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

2. 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

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

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

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

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

7. But it comes as no surprise that they are not so sure when it comes to the areas of the other plane figures. First – the Parallelogram h b 7 Lesley University Copyright Steve Yurek January 9, 2014

8. b h 8

9. b h 9 h

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

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

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

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

14. The Blue area is still the same size, but it’s just in the shape of a rectangle now, so A = bh once again b h 14 Lesley University Copyright Steve Yurek January 9, 2014

15. But it comes as no surprise that they are not so sure when it comes to the areas of the other plane figures. Next a Triangle h b 15 Lesley University Copyright Steve Yurek January 9, 2014

16. But it comes as no surprise that they are not so sure when it comes to the areas of the other plane figures. Next a Triangle h b Not a real stretch to see that for a triangle A = ½ bh h 16 Lesley University Copyright Steve Yurek January 9, 2014

17. 17

18. Lesley University Copyright Steve Yurek January 9, 2014 18

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21. Lesley University Copyright Steve Yurek January 9, 2014 21

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24. Lesley University Copyright Steve Yurek January 9, 2014 24

25. 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.

26. Lesley University Copyright Steve Yurek January 9, 2014 26

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33. Lesley University Copyright Steve Yurek January 9, 2014 33

34. 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

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

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

37. a b h

38. a b h

39. a b h

40. a b h

41. a b h b a

42. 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

43. 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

44. 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

45. 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

46. 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

47. Consider any right triangle

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

49. 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

50. 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

51. 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

52. 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

53. 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

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

55. a a b b c c 1 1 2 2

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

57. a a b b c c 1 1 2 2

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

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

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

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

62. 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

63. 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

64. 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

65. 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

66. 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

67. 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

68. 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

69. 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

70. 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

71. 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

72. 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

73. 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

74. 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

75. 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

76. 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

77. 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

78. 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

79. 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

80. 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

81. 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

82. 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

83. 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

84. 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

85. 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

86. 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

87. 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

88. 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

89. 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

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

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

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

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

94. a 94 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

99. 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

100. 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

101. 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

102. 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

103. 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

104. 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

105. 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

106. 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

107. 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

108. 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

109. 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

110. 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

111. 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

112. 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

113. 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

114. 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

115. 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

116. 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

117. 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

118. 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

119. 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

120. 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

121. 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

122. 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

123. 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

124. 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

125. 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

126. 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

127. 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

128. 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. = 129 Lesley University Copyright Steve Yurek January 9, 2014

130. = 130 Lesley University Copyright Steve Yurek January 9, 2014

131. = 131 Lesley University Copyright Steve Yurek January 9, 2014

132. = 132 Lesley University Copyright Steve Yurek January 9, 2014

133. = 133 Lesley University Copyright Steve Yurek January 9, 2014

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

135. 135 Lesley University Copyright Steve Yurek January 9, 2014

136. 136 Lesley University Copyright Steve Yurek January 9, 2014

137. 137 Lesley University Copyright Steve Yurek January 9, 2014

138. 138 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

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

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

145. 145 Lesley University Copyright Steve Yurek January 9, 2014

146. 146 Lesley University Copyright Steve Yurek January 9, 2014

147. 147 Lesley University Copyright Steve Yurek January 9, 2014

148. 148 Lesley University Copyright Steve Yurek January 9, 2014

149. 149 Lesley University Copyright Steve Yurek January 9, 2014

150. 150 Lesley University Copyright Steve Yurek January 9, 2014

151. 151 Lesley University Copyright Steve Yurek January 9, 2014

152. 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

153. 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

154. 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

155. 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

156. 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. 157 Lesley University Copyright Steve Yurek January 9, 2014

158. Sarah Jane bought cartridges for the various printers in her office and spent a total of $350 for each type. She paid $25 for each of the cartridges for the black & white printers for the staff and $70 for each of the cartridges for her supervisor’s color printer. What was the average cost of a single cartridge? 158 Lesley University Copyright Steve Yurek January 9, 2014

