atmim winter conference boston college january 9, 2014 steve yurek cambridge, massachusetts...

Lesley University Copyright Steve Yurek January 9, 2014

ATMIM Winter ConferenceBoston CollegeJanuary 9, 2014

Steve YurekCambridge, Massachusetts

[email protected]

1

Lessons from the “Lowly” Trapezoid


2

Every Middle Schooler knows the formula for the area of a rectangle

A = bh

h

b


3

And the right triangle

A = bh

h

b


4


A = bh

h

b


5


A = bh

h

b


6


A = ½ bh

h

b


7

But it comes as no surprise that they are not so sure when it comes to the areas of the other plane figures. First – the Parallelogram

h

b


8

b

h


9

b

hh


10

b

hh h

x y

y x


11

b

hh h

x y

y x

x


12

b

hh h


13

b

h


14

The Blue area is still the same size, but it’s just in the shape of a rectangle now, so A = bh once again

b

h


15

But it comes as no surprise that they are not so sure when it comes to the areas of the other plane figures. Next a Triangle

h

b


16

But it comes as no surprise that they are not so sure when it comes to the areas of the other plane figures. Next a Triangle

h

b

Not a real stretch to see that for a triangle A = ½ bh

h


17

And Now the Reason Why We’re Here!!!


18

Let’s look at all we ever had to know

about the Lowly Trapezoid


19


20


21


22


23


24


25

The sum of all 3 shapes will yield the formula for the area of a trapezoid.


26

Or


27


28


29


30


31


32


33


34

Subtract the sum of the 2 triangles from the outer rectangle and you have the formula for the area of a trapezoid


35

So this is starting from the EASY shapesand building to the COMPLEX shapes.


36

Let’s reverse it and see where it leads.

a

b

h

a

b

h

b

a

So the blue area is that of a parallelogram and the area is:

A = bh = hbA = h(a + b)

a

b

h

b

a b

a

But the area of the original trapezoid is half the area of the parallelogram, so the area of the trapezoid must be:

A = ½ h(a + b)

a

b

h

b

a


A = ½ h(a + b)

a

b

h

b

aAnd this leads us --- where???


A = ½ h(a + b)

a

b

h

b

aAnd this leads us --- where???

Click here to see where “where” is

Consider any right triangle


Label the legs as “a” & “b”, with hypotenuse “c”

a

b

c



Label the acute angles as “1” “2”

a

b

c

a

c

1

2




Now make a copy of the triangle

a

b

c

a

c

1

2





Then rotate and translate until it looks like this

a

b

c

a

c

1

2





Then rotate and translate until it looks like this, then translate it to the upper vertex

a

b

c

a

c

1

2

a

b

c

1

2

Let’s insert the notation into their proper places

a

a

b

b

c

c

1

1

2

2

a

a

b

b

c

c

1

1

2

2 Now draw the line segment as indicated

a

a

b

b

c

c

1

1

2

2

a

a

b

b

c

c

1

1

2

2The quadrilateral is a _________.

a

a

b

b

c

c

1

1

2

2The quadrilateral is a trapezoid.

a

a

b

b

c

c

1

1

2

2The quadrilateral is a trapezoid. WHY?

a

a

b

b

c

c

1

1

2


What is the measure in angle 3?

3

a

a

b

b

c

c

1

1

2


What is the measure in angle 3? WHY?

3

a

a

b

b

c

c

1

1

2



The area of the trapezoid is given by A = ½ h (b1 + b2)

3

a

a

b

b

c

c

1

1

2



The area of the trapezoid is given by A = ½ h (b1 + b2) OR

A = ½ (a + b) (a + b)

3

a

a

b

b

c

c

1

1

2




A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2)

3

a

a

b

b

c

c

1

1

2




A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2

3

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab,

3

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab, for a total area of ab 3

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab, for a total area of ab So the area of the large right triangle on the right must be determined by

3

c

c

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab, for a total area of ab So the area of the large right triangle on the right must be determined by subtracting the total of the areas of the 2 smaller right triangles from the area of the trapezoid.

