1. Lesson 9-1: Area of 2-D Shapes 1 Part 1 Area of 2-D Shapes

2. Lesson 9-1: Area of 2-D Shapes 2 Squares and Rectangles s s A = s² 6 6 A = 6² = 36 sq. units L W A = LW 12 5 A = 12 x 5 = 60 sq. units Example: Area of Rectangle: A = LW Area of Square: A = s²

3. Lesson 9-1: Area of 2-D Shapes 3 Circles and Sectors r 9 cm A =  (9)² = 81  sq. cm Area of Circle: A =  r² arc r B C A 120° Example: 9 cm

4. Lesson 9-1: Area of 2-D Shapes 4 Triangles and Trapezoids h h h bb b1b1 b2b2 h is the distance from a vertex of the triangle perpendicular to the opposite side. h is the distance from b1 to b2, perpendicular to each base

5. Lesson 9-1: Area of 2-D Shapes 5 Example: Triangles and Trapezoids 7 6 8 12 6

6. Lesson 9-1: Area of 2-D Shapes 6 Parallelograms & Rhombi Area of Parallelogram: A = b h 6 9 A = 9 x 6 = 54 sq. units 8 10 A = ½ (8)(10) = 40 sq units h b Example:

7. Lesson 9-1: Area of 2-D Shapes 7 Area of Regions 8 10 12 414 8 The area of a region is the sum of all of its non-overlapping parts. A = ½(8)(10) A= 40 A = (12)(10) A= 120 A = (4)(8) A=32 A = (14)(8) A=112 Area = 40 + 120 + 32 + 112 = 304 sq. units

8. Lesson 9-2: Prisms & Pyramids 8 Prisms and Pyramids Part 2

9. Lesson 9-2: Prisms & Pyramids 9 Right Prism Lateral Area of a Right Prism (LA) = ph Surface Area (SA) = ph + 2B = [Lateral Area + 2 (area of the base)] Volume of a Right Prism (V )= Bh (P = perimeter of the base, h = height of prism, B = base area) h h h Triangular Right Prism

10. Lesson 9-2: Prisms & Pyramids 10 Examples: 5 4 8 perimeter of base = 2(5) + 2(4) = 18 B = 5 x 4 = 20 L. A.= 18 x 8 = 144 sq. units S.A. = 144 + 2(20) = 184 sq. units V = 20 x 8 = 160 cubic units h = 8 6 8 5 4 4 perimeter of base = 6 + 5 + 8 = 19 L. A. = 19 x 4 = 76 sq. units B = ½ (6)(4) = 12 S. A. = 76 + 2(12) = 100 sq. units V = 12 x 4 = 48 cubic units h = 4

11. Lesson 9-2: Prisms & Pyramids11 Regular Pyramids A regular pyramid has a base which is always a regular polygon. The lateral faces all intersect at a point called the vertex and form triangles. The altitude is a segment from the vertex perpendicular to the base. The slant height is the height of a lateral face. Lateral side vertex altitude Slant height Base

12. Lesson 9-2: Prisms & Pyramids 12 Regular Pyramid ( p = perimeter of the base, l = slant height, B = base area) Lateral area = ½ (13)(40) = 260 sq. units Perimeter = (2 x 10) + (2 x 10) = 40 Slant height l = 13 ; Height h = 12 The lateral area of a regular pyramid (L.A.)= ½ lp Surface area of a regular pyramid (S.A.) = ½ lp + B The volume of a right pyramid (V)= ⅓ Bh Area of base = 10 x 10 = 100 sq. units Surface area = 260 + 100 = 360 sq. units Volume = ⅓ ⅓⅓ ⅓ (100)(12) = 400 cubic units 10 13 12 Example:

13. ** To find the lateral area of a non-regular pyramid, find the area of each face separately. Lesson 9-2: Prisms & Pyramids Find the slant height of the front face and the side face. l 2 = 3 2 + 9 2 m 2 = 5 2 + 9 2 l 2 = 9 + 81 = 90m 2 = 25 + 81 = 106 l = 9.4m = 10.3 Since the front and back faces are the same, and the two side faces are the same, add the areas of the four faces. Lateral area = 2(47) + 2(30.9) = 155.8 Continue with the formula for surface area and volume. Non-regular Pyramids Area of front face = ½ (10)(9.4) = 47 Area of side face = ½ (6)(10.3) = 30.9 6 10 9 m l 53

14. Lesson 9-2: Prisms & Pyramids 14 Prism and Pyramids Formulas Prisms: Lateral Area: L.A. = ph (p = perimeter, h = height) Surface Area: S.A. = ph + 2B (B = area of base) Volume: V = Bh Regular Pyramids: Lateral Area: L.A. = ½ lp (p = perimeter, l = slant height) Surface Area: S.A. = ½ lp + B (B = area of base) Volume: V = ⅓ Bh ( B = area of base, h = height)

