ch11-1: euler’s formula geometry b-ch11 2 surface area …€¦ · · 2018-04-26ch11-1:...
TRANSCRIPT
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MRS . KUMMER
FALL , 201 1 - 2012
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!)
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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.
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CH11-1: Euler’s Formula
Euler’s Formula:
F + V = E +2
F = # Faces
V=# Vertices
E = # Edges
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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
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CH11-1: Euler’s Formula
Now, you do
#s 1-9 on page 601!
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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
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11-2 Surface Areas of Prisms and Cylinders
Let’s go make a model!
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CH11-2: Surface Area of Prisms and Cylinders
Lateral Area of a Prism
LA = Sum of areas of all sides
h
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CH11-2: Surface Area of Prisms and Cylinders
Surface Area of a Prism
SA = LA +2∙Ab
LA = Sum of areas of sides
Ab = Area of Base Shape
h
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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??
c2 = a2 + b2
c2 = 52 + 82
c2 = 25 + 64
c2 = 89
c = 9.43 in.
18 in.
10 in.h = 8
in.
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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 in2
18 in.
10 in.h = 8
in.
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CH11-2: Surface Area of Prisms and Cylinders
Now, we can find SA:
SA = LA +2∙Ab
LA = Sum of areas of sides
Ab = 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 in2
SA = LA + 2∙Ab
SA = 518.4 + 2(40) in2
SA = 598.4 in2
18 in.
10 in.h = 8
in.
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CH11-2: Surface Area of Prisms and Cylinders
Now, you do EVENS
8, 10, & 12
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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 ∙ r2
h
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CH11-2: Surface Area of Prisms and Cylinders
Surface Area of a Prism
SA = LA +2∙Ab
LA = Sum of areas of sides
Ab = Area of Base Shape
h
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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 ∙ r2
SA = 2∙π ∙ 1m ∙ 3m+2 ∙ π ∙ (1m)2
SA = 6π +2π
SA = 8π
1m
3m
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CH11-2: Surface Area of Prisms and Cylinders
Now, you do EVENS
2,4,6,14
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CH11-3: Surface Area of Pyramids and Cones
Lateral Area of a Pyramid
LA = ½∙llll····PBase
LA = Lateral Area
LLLL = slant height
PBase=Perimeter of Base Shape
h
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llll
CH11-3: Surface Area of Pyramids and Cones
Surface Area of a Pyramid
SA = LA + ABase
LA = Lateral Area
Abase = Area of Base Shape
h
20
llll
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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∙llll····Pbase
LA = ½∙llll····(12+12+12+12)
How do we find slant height, llll?
21 ft
21
llll
12 ft12 ft
CH11-3: Surface Area of Pyramids and Cones
Yes! Pythagorean Theorem
c2 = a2+b2
c2 = 212+62
c2 = 441+36
c2 = 477
√c2 = √477
c= 21.84
Now, we can find Lateral Area, LA
21 ft
22
llll
12 ft12 ft
CH11-3: Surface Area of Pyramids and Cones
LA = ½∙llll····(12+12+12+12)
And we now know c=llll=21.84
LA = ½∙21.84····(12+12+12+12)
LA = 524.2 ft2
What shape is the base?
Square
SA = LA + Abase
SA = 524.2 + (12)2
SA = 668.2 ft2
21 ft
23
llll
12 ft12 ft
CH11-3: Surface Area of Pyramids and Cones
Now, you do 7, 9, and 11
21 ft
24
llll
12 ft12 ft
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CH11-3: Surface Area of Pyramids and Cones
Lateral Area of a Cone
LA = π∙r∙llll
LA = Lateral Area
LLLL = slant height
r=radius of base
25
r
llll
CH11-3: Surface Area of Pyramids and Cones
Surface Area of a Cone
SA = π∙r∙llll+πr2
SA = Surface Area
LLLL = slant height
r=radius of base
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r
llll
CH11-3: Surface Area of Pyramids and Cones
Now, find Lateral Area
c2=a2+b2
c2=202+92
c2=400+81
c2=481
√c2=√481
c=21.9
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9cm
20cmllll
CH11-3: Surface Area of Pyramids and Cones
Now, find Lateral Area
Now, find Lateral Area
LA = π∙r∙llll
LA = π∙(9)∙(21.9)
LA = 618.9 cm2
SA = π∙r∙llll+πr2
SA = π∙9∙(21.9)+π92
SA = 197.1π+81π
SA = 278.1π cm2
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9cm
20cmllll
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CH11-3: Surface Area of Pyramids and Cones
Now, you do 1, 3, and 5
21 ft
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llll
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
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CH11-4 Volumes of Prisms & Cylinders
Volume of a Cylinder
V=π•r2•h
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r
h
CH11-4 Volumes of Prisms & Cylinders
Ex.1 Find the volume of each cylinder to the nearest tenth.
V=π•r2•h
V = π•(2m)2•(2m)
V = 25.1 m3
4m
2m
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CH11-4 Volumes of Prisms & Cylinders
Now, you do
EVENS 2,4, 6
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CH11-4 Volumes of Prisms & Cylinders34
Volume of a Prism
V=Ab•h
Ab=Area of Base Shape
h=height of prism
18
in.
10
in.
h =
8 in.
CH11-4: Volume of Prisms & Cylinders
Ex.1 Find the volume of each prism.
First, what shape is the base?
Triangle, so
Ab = ½∙b∙h
Ab = ½∙(10in)∙(8in)
Ab = 40 in2
V= Ab•h
V = (40in2) •(18in)
V = 720 in3
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18 in.
10 in.h = 8
in.
CH11-4 Volumes of Prisms & Cylinders
Now, you do
EVENS 8 -14
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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
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h
CH11-5: Volumes of Pyramids and Cones
Ex.1 Find the volume of each
pyramid.
What shape is the base?
Square, so
Ab = s2
Ab = (12cm)2
Ab = 144cm2
V = 1/3∙Ab∙h
V = 1/3∙(144cm2)∙(21cm)
V = 1008 cm3
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21 ft
llll
12 ft12 ft
CH11-5: Volumes of Pyramids and Cones
Now, you do #S,
1,3,5,13&15
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CH11-5: Volumes of Pyramids and Cones
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
llll
h
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CH11-5: Volumes of Pyramids and Cones
Ex.2 Find the Volume of the cone.
V = 1/3∙π∙r2∙h
V = 1/3∙π∙(9cm)2∙(20cm)
V = 1696.5 cm3
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9cm
20cm
llll
CH11-5: Volumes of Pyramids and Cones
Now, you do #s 7,9, & 11
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11-6 Surface Areas and Volumes of Spheres43
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= 314.2 m2
b)
V= 4/3∙π∙r3
V= 4/3∙π∙(5m)3 V= 1.3333∙π∙(5m)3
V = 523.6 m3
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10 m
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CH11-6: Surface Areas and Volumes of Spheres
Now, you do EVENS (2-18)!
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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.
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SA1 = a2
SA2 b2
V1 = a3
V2 b3
a
b = similarity ratio
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11-7 Areas & Volumes of Similar Solids
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=a
4 b
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11-7 Areas & Volumes of Similar Solids
SA = 9SA = 16
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Now, you do ODDS
1-11
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11-7 Areas & Volumes of Similar Solids
Woooohoooo! We’re done with CH11!!!!!!
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11-7 Areas & Volumes of Similar Solids