3

c

c

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab, for a total area of ab So the area of the large right triangle on the right must be determined by subtracting the total of the areas of the 2 smaller right triangles from the area of the trapezoid. In other words, area of this triangle is (½ a2 + ab + ½ b2)

3

c

c

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab, for a total area of ab So the area of the large right triangle on the right must be determined by subtracting the total of the areas of the 2 smaller right triangles from the area of the trapezoid. In other words, area of this triangle is (½ a2 + ab + ½ b2) – ab =

3

c

c

a

a

b

b

c

c

1

1

2





The area of the original triangle and its duplicate are each ½ ab, for a total area of ab So the area of the large right triangle on the right must be determined by subtracting the total of the areas of the 2 smaller right triangles from the area of the trapezoid. In other words, area of this triangle is (½ a2 + ab + ½ b2) – ab = ½ a2 + ½ b2

3

c

c

a

a

b

b

c

c

1

1

2

2So we now know that the area of the right triangle on the right has an area of ½ a2 + ½ b2.

3

c

c

a

a

b

b

c

c

1

1

2

2So we now know that the area of the right triangle on the right has an area of ½ a2 + ½ b2.But since the measure of the length of each of the legs in this triangle is “c” ,

3

c

c

a

a

b

b

c

c

1

1

2

2So we now know that the area of the right triangle on the right has an area of ½ a2 + ½ b2.But since the measure of the length of each of the legs in this triangle is “c” , then its area is: A = ½ c2

3

c

c

a

a

b

b

c

c

1

1

2

2So we now know that the area of the right triangle on

the right has an area of ½ a2 + ½ b2.But since the measure of the length of each of the legs

in this triangle is “c” , then its area is: A = ½ c2

3

We now have 2 expressions that measure the same value, so they must be equal to themselves, namely:

c

c

a

a

b

b

c

c

1

1

2


3

We now have 2 expressions that measure the same value, so they must be equal to themselves, namely:½ c2 = ½ a2 + ½ b2

c

c

a

a

b

b

c

c

1

1

2


3

We now have 2 expressions that measure the same value, so they must be equal to themselves, namely:½ c2 = ½ a2 + ½ b2 OR

c

c

a

a

b

b

c

c

1

1

2


3

We now have 2 expressions that measure the same value, so they must be equal to themselves, namely:½ c2 = ½ a2 + ½ b2 OR c2 = a2 + b 2

c

c

a

a

b

b

c

c

1

1

2


3


Q

c

c

a

a

b

b

c

c

1

1

2


3


QE

c

c

a

a

b

b

c

c

1

1

2


3


QED

c

c

a

a

b

b

c

c

1

1

2


3


This proof has ties to the U.S. House of Representatives because it is the handiwork of President James Garfield, who was a member of the House at the time.

c

c


90

Some Unexpectedness

Thanks to Alfred S. Posamentier: The Glorius Golden Ratio


91

Some Unexpectedness


92

Some Unexpectedness


93

Some Unexpectedness


94

a


95

a

aaa


96

a

aaa

Draw a segment (B)// base

so that the 2 areas are equal


97

a

aaa

Draw a segment (B) // base

so that the 2 areas are equalB


98

a

aaa


so that the 2 areas are equalB


99

a

aaa



h1

h2

B

Area top


100

a

aaa



h1

h2

B

1

1Area ( )

2top h


101

a

aaa



h1

h2

B

1

1Area ( )( B)

2top h a


102

a

aaa



h1

h2

B

1

1Area ( )( B)

2

Area =

top

bottom

h a


103

a

aaa



h1

h2

B

1

2

1Area ( )( B)

2

1Area = ( )

2

top

bottom

h a

h


104

a

aaa



h1

h2

B

1

2

1Area ( )( B)

2

1Area = ( )(B + 3 )

2

top

bottom

h a

h a


105

a

aaa



h1

h2

B

1

2

.......