15. Lesson 9-3: Cylinders and Cones 15 Part 3 Cylinders and Cones

16. Lesson 9-3: Cylinders and Cones 16 Cylinders Surface Area (SA) = 2B + LA = 2πr ( r + h ) Cylinders are right prisms with circular bases. Therefore, the formulas for prisms can be used for cylinders. Volume (V) = Bh = The base area is the area of the circle: The lateral area is the area of the rectangle: 2πrh h 2πr h Formulas: S.A. = 2πr ( r + h ) V =

17. Lesson 9-3: Cylinders and Cones 17 Example For the cylinder shown, find the lateral area, surface area and volume. L.A.= 2πrh L.A.= 2π(3)(4) L.A.= 24π sq. cm. 4 cm 3 cm S.A.= 2πr 2 + 2πrh S.A.= 2π(3) 2 + 2π(3)(4) S.A.= 18π +24π S.A.= 42π sq. cm. V = πr 2 h V = π(3) 2 (4) V = 36π

18. Lesson 9-3: Cylinders and Cones 18 Cones Surface Area (SA) = B + LA = π r (r + l) Cones are right pyramids with a circular base. Therefore, the formulas for pyramids can be used for cones. Volume (V) = Lateral Area (LA) = π r l, where l is the slant height. The base area is the area of the circle: Notice that the height (h) (altitude), the radius and the slant height create a right triangle. l r h Formulas: S.A. = π r ( r + l ) V =

19. Lesson 9-3: Cylinders and Cones 19 Example: For the cone shown, find the lateral area surface area and volume. L.A.= πrl Note: We must use the Pythagorean theorem to find l. L.A.= π(6)(10) L.A.= 60π sq. cm. 6 cm 8 6 2 +8 2 = l 2 10 S.A.= πr (r + l ) S.A.= π6 (6 + 10) S.A.= 6π (16) S.A.= 96π sq. cm. V= 96π cubic cm.

20. Lesson 9-4: Spheres 20 Part 4 Spheres

21. Lesson 9-4: Spheres 21 Spheres A sphere is formed by revolving a circle about its diameter. In space, the set of all points that are a given distance from a given point, called the center. Definition:

22. Lesson 9-4: Spheres 22 Spheres – special segments & lines Radius: A segment whose endpoints are the center of the sphere and a point on the sphere. Chord: A segment whose endpoints are on the sphere. Diameter: A chord that contains the sphere’s center. Tangent: A line that intersects the sphere in exactly one point. Radius Chord Diameter Tangent

23. Lesson 9-4: Spheres 23 Surface Area & Volume of Sphere Volume (V) = Surface Area (SA) = 4 π r 2 Example:Find the surface area and volume of the sphere. 12 cm

24. Lesson 9-4: Spheres 24 Great Circle & Hemisphere Great Circle: For a given sphere, the intersection of the sphere and a plane that contains the center of the sphere. Hemisphere: One of the two parts into which a great circle separates a given sphere. Great Circle Hemispher e

25. Lesson 9-4: Spheres 25 Surface Area & Volume of Hemisphere Find the surface area and volume of the following solid (Hemisphere). r= 12 cm

26. Lesson 9-5: Similar Solids 26 Part 5 Similar Solids

27. Lesson 9-5: Similar Solids 27 Similar Solids Two solids of the same type with equal ratios of corresponding linear measures (such as heights or radii) are called similar solids.

28. Lesson 9-5: Similar Solids 28 Similar Solids Similar solids NOT similar solids

29. Lesson 9-5: Similar Solids 29 Similar Solids & Corresponding Linear Measures To compare the ratios of corresponding side or other linear lengths, write the ratios as fractions in simplest terms. 12 3 6 8 2 4 Length: 12 = 3 width: 3 height: 6 = 3 8 2 2 4 2 Notice that all ratios for corresponding measures are equal in similar solids. The reduced ratio is called the “scale factor”.

30. Lesson 9-5: Similar Solids 30 16 12 8 6 9 All corresponding ratios are equal, so the figures are similar Are these solids similar? Example: Solution:

31. Lesson 9-5: Similar Solids 31 8 18 4 6 Corresponding ratios are not equal, so the figures are not similar. Are these solids similar? Solution: Example:

32. Lesson 9-5: Similar Solids 32 10 4 8 Similar Solids and Ratios of Areas If two similar solids have a scale factor of a : b, then corresponding areas have a ratio of a 2 : b 2. This applies to lateral area, surface area, or base area. Surface Area = base + lateral = 40 + 108 = 148 5 2 4 3.5 Surface Area = base + lateral = 10 + 27 = 37 Ratio of sides = 2: 1 Ratio of surface areas: 148:37 = 4:1 = 2 2 : 1 2 7

33. Lesson 9-5: Similar Solids 33 9 15 Similar Solids and Ratios of Volumes If two similar solids have a scale factor of a : b, then their volumes have a ratio of a 3 : b 3. 6 10 Ratio of heights = 3:2 V =  r 2 h =  (9 2 ) (15) = 1215V=  r 2 h =  (6 2 )(10) = 360 Ratio of volumes: 1215:360 = 27:8 = 3 3 : 2 3