12.2 – Surface Area of Prisms And Cylinders. Polyhedron with two parallel, congruent bases Named after its base Prism:

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12.2 – Surface Area of Prisms And Cylinders

Polyhedron with two parallel, congruent bases Named after its base Prism:

Surface area: Sum of the area of each face of solid

Lateral area: Area of each lateral face

Right Prism: Each lateral edge is perpendicular to both bases

Oblique Prism: Each lateral edge is NOT perpendicular to both bases

Cylinder: Prism with circular bases

Net: Two-dimensional representation of a solid

Surface Area of a Right Prism: SA = 2B + PH B = area of one base P = Perimeter of one base H = Height of the prism H

Surface Area of a Right Cylinder: H SA = 2B + PH

1. Name the solid that can be formed by the net. Cylinder

1. Name the solid that can be formed by the net. Triangular prism

1. Name the solid that can be formed by the net. rectangular prism

2. Find the surface area of the right solid. SA = 2B + PH SA = 2(30) + (22)(7) B = bh B = (5)(6) B = 30 P = P = 22 SA = SA = 214m2m2

2. Find the surface area of the right solid. SA = 2B + PH SA = 2(30) + (30)(10) P = P = 30 SA = SA = 100cm 2 c 2 = a 2 + b 2 c 2 = (5) 2 + (12) 2 c 2 = c 2 = 169 c = 13

2. Find the surface area of the right solid. cm 2

2. Find the surface area of the right solid. in 2 144in

3. Solve for x, given the surface area. SA = 2B + PH 142 = 2(5x) + (2x + 10)(7) B = bh B = 5x P = 5 + x x P = 2x = 10x + 14x = 24x = 24x 3ft = x

3. Solve for x, given the surface area.

, 5, 6, 7-15 odd HW Problems #5 SA = 2B + PH SA = 2(4156.8) + (240)(80) P = 40  6 P = 240 SA = SA = ft 2