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Solution Manual for Mathematical Reasoning for Elementary Teachers, 7th Edition

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    Guide for Activities 1 Chapter 1 Thinking Critically Activity 1 Box 1: UNDERSTAND THE PROBLEM 1. Answers will vary. 2. Sample Responses: Shapes are congruent if they have exactly the same shape and size. Shapes are congruent if they can be made to match exactly by placing one on top of the other. 3. a. b. 3 ways 4. Answers will vary. 5. Answers will vary. Box 2: DEVISE A PLAN 1. Squares could be moved around to try to construct the staircases or they could be sketched on graph paper. 2. Simplify the problem, look for a pattern, and make a table 3. Answers will vary. Box 4: LOOK BACK 3. a. 3 ways b. Since the two staircases must fit together to form a rectangle with area 120, find the dimensions of all rectangles with integral dimensions and an area of 120. The possible dimensions are 1 × 120, 2 × 60, 3 × 40, 4 × 30, 5 × 24, 6 × 20, 8 × 15, 10 × 12. To form a staircase, both dimensions must be greater than 1, and one dimension must be even and the other odd. This leaves 3 possibilities, a 3 × 40 rectangle, a 5 × 24 rectangle, and an 8 × 15 rectangle. The 3 × 40 rectangle can be divided into two 3-step staircases. The bottom step is 21 squares long, the middle step is 20 steps long, and the top step is 19 squares long. The 5 × 24 rectangle can be divided into two 5-step staircases. The bottom step is 14 squares long, the next step is 13 squares long, the middle step is 12 steps long, the fourth step is 11 squares long and the top step is 10 squares long. The 8 × 15 rectangle can be divided into two 8-step staircases. The steps are 11, 10, 9, 8, 7, 6, 5, and 4 steps long. 4. To determine whether n squares can be used to construct two congruent staircases, find a pair of factors of n . There is a pair of staircases for each factor pair in which both factors are greater than 1 and one factor is odd and the other even.

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    2 Chapter 1 • Thinking Critically Activity 2 Box 1 1. 2. 3. 4. 6, 3, 4, 5, 6, 7 5. 14, 17, 20, 23, 26, 29 6. 25, 18, 11, 4, –3, –10 7. 15, 21, 28, 36, 45, 55 8. 28, 39, 52, 67, 84, 103 9. 32, 64, 128, 256, 512, 1024 10. 9, 3, 1, 1 3 , 1 9 , 1 27 11. 34, 55, 89, 144, 233, 377 12. Each term of the sequence in Exercise 5 is the number of squares in the corresponding term of the sequence in Exercise 1. Similarly, each term of the sequence in Exercise 7 is the number of triangles in the corresponding term of the sequence in Exercise 3. 13. a. The number of triangles in each term of the sequence in Exercise 3 is half the number of squares in the corresponding term of the sequence in Exercise 2. b. Sample Response: (Continued on next page.) ,

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    Guide for Activities 3 Activity 2 Continued Box 2 1. 7, 12, 17, 22, 27, 32, 37, 42, 47 2. 2, 5, 8, 11, 14, 17, 20, 23, 26 Each term of the sequence in Exercise 1 on page 5 is a rectangular array 3 squares tall and n squares long (where n is the term number) with the top right-hand square removed. Thus the number of squares used to form each term is 1 less than 3 times n . 3. 44, 41, 38, 35, 32, 29, 26, 23, 20 4. 7, 13, 23, 37, 55, 77, 103, 133, 167 5. 2, 6, 12, 20, 30, 42, 56, 72, 90 Each term of the sequence in Exercise 2 on page 5 is a rectangular array of squares. The height of the rectangle is equal to the term number and its length is 1 more than the term number. Since the area of the rectangle is the length times the width, the number of squares in the array is the term number times the term number plus 1 . That is, the term number times the next term number . 6. In Exercises 1–3, the difference between successive terms is a constant. In Exercise 1 the difference is always 5, in Exercise 2 it is always 3, and in Exercise 3 it is –3. In each case, the difference is the constant by which the term number was multiplied. The differences between successive terms in Exercises 4 and 5 are not constant. Note that, in these cases, the rule for generating the sequence does not involve multiplying the term number by a constant. 7. Answers will vary. One possibility is to look for a pattern in the differences between successive terms and to continue that pattern to find the next five terms in the sequence. For example, in Exercise 5 the differences are the sequence of even numbers 4, 6, 8, 10, 12, 14, 16, 18… Thus, the next five terms would be 90 + 20 = 110, 110 + 22 = 132, 132 + 24 = 156, 156 + 26 = 182, and 182 + 28 = 210. 8. 2, 4, 8, 16, 30, 52 , 84 , 128 , 186, 260 9. The first four terms of the sequences are the same, but from there on they are different. So no matter how many of the initial terms of a sequence you may know, there may be more than one way to extend the sequence.

