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ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts syurek@lesley.edu 1 Lesley University Copyright Steve Yurek January 9, 2014

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Every Middle Schooler knows the formula for the area of a rectangle A = bh h b 2 Lesley University Copyright Steve Yurek January 9, 2014

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And the right triangle A = bh h b 3 Lesley University Copyright Steve Yurek January 9, 2014

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And the right triangle A = bh h b 4 Lesley University Copyright Steve Yurek January 9, 2014

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And the right triangle A = bh h b 5 Lesley University Copyright Steve Yurek January 9, 2014

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And the right triangle A = ½ bh h b 6 Lesley University Copyright Steve Yurek January 9, 2014

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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 7 Lesley University Copyright Steve Yurek January 9, 2014

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b h 8

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b h 9 h

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b h 10 Lesley University Copyright Steve Yurek January 9, 2014 hh x y y x

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b h 11 Lesley University Copyright Steve Yurek January 9, 2014 hh x y y x x

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b h 12 Lesley University Copyright Steve Yurek January 9, 2014 hh

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b h 13 Lesley University Copyright Steve Yurek January 9, 2014

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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 14 Lesley University Copyright Steve Yurek January 9, 2014

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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 15 Lesley University Copyright Steve Yurek January 9, 2014

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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 16 Lesley University Copyright Steve Yurek January 9, 2014

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Lesley University Copyright Steve Yurek January 9, 2014 24

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Lesley University Copyright Steve Yurek January 9, 2014 25 The sum of all 3 shapes will yield the formula for the area of a trapezoid.

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Lesley University Copyright Steve Yurek January 9, 2014 26

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Lesley University Copyright Steve Yurek January 9, 2014 32

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Lesley University Copyright Steve Yurek January 9, 2014 33

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Lesley University Copyright Steve Yurek January 9, 2014 34 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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Lesley University Copyright Steve Yurek January 9, 2014 35 So this is starting from the EASY shapes and building to the COMPLEX shapes.

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Lesley University Copyright Steve Yurek January 9, 2014 36 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 = hb A = 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 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 Click here to see where “where” is

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Consider any right triangle

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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 2 The quadrilateral is a _________.

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid.

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY?

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a a b b c c 1 1 2 2 What is the measure in angle 3? 3

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a a b b c c 1 1 2 2 The 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 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 2 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 2 The area of the original triangle and its duplicate are each ½ ab, 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 2 The area of the original triangle and its duplicate are each ½ ab, 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 2 The area of the original triangle and its duplicate are each ½ ab, 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 2 The area of the original triangle and its duplicate are each ½ ab, for a total area of ab 3

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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 a 2 + ab + ½ b 2 ) 3 c c

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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 a 2 + ab + ½ b 2 ) 3 c c

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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 a 2 + ab + ½ b 2 ) – ab = 3 c c

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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 a 2 + ab + ½ b 2 ) – ab = 3 c c

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a a b b c c 1 1 2 2 The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? The area of the trapezoid is given by A = ½ h (b 1 + b 2 ) OR A = ½ (a + b) (a + b) A = ½ (a 2 + 2ab + b 2 ) A = ½ a 2 +ab + ½ b 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 a 2 + ab + ½ b 2 ) – ab = ½ a 2 + ½ b 2 3 c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. 3 c c

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a a b b c c 1 1 2 2 But since the measure of the length of each of the legs in this triangle is “c”, 3 c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is : A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 OR c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 OR c 2 = a 2 + b 2 c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 OR c 2 = a 2 + b 2 Q c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 OR c 2 = a 2 + b 2 QE c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 OR c 2 = a 2 + b 2 QED c c

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a a b b c c 1 1 2 2 So we now know that the area of the right triangle on the right has an area of ½ a 2 + ½ b 2. But since the measure of the length of each of the legs in this triangle is “c”, then its area is: A = ½ c 2 3 We now have 2 expressions that measure the same value, so they must be equal to themselves, namely: ½ c 2 = ½ a 2 + ½ b 2 OR c 2 = a 2 + 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

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Some Unexpectedness 90 Lesley University Copyright Steve Yurek January 9, 2014 Thanks to Alfred S. Posamentier: The Glorius Golden Ratio

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Some Unexpectedness 91 Lesley University Copyright Steve Yurek January 9, 2014

