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Lesley University Copyright Steve Yurek January 9, 2014
ATMIM Winter ConferenceBoston CollegeJanuary 9, 2014
Steve YurekCambridge, Massachusetts
1
Lessons from the “Lowly” Trapezoid
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Every Middle Schooler knows the formula for the area of a rectangle
A = bh
h
b
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And the right triangle
A = bh
h
b
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And the right triangle
A = bh
h
b
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And the right triangle
A = bh
h
b
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And the right triangle
A = ½ bh
h
b
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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
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b
h
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b
hh
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b
hh h
x y
y x
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b
hh h
x y
y x
x
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b
hh h
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b
h
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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
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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
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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
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And Now the Reason Why We’re Here!!!
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Let’s look at all we ever had to know
about the Lowly Trapezoid
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The sum of all 3 shapes will yield the formula for the area of a trapezoid.
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Or
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Subtract the sum of the 2 triangles from the outer rectangle and you have the formula for the area of a trapezoid
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So this is starting from the EASY shapesand building to the COMPLEX shapes.
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Let’s reverse it and see where it leads.
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a
b
h
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a
b
h
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a
b
h
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a
b
h
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a
b
h
b
a
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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
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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
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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
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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
aAnd this leads us --- where???
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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
aAnd this leads us --- where???
Click here to see where “where” is
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Consider any right triangle
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Consider any right triangle
Label the legs as “a” & “b”, with hypotenuse “c”
a
b
c
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Consider any right triangle
Label the legs as “a” & “b”, with hypotenuse “c”
Label the acute angles as “1” “2”
a
b
c
a
c
1
2
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Consider any right triangle
Label the legs as “a” & “b”, with hypotenuse “c”
Label the acute angles as “1” “2”
Now make a copy of the triangle
a
b
c
a
c
1
2
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Consider any right triangle
Label the legs as “a” & “b”, with hypotenuse “c”
Label the acute angles as “1” “2”
Now make a copy of the triangle
a
b
c
a
c
1
2
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Consider any right triangle
Label the legs as “a” & “b”, with hypotenuse “c”
Label the acute angles as “1” “2”
Now make a copy of the triangle
Then rotate and translate until it looks like this
a
b
c
a
c
1
2
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Consider any right triangle
Label the legs as “a” & “b”, with hypotenuse “c”
Label the acute angles as “1” “2”
Now make a copy of the triangle
Then rotate and translate until it looks like this, then translate it to the upper vertex
a
b
c
a
c
1
2
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a
b
c
1
2
Let’s insert the notation into their proper places
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a
a
b
b
c
c
1
1
2
2
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a
a
b
b
c
c
1
1
2
2 Now draw the line segment as indicated
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a
a
b
b
c
c
1
1
2
2
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a _________.
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid.
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3?
3
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
3
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2)
3
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a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b)
3
![Page 65: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/65.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b)
3
![Page 66: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/66.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2)
3
![Page 67: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/67.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
3
![Page 68: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/68.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
The area of the original triangle and its duplicate are each ½ ab,
3
![Page 69: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/69.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
The area of the original triangle and its duplicate are each ½ ab,
3
![Page 70: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/70.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
The area of the original triangle and its duplicate are each ½ ab,
3
![Page 71: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/71.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
The area of the original triangle and its duplicate are each ½ ab, for a total area of ab 3
![Page 72: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/72.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 73: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/73.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 74: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/74.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 75: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/75.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 76: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/76.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 77: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/77.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 78: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/78.jpg)
a
a
b
b
c
c
1
1
2
2The quadrilateral is a trapezoid. WHY?
What is the measure in angle 3? WHY?
