Warm-Up Geometry 1st Hour – Unit 10 Test Scores 103 94 87 77 65 103 94 87 77 65 102 93 83 76 56 100 93 82 75 37 97 90 82 75 97 89 82 72 Mean = _______ Median = _______ Mode = _______ Range = ________

Warm-Up Geometry 2nd Hour – Unit 10 Test Scores 103 92 86 65 31 103 92 86 65 31 103 89 85 56 99 87 83 52 96 87 82 35 92 86 79 34 Mean = _______ Median = _______ Mode = _______ Range = ________

Warm-Up Geometry 7th Hour – Unit 10 Test Scores 104 101 93 80 37 104 101 93 80 37 104 100 93 77 26 102 99 90 75 101 97 83 69 101 94 80 66 Mean = _______ Median = _______ Mode = _______ Range = ________

Surface Area of Prisms and Cylinders Section 12 – 2 Surface Area of Prisms and Cylinders

Vocabulary Polyhedron: A solid that is bounded by polygons, called faces, that enclose a single region of space. Prism: A polyhedron with two congruent faces, called bases, that lie in parallel planes. Lateral Faces: The faces of a prism that are parallelograms formed by connecting the corresponding vertices of the bases of the prism. Lateral Edges: The segments connecting the corresponding vertices of the bases of a prism.

Vocabulary Lateral Area: The sum of the areas of the faces of a polyhedron or other solid. Surface Area: The sum of the areas of the lateral faces of a polyhedron or other solid with one or two bases.

Surface Area of a Prism: Theorem 12-2 Surface Area of a Prism: The surface area SA of a right prism is SA = LA + 2B LA = Ph SA = Ph + 2B where B is the area of the base, P is the perimeter of a base and h is the height.

Surface Area of a Right Cylinder: Theorem 12-3 Surface Area of a Right Cylinder: The surface area SA of a right cylinder is SA = 2Πr2 + 2Πrh r is the radius of a base and h is the height.

Example 1 LA = Ph LA = 22(6) LA = 132 cm2 SA = LA + 2B Find the surface area of a rectangular prism. LA = Ph LA = 22(6) 3 cm LA = 132 cm2 8 cm 6 cm SA = LA + 2B SA = 132 + 2(8)(3) SA = 132 + 48 SA = 180 cm2

Example 2 SA = 2Πr2 + 2Πrh SA = 2(3.14)(4)2 + 2(3.14)(4)(36) Find the surface area of the cylinder. 4 in SA = 2Πr2 + 2Πrh SA = 2(3.14)(4)2 + 2(3.14)(4)(36) 36 in SA = 100.48 + 904.32 SA = 1004.8 in2

Example 3 SA = 2Πr2 + 2Πrh 262.64 = 90.6832 + 23.864(h) Find the height of the right cylinder, which has a surface area of 262.64 cm2. 3.8 SA = 2Πr2 + 2Πrh 262.64 = 2(3.14)(3.8)2 + 2(3.14)(3.8)(h) h 262.64 = 90.6832 + 23.864(h) 171.9568 = 23.864h h = 7.2 cm

Homework Section 12-2 Page 806 – 809 3, 6, 9 – 11, 13, 14, 17, 18, 20, 22, 28