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Chapter 11 Section 11.1 – Space Figures and Cross Sections

Chapter 11 Section 11.1 – Space Figures and Cross Sections. Objectives: To recognize polyhedra and their parts To visualize cross sections of space figures. Polyhedron  a three-dimensional figure whose surfaces are polygons Face  name for a side of a polyhedron

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Chapter 11 Section 11.1 – Space Figures and Cross Sections

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  1. Chapter 11Section 11.1 – Space Figures and Cross Sections Objectives: To recognize polyhedra and their parts To visualize cross sections of space figures

  2. Polyhedron  a three-dimensional figure whose surfaces are polygons • Face  name for a side of a polyhedron • Edge  a segment that is formed by the intersection of two faces • Vertex  a point where three or more edges intersect

  3. Faces Vertex Edge

  4. A • Ex: • Identify the vertices/edges/faces of the figure D B C E H F G

  5. Euler’s Formula • The numbers of faces (F), vertices (V), and edges (E) of a polyhedron are related by the formula: F + V = E + 2

  6. Ex: Using Euler’s Formula • Use Euler’s Formula to find the number of edges on a polyhedron with eight triangular faces.

  7. Cross-Section  the intersection of a solid and a plane. You can think of it as a very thin slice of the solid. • Examples of Cross Sections • One slice of bread in a loaf • CAT Scans and MRI’s

  8. Homework #25 • Due Tuesday (April 09) • Page 601 • # 1 – 19 all

  9. Section 11.2 – Surface Areas of Prisms and Cylinders • Objectives: To find the surface area of a prism To find the surface area of a cylinder

  10. Prism  a polyhedron with exactly two congruent, parallel faces called bases. The other faces of the prism are called lateral faces. • Altitude  of a prism is a perpendicular segment that joins the planes of the bases. • Height  of the prism is the length of an altitude.

  11. Lateral Area  of a prism is the sum of the areas of the lateral faces. • Surface Area  the sum of the lateral area of the area of the two bases.

  12. Theorem 11.1 – Lateral and Surface Areas of a Prism • The lateral area of a right prism is the product of the perimeter of the base and the height. L.A. = p · h • The surface area of a right prism is the sum of the lateral area and the areas of the two bases. S.A. = L.A. + 2B

  13. Cylinder  has two congruent parallel bases, just like a prism. However, the bases of a cylinder are circles. • Altitude  of a cylinder is a perpendicular segment that joins the planes of the bases • Height  of a cylinder is the length of an altitude

  14. Lateral Area  visualize “unrolling” the curved surface of the cylinder. Imagine taking the label off of a water bottle. The area of the resulting rectangle is the lateral area. • Surface Area  of a cylinder is the sum of the lateral area and the areas of the two circular bases.

  15. Theorem 11.2 – Lateral and Surface Areas of a Cylinder • The lateral area of a right cylinder is the product of the circumference of the base and the height of the cylinder. L.A. = 2Πrh-or- L.A. =Πdh • The surface area of a right cylinder is the sum of the lateral area and the areas of the two bases. S.A. = L.A. + 2B -or- S.A. = 2Πrh + 2Π

  16. Homework # 26 • Due Wednesday (April 10) • Page 611 – 612 • #1 – 19 all

  17. Section 11.3 – Surface Areas of Pyramids and Cones • Objectives: To find the surface area of a pyramid To find the surface area of a cone

  18. Pyramid  a polyhedron in which one face (the base) can be any polygon and the other faces (lateral faces) are triangles that meet at a common vertex (vertex of the pyramid). • Altitude  of a pyramid is the perpendicular segment from the vertex to the plane of the base. • Height  length of the altitude

  19. Regular Pyramid  a pyramid whose base is a regular polygon and whose lateral faces are congruent isosceles triangles. • Slant height (l)  the length of the altitude of a lateral face of the pyramid.

  20. Lateral Area  of a pyramid is the sum of the areas of the congruent lateral faces. • Surface Area  of a pyramid is the sum of the lateral area and the area of its base

  21. Theorem 11.3 – Lateral and Surface Areas of a Regular Pyramid • The lateral area of a regular pyramid is half the product of the perimeter of the base and the slant height. L.A. = p · l • The surface area of a regular pyramid is the sum of the lateral area and the area of the base. S.A. = L.A. + B

  22. Cone  has a pointed top like a pyramid, but its base is a circle • Altitude  a perpendicular segment from the vertex of the cone to the center of its base • Height  the length of the altitude • Slant Height  the distance from the vertex to a point on the edge of the base

  23. Lateral Area  as with a pyramid, it is the circumference of the base times the slant height. • Surface Area  similar to a pyramid as well, it is the sum of the lateral area and the area of the base.