The area of the trapezoid is given by A = ½ h (b1 + b2) OR
A = ½ (a + b) (a + b) A = ½ (a2 + 2ab + b2) A = ½ a2 +ab + ½ b2
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
![Page 79: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/79.jpg)
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
![Page 80: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/80.jpg)
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
![Page 81: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/81.jpg)
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
![Page 82: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/82.jpg)
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
![Page 83: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/83.jpg)
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:½ c2 = ½ a2 + ½ b2
c
c
![Page 84: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/84.jpg)
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:½ c2 = ½ a2 + ½ b2 OR
c
c
![Page 85: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/85.jpg)
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:½ c2 = ½ a2 + ½ b2 OR c2 = a2 + b 2
c
c
![Page 86: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/86.jpg)
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:½ c2 = ½ a2 + ½ b2 OR c2 = a2 + b 2
Q
c
c
![Page 87: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/87.jpg)
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:½ c2 = ½ a2 + ½ b2 OR c2 = a2 + b 2
QE
c
c
![Page 88: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/88.jpg)
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:½ c2 = ½ a2 + ½ b2 OR c2 = a2 + b 2
QED
c
c
![Page 89: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/89.jpg)
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:½ c2 = ½ a2 + ½ b2 OR c2 = a2 + b 2
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
![Page 90: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/90.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
90
Some Unexpectedness
Thanks to Alfred S. Posamentier: The Glorius Golden Ratio
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Lesley University Copyright Steve Yurek January 9, 2014
91
Some Unexpectedness
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Lesley University Copyright Steve Yurek January 9, 2014
92
Some Unexpectedness
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Lesley University Copyright Steve Yurek January 9, 2014
93
Some Unexpectedness
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Lesley University Copyright Steve Yurek January 9, 2014
94
a
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Lesley University Copyright Steve Yurek January 9, 2014
95
a
aaa
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Lesley University Copyright Steve Yurek January 9, 2014
96
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
![Page 97: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/97.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
97
a
aaa
Draw a segment (B) // base
so that the 2 areas are equalB
![Page 98: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/98.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
98
a
aaa
Draw a segment (B)// base
so that the 2 areas are equalB
![Page 99: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/99.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
99
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
Area top
![Page 100: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/100.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
100
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
1Area ( )
2top h
![Page 101: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/101.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
101
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
1Area ( )( B)
2top h a
![Page 102: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/102.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
102
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
1Area ( )( B)
2
Area =
top
bottom
h a
![Page 103: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/103.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
103
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
1Area ( )( B)
2
1Area = ( )
2
top
bottom
h a
h
![Page 104: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/104.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
104
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
1Area ( )( B)
2
1Area = ( )(B + 3 )
2
top
bottom
h a
h a
![Page 105: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/105.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
105
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
.......
.......
h
h
1
2
1Area ( )( B)
2
1Area = ( )(B + 3 )
2
top
bottom
h a
h a
![Page 106: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/106.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
106
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 107: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/107.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
107
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
3h B a
h B a
Now consider the 2 similar triangles on the left
![Page 108: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/108.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
108
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
3h B a
h B a
Now consider the 2 similar triangles on the left
![Page 109: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/109.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
109
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
3h B a
h B a
Now consider the 2 similar triangles on the left
![Page 110: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/110.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
110
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
3h B a
h B a
![Page 111: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/111.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
111
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
1
2
3h B a
h B a
![Page 112: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/112.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
112
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
2
B a
1
2
3h B a
h B a
![Page 113: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/113.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
113
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
2
B a
1
2
3h B a
h B a
1
h
![Page 114: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/114.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
114
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
2
B a
1
2
3h B a
h B a
1
1 2
h
h h
![Page 115: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/115.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
115
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
2
B a
1
2
3h B a
h B a
1
1 2
h
h h
![Page 116: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/116.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
116
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
2
B a
1
2
3h B a
h B a
1
1 2
2B a
h
h h
![Page 117: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/117.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
117
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B
2
B a
1
2
3h B a
h B a
1
1 2
2B a
h
h h a
![Page 118: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/118.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
118
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B2
B a
1
2
3h B a
h B a
1
1 2
2B a
h
h h a
![Page 119: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/119.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
119
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 120: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/120.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
120
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B2
B a
1
2
3h B a
h B a
1
1 2
2B a
h
h h a
![Page 121: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/121.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
121
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 122: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/122.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
122
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 123: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/123.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
123
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 124: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/124.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
124
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 125: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/125.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
125
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 126: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/126.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
126
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
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
![Page 127: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/127.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
127
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B2
B a
2B a
a
3
2 4
B a
B a
=
![Page 128: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/128.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
128
a
aaa
Draw a segment (B)// base
so that the 2 areas are equal
h1
h2
B2
B a
2B a
a
3
2 4
B a
B a
=
So now we can solve for B
![Page 129: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/129.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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2B a
a
3
2 4
B a
B a
=
![Page 130: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/130.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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2B a
a
3
2 4
B a
B a
=
( - )( 2 ) ( 3 )B a B a a B a
![Page 131: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/131.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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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
![Page 132: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/132.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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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
![Page 133: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/133.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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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
5B a
![Page 134: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/134.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
134
Now for one more substitution
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
5B a
![Page 135: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/135.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
135
1
2
h +3aSince = and =a 5, then
h +a
BB
B
![Page 136: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/136.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
136
1
2
h +3aSince = and =a 5, then
h +a
BB
B
1
2
5 3
5
h a a
h a a
![Page 137: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/137.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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1
2
h +3aSince = and =a 5, then
h +a
BB
B
1
2
5 3
5
h a a
h a a
![Page 138: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/138.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
138
1
2
h +3aSince = and =a 5, then
h +a
BB
B
1
2
5 3
5
h a a
h a a
1
2
5 3
5 1
h
h
![Page 139: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/139.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
139
1
2
h +3aSince = and =a 5, then
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
![Page 140: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/140.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
140
1
2
h +3aSince = and =a 5, then
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
1
2
h
h
![Page 141: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/141.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
141
1
2
h +3aSince = and =a 5, then
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
1
2
1 5
2
h
h
![Page 142: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/142.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
142
1
2
h +3aSince = and =a 5, then
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
1
2
1 5
2
h
h
= ?
