Lesson 12-1, 2, 7 & 13-2

1. Lesson 12-1, 2, 7 & 13-2 3 D FiguresNetsSpheres

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 Triangular Prism Pentagonal Prism Prism – a polyhedron with two parallel congruent faces called bases. Other faces areparallelograms. Faces (sides) Rectangular Prism Vertexes (corner pts) Edges (lines between vertexes) Base (front and back)

6. l h h r B l h r r Other 3d Figures Pyramid (Square) Cylinder Cone Sphere 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 l – slant height

7. Nets Triangular Prism Cylinder 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 h h r C Square Prism (Cube) Surface Area – Sum of each area of the faces of the solid

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

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

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

11. Spheres – Surface Area & Volume r 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. Sphere V = 4/3 •π• r3 SA = 4π• r2 Sphere – All points equal distant from a center point in 3-space

12. Example 1: Find the surface area and the volume of the sphere to the right 10 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: 16 Find the surface area and the volume of the hemi-sphere to the right ½ of a sphere’s SA is just ½ SA = ½ 4πr²  NO! We need to include the newly exposed “flat surface” SA of a hemisphere = ½ 4π(r)² + π(r)² = 2πr² + πr² = 3πr² SA = 3πr² = 3π(8)² = 192π = 603.19 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

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 Substitution Use a calculator. Answer: The surface area is approximately 136.1 sq inches.

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 ≈ 3.8 Next, find the surface area of the sphere. Surface area of a sphere Answer: The surface area is approximately 183.3 sq inches.

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

18. Use a calculator. C 25 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. Now find the volume. Volume of a sphere Answer: The volume of the sphere is approximately 263.9 cm³.

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