ANSWERS FOR ONE-PAGE MATH ACTIVITIES Math Activity 1.1. Three moves are required for 1 peg on each side and 8 moves are required for pegs on each side.. For pegs on each side the minimum number of moves is 15 and the strategy is. Here are a few patterns. R GG RRR GGG RRR GG R I. The strategies have patterns in the numbers of letters R and G. Consider the following strategies: pegs on a side: R GG RR GG R pegs on a side: R GG RRR GGG RRR GG R II. Each sequence of letters in the strategies is symmetric about the center. III. The center group of letters in the letter sequence has the same number of letters as pegs on a side and the color of the center group is red if the number of pegs on a side is even, and green if the number of pegs on a side is odd. IV. The total number of moves for two pegs on a side: 1++++1 The total number of moves for three pegs on a side: 1++++++1 Predicting the number of moves for pegs on a side: Using pattern IV, the total number of moves is 1++++++++1; and extending pattern I, the next letter sequence is R GG RRR GGGG RRRR GGGG RRR GG R Notice that for pegs on a side, the center group of numbers is red. 5. For five pegs on each side, the minimum number of moves is 5. Math Activity 1. 1. a. Pattern: Starting with a trapezoid, each successive figure is obtained by alternately adding a pair of parallelograms (one at each end) or a pair of trapezoids (one at each end). b. 9 trapezoids and 10 parallelograms c. Each odd numbered figure in the sequence will have trapezoids on its ends. The 5th figure will have 5 trapezoids and parallelograms. d. Here are two of several ways to determine the number of pattern blocks in any figure. (1) After the first figure, blocks are added each time. Therefore, the 75th figure has 1 + 7 = 19 blocks. () Each odd-numbered figure has that many trapezoids and one less parallelogram. So the 75th figure has 75 + 7 = 19 pattern blocks.. a. Pattern: Starting with a hexagon, first a triangle is attached and then a square. The pattern begins repeating after the third term. b. The 0th figure would look like 7 copies of the rd figure, in a row, with the square removed from the right end.
c. The 17th figure would look like 58 copies of the rd figure, in a row, and contain 58 hexagons, 58 triangles, and 58 squares. The 175th will have 1 more hexagon than the 17th and the 176th will have 1 more hexagon and 1 more triangle than the 17th.. a. After the first parallelogram you alternately attach triangles (two) and parallelograms (one). Or, one parallelogram is added to odd numbered figures and two triangles are added to even numbered figures. The 0th will look like 5 copies of the th figure and contain 0 triangles and 10 parallelograms. b. The 5th figure will look like 11 copies of the th figure with an additional parallelogram on the end. It will have 67 pattern blocks. c. The 58th figure. A figure with 87 pattern blocks will look like 1 copies of figure (8 blocks) with 1 parallelogram and triangles attached at the end. The figure that has 1 copies of figure is the 56th so attaching 1 parallelogram and triangles brings us to the 58th. Math Activity 1. 1. a. Figure number 1 5 Green tile 5 8 1 18 Yellow tile 8 1 18 b. 61 green and 60 yellow c. 11 tiles d. There are many descriptions. Here are two. Place a green tile on the table to mark the lower right corner of the square figure. To its left alternately place yellow and green tiles until there are a total of 6 tiles in that row. Above each tile in that row alternately place a green or yellow tile until each column has a total of 6 tiles. Or, make a 6 by 6 checkerboard from green and yellow tiles with green tiles down the main diagonal from upper left to lower right and yellow tiles down the other diagonal. Each of the 6 columns will have 1 yellow and 1 green for a total of 8 yellow and 8 green in the 6th figure.. a. Figure number 1 5 10 5 Red tile 1 1 6 6 15 5 5 Blue tile 0 10 10 55 00 Starting from the bottom right corner the red tiles form an L-shape pattern of 1 + 5 + 9 + 1 +... tiles, and the blue tiles form an L-shape pattern of + 7 + 11 + 15 +... tiles. b. Figure number 1 5 10 5 Yellow tile 6 1 18 0 60 150 Red tile 0 10 1 6 171 1176 Notice the relationships between the dimensions of each rectangle and its colored parts: Figure number 1 Dimensions 5 7 5 9 No. of yellows + 1 + 1 + 5 + 5 + + 7 + 7 + + 9 + 9 No. of reds 0 1 5 7
