1. WINTER, 2011 Geometry B-CH11 Surface Area and Volume

2. CH11-1: Euler’s Formula Background: This Swiss guy, Leonhard Euler lived from 1707 to 1783 doing most of his work in Berlin, Germany. In 1735 through 1766 he went totally blind. He dictated his formulas and mathematical papers to an assistant and did most of his calculations in his head! His formulas and work served as a basis of advanced math topics like differential equations (math beyond Calculus!) 2

3. CH11-1: Euler’s Formula Vocabulary: Euler developed a formula to help analyze various polyhedrons. Polyhedron: a 3D shape with a surface of a polygon (ex. Volleyball). Face: The polygon is called the face. Edge: Segment that is formed by the intersection of two faces. Vertex: a point where 3 or more edges intersect. 3

4. CH11-1: Euler’s Formula Euler’s Formula: F + V = E +2 F = # Faces V=# Vertices E = # Edges 4

5. CH11-1: Euler’s Formula Ex.1 Use Euler’s Formula to find the missing number for the polyhedron. #Faces: 6 #Edges: 12 #Vertices=? F+V = E +2 6+V = 12+2 6+V = 14 -6= -6 V = 8 5

6. CH11-1: Euler’s Formula Now, you do 1,2, and 3 in 5 minutes! 6

7. CH11-2 Surface Areas of Prisms and Cylinders Surface Area (SA): Sum of the areas of all the faces of a 3D object. Lateral Area (LA): Sum of the areas of all the faces EXCEPT THE TOP AND THE BOTTOM. The word lateral means “Side.” Since it is still a calculation of area, the units for Surface Area and Lateral Area are still squared units. Ex. m 2, cm 2, in 2 7

8. 11-2 Surface Areas of Prisms and Cylinders Let’s go make a model! 8

9. CH11-2: Surface Area of Prisms and Cylinders Lateral Area of a Prism LA = Sum of areas of all sides h 9

10. CH11-2: Surface Area of Prisms and Cylinders Surface Area of a Prism SA = LA +2∙A b LA = Sum of areas of sides A b = Area of Base Shape h 10

11. CH11-2: Surface Area of Prisms and Cylinders Ex.1 Find a) the lateral area and b) the surface area of each prism. DON’T FORGET YOUR UNITS! LA = Sum of area of sides What shape are the sides? Rectangles What can we use to find missing sides?? c 2 = a 2 + b 2 c 2 = 5 2 + 8 2 c 2 = 25 + 64 c 2 = 89 c = 9.43 in. 18 in. 10 in. h = 8 in. 11

12. CH11-2: Surface Area of Prisms and Cylinders Now, we can find LA: LA =Sum of areas of all sides What side shape do we have? Rectangle Area of Rectangle = b∙h Area of rectangle 1 = 10∙18 Area of rectangle 2= 9.4∙18 Area of rectangle 3 = 9.4∙18 LA=Area1+Area2+Area3 LA = 180+169.2+169.2 LA = 518.4 in 2 18 in. 10 in. h = 8 in. 12

13. CH11-2: Surface Area of Prisms and Cylinders Now, we can find SA: SA = LA +2∙A b LA = Sum of areas of sides A b = Area of Base Shape What base shape do we have? Triangle Area of Triangle = ½ ∙b∙h Area of Triangle = ½ ∙(10in)(8in) Area of Triangle = 40 in 2 SA = LA + 2∙A b SA = 518.4 + 2(40) in 2 SA = 598.4 in 2 18 in. 10 in. h = 8 in. 13

14. CH11-2: Surface Area of Prisms and Cylinders Now, you do EVENS 8, 10, & 12 14

15. CH11-2: Surface Area of Prisms and Cylinders Lateral Area of a Cylinder LA = 2∙ π ∙ r ∙ h LA = Lateral Area r = radius of circle h= height Surface Area of a Cylinder SA = LA +2 ∙ π ∙ r 2 h 15