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    4 Chapter 1 • Thinking Critically Beginning of month Number of newborn pairs Number of 1-month- old pairs Number of pairs 2 months old or older Total number of pairs 1 1 0 0 1 2 0 1 0 1 3 1 0 1 2 4 1 1 1 3 5 2 1 2 5 6 3 2 3 8 7 5 3 5 13 Activity 3 Box 1 1. 2. a. The number of 1-month old pairs in month N ( N ≥ 2) is equal to the number of newborn pairs in month N – 1. b. The number of pairs 2 months old or older in month N ( N ≥ 2) is equal to the sum of the number of 1-month old pairs and the number of pairs 2-months old or older in month N – 1. c. From the second month on, the number of newborn pairs each month is equal to the number of pairs 2 months old or older. Box 2: CREATING A SPREADSHEET 2. a. 1 was added to the 1 in cell A2, so a 2 appears in cell A3. b. The formula “= A3 + 1” is copied into cell A4, the formula “= A4 + 1” is copied into cell A5, and so on. Thus for each cell A4 through A14, 1 is added to the number in the preceding cell. The result is that the numbers 3 through 13 appear in cells A4 through A14 respectively. c. To find the number of pairs after one year, you must extend the table to the beginning of the 13th month. 3. a. The content of cell B2 was copied into cell C3, so a 1 appears in cell C3. 7. 233 pairs 8. From the third term on, each term equals the sum of the two preceding terms. Month Newborn 1 Month Old 2 Months Old Total Pairs 1 1 0 0 1 2 0 1 0 1 3 1 0 1 2 4 1 1 1 3 5 2 1 2 5 6 3 2 3 8 7 5 3 5 13 8 8 5 8 21 9 13 8 13 34 10 21 13 21 55 11 34 21 34 89 12 55 34 55 144 13 89 55 89 233 (Continued on next page.)

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    Guide for Activities 5 Activity 3 Continued Box 3: Looking for Patterns Fibonacci #s Sums Fib. # – 1 Squares Sum of Squares Products Quotients 1 1 0 1 1 1 1 1 2 0 1 2 2 2 2 4 1 4 6 6 1.5 3 7 2 9 15 15 1.666666667 5 12 4 25 40 40 1.6 8 20 7 64 104 104 1.625 13 33 12 169 273 273 1.615384615 21 54 20 441 714 714 1.619047619 34 88 33 1156 1870 1870 1.617647059 55 143 54 3025 4895 4895 1.618181818 89 232 88 7921 12816 12816 1.617977528 144 376 143 20736 33552 33552 1.618055556 233 609 232 54289 87841 87841 1.618025751 377 986 376 142129 229970 229970 1.618037135 610 1596 609 372100 602070 602070 1.618032787 987 2583 986 974169 1576239 1576239 1.618034448 1597 4180 1596 2550409 4126648 4126648 1.618033813 2584 6764 2583 6677056 10803704 10803704 1.618034056 4181 10945 4180 17480761 28284465 28284465 1.618033963 6765 17710 6764 45765225 74049690 0 0 1. c. The first sum is 1 less than the third Fibonacci number, the second sum is 1 less than the fourth Fibonacci number, and so on—the n th sum is 1 less than Fibonacci number n + 2. 2. c. The numbers in the two columns are the same. d. The sum of the squares of the first n Fibonacci numbers equals the product of Fibonacci number n and Fibonacci number n + 1. 3. The quotients approach 1.61803.

