1. Unit 9: Solids

2. A polyhedron is a solid that is bounded by polygons called faces, that enclose a region of space. An edge of a polyhedron is a line segment formed by the intersection of two faces.

3. A vertex of a polyhedron is a point where three or more edges meet. The plural of polyhedron is polyhedra or polyhedrons.

4. Example Decide whether the solid is a polyhedron. If so, count the number of faces, vertices, and edges of the polyhedron.

5. a.This is a polyhedron. It has 5 faces, 6 vertices, and 9 edges. b.This is not a polyhedron. Some of its faces are not polygons. c.This is a polyhedron. It has 7 faces, 7 vertices, and 12 edges.

6. Types of Solids

7. Prisms

8. A prism is a polyhedron with two congruent faces, called bases. The other faces, called lateral faces, are parallelograms. The segments connecting these vertices are lateral edges.

9. Prisms are classified by the shape of their bases. The surface area of any polyhedron is the sum of the areas of its faces.

10. Find the surface area of a right rectangular prism with a height of 8 inches, a length of 3 inches, and a width of 5 inches.

12. Nets Imagine that you cut some edges of a right hexagonal prism and unfolded it. The two-dimensional representation of all of the faces is called a NET. Nets make it easier to identify the faces

13. Surface Area of a Prism

14. Use the formula to find the exact surface area of the triangular prism

15. Volume of a Prism The volume, V, of a prism is V=Bh, where B is the area of the base and h is the height of the prism.

16. Example Find the volume of the right triangular prism. V = BhVolume of a prism formula A = ½ bhArea of a triangle A = ½ (3)(4)Substitute values A = 6 cm 2 Multiply values -- base V = (6)(2) Substitute values V = 12 cm 3 Multiply values & solve

17. Do Now

18. Page 524 #2-5

19. Surface Area – Page 466 #1-3

20. Volume - Page 533 #1-3

21. Cylinder

22. A cylinder is a solid with congruent circular bases that lie in parallel planes. The height of a cylinder is the perpendicular distance between its bases.

23. Surface Area of Cylinders The surface area of a cylinder is the same formula as prism SA = 2B + ph Since the bases are both circles we can say  SA = 2  r 2 + 2  rh

24. Find the surface area of the cylinder.

25. Find the height of a cylinder which has a radius of 6.5 centimeters and a surface area of 592.19 square centimeters.

26. Volume of a Cylinder The volume, V, of a cylinder is V=Bh, where B is the area of a base, h is the height. Since the base is a circle V=  r 2 h

27. Ex. 2: Finding Volumes Find the volume of the right cylinder. V = BhVolume of a prism formula A =  r 2 Area of a circle A =  8 2 Substitute values A = 64  in. 2 Multiply values -- base V = 64  (6) Substitute values V = 384  in. 3 Multiply values & solve V = 1206.37 in. 3 Simplify

28. Use the measurements given to solve for x.

29. SA - Page 466 #6, 7, 9

30. Volume - Page 534 #4-6

31. Pyramids

32. A pyramid is a polyhedron in which the base is a polygon and the lateral faces are triangles with a common vertex. The altitude or height of a pyramid is the perpendicular distance between the base and the vertex.

33. A regular pyramid has a regular polygon for a base and its height meets the base at its center. The slant height, l, of a regular pyramid is the altitude (height) of any lateral face (triangle).

34. Surface Area of Pyramids SA = the sum of the areas of all faces Examples 1) Find the surface area of the square pyramid

35. 2)Find the exact surface area of the square pyramid

36. The volume, V, of a pyramid is V = ⅓ Bh, where B is the area of the base and h is the height of the pyramid.

37. 3) Find the volume of the pyramid

38. Cones

39. A cone is a nonpolyhedron that has a circular base and a vertex that is NOT in the same plane as the base. The altitude, or height, is the perpendicular distance between the vertex and the base. The slant height is the distance between the vertex and a point on the base edge.

40. Surface Area of a Cone SA =  r 2 +  r l where r is the radius of the base and l is the slant height

41. To find the surface area of the right cone shown, use the formula for the surface area. S =  r 2 +  r l Write formula S =  4 2 +  (4)(6) Substitute S = 16  + 24  Simplify S = 40  Simplify  The surface area is 40  square inches or about 125.7 square inches.

42. Volume of a Cone The volume of a cone is V = Bh. Since the base is a circle  V =  r 2 h

43. Find the volume of the cone. V = ⅓ πr 2 h To find h 6 2 + h 2 = 10 2 h = 8 Therefore, V = ⅓ πr 2 h  V = ⅓ π(6) 2 (8) = 96 π

44. Ex. 3: Using the Volume of a Cone

45. Spheres

46. A circle is defined as all of the points in a plane that are a given distance from a point. A sphere is all points in space that are a given distance from a point. The point is called the center of the sphere. A radius of a sphere is a segment from the center to a point on the sphere.

47. Surface Area of a Sphere SA = 4  r 2, where r is the radius of the sphere

48. Volume of a Sphere,where r is the radius