16. CH11-2: Surface Area of Prisms and Cylinders Surface Area of a Prism SA = LA +2∙A b LA = Sum of areas of sides A b = Area of Base Shape h 16

17. CH11-2: Surface Area of Prisms and Cylinders Ex.2 Find the surface area of each cylinder in terms of π. SA = 2∙ π ∙ r ∙ h + 2 ∙ π ∙ r 2 SA = 2∙ π ∙ 1m ∙ 3m +2 ∙ π ∙ ( 1m) 2 SA = 6 π +2π SA = 8π 1m 3m 17

18. CH11-2: Surface Area of Prisms and Cylinders Now, you do EVENS 2,4,6,14 18

19. CH11-3: Surface Area of Pyramids and Cones Lateral Area of a Pyramid LA = ½∙ l ∙ P Base LA = Lateral Area L = slant height P Base =Perimeter of Base Shape h 19 l

20. CH11-3: Surface Area of Pyramids and Cones Surface Area of a Pyramid SA = LA + A Base LA = Lateral Area A base = Area of Base Shape h 20 l

21. CH11-3: Surface Area of Pyramids and Cones Ex.1 Find a) the lateral area and b) the surface area of the shape provided. First, the Lateral Area What shape is the base? Square, so LA = ½∙ l ∙ P base LA = ½∙ l ∙ (12+12+12+12) How do we find slant height, l ? 21 ft 21 l 12 ft

22. CH11-3: Surface Area of Pyramids and Cones Yes! Pythagorean Theorem c 2 = a 2 +b 2 c 2 = 21 2 +6 2 c 2 = 441+36 c 2 = 477 √ c 2 = √ 477 c= 21.84 Now, we can find Lateral Area, LA 21 ft 22 l 12 ft

23. CH11-3: Surface Area of Pyramids and Cones LA = ½∙ l ∙ (12+12+12+12) And we now know c= l =21.84 LA = ½∙21.84 ∙ (12+12+12+12) LA = 524.2 ft 2 What shape is the base? Square SA = LA + A base SA = 524.2 + (12) 2 SA = 668.2 ft 2 21 ft 23 l 12 ft

24. CH11-3: Surface Area of Pyramids and Cones Now, you do 7, 9, and 11 21 ft 24 l 12 ft

25. CH11-3: Surface Area of Pyramids and Cones Lateral Area of a Cone LA = π∙r∙ l LA = Lateral Area L = slant height r=radius of base 25 r l

26. CH11-3: Surface Area of Pyramids and Cones Surface Area of a Cone SA = π∙r∙ l +πr 2 SA = Surface Area L = slant height r=radius of base 26 r l

27. CH11-3: Surface Area of Pyramids and Cones Now, find Lateral Area c 2 =a 2 +b 2 c 2 =20 2 +9 2 c 2 =400+81 c 2 =481 √c 2 =√481 c=21.9 27 9cm 20cm l

28. CH11-3: Surface Area of Pyramids and Cones Now, find Lateral Area LA = π∙r∙ l LA = π∙(9)∙(21.9) LA = 618.9 cm 2 SA = π∙r∙ l +πr 2 SA = π∙9∙ (21.9)+π9 2 SA = 197.1π+81π SA = 278.1π cm 2 28 9cm 20cm l

29. CH11-3: Surface Area of Pyramids and Cones Now, you do 1, 3, and 5 21 ft 29 l 12 ft

30. CH11-4 Volumes of Prisms & Cylinders Background: Often, it is useful to know the amount of the inside of a 3D shape. For example, out in Milan Dragway, there are large plastic cylinders for recycling used oil. Vocabulary: Volume, V: Product of any 3 dimensions. Measures an objects INTERIOR PLUS DEPTH and has cubed units. Ex. m 3, cm 3, ft 3 30