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    6 Chapter 1 • Thinking Critically Activity 4 Box 1 Term Number 1 2 3 4 5 6 7 8 Term 4 11 18 25 32 39 46 53 Difference 7 7 7 7 7 7 7 Constant difference = 7 Term Constant What To Number Difference Was Done? Get 1 ´ 7 → 7 ___–3___ = 4 First Term 2 ´ 7 → __14_ ___–3___ = 11 Second Term 3 ´ __7__ → __21_ ___–3___ = __18_ Third Term 10 ´ __7__ → __70_ ___–3___ = __67_ Tenth Term 50 ´ __7__ → _350_ ___–3___ = _347_ Fiftieth Term Each term is 3 less than 7 times the term number, Rule: n th term = 7 n – 3 Box 2 Rule 25th 100th Term Term 1. 25, 29, 33, 37 4 T + 5 105 405 2. 30, 37, 44, 51 7 T – 5 170 695 3. 5, 7, 9, 11 2 T – 5 45 195 4. 90, 88, 86, 84 –2 T + 100 50 –100 5. 65, 62, 59, 56 –3 T + 80 5 –220

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    Guide for Activities 7 Activity 5 2. Number of People 2 4 6 8 10 2 N Number of Pairs 1 2 3 4 5 N Min Number of Moves 3 8 15 24 35 N 2 + 2 N Sequence of Moves RLR RLLRRL LR RLLRRR LLLRRR LLR RLLRRR LLLLRR RRLLLL RRRLLR RLLRRR LLLLRR RRRLLL LLRRRR RLLLLR RRLLR 1R, 2L, 3R, ... , ( N – 1)L, N R, N L, N R, ( N – 1)L, ... , 3R, 2L, 1R if N is odd 1R, 2L, 3R, ... , ( N – 1)R, N L, N R, N L, ( N – 1)R, ... , 3R, 2L, 1R if N is even 4. 12 people 5. 255 moves 6. Number of People 1 3 5 7 9 N Min Number of Moves 1 5 11 19 29 ( N 2 + 4 N – 1)/4 Sequence of Moves R RLRRL RLLRRR LLRRL RLLRRR LLLRRR RLLLRR L RLLRRR LLLLRR RRRLLL LRRRRL LLRRL The Legend of the Tower of Brahma 1. Number of Disks 1 2 3 4 5 n Number of Moves 1 3 7 15 31 2 n – 1 2. a. 2 10 – 1 = 17 min 3 sec b. 2 30 – 1 ≈ 34 years c. 2 50 – 1 ≈ 36 million years d. 2 64 – 1 ≈ 585 billion years Activity 6 Box 1 1. The results for Steps 1-5 will vary depending on the number selected by each person. 2. Sample Response: 10 because that was the answer I got for Persons 1 through 5. 3. Step 1: n Step 2: 3 n Step 3: 3 n + 30 Step 4: (3 n + 30) ÷ 3 = n + 10 Step 5: n + 10 – n The result is 10. (Continued on next page.)

