WINTER, 2011 Geometry B-CH11 Surface Area and Volume

WINTER, 2011

Geometry B-CH11 Surface Area and Volume

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


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


Euler’s Formula: F + V = E +2

F = # FacesV=# VerticesE = # Edges

4


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


Now, you do

1,2, and 3 in 5 minutes!

6

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. m2, cm2, in2

7

11-2 Surface Areas of Prisms and Cylinders

Let’s go make a model!

8

CH11-2: Surface Area of Prisms and Cylinders

Lateral Area of a Prism

LA = Sum of areas of all sides

h

9


Surface Area of a Prism

SA = LA +2∙Ab

LA = Sum of areas of sidesAb = Area of Base Shape

h

10


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 sidesWhat shape are the sides?RectanglesWhat can we use tofind missing sides?? c2 = a2 + b2

c2 = 52 + 82

c2 = 25 + 64 c2 = 89 c = 9.43 in.

18 in.

10 in.h = 8

in.

11


Now, we can find LA:LA =Sum of areas of all sidesWhat side shape do we have?RectangleArea of Rectangle = b∙hArea of rectangle 1 = 10∙18Area of rectangle 2= 9.4∙18Area of rectangle 3 = 9.4∙18LA=Area1+Area2+Area3LA = 180+169.2+169.2LA = 518.4 in2

18 in.

10 in.h = 8

in.

12


Now, we can find SA:SA = LA +2∙Ab


What base shape do we have?TriangleArea of Triangle = ½ ∙b∙hArea of Triangle = ½ ∙(10in)(8in)Area of Triangle = 40 in2

SA = LA + 2∙Ab

SA = 518.4 + 2(40) in2

SA = 598.4 in2

18 in.

10 in.h = 8

in.

13


Now, you do EVENS

8, 10, & 12

14


Lateral Area of a Cylinder

LA = 2∙π ∙ r ∙ hLA = Lateral Area r = radius of circleh= height

Surface Area of a CylinderSA = LA +2 ∙ π ∙ r2

h

15


Surface Area of a Prism

SA = LA +2∙Ab


h

16


Ex.2 Find the surface area of each cylinder in terms of π.

SA = 2∙π ∙ r ∙ h + 2 ∙ π ∙ r2

SA = 2∙π ∙ 1m ∙ 3m+2 ∙ π ∙ (1m)2

SA = 6π +2πSA = 8π

1m

3m

17


Now, you do EVENS

2,4,6,14

18

CH11-3: Surface Area of Pyramids and Cones

Lateral Area of a Pyramid

LA = ½∙l∙PBase

LA = Lateral Area

L = slant heightPBase=Perimeter of Base Shape

h

19

l


Surface Area of a Pyramid

SA = LA + ABase

LA = Lateral Area

Abase = Area of Base Shape

h

20

l


Ex.1 Find a) the lateral areaand b) the surface area of the shape provided.First, the Lateral AreaWhat shape is the base?Square, so

LA = ½∙l∙Pbase

LA = ½∙l∙(12+12+12+12)

How do we find slant height, l?

21 ft

21

l

12 ft12 ft


Yes! Pythagorean Theoremc2 = a2+b2

c2 = 212+62

c2 = 441+36c2 = 477√c2 = √477c= 21.84

Now, we can find Lateral Area, LA

21 ft

22

l

12 ft12 ft


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 ft2

What shape is the base?SquareSA = LA + Abase

SA = 524.2 + (12)2

SA = 668.2 ft2

21 ft

23

l

12 ft12 ft


Now, you do 7, 9, and 11

21 ft

24

l

12 ft12 ft


Lateral Area of a Cone

LA = π∙r∙lLA = Lateral Area

L = slant heightr=radius of base

25

r

l


Surface Area of a Cone

SA = π∙r∙l+πr2

SA = Surface Area

L = slant heightr=radius of base

26

r

l


Now, find Lateral Area

c2=a2+b2

c2=202+92

c2=400+81c2=481√c2=√481c=21.9

27

9cm

20cm

l


Now, find Lateral Area

Now, find Lateral AreaLA = π∙r∙lLA = π∙(9)∙(21.9)LA = 618.9 cm2

SA = π∙r∙l+πr2

SA = π∙9∙(21.9)+π92

SA = 197.1π+81πSA = 278.1π cm2

28

9cm

20cm

l


Now, you do 1, 3, and 5

21 ft

29

l

12 ft12 ft

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. m3, cm3, ft3

30


Volume of a Cylinder

V=π•r2•h

31

r

h


Ex.1 Find the volume of each cylinder to the nearest tenth.

V=π•r2•hV = π•(2m)2•(2m)V = 25.13 m3

4m

2m

32


Now, you do

EVENS 2,4, 6

33

CH11-4 Volumes of Prisms & Cylinders34

Volume of a Prism

V=Ab•h

Ab=Area of Base Shape

h=height of prism

h

CH11-4: Volume of Prisms & Cylinders

Ex.1 Find the volume of each prism. First, what shape is the base?Square, soAb = s2

Ab = (12ft)2

Ab = 144ft2

V= Ab•h

V = (144ft2) •(21 ft)V = 3024 ft3

35

21 ft

l

12 ft12 ft


Now, you do

EVENS 8 -14

36

CH11-5: Volumes of Pyramids and Cones

Volume of a Pyramid

V = 1/3∙Ab∙h

Ab = Area of Base Shape

h= height of pyramid

h

37

l


Ex.1 Find the volume of each pyramid.What shape is the base?Square, soAb = s2

Ab = (54cm)2

Ab = 2916 cm2

V = 1/3∙Ab∙h

V = 1/3∙(2916cm2)∙(45cm)V = 43470 cm3

45 cm

38

l

54 cm54 cm


Now, you do EVENS 2,4,6

39


Volume of a Cone

V = 1/3∙π∙r2∙h

r = radius h=height of cone(use Pythagorean

Theorem or Trig to find h)

40

r

l


Ex.2 Find the Volume of the cone.

V = 1/3∙π∙r2∙hV = 1/3∙π∙(9cm)2∙(20cm)V = 1696.5 cm3

41

9cm

20cm

l


Now, you do EVENS 8,10,12

42

11-6 Surface Areas and Volumes of Spheres

43

Surface Area of a Sphere

SA= 4∙π∙r2

r = radius r

Volume of a Sphere

V= 4/3∙π∙r3

r = radius

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∙r2

SA= 4∙π∙(5m)2

SA= 314.6 m2

b)V= 4/3∙π∙r3

V= 4/3∙π∙(5m)3

V = 523.6 m3

44

10 m

CH11-6: Surface Areas and Volumes of Spheres

Now, you do ODDS 1-11

45

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

SA1 = a2

SA2 b2

V1 = a3

V2 b3

47


a

b

How To Use It:Ex.1 For each pair of similar figures, find the

similarity ratios of the smaller to the larger shape.

SA1 = a2

SA2 b2

9 = a2

16 b2

√ 9 = √a2

√16 √b2

3=a4 b

48


SA = 9SA = 16

Now, you do ALL

1-6

49


Woooohoooo! We’re done with CH11!!!!!!

50


Documents

WINTER, 2011 Geometry B-CH11 Surface Area and Volume