31. CH11-4 Volumes of Prisms & Cylinders Volume of a Cylinder V= π r 2 h 31 r h

32. CH11-4 Volumes of Prisms & Cylinders Ex.1 Find the volume of each cylinder to the nearest tenth. V= π r 2 h V = π( 2m) 2 (2m) V = 25.13 m 3 4m 2m 32

33. CH11-4 Volumes of Prisms & Cylinders Now, you do EVENS 2,4, 6 33

34. CH11-4 Volumes of Prisms & Cylinders 34 Volume of a Prism V=A b h A b =Area of Base Shape h=height of prism h

35. CH11-4: Volume of Prisms & Cylinders Ex.1 Find the volume of each prism. First, what shape is the base? Square, so A b = s 2 A b = (12ft) 2 A b = 144ft 2 V= A b h V = (144ft 2 ) (21 ft) V = 3024 ft 3 35 21 ft l 12 ft

36. CH11-4 Volumes of Prisms & Cylinders Now, you do EVENS 8 -14 36

37. CH11-5: Volumes of Pyramids and Cones Volume of a Pyramid V = 1/3∙A b ∙h A b = Area of Base Shape h= height of pyramid h 37 l

38. CH11-5: Volumes of Pyramids and Cones Ex.1 Find the volume of each pyramid. What shape is the base? Square, so A b = s 2 A b = (54cm) 2 A b = 2916 cm 2 V = 1/3∙A b ∙h V = 1/3∙(2916cm 2 )∙(45cm) V = 43470 cm 3 45 cm 38 l 54 cm

39. CH11-5: Volumes of Pyramids and Cones Now, you do EVENS 2,4,6 39

40. CH11-5: Volumes of Pyramids and Cones Volume of a Cone V = 1/3∙π∙r 2 ∙h r = radius h=height of cone (use Pythagorean Theorem or Trig to find h) 40 r l

41. CH11-5: Volumes of Pyramids and Cones Ex.2 Find the Volume of the cone. V = 1/3∙π∙r 2 ∙h V = 1/3∙π∙(9cm) 2 ∙(20cm) V = 1696.5 cm 3 41 9cm 20cm l

42. CH11-5: Volumes of Pyramids and Cones Now, you do EVENS 8,10,12 42

43. 11-6 Surface Areas and Volumes of Spheres 43 Surface Area of a Sphere SA= 4∙π∙r 2 r = radius r Volume of a Sphere V= 4/3∙π∙r 3 r = radius

44. 11-6 Surface Areas and Volumes of Spheres Ex.1 Find a) the Surface Area and b) Volume of the sphere. Round your answers to the nearest tenth. a) SA= 4∙π∙r 2 SA= 4∙π∙(5m) 2 SA= 314.6 m 2 b) V= 4/3∙π∙r 3 V= 4/3∙π∙(5m) 3 V = 523.6 m 3 44 10 m

45. CH11-6: Surface Areas and Volumes of Spheres Now, you do ODDS 1-11 45

46. 11-7 Areas & Volumes of Similar Solids Background: Sometimes, you don’t have all the dimensions of all sides for your shapes. So, if you know the surface areas or volumes, you can make a proportion to figure it out. Vocabulary: Surface Area: The total of any shape rolled out flat. Proportion: Two ratios set equal to each other. 46

47. SA 1 = a 2 SA 2 b 2 V 1 = a 3 V 2 b 3 47 11-7 Areas & Volumes of Similar Solids a b

48. How To Use It: Ex.1 For each pair of similar figures, find the similarity ratios of the smaller to the larger shape. SA 1 = a 2 SA 2 b 2 9 = a 2 16 b 2 √ 9 = √a 2 √16 √b 2 3=a3=a 4 b 48 11-7 Areas & Volumes of Similar Solids SA = 9 SA = 16

49. Now, you do ALL 1-6 49 11-7 Areas & Volumes of Similar Solids

50. Woooohoooo! We’re done with CH11!!!!!! 50 11-7 Areas & Volumes of Similar Solids