1. Lesson 12-1, 2, 7 & 13-2 3 D Figures Nets Spheres

2. Objectives Find the nets of 3-dimensional objects Find the Surface Area of Spheres Find the Volume of Spheres

3. Vocabulary Orthogonal Drawing – Two-dimensional view from top, left, front and right sides Corner View – View of a figure from a corner Perspective View – same as a corner view Polyhedron – A solid with all flat surfaces that enclose a single region of space Face – Flat surface of a polyhedron Edges – Line segments where two faces intersect (edges intersect at a vertex) Bases – Two parallel congruent faces Prism – Polyhedron with two bases Regular Prism – Prism with bases that are regular polygons

4. Vocabulary Cont Pyramid – Polyhedron with all faces (except for one base) intersecting at one vertex Regular Polyhedron – All faces are regular congruent polygons and all edges congruent Platonic Solids – The five types of regular polyhedra (named after Plato) Cylinder – Solid with congruent circular bases in a pair of parallel planes Cone – Solid with a circular base and a vertex (where all “other sides” meet) Sphere – Set of points in space that are a given distance from a given point (center) Cross Section – Intersection of a plane and a solid Reflection Symmetry – Symmetry with respect to different planes (instead of lines)

5. Prisms & 3d-Terms Faces (sides) Vertexes (corner pts) Edges (lines between vertexes) Base (front and back) Prism – a polyhedron with two parallel congruent faces called bases. Other faces are parallelograms. Triangular Prism Pentagonal Prism Rectangular Prism

6. Other 3d Figures Pyramid (Square)Cylinder h r h B l h r l ConeSphere r Pyramid – A polyhedron with all faces (except the base) intersecting at one vertex. Named for their bases (which can be any polygon). Cylinder – A solid with circular congruent bases in two parallel planes (a can). Cone – A solid with circular base and a vertex. Sphere – All points equal distant from a center point in 3- space l – slant height

7. Nets Triangular Prism Square Prism (Cube) Cylinder h r h C Nets – cut a 3d figure on its edges and lay it flat. It can be folded into the shape of the 3d figure with no overlap Surface Area – Sum of each area of the faces of the solid

8. Example 1 h r l Which of the following represents the net of the cone above? D.C.B.A. D.

9. Example 2 Which of the following represents the net of the cylinder above? h r D.C.B.A. D.

10. Example 3 Which of the following represents the net of the triangular prism above? l b h c c D.C.B.A.

11. Spheres – Surface Area & Volume V = 4/3 π r 3 Sphere r SA = 4 π r 2 Sphere – All points equal distant from a center point in 3-space Circles – Intersection between a plane and a sphere Great Circles – Intersections between a plane passing through the center of the sphere and the sphere. Great circles have the same center as the sphere. The shortest distance between two points on a sphere lie on the great circle containing those two points. Hemisphere – a congruent half of a sphere formed by a great circle. Surface areas of hemispheres are half of the SA of the sphere and the area of the great circle. Volumes of hemispheres are half of the volume of the sphere.

12. Example 1: 10 Find the surface area and the volume of the sphere to the right SA = 4πr²  need to find r SA = 4π(10)² = 400π = 1256.64 V= 4/3πr³  need to find r V= 4/3π(10)³ = 4000π/3 = 4188.79

13. Example 2: Find the surface area and the volume of the sphere to the right 18 SA = 4πr²  need to find r r = ½ d = ½(18) = 9 SA = 4π(9)² = 324π = 1017.88 V= 4/3πr³  need to find r V= 4/3π(9)³ = 2916π/3 = 3053.63

14. Example 3: Find the surface area and the volume of the hemi-sphere to the right 16 ½ of a sphere’s SA is just ½ SA = ½ 4πr² SA of a hemisphere = ½ 4π(r)² + π(r)² = 2πr² + πr² = 3πr² Volume of ½ a sphere ½ V= ½ (4/3πr³) = 2/3πr³  need to find r V= 2/3π(8)³ = 1024π/3 = 1072.33  NO! We need to include the newly exposed “flat surface” SA = 3πr² = 3π(8)² = 192π = 603.19

15. Example 7-2b Find the surface area of a hemisphere with a radius of 3.8 inches. A hemisphere is half of a sphere. To find the surface area, find half of the surface area of the sphere and add the area of the great circle. Surface area of a hemisphere Use a calculator. Answer: The surface area is approximately 136.1 sq inches. Substitution

16. Example 7-3a Find the surface area of a ball with a circumference of 24 inches to determine how much leather is needed to make the ball. First, find the radius of the sphere. Circumference of a circle Next, find the surface area of the sphere. Surface area of a sphere Answer: The surface area is approximately 183.3 sq inches. ≈ 3.8

17. Example 3-1a Find the volume of the sphere to the nearest tenth. Answer:The volume of the sphere is approximately 14,137.2 cubic centimeters. Volume of a sphere r = 15 Use a calculator.

18. Example 3-1b Find the volume of the sphere to the nearest tenth. First find the radius of the sphere. Circumference of a circle Solve for r. C 25 Answer: The volume of the sphere is approximately 263.9 cm³. Volume of a sphere Use a calculator. Now find the volume.

19. Summary & Homework Summary: –Nets Every 3-dimensional solid can be represented by one or more 2-dimensional nets (cardboard cutouts) Area of a net is the same as the surface area of the solid –Sphere Volume: V = 4/3πr³ Surface Area: SA = 4πr² –Hemi-sphere Volume: ½(sphere) = 2/3 πr³ Surface Area: SA = 3πr² A hemi-sphere’s surface consists of ½ SA of a sphere plus area of exposed circle Homework: –pg 704-705; 9-18