.......

h

h

1

2

1Area ( )( B)

2

1Area = ( )(B + 3 )

2

top

bottom

h a

h a


106

a

aaa



h1

h2

B

1

2

3h B a

h B a

1

2

1Area ( )( B)

2

1Area = ( )(B + 3 )

2

top

bottom

h a

h a


107

a

aaa



h1

h2

B

1

2

3h B a

h B a

Now consider the 2 similar triangles on the left


108

a

aaa



h1

h2

B

1

2

3h B a

h B a



109

a

aaa



h1

h2

B

1

2

3h B a

h B a



110

a

aaa



h1

h2

B

1

2

3h B a

h B a


111

a

aaa



h1

h2

B

1

2

3h B a

h B a


112

a

aaa



h1

h2

B

2

B a

1

2

3h B a

h B a


113

a

aaa



h1

h2

B

2

B a

1

2

3h B a

h B a

1

h


114

a

aaa



h1

h2

B

2

B a

1

2

3h B a

h B a

1

1 2

h

h h


115

a

aaa



h1

h2

B

2

B a

1

2

3h B a

h B a

1

1 2

h

h h


116

a

aaa



h1

h2

B

2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h


117

a

aaa



h1

h2

B

2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a


118

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a


119

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

Let’s see what can be done with proportions


120

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a


121

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

1

1 2

h

h h


122

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

1

1 2

h

h h

3

( 3 ) ( )

B a

B a B a


123

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

1

1 2

h

h h

3

( 3 ) ( )

B a

B a B a

3

2 4

B a

B a


124

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

1

1 2

h

h h

3

( 3 ) ( )

B a

B a B a

3

2 4

B a

B a


125

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

1

1 2

h

h h

3

( 3 ) ( )

B a

B a B a

3

2 4

B a

B a


126

a

aaa



h1

h2

B2

B a

1

2

3h B a

h B a

1

1 2

2B a

h

h h a

1

1 2

h

h h

3

( 3 ) ( )

B a

B a B a

3

2 4

B a

B a


127

a

aaa



h1

h2

B2

B a

2B a

a

3

2 4

B a

B a

=


128

a

aaa



h1

h2

B2

B a

2B a

a

3

2 4

B a

B a

=

So now we can solve for B


129

2B a

a

3

2 4

B a

B a

=


130

2B a

a

3

2 4

B a

B a

=

( - )( 2 ) ( 3 )B a B a a B a


131

2B a

a

3

2 4

B a

B a

=

( - )( 2 ) ( 3 )B a B a a B a 2 2 22 3B aB a aB a


132

2B a

a

3

2 4

B a

B a

=

( - )( 2 ) ( 3 )B a B a a B a 2 2 22 3B aB a aB a 2 25B a


133

2B a

a

3

2 4

B a

B a

=


5B a


134

Now for one more substitution

2B a

a

3

2 4

B a

B a

=


5B a


135

1

2

h +3aSince = and =a 5, then

h +a

BB

B


136

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a


137

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a


138

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a

1

2

5 3

5 1

h

h


139

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a

1

2

5 3

5 1

h

h

After we rationalize the denominators, we get


140

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a

1

2

5 3

5 1

h

h


1

2

h

h


141

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a

1

2

5 3

5 1

h

h


1

2

1 5

2

h

h


142

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a

1

2

5 3

5 1

h

h


1

2

1 5

2

h

h

= ?


143

1

2


h +a

BB

B

1

2

5 3

5

h a a

h a a

1

2

5 3

5 1

h

h


1

2

1 5

2

h

h

= Ø


144

Thanks to Alfred S. Posamentier: The Glorius Golden Ratio


145


146


147


148


149


150


151


152

Can you find any relationship between any of the sides of this trapezoid?Specifically between sides b & c?


153



154


b


155


b

c


156


b

c


157

Let’s look for some

added value


158

Sarah Jane bought cartridges for the various printers in her office and spent a total of $350 for each type. She paid $25 for each of the cartridges for the black & white printers for the staff and $70 for each of the cartridges for her supervisor’s color printer. What was the average cost of a single cartridge?