159. 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

160. 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

161. 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

162. Sarah Jane bought cartridges for the various printers in her office and spent a total of $350 for each type. She paid $25 for each of the cartridges for the black & white printers for the staff and $70 for each of the cartridges for her supervisor’s color printer. What was the average cost of a single cartridge? Did you get $36.84 ? 162 Lesley University Copyright Steve Yurek January 9, 2014

163. Questions such as these involve the Harmonic Mean. There are 3 ways to compute the Harmonic Mean of a & b. 1.Arithmetic Sense We just did that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 163 Lesley University Copyright Steve Yurek January 9, 2014

164. 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

165. Questions such as these involve the Harmonic Mean. There are 3 ways to compute the Harmonic Mean of a & b. 1.Arithmetic Sense: We just did that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 165 Lesley University Copyright Steve Yurek January 9, 2014

166. Questions such as these involve the Harmonic Mean. There are 3 ways to compute the Harmonic Mean of a & b. 1.Arithmetic Sense: We just did that 2.Definition: The reciprocal of the arithmetic mean of the reciprocals 3.Formula: 166 Lesley University Copyright Steve Yurek January 9, 2014

167. 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

168. 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

169. 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

170. 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

171. 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

172. 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

173. 173 Lesley University Copyright Steve Yurek January 9, 2014

174. 174 Lesley University Copyright Steve Yurek January 9, 2014

175. 175 Lesley University Copyright Steve Yurek January 9, 2014

176. 176 Lesley University Copyright Steve Yurek January 9, 2014

177. Sarah Jane bought cartridges for the various printers in her office and spent a total of $350 for each type. She paid $25 for each of the cartridges for the black & white printers for the staff and $70 for each of the cartridges for her supervisor’s color printer. What was the average cost of a single cartridge? Sarah Jane and her PrintersSarah Jane and her Printers 177 Lesley University Copyright Steve Yurek January 9, 2014

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

179. 179 Lesley University Copyright Steve Yurek January 9, 2014

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

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

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

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

184. 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

185. 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

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

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

188. Lesley University Copyright Steve Yurek January 9, 2014 188

189. Lesley University Copyright Steve Yurek January 9, 2014 189

190. Lesley University Copyright Steve Yurek January 9, 2014 190

191. 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:

192. We can determine the formula for the sum of the squares of the first n integers in many ways: Finite Differences and Mathematical Induction involve only algebra, but let’s take a look at this: Lesley University Copyright Steve Yurek January 9, 2014 192

193. Lesley University Copyright Steve Yurek January 9, 2014 193

194. Lesley University Copyright Steve Yurek January 9, 2014 194

195. Lesley University Copyright Steve Yurek January 9, 2014 195

196. Lesley University Copyright Steve Yurek January 9, 2014 196

197. Lesley University Copyright Steve Yurek January 9, 2014 197

198. Lesley University Copyright Steve Yurek January 9, 2014 198

199. Lesley University Copyright Steve Yurek January 9, 2014 199

200. Lesley University Copyright Steve Yurek January 9, 2014 200

201. Lesley University Copyright Steve Yurek January 9, 2014 201

202. Lesley University Copyright Steve Yurek January 9, 2014 202 N

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

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

205. Lesley University Copyright Steve Yurek January 9, 2014 205

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

207. Or Lesley University Copyright Steve Yurek January 9, 2014 207

208. Lesley University Copyright Steve Yurek January 9, 2014 208

209. Lesley University Copyright Steve Yurek January 9, 2014 209 n

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

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

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

213. Or Lesley University Copyright Steve Yurek January 9, 2014 213

214. Lesley University Copyright Steve Yurek January 9, 2014 214

215. Lesley University Copyright Steve Yurek January 9, 2014 215 n

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

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

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

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

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

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

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

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

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

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

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 226

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 227

228. 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

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 229

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 230

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 231

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 232

233. 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

234. Lesley University Copyright Steve Yurek January 9, 2014 234

235. Lesley University Copyright Steve Yurek January 9, 2014 235