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    8 Chapter 1 • Thinking Critically Activity 6 Continued Box 3 1. Answers will vary. 2. The display on the calculator shows the month and day of a person’s birthday. For example, 1021 for October 21. 3. Step 1 m Step 2: 5 m Step 3: 5 m + 20 Step 4: 20 m + 80 Step 5: 20 m + 73 Step 6: 100 m + 365 Step 7: 100 m + 365 + d Step 8: 100 m + d In Step 8, multiplying the month, m , by 100 moves the month to the hundreds place in the final answer. If m is a single digit number, that digit will be in the hundreds place of the answer. If m is a two digit number, its tens digit will be in the thousands place of the answer and its units digit will be in the hundreds place. When the day, d , is added, the digits of d will appear in the tens and units places of the final answer. Activity 7 2. a. 1 Subtracting 1 leaves 4 on the display. If Player A subtracts 1, that leaves 3 on the display and B wins by subtracting 3. If A subtracts 2, that leaves 2 on the display and B wins by subtracting 2. If A subtracts 3, it leaves 1 on the display and B wins by subtracting 1. b. To be certain to win, you want to be the first player. At the end of your turn you want to leave 16, 12, 8, or 4 on the display. 3. a. In this game you want to be the second player. At the end of your turn you want to leave 20, 15, 10, or 5 on the display. b. In this game you want to be the second player. At the end of your turn you want to leave 40, 30, 20, or 10 on the display. Activity 8 Box 1 1. 1580 2. Sample Response: From clues b and d, I knew the number had to be a multiple of 10. These numbers are easy to pick out since the must have a 0 in the units place. Box 2 1. 153 (or 1 if it isn’t a multi-digit number) 2. Sample Response: Since 7 3 = 343, from the last two clues I knew the digits in the number had to be less than 7. From the second clue, I knew the digits could only be 1, 3, or 5. Box 3 25 cards or 85 cards

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    Guide for Activities 9 Activity 9 Box 1 1. No. If there were green gumballs in the jar, it would be labeled correctly, but no jar has a correct label. 2. Red 3. No. If so, the jar labeled GREEN would have GREEN gumballs, but all labels are incorrect. 4. The correct label for the jar labeled Red - Green is Red, the correct label for the jar labeled Red is Green, and the correct label for the jar labeled Green is Red - Green. Box 2 Greatest amount = $1.19 Coins = 1 half dollar, 1 quarter, 4 dimes, 4 pennies Box 3 1. Alabama Alaska Oklahoma Minnesota Camellia Forget Me Not Mistletoe Lady's slipper Yellowhammer Willow ptarmigan Flycatcher Loon 2. Clues b, d, and c are the keys to solving the problem. From b, we know that Alaska and forget me nots go together. From d, we know that Alabama and yellowhammer go together. From c, either camellia or lady 's slipper go with Minnesota. If camellia and Minnesota go together, then the state bird can't be the willow ptarmigan (clue e) or the loon (clue c) so it must be the flycatcher. But this would contradict clue a, so Minnesota, lady's slipper, and loons must go together. Box 4 Two answers are possible. Freddie or Susie can be in either First or Fourth place. First Second Third Fourth Yellow Purple Green Red Freddie or Susie Liz Joe Susie or Freddie

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    Guide for Activities 49 Chapter 10 Measurement: Length, Area, and Volume Activity 1 Box 1 1. Answers will vary. 2. Answers will vary. Box 2 1. Answers will vary. 2. Answers will vary. 3. a. Answers will vary. b. Answers will vary. c. On the geo-board construct a right triangle with a horizontal base of 3 and a vertical altitude of 2. The area is 3 square units. Pick up the geoband at the intersection of the hypotenuse and the altitude. Move the band one peg to the right. Continue stretching the band, always moving it one peg to the right so that the triangle has a base of 3 units and an altitude of 2 units. The perimeter will increase without bound, but the area will always be 3 square units. Thus it is possible to obtain a perimeter of approximately100, 1000, or any large number. Activity 2 Box 1 1. a rectangle 2. Sample Response: The angles are all right angles. 3. They base of the parallelogram has the same length as the base of the rectangle and the altitude of the parallelogram equals the altitude of the rectangle. 4. Answers will vary. 5. The areas are equal. 6. a. Count the number of square units in the rectangle. b. Use the formula, A = base × altitude. Box 2 1. Answers will vary. 2. The areas are equal. 3. A = base × altitude.

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    50 Chapter 10 • Measurement: Length, Area, and Volume Activity 3 Box 1 1. a parallelogram 2. Sample Response: Sides MI and EK are parallel and congruent. 3. A = 12 4. 1 2 5. A = 6 Box 2 1. The area of the triangle is one-half the area of the parallelogram. 2. The base and altitude of the triangle are the same as the base and altitude of the parallelogram. 3. A = b × a 4. A = 1 2 b × a Activity 4 Box 1 1. a parallelogram 2. ML is parallel to JK . 3. Answers will vary. 4. The area of the trapezoid is one-half the area of the parallelogram. 5. Answers will vary. Box 2 1. The area of the trapezoid is one-half the area of the parallelogram. 2. The measure of the base of the parallelogram is equal to the sum of the measures of the bases of the trapezoid. 3. A (parallelogram) = b x a , so A (trapezoid) = 1 2 ( b 1 + b 2 ) a .