![Page 143: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/143.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
143
1
2
h +3aSince = and =a 5, then
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
1
2
1 5
2
h
h
= Ø
![Page 144: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/144.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
144
Thanks to Alfred S. Posamentier: The Glorius Golden Ratio
![Page 145: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/145.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
145
![Page 146: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/146.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
146
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Lesley University Copyright Steve Yurek January 9, 2014
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Lesley University Copyright Steve Yurek January 9, 2014
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![Page 152: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/152.jpg)
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Can you find any relationship between any of the sides of this trapezoid?Specifically between sides b & c?
![Page 153: ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek.](https://reader038.fdocuments.us/reader038/viewer/2022110304/551bf452550346ad4f8b45b2/html5/thumbnails/153.jpg)
Lesley University Copyright Steve Yurek January 9, 2014
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Can you find any relationship between any of the sides of this trapezoid?Specifically between sides b & c?
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Can you find any relationship between any of the sides of this trapezoid?Specifically between sides b & c?
b
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Can you find any relationship between any of the sides of this trapezoid?Specifically between sides b & c?
b
c
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Can you find any relationship between any of the sides of this trapezoid?Specifically between sides b & c?
b
c
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Let’s look for some
added value
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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?
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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?
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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?
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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?
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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 ?
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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:
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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:
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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:
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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:
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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:
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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:
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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:
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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:
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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:
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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: 2ab
HMa b
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11 125 70
2
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11 125 7036.84210526
2
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2*25*70
25 70
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2*25*7034.84210526
25 70
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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
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Here’s another “unexpected “
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80’
50’30’
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80’
50’30’
E
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80’
50’30’
E
?
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80’
50’30’
E
18.75
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80’
50’30’
E
18.75
How should the pole(s) be moved so that E will eventually be 20 feet
above the ground?
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80’
50’30’
E
18.75
As it turns out, this is very cool --- watch this
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Where Can This Lead Us and our
Students?
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Perhaps from 2D to 3D
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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
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• 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:
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N
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N + 1
N
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N + 1N + 1
N
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N + 1N + 1
N
11 1 1
2V n n n n n
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Or
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n
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n
n
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n
n + 1n
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n
n + 1n
11 1
2V n n n n n
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Or
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n
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n + 1 n
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n + 1 n
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n + 1 n
n + ½
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n + 1 n
n + ½
11
2V n n n
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So Let’s Compare the Values of:
11 1 1
2V n n n n n
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So Let’s Compare the Values of:
11 1 1
2V n n n n n
11 1
2V n n n n n
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So Let’s Compare the Values of:
11 1 1
2V n n n n n
11 1
2V n n n n n
11
2V n n n
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They all Simplify to:3 22 3
2
n n nV
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They all Simplify to:3 22 3
2
n n nV
22 3 1
2
n n nV
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They all Simplify to:3 22 3
2
n n nV
22 3 1
2
n n nV
1 (2n 1)
2
n nV
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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:
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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:
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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:
2
1
( 1)(2 1)
6
n
i
n n ni
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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:
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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:
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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:
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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:
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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:
1 ( )
3V h A B AB
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Thanks for Being HereEnjoy the Remainder
of the Conference
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Thanks for Being HereEnjoy the Remainder
of the ConferenceDownload these
Powerpoint Slides atwww.atmim.net