1. 12.2 – Surface Area of Prisms And Cylinders

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

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

4. Lateral area: Area of each lateral face

5. Right Prism: Each lateral edge is perpendicular to both bases

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

7. Cylinder: Prism with circular bases

8. Net: Two-dimensional representation of a solid

9. 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

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

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

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

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

14. 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 = 5 + 6 + 5 + 6 P = 22 SA = 60 + 154 SA = 214m2m2

15. 2. Find the surface area of the right solid. SA = 2B + PH SA = 2(30) + (30)(10) P = 5 + 12 + 13 P = 30 SA = 60 + 40 SA = 100cm 2 c 2 = a 2 + b 2 c 2 = (5) 2 + (12) 2 c 2 = 25 + 144 c 2 = 169 c = 13

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

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

18. 3. Solve for x, given the surface area. SA = 2B + PH 142 = 2(5x) + (2x + 10)(7) B = bh B = 5x P = 5 + x + 5 + x P = 2x + 10 142 = 10x + 14x + 70 142 = 24x + 70 72 = 24x 3ft = x

19. 3. Solve for x, given the surface area.

20. 12.2806-8083, 5, 6, 7-15 odd HW Problems #5 SA = 2B + PH SA = 2(4156.8) + (240)(80) P = 40  6 P = 240 SA = 8313.6 + 19200 SA = 27513.6 ft 2