: 9  7 = 63 63  2 = 126 Top: 9  5 = 45 45  2 = 90 Side: 7  5 = 35 35  2 = 70 S = = 286 Add the areas of each face. The surface area is 286 cm2.

11 Surface Area 10-7 Try This: Example 1A
Course 1 10-7 Surface Area Try This: Example 1A Find the surface area S of the prism. A. Method 1: Use a net. A 3 in. 3 in. 6 in. 6 in. 3 in. 3 in. 6 in. 11 in. 11 in. B C D E F 3 in. Draw a net to help you see each face of the prism. Use the formula A = lw to find the area of each face.

12 Surface Area 10-7 Try This: Example 1A A: A = 6  3 = 18 A 3 in.
Course 1 10-7 Surface Area Try This: Example 1A A: A = 6  3 = 18 A 3 in. B: A = 11  6 = 66 3 in. 6 in. 6 in. 3 in. C: A = 11  3 = 33 11 in. D: A = 11  6 = 66 B C D E E: A = 11  3 = 33 F 3 in. F: A = 6  3 = 18 Add the areas of each face. S = = 234 The surface area is 234 in2.

13 Surface Area 10-7 Try This: Example 1B
Course 1 10-7 Surface Area Try This: Example 1B Find the surface area S of each prism. B. Method 2: Use a three-dimensional drawing. top side front 8 cm 10 cm 6 cm Find the area of the front, top, and side, and multiply each by 2 to include the opposite faces.

14 Try This: Example 1B Continued
Course 1 10-7 Surface Area Try This: Example 1B Continued top side front 8 cm 10 cm 6 cm Front: 10  8 = 80 80  2 = 160 Top:  6 = 60 60  2 = 120 Side:  6 = 48 48  2 = 96 S = = 376 Add the areas of each face. The surface area is 376 cm2.

15 Course 1 10-7 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 Additional Example 2: Finding the Surface Area of a Pyramid
Course 1 10-7 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 = s2 + 4  ( bh) 1 2 __ S =  (  7  8) 1 2 __ Substitute. S =  28 S = S = 161 The surface area is 161 ft2.

17 Surface Area 10-7 Try This: Example 2
Course 1 10-7 Surface Area Try This: Example 2 Find the surface area S of the pyramid. S = area of square + 4  (area of triangular face) 10 ft 5 ft S = s2 + 4  ( bh) 1 2 __ 5 ft S =  (  5  10) 1 2 __ Substitute. 10 ft S =  25 5 ft S = S = 125 The surface area is 125 ft2.

18 Course 1 10-7 Surface Area The surface area of a cylinder equals the sum of the area of its bases and the area of its curved surface. To find the area of the curved surface of a cylinder, multiply its height by the circumference of the base. Helpful Hint

19 Additional Example 3: Finding the Surface Area of a Cylinder
Course 1 10-7 Surface Area Additional Example 3: Finding the Surface Area of a Cylinder Find the surface area S of the cylinder. Use 3.14 for , and round to the nearest hundredth. ft S = area of lateral surface + 2  (area of each base) S = h  (2r) + 2  (r2) Substitute. S = 7  (2    4) + 2  (  42)

20 Additional Example 3 Continued
Course 1 10-7 Surface Area Additional Example 3 Continued Find the surface area S of the cylinder. Use 3.14 for , and round to the nearest hundredth. S = 7  8 + 2  16 S  7  8(3.14) + 2  16(3.14) Use 3.14 for . S  7   50.24 S  S  The surface area is about ft2.

21 Surface Area 10-7 Try This: Example 3
Course 1 10-7 Surface Area Try This: Example 3 Find the surface area S of the cylinder. Use 3.14 for , and round to the nearest hundredth. 6 ft 9 ft S = area of lateral surface + 2  (area of each base) S = h  (2r) + 2  (r2) Substitute. S = 9  (2    6) + 2  (  62)

22 Try This: Example 3 Continued
Course 1 10-7 Surface Area Try This: Example 3 Continued Find the surface area S of the cylinder. Use 3.14 for , and round to the nearest hundredth. S = 9  12 + 2  36 S  9  12(3.14) + 2  36(3.14) Use 3.14 for . S  9   S  S  565.2 The surface area is about ft2.

23 Insert Lesson Title Here
Course 1 10-7 Surface Area Insert Lesson Title Here Lesson Quiz Find the surface area of each figure. Use 3.14 for . 1. rectangular prism with base length 6 ft, width 5 ft, and height 7 ft 2. cylinder with radius 3 ft and height 7 ft 3. Find the surface area of the figure shown. 214 ft2 188.4 ft2 208 ft2


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