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Some Unexpectedness 92 Lesley University Copyright Steve Yurek January 9, 2014

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Some Unexpectedness 93 Lesley University Copyright Steve Yurek January 9, 2014

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a 94 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa 95 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal 96 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal B 97 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal B 98 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 99 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 100 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 101 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 102 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 103 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 104 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 105 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 106 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B Now consider the 2 similar triangles on the left 107 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B Now consider the 2 similar triangles on the left 108 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B Now consider the 2 similar triangles on the left 109 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 110 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 111 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 112 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 113 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 114 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 115 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 116 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 117 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 118 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 119 Lesley University Copyright Steve Yurek January 9, 2014 Let’s see what can be done with proportions

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 120 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 121 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 122 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 123 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 124 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 125 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B 126 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B = 127 Lesley University Copyright Steve Yurek January 9, 2014

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a aaa Draw a segment (B) // base so that the 2 areas are equal h1h1 h2h2 B = So now we can solve for B 128 Lesley University Copyright Steve Yurek January 9, 2014

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= 129 Lesley University Copyright Steve Yurek January 9, 2014

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= 130 Lesley University Copyright Steve Yurek January 9, 2014

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= 131 Lesley University Copyright Steve Yurek January 9, 2014

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= 132 Lesley University Copyright Steve Yurek January 9, 2014

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= 133 Lesley University Copyright Steve Yurek January 9, 2014

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Now for one more substitution = 134 Lesley University Copyright Steve Yurek January 9, 2014

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135 Lesley University Copyright Steve Yurek January 9, 2014

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136 Lesley University Copyright Steve Yurek January 9, 2014

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137 Lesley University Copyright Steve Yurek January 9, 2014

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138 Lesley University Copyright Steve Yurek January 9, 2014

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After we rationalize the denominators, we get 139 Lesley University Copyright Steve Yurek January 9, 2014

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After we rationalize the denominators, we get 140 Lesley University Copyright Steve Yurek January 9, 2014

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After we rationalize the denominators, we get 141 Lesley University Copyright Steve Yurek January 9, 2014

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After we rationalize the denominators, we get = ? 142 Lesley University Copyright Steve Yurek January 9, 2014

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After we rationalize the denominators, we get = Ø 143 Lesley University Copyright Steve Yurek January 9, 2014

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144 Lesley University Copyright Steve Yurek January 9, 2014 Thanks to Alfred S. Posamentier: The Glorius Golden Ratio

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145 Lesley University Copyright Steve Yurek January 9, 2014

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146 Lesley University Copyright Steve Yurek January 9, 2014

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147 Lesley University Copyright Steve Yurek January 9, 2014

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148 Lesley University Copyright Steve Yurek January 9, 2014

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150 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? 152 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? 153 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? 154 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? 155 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? 156 Lesley University Copyright Steve Yurek January 9, 2014

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157 Lesley University Copyright Steve Yurek January 9, 2014

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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? 158 Lesley University Copyright Steve Yurek January 9, 2014

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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? 159 Lesley University Copyright Steve Yurek January 9, 2014

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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? 160 Lesley University Copyright Steve Yurek January 9, 2014

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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? 161 Lesley University Copyright Steve Yurek January 9, 2014

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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 ? 162 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 163 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 164 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 165 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 166 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 167 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 168 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 169 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 170 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 171 Lesley University Copyright Steve Yurek January 9, 2014

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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 that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 172 Lesley University Copyright Steve Yurek January 9, 2014

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173 Lesley University Copyright Steve Yurek January 9, 2014

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174 Lesley University Copyright Steve Yurek January 9, 2014

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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 PrintersSarah Jane and her Printers 177 Lesley University Copyright Steve Yurek January 9, 2014

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Here’s another “unexpected “ 178 Lesley University Copyright Steve Yurek January 9, 2014

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179 Lesley University Copyright Steve Yurek January 9, 2014

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80’ 50’ 30’ 180 Lesley University Copyright Steve Yurek January 9, 2014

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80’ 50’ 30’ E 181 Lesley University Copyright Steve Yurek January 9, 2014

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80’ 50’ 30’ E ? 182 Lesley University Copyright Steve Yurek January 9, 2014

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80’ 50’ 30’ E 18.75 183 Lesley University Copyright Steve Yurek January 9, 2014