Math Activity.1 1. There are many ways to continue the sequence that has been started. Here are three ways which use different attribute pieces. (1) SYS, LYS, LYC, SYC, SRC () LYH, LYT, LBT, SBT, SBC () SBH, SBT, LBT, LRT, LRH. There are many ways to form a two-difference train that starts with the four given pieces. Here is one way that uses all the pieces and is circular. That is, the last piece differs from the starting piece by exactly two attributes. LYH, LRS, SBS, SRH, LBH, LRT, SYT, SBC, LBS, SYS, LYC, SRC, LRH, SBH, SRS, LYS, LRC, LBT, SRT, SYH, SBT, LYT, LBC, SYC. Here are the pieces in each region: Region 1, LBS, SBS, LBC, SBC, LBT, SBT; Region, LBH, SBH; Region, LRH, SRH, LYH, SYH; Outside, LRS, SRS, LRT, SRT, LRC, SRC, LYS, SYS, LYT, SYT, LYC, SYC. Here are all pieces in each region: Region 1, SBH, SBC, SBS; Region, SBT; Region, SYT, SRT; Region, SRH, SYH, SRS, SYS, SRC, SYC Region 5, LBT; Region 6, LRT, LYT; Region 7, LRH, LYH, LRS, LYS, LRC, LYC; Outside, LBS, LBH, LBC Math Activity. 1. a. 1 b. c. d. = 1 As the lines get steeper, the slopes increase.. slope, 1 slope, slope, slope,. length of length of length of 1 length of 5 slope of 0 undefined slope undefined slope slope of /. From least to greatest the slopes are: 0 1 1 1 1
Math Activity. 1. Player A's number does not contain the digits 1,, or.. Player A's number contains of the digits, 5, or 6, but no digit is in the correct place value position.. Since of the digits are correct but in the wrong position, switching positions will give us more information.. Now player B knows that one of the digits 6 or is in the correct position, but cannot be sure which one. 5 may still be the correct digit but in the incorrect position. Math Activity.1 1. A square made of 5 long-flats. long-flats flats longs units a. 1 b. 1 0 c. 0. a. 0 units b. 61 units c. 58 units d. 18 units. a. 11three b. 68 units
Math Activity. 1. flats, 1 long, 1 unit. Player 1: 1 long-flat Player : flats, longs Player : flats, longs, units Player 1 won the game.. After 6 turns the player had flats, 1 long, and units. The player will need 1 flat, longs and units to obtain one long-flat.. 1 flat, longs and units Math Activity. 1. 1 long-flat, flats, 1 long, units. a. 1five b. 101five c. 1five. a. 1 flat, longs, 0 units 10five. b. 1 long-flat, flats, longs, 0 units 10five c. Any group of five of the same size base five pieces can be traded for one of the next larger base five pieces. So when an arbitrary collection of base five pieces is multiplied by five, every single piece in the collection will be replaced by the next larger piece. In particular, each unit will become a long when multiplied by five so there will be no units in the product. 10 four 1 four = 10 four 10 four four = 0 four 5. Base ten: 6ten 7ten = 8ten Math Activity.
1. The minimal collection consists of longs and 1 unit.. a. 1five b. 11five c. five. a. 0five b. 1five c. five d. five. a. Each person would receive 1 units. The sketch will have 1 groups of 1 units each. b. Fourteen people would receive 1 units each. The sketch will have 1 groups of 1 units each. c. Yes. The sketch in a illustrates dividing 168 objects into 1 equal groups. The sketch in activity b illustrates the number of times that 1 objects can be subtracted from 168 objects. Math Activity.1 1. a. A three digit base five numeral indicates the number of flats, longs and units in the collection for the number. When a flat is divided into equal parts there is one unit remaining and when a long is divided into equal parts there is also 1 unit remaining. So the sum of the digits in the numeral indicates the total number of units remaining in the collection after flats and longs have each been divided into equal parts. b. When the units in a base five long-flat are divided into equal parts there is one unit remaining. (In fact this will be true for any base five piece larger than a unit.) 1five is divisible by because 1 + + + is divisible by. A base five number is divisible by if the sum of its digits is divisible by.. a. A number is divisible by three if the sum of its digits is divisible by three. Also a number is divisible by nine if the sum of its digits is divisible by nine. b. Since divides 9, a number divisible by 9 is divisible by. However, the reverse is not true. For example 75 is divisible by but not by 9.. The following base ten pieces show that has a remainder of. Since each flat can be divided into equal parts, as well as the base pieces for all higher powers of ten, to determine if a number is divisible by, it is only necessary to determine if the value of the longs and units is divisible by. that is, if the number represented by the tens and units digits of a number is divisible by, then the number will be divisible by. Math Activity. 1. a. 0th is blue; 5th is yellow b. Divide the figure number by. If there is no remainder the color is blue. If the remainder is 1,, or the color is red, green or yellow, respectively.
