1 Ch10 L7
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1 Ch10 L7

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1 Ch10 L7 1 Ch10 L7 Presentation Transcript

  • Warm Up Lesson Presentation Problem of the Day 10-7 Surface Area Course 1
  • 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 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 ft 2 or a square with perimeter of 84 ft? A square with a perimeter of 84 ft; by 12 ft Course 1 10-7 Surface Area
  • Learn to find the surface areas of prisms, pyramids, and cylinders . Course 1 10-7 Surface Area
  • Vocabulary surface area net Insert Lesson Title Here 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. 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. 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 S = 10 + 60 + 24 + 60 + 24 + 10 = 188 The surface area is 188 in 2 . Add the areas of each face. 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. Course 1 10-7 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 = 286 Add the areas of each face. The surface area is 286 cm 2 . Course 1 10-7 Surface Area
  • Try This : Example 1A 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. 3 in. 11 in. 6 in. 11 in. 6 in. 6 in. 3 in. 3 in. 3 in. 3 in. A B C D E F Course 1 10-7 Surface Area
  • Try This : 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 The surface area is 234 in 2 . 11 in. 6 in. 6 in. 3 in. 3 in. 3 in. 3 in. A B C D E F Add the areas of each face. 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. 6 cm 10 cm 8 cm top front side Find the area of the front, top, and side, and multiply each by 2 to include the opposite faces. Course 1 10-7 Surface Area
  • Try This : Example 1B Continued Front : 10  8 = 80 Top : 10  6 = 60 Side : 8  6 = 48 80  2 = 160 60  2 = 120 48  2 = 96 S = 160 + 120 + 96 = 376 Add the areas of each face. The surface area is 376 cm 2 . 6 cm 10 cm 8 cm top front side 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. 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 = 49 + 4  28 S = 49 + 112 S = 161 The surface area is 161 ft 2 . Substitute. S = s 2 + 4  ( bh ) 1 2 __ S = 7 2 + 4  (  7  8 ) 1 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) S = 25 + 4  25 S = 25 + 100 S = 125 The surface area is 125 ft 2 . 5 ft 5 ft 10 ft 10 ft 5 ft Substitute. S = s 2 + 4  ( bh ) 1 2 __ S = 5 2 + 4  (  5  10 ) 1 2 __ 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 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. S = area of lateral surface + 2  (area of each base) Substitute. S = h  (2  r ) + 2  (  r 2 ) S = 7  (2    4 ) + 2  (   4 2 ) ft 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( 3.14 ) + 2  16( 3.14 ) S  7  25.12 + 2  50.24 The surface area is about 276.32 ft 2 . Use 3.14 for  . S  175.84 + 100.48 S  276.32 S = 7  8  + 2  16  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. S = area of lateral surface + 2  (area of each base) Substitute. S = h  (2  r ) + 2  (  r 2 ) S = 9  (2    6 ) + 2  (   6 2 ) 6 ft 9 ft 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( 3.14 ) + 2  36( 3.14 ) S  9  37.68 + 2  113.04 The surface area is about 565.2 ft 2 . Use 3.14 for  . S  339.12 + 226.08 S  565.2 S = 9  12  + 2  36  Course 1 10-7 Surface Area
  • 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. Insert Lesson Title Here 214 ft 2 188.4 ft 2 208 ft 2 Course 1 10-7 Surface Area