Download presentation

Presentation is loading. Please wait.

Published byAlfredo Marsh Modified over 2 years ago

1
ATMIM Winter Conference Boston College January 9, 2014 Steve Yurek Cambridge, Massachusetts 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,

19
Lesley University Copyright Steve Yurek January 9,

20
Lesley University Copyright Steve Yurek January 9,

21
Lesley University Copyright Steve Yurek January 9,

22
Lesley University Copyright Steve Yurek January 9,

23
Lesley University Copyright Steve Yurek January 9,

24
Lesley University Copyright Steve Yurek January 9,

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

26
Lesley University Copyright Steve Yurek January 9,

27
Lesley University Copyright Steve Yurek January 9,

28
Lesley University Copyright Steve Yurek January 9,

29
Lesley University Copyright Steve Yurek January 9,

30
Lesley University Copyright Steve Yurek January 9,

31
Lesley University Copyright Steve Yurek January 9,

32
Lesley University Copyright Steve Yurek January 9,

33
Lesley University Copyright Steve Yurek January 9,

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

36
Lesley University Copyright Steve Yurek January 9, 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

56
a a b b c c Now draw the line segment as indicated

57
a a b b c c

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

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

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

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

62
a a b b c c The quadrilateral is a trapezoid. WHY? What is the measure in angle 3? WHY? 3

63
a a b b c c 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 Lesley University Copyright Steve Yurek January 9, 2014

184
80’ 50’ 30’ E 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 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,

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

188
Lesley University Copyright Steve Yurek January 9,

189
Lesley University Copyright Steve Yurek January 9,

190
Lesley University Copyright Steve Yurek January 9,

191
Lesley University Copyright Steve Yurek January 9, 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,

193
Lesley University Copyright Steve Yurek January 9,

194
Lesley University Copyright Steve Yurek January 9,

195
Lesley University Copyright Steve Yurek January 9,

196
Lesley University Copyright Steve Yurek January 9,

197
Lesley University Copyright Steve Yurek January 9,

198
Lesley University Copyright Steve Yurek January 9,

199
Lesley University Copyright Steve Yurek January 9,

200
Lesley University Copyright Steve Yurek January 9,

201
Lesley University Copyright Steve Yurek January 9,

202
Lesley University Copyright Steve Yurek January 9, N

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

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

205
Lesley University Copyright Steve Yurek January 9,

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

207
Or Lesley University Copyright Steve Yurek January 9,

208
Lesley University Copyright Steve Yurek January 9,

209
Lesley University Copyright Steve Yurek January 9, n

210
Lesley University Copyright Steve Yurek January 9, n n

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

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

213
Or Lesley University Copyright Steve Yurek January 9,

214
Lesley University Copyright Steve Yurek January 9,

215
Lesley University Copyright Steve Yurek January 9, n

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

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

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

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

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

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

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

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

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

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

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,

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,

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,

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,

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,

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,

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,

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,

234
Lesley University Copyright Steve Yurek January 9,

235
Lesley University Copyright Steve Yurek January 9,

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google