# ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts 1 Lesley University Copyright Steve Yurek.

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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

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

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

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

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

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

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

b h 8

b h 9 h

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

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

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

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

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

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

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

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

Lesley University Copyright Steve Yurek January 9, 2014 36 Let’s reverse it and see where it leads.

a b h

a b h

a b h

a b h

a b h b a

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

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

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

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

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

Consider any right triangle

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

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

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

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

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

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

a b c 1 2 Let’s insert the notation into their proper places

a a b b c c 1 1 2 2

a a b b c c 1 1 2 2 Now draw the line segment as indicated

a a b b c c 1 1 2 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a 94 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

= 129 Lesley University Copyright Steve Yurek January 9, 2014

= 130 Lesley University Copyright Steve Yurek January 9, 2014

= 131 Lesley University Copyright Steve Yurek January 9, 2014

= 132 Lesley University Copyright Steve Yurek January 9, 2014

= 133 Lesley University Copyright Steve Yurek January 9, 2014

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

135 Lesley University Copyright Steve Yurek January 9, 2014

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138 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

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

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

145 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

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

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

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

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

157 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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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

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

179 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

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

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

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

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

Lesley University Copyright Steve Yurek January 9, 2014 188

Lesley University Copyright Steve Yurek January 9, 2014 189

Lesley University Copyright Steve Yurek January 9, 2014 190

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:

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

Lesley University Copyright Steve Yurek January 9, 2014 193

Lesley University Copyright Steve Yurek January 9, 2014 194

Lesley University Copyright Steve Yurek January 9, 2014 195

Lesley University Copyright Steve Yurek January 9, 2014 196

Lesley University Copyright Steve Yurek January 9, 2014 197

Lesley University Copyright Steve Yurek January 9, 2014 198

Lesley University Copyright Steve Yurek January 9, 2014 199

Lesley University Copyright Steve Yurek January 9, 2014 200

Lesley University Copyright Steve Yurek January 9, 2014 201

Lesley University Copyright Steve Yurek January 9, 2014 202 N

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

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

Lesley University Copyright Steve Yurek January 9, 2014 205

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

Or Lesley University Copyright Steve Yurek January 9, 2014 207

Lesley University Copyright Steve Yurek January 9, 2014 208

Lesley University Copyright Steve Yurek January 9, 2014 209 n

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

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

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

Or Lesley University Copyright Steve Yurek January 9, 2014 213

Lesley University Copyright Steve Yurek January 9, 2014 214

Lesley University Copyright Steve Yurek January 9, 2014 215 n

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Lesley University Copyright Steve Yurek January 9, 2014 234

Lesley University Copyright Steve Yurek January 9, 2014 235

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