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    Guide for Activities 51 Activity 5 Box 1 1. m ∠ X = 56°, m ∠ Y = 90°, m ∠ Z = 34°, 2. right 3. Area of Square A + Area of Square B = Area of Square C . Box 2 1. Area of Square A + Area of Square B = Area of Square C . 2. right Activity 6 Box 1 1. The area of the square is 64. 2. The area of the rectangle is 65. 3. The area of the other figure is 63. 4. AB + BC > AC. This shows that AC is not a straight line, and the parts do nit fit together as shown in the drawing. Box 2 The next four Fibonacci numbers are 13, 21, 34, and 55. Box 3 1. The area of the square is 169 2. The area of the rectangle is 168. 3. The area of the other figure is 170. (Continued on next page.)

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    52 Chapter 10 • Measurement: Length, Area, and Volume Activity 6 Continued Box 4 1. 21 2. Divide two sides of the square into segments of 8 and 13 3. The area of the rectangle is 442. The area of the other figure is 440. EXTENSIONS For any Fibonacci number n , the area of the square is n 2 . The area of the related rectangle formed by partitioning the square and rearranging the pieces is the product of the preceding Fibonacci number and the next Fibonacci number. If the area of the rectangle is one more than the area of the square, the area of the other figure is one less. If the area of the rectangle is one less than the area of the square, the area of the other figure is one more. Activity 7 Box 1 Area of the Largest Square Area of the Smallest Square Area of the Third Square Sum of the Areas of the Two Smaller Squares Is the Triangle Acute, Right, or Obtuse? 64 9 36 45 obtuse 100 36 64 100 right 36 16 25 41 acute The remaining entries in the table will vary. Box 2 Entries in the tables will vary. a. acute b. right c. obtuse Activity 8 Box 1 1. a. 9, 6, 3, 2, 1 21 b. 6, 6, 5, 4, 3, 3, 2, 2, 1, 1 33 c. 12, 12, 10, 7, 4, 2 47 2. a. 24 b. 27 c. 25 3. a. 45 b. 60 c. 72 (Continued on next page.)

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    Guide for Activities 53 Activity 8 Continued Box 2 Length Width Height Number of Cubes 3 5 2 30 4 6 5 120 4 4 4 64 7 8 3 168 6 5.5 10 330 14 3 6.5 273 15 9 5 675 8.7 13 5 565.5 1. V = lwh 2. Number of cubes = area of the base × height 3. a. A 1 = 2 × 5 = 10 A 2 = 5 × 8 = 40 A 3 = 2 × 8 = 16 b. V = 10 × 8 = 80 V = 40 × 2 = 80 V = 16 × 5 = 80 4. Volume of cylinder = area of the base × height Activity 9 Box 1 1. They are equal. 2. They are equal. 3. Answers will vary. 4. 3 5. 1 3 6. Volume of a pyramid = 1 3 × area of the base × altitude Box 2 1. 1 3 2. Volume of a cone = 1 3 × area of the base × altitude or 1 3 ( π r 2 ) h

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    54 Chapter 10 • Measurement: Length, Area, and Volume Activity 10 Box 1 1. trapezoidal prism 2. square pyramid 3. rectangular prism 4. triangular prism Box 2 Since measures are not accurate and printing of the nets may reduce or enlarge the figures, all answers are approximate. Measures of surface area are in cm 2 and measures of volume are in cm 3 . 1. S.A. ≈ 87 V ≈ 51 2. S.A. ≈ 50 V ≈ 18 3. S.A. ≈ 117 V ≈ 83 4. S.A. ≈ 133 V ≈ 72 EXTENSION Volume of the prism is 54. Volume of each pyramid is 18.

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