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80’ 50’ 30’ E 18.75 184 How should the pole(s) be moved so that E will eventually be 20 feet above the ground? Lesley University Copyright Steve Yurek January 9, 2014

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80’ 50’ 30’ E 18.75 As it turns out, this is very cool --- watch this 185 Lesley University Copyright Steve Yurek January 9, 2014

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Where Can This Lead Us and our Students? Lesley University Copyright Steve Yurek January 9, 2014 186

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Perhaps from 2D to 3D Lesley University Copyright Steve Yurek January 9, 2014 187

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Lesley University Copyright Steve Yurek January 9, 2014 188

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Lesley University Copyright Steve Yurek January 9, 2014 189

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Lesley University Copyright Steve Yurek January 9, 2014 190

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Lesley University Copyright Steve Yurek January 9, 2014 191 It turns out that a crucial component of determining the volume of a right rectangular pyramid is to determine the formula for the following series:

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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: Lesley University Copyright Steve Yurek January 9, 2014 192

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Lesley University Copyright Steve Yurek January 9, 2014 202 N

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Lesley University Copyright Steve Yurek January 9, 2014 203 N + 1 N

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Lesley University Copyright Steve Yurek January 9, 2014 204 N + 1 N

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Lesley University Copyright Steve Yurek January 9, 2014 205

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Lesley University Copyright Steve Yurek January 9, 2014 206 N + 1 N

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Or Lesley University Copyright Steve Yurek January 9, 2014 207

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Lesley University Copyright Steve Yurek January 9, 2014 208

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Lesley University Copyright Steve Yurek January 9, 2014 209 n

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Lesley University Copyright Steve Yurek January 9, 2014 210 n n

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Lesley University Copyright Steve Yurek January 9, 2014 211 n n + 1 n

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Lesley University Copyright Steve Yurek January 9, 2014 212 n n + 1 n

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Or Lesley University Copyright Steve Yurek January 9, 2014 213

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Lesley University Copyright Steve Yurek January 9, 2014 214

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Lesley University Copyright Steve Yurek January 9, 2014 215 n

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Lesley University Copyright Steve Yurek January 9, 2014 216 n + 1 n

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Lesley University Copyright Steve Yurek January 9, 2014 217 n + 1 n

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Lesley University Copyright Steve Yurek January 9, 2014 218 n + 1 n n + ½

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Lesley University Copyright Steve Yurek January 9, 2014 219 n + 1 n n + ½

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So Let’s Compare the Values of: Lesley University Copyright Steve Yurek January 9, 2014 220

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So Let’s Compare the Values of: Lesley University Copyright Steve Yurek January 9, 2014 221

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So Let’s Compare the Values of: Lesley University Copyright Steve Yurek January 9, 2014 222

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They all Simplify to: Lesley University Copyright Steve Yurek January 9, 2014 223

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They all Simplify to: Lesley University Copyright Steve Yurek January 9, 2014 224

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They all Simplify to: Lesley University Copyright Steve Yurek January 9, 2014 225

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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 1 st “n” squares”, can be represented by: Lesley University Copyright Steve Yurek January 9, 2014 226

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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 1 st “n” squares”, can be represented by: Lesley University Copyright Steve Yurek January 9, 2014 227

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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 1 st “n” squares”, can be represented by: Lesley University Copyright Steve Yurek January 9, 2014 228

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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: Lesley University Copyright Steve Yurek January 9, 2014 229

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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: Lesley University Copyright Steve Yurek January 9, 2014 230

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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: Lesley University Copyright Steve Yurek January 9, 2014 231

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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: Lesley University Copyright Steve Yurek January 9, 2014 232

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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: Lesley University Copyright Steve Yurek January 9, 2014 233

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Lesley University Copyright Steve Yurek January 9, 2014 235

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2 20 107 1 106 13 19 21 25 22/24 3 12 14 23 4 11 15 26 27 30 5 7 6 9 16 18 28 104 29 8/17 32/109 100 99 105 108 31 34 40/42 41 98 68 93 90/92 91 81 80.

2 20 107 1 106 13 19 21 25 22/24 3 12 14 23 4 11 15 26 27 30 5 7 6 9 16 18 28 104 29 8/17 32/109 100 99 105 108 31 34 40/42 41 98 68 93 90/92 91 81 80.

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