159



160



161



162

Sarah Jane bought cartridges for the various printers in her office and spent a total of $350 for each type. She paid $25 for each of the cartridges for the black & white printers for the staff and $70 for each of the cartridges for her supervisor’s color printer. What was the average cost of a single cartridge? Did you get $36.84 ?


163

Questions such as these involve the Harmonic Mean. There are 3 ways to compute the Harmonic Mean of a & b.1. Arithmetic Sense We just did that2. Definition: The reciprocal of the

arithmetic mean of the reciprocals3. Formula:


164




165

Questions such as these involve the Harmonic Mean. There are 3 ways to compute the Harmonic Mean of a & b.1. Arithmetic Sense: We just did that2. Definition: The reciprocal of the



166

Questions such as these involve the Harmonic Mean. There are 3 ways to compute the Harmonic Mean of a & b.1. Arithmetic Sense: We just did that2. Definition: The reciprocal of the



167




168




169




170




171




172


arithmetic mean of the reciprocals3. Formula: 2ab

HMa b


173

11 125 70

2


174

11 125 7036.84210526

2


175

2*25*70

25 70


176

2*25*7034.84210526

25 70


177

Sarah Jane bought cartridges for the various printers in her office and spent a total of $350 for each type. She paid $25 for each of the cartridges for the black & white printers for the staff and $70 for each of the cartridges for her supervisor’s color printer. What was the average cost of a single cartridge? Sarah Jane and her Printers


178

Here’s another “unexpected “


179


180

80’

50’30’


181

80’

50’30’

E


182

80’

50’30’

E

?


183

80’

50’30’

E

18.75


184

80’

50’30’

E

18.75

How should the pole(s) be moved so that E will eventually be 20 feet

above the ground?


185

80’

50’30’

E

18.75

As it turns out, this is very cool --- watch this


186

Where Can This Lead Us and our

Students?


187

Perhaps from 2D to 3D


188


189


190


191

It turns out that a crucial component of determining the

volume of a right rectangular pyramid is todetermine the formula for the following series:

2 2 2 2 21 2 3 4 ... n


192

• We can determine the formula for the sum of the squares of the first n integers in many ways: Finite Differences and Mathematical Induction involve only algebra, but let’s take a look at this:


193


194


195


196


197


198


199


200


201


202

N


203

N + 1

N


204

N + 1N + 1

N


205


206

N + 1N + 1

N

11 1 1

2V n n n n n


207

Or


208


209

n


210

n

n


211

n

n + 1n


212

n

n + 1n

11 1

2V n n n n n


213

Or


214


215

n


216

n + 1 n


217

n + 1 n


218

n + 1 n

n + ½


219

n + 1 n

n + ½

11

2V n n n


220

So Let’s Compare the Values of:

11 1 1

2V n n n n n


221


11 1 1

2V n n n n n

11 1

2V n n n n n


222


11 1 1

2V n n n n n

11 1

2V n n n n n

11

2V n n n


223

They all Simplify to:3 22 3

2

n n nV


224


2

n n nV

22 3 1

2

n n nV


225


2

n n nV

22 3 1

2

n n nV

1 (2n 1)

2

n nV


226

But this represents the volume of all 3 sets of shapes combined, meaning that each original set, or sum of the 1st “n” squares”, can be represented by:


227



228


2

1

( 1)(2 1)

6

n

i

n n ni


229

By the way

Lest we forget our target:The volume of a frustrum of a square pyramid with larger base of area “B”, smaller base of area “A” and height “h” is given by:


230

By the way



231

By the way



232

By the way



233

By the way


1 ( )

3V h A B AB


234

Thanks for Being HereEnjoy the Remainder

of the Conference


235

Thanks for Being HereEnjoy the Remainder

of the ConferenceDownload these

Powerpoint Slides atwww.atmim.net

atmim winter conference boston college january 9, 2014 steve yurek cambridge, massachusetts...

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