  24. Theorem 11.4 – Lateral and Surface Areas of a Cone • The lateral area of a right cone is half the product of the circumference of the base and the slant height. L.A. = · 2Πr ·l -or- L.A. = Πrl • The surface area of a right cone is the sum of the lateral area and the area of the base. S.A. = L.A. + B

  25. Homework #27 • Due Thurs/Fri (Apr 11/12) • Page 620 – 621 • # 1 – 21 all

  26. Section 11.4 – Volumes of Prisms and Cylinders • Objectives: To find the volume of a prism To find the volume of a cylinder

  27. Volume  the space that a figure occupies. It is measured in cubic units (, ). **Notice that volume and surface area are different. Surface area only includes the area of the container (empty soda can). Volume includes everything inside the container (full can of soda).**

  28. Theorem 11.5 – Cavalieri’s Principle • If two space figures have the same height and the same cross-sectional area at every level, then they have the same volume.

  29. Theorem 11.6 – Volume of a Prism • The volume of a prism is the product of the area of a base and the height of the prism. V = B · h h B

  30. Theorem 11.7 – Volume of a Cylinder • The volume of a cylinder is the product of the area of the base and the height of the cylinder. V = B ·h -or- V = Πh r B h

  31. Composite Space Figure  a 3D figure that is the combination of two or more simpler figures. • Think of a rocket. It is composed of a conical top and a cylindrical body. The composite volume would be the volume of the cone added to the volume of the cylinder.

  32. Section 11.5 – Volumes of Pyramids and Cones • Objectives: To find the volume of a pyramid To find the volume of a cone

  33. Theorem 11.8 – Volume of a Pyramid • The volume of a pyramid is one third the product of the area of the base and the height of the pyramid. V = B · h

  34. Theorem 11.9 – Volume of a Cone • The volume of a cone is one third the product of the area of the base and the height of the cone. V = B · h -or- V = h

  35. Homework #28 • Due Monday (April 15) • Page 627 – 628 • # 1 – 19 odd • Homework #29 • Due Monday (April 15) • Page 634 – 635 • # 1 – 19 odd Quiz Tuesday

  36. Section 11.6 – Surface Areas and Volumes of Spheres • Objectives: To find the surface area and volume of a sphere

  37. Sphere  the set of all points in space equidistant from a given point called the center. • Radius  a segment that has one endpoint at the center and the other endpoint on the sphere • Diameter  a segment passing through the center with endpoints on the sphere

  38. When a plane and a sphere intersect in more than one point, the intersection is a circle. If the center of the circle is also the center of the sphere, the circle is called a great circle. • Circumference  of the great circle is the same as the sphere • Hemispheres  two equal halves of a sphere. These are created by a great circle.

  39. Theorem 11.10 – Surface Area of a Sphere • The surface area of a sphere is four times the product of Π and the square of the radius of the sphere. S.A. = 4Π

  40. Theorem 11.11 – Volume of a Sphere • The volume of a sphere is four thirds the product of Π (pi) and the cube of the radius of the sphere. V = Π

  41. Ex: Earth’s equator is about 24,902 mi long. Approximate the surface area of Earth by finding the surface area of a sphere with circumference 24,902 mi. • Ex: The volume of a sphere is 4200 . Find the surface area to the nearest tenth.

  42. Homework #30 • Due Monday (April 22) • Page 640 – 641 • # 1 – 21 all

  43. Section 11.7 – Areas and Volumes of Similar Solids • Objectives: To find relationships between the ratios of the areas and volumes of similar solids

  44. Similar Solids  have the same shape, and all their corresponding dimensions are proportional. • Similarity Ratio  the ratio of corresponding linear dimensions of two similar solids. **Any two cubes are similar and any two spheres are similar**

  45. Ex: Identifying Similar Solids • Are the following figures similar? 3 6 2 3 4 6 6 5 12 10

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