c. red green yellow blue Figure 0 5 50 55 60 Figure 5 15 16 171 1 This chart suggests one way to organize the data and look for patterns. Figure red green yellow blue 1 1 1 1 1 5 1+5 6 1+5 +6 7 1+5 +6 +7 8 1+5 +6 +7 +8 9 1+5+9 +6 +7 +8 a. One diagonal is always yellow; the colors are symmetric about the yellow diagonal so there are the same number of green, blue and red on each side of the yellow diagonal; as the squares get larger one yellow is added to the diagonal and the newly created space in each step is filled in rotation by green, blue and red tile. b. yellow green blue red Figure 10 10 0 6 Figure 1 1 5 60 Math Activity 5.1 1. a. The answers will vary. b. You get the opposite value.. a. - - 7 + 5 = - b. 1 + - 9 =
c. - 1-7 + - 6 = - 1 d. 0-8 + 8 = 0. a. - 7 b. - 1 c. d. - 8 - = - 7 - - = - 1 - - 6 = - 5 = - 8 remove black tiles remove red tiles remove 6 red tiles remove 5 black tiles. Add black and red tile to the collection. To subtract you add black and red and then withdraw black. This has the same effect as adding red. Math Activity 5. 1. a. 1 = = 6 = 6 1 6 = = 8 1 1 = 1 9 1 = 1 = 6 = 1 b. Whole bars: 6 6 1 1 Zero bars: 0 0 0 0 6 0 1. = 1 6 = 1 6 = 1 6 = 1 = 1 6 1 = 1 1 = 1 6 1 = 1 8 1 = 9 1 = 10 1 = 5 6. a. 7 1 b. 1 or 1 c. 5 6 or 10 1 d. or 9 1
. a. b. 5. Common denominator is the concept that is being modeled by this activity. Math Activity 5. 1. a. 5 1 + = 11 1. a. 5 6-1 = 6. Answers will vary b. 1 6 + = 5 6 b. 7 10 - = 5 10 c. 1 + 1 = c. 5 1-1 = 1
. 5. a. 8 10 5 = b. 5 6 1 = 5 c. 5 1 1 = 1 Math Activity 6.1 1. a. Ten of any one of the pieces has the value of the next larger piece; the pattern of the shapes alternates between non-square rectangle and square; there are 100 one-hundredth squares in the unit square and 1000 one-thousandth regions in the unit square. b. The second row shows the least number of pieces. units tenths hundredths thousandths 1 1 1 5 0 15 0 0 15 0 0 0 150 1 0 5 1 1 1 1 1 10
. a. b. c.. a. b. c. Math Activity 6. 1. a..76 b..65 c.. d..6
. a.. b..09 c..0. a. b. 18 c. 11 Math Activity 6. 1. a. b.c. d. 1 7% % % 16%. 1 small squares (1% of 00) represents 1 = ` 8 small squares (8% of 00) represents years. One square represents 1.8 skateboards so 5% (5 small squares) represents 5 1.8 = 81 skateboards. Or, 10 squares represent 18 skateboards so 5 squares represent.5 18 = 81. Fifteen (15) squares represent 7 skateboards.
. If 8 squares represent 70 teachers then 1 square represents.5 teachers and there must be 100.5 = 50 teachers in the district. 5. One small square (1%) represents 158 1 = 1 cars; 15% represents 180 cars; and 100% represents 100 cars. Math Activity 6. 1. 6 1 square units. Here are some possible solutions.. Here are some possibilities.. 5. Comparing the length of a side of the first square above to the length of a side of the second square shows that 8 is times.
Math Activity 7.1 1. a. Each lettered square does not have the same chance of receiving a tile. Theoretically, the numbers of tiles that would land on each of the letters are: A, ; B, 8; C, 1; D, 8; and E, c. Squares B, C, and E are the most likely to have tiles land on them and squares A and E are the least likely.. a. 1 path each to points A and E paths each to points B and D 6 paths to point C b. 6.5% to A 5% to B 7.5% to C 5% to D 6.5% to E c. 8 in column A in column B 8 in column C in column D 8 in column E Math Activity 7. 1. First way: Take 6 tiles off the top of the column with 15 tiles so both columns now have the same height. Divide the difference by and put these tiles on top of each column. Second way: Count all the tiles and divide equally into columns. 15 9 a. First way: 15-9 = 6 is the difference; (15-9) = is half the difference; and 9 + = 1 is the average. Second way: 15 + 9 = is total; and = 1 is average b. One way is y + x - y Another is (x + y). A third is x (x - y) c. y + x y = y + x y = y + x y = x + y. a. 16 and b. 10 and 6. Only option will work and only if the score is 1 or better.
Math Activity 7. 1. a,b. Answers will vary. The smallest possible number of spins is. c,e. After a large number of experiments the average should be between 8 and 9 yogurts. The theoretical solution to this problem is 8 1 yogurts. ( + + + 1 = 81 ) d. A person might get lucky and obtain all four types of covers by buying fewer than 8 yogurts. However, on the average a person must buy between 8 and 9 yogurts.. For a large number of experiments, the total number of spins should be between 7 and 8. The theoretical solution to this problem is 7 1 people. (The probability of selecting a person with type A blood is 5 and so 5 = 1 people must be selected on the average to obtain one person with type A blood. Thus 1 average to obtain people with type A blood.) = 71 people need to be selected on the Math Activity 8.1 1. The chances are better than 60% that at least one sixth-grader will be chosen. The theoretical probability that at least one sixth-grader will be chosen is.6 to two decimal places.. b. It is unlikely that when six cars are selected they will all have different inspection months (about a % chance). c. Increase the number of experiments. Math Activity 8. 1. b. 1 1 1 6 8 6 9 1 8 1 16 c. Out of all 16 outcomes occurs most frequently ( times) so over a large number of spins one would expect to come up most often. The numbers 1, 9, and 16 each occur once in the 16 possible outcomes so one would expect them to be less likely to reach the finish line first. d. Because points are awarded for the total number of Xs, this is a fair game. The numbers 1,,, and constitute 50% of the total outcomes and the numbers 6, 8, 9, 1, and 16 the other 50%.
. There are 15 different products. If player A uses the products 1,,,, 5, 6, and player B uses products 8, 9, 10, 1, 15, 16, 18, 0,, then it will be a fair game, because the multiplication table contains products and each player has 1 of these. Math Activity 9.1 1. The triangle has three 60 degree interior angles. The square has four 90 degree interior angles. The hexagon has six 10 degree interior angles The trapezoid has two 60 degree and two 10 degree interior angles. The blue parallelogram has two 60 degree and two 10 degree interior angles. The white parallelogram has two 0 degree and two 150 degree interior angles.. a. The sum of the measures of the interior angles in any 6-sided polygon is 70 degrees. b. Yes, the sum of the measures of the interior angles of this hexagon is 70 degrees.. The sum of the measures of the interior angles of this pentagon and any pentagon is 50 degrees.. The sum of the measures of the interior angles of an 8-sided polygon is 1080 degrees. The sum of the measures of an n-sided polygon is (n-) 180 degrees. Math Activity 9. 1. Answers will vary.. A regular triangle, a square, or a regular hexagon. Different vertices have different numbers of shapes surrounding them. For example, one vertex has a code of,,,,6 while another has code,,6,6.. There are 6 different possibilities having the following codes: 8, 8, ; 1, 1, ;,,,, ;,,,, ;,,,, 6; and 6,, 6,. 5. One of them has code,,, 6 and the other has code 1, 6,. 6. There are no semiregular tessellations that use four regular polygons because each vertex in a semiregular tessellation would need to be the vertex of at least one vertex angle from each of the four regular polygons and the sum of the measures of four such angles would be greater than 60 degrees.
Math Activity 9. 1. There are many patterns. Here are two:. Answers will vary.. n +. There are many possibilities for nets. Volume Volume 1 Volume Volume Surface Area 1 Surface Area Surface Area 18 Surface Area 18 Math Activity 9. 1. There are many solutions. Here is one of each:
. a. 60 degree rotational symmetry b. rotational symmetries of 60, 10, 180, 0, 00, and 60 degrees c. rotational symmetries of 90, 180, 70, and 60 degrees d. rotational symmetries of 180 and 60 degrees. Here is one example of each: a. b. c.. a. The 6 ways this can be done to obtain a line of symmetry are shown here. b. The ways this can be done to obtain rotation symmetry are shown here.
Math Activity 10.1 1. Square, units; triangle, units; trapezoid, 5 units; hexagon, 6 units; blue parallelogram, units; white parallelogram, units.. Other answers are possible. a. b. c.. perimeter, 6 perimeter, 10 perimeter, 10 9 units 10 units 11 units 1 units 1 units 1 units 15 units. Maximum Minimum Triangle 10 White Parallelogram 1 Hexagon 6 18 5. To obtain a figure with the least perimeter, make the figure as compact as possible by adjoining as many sides as possible. To obtain a figure with the greatest perimeter, separate the blocks so no sides overlap. 6. a. 61 units b. n - 1 triangles c. (n - 1) + units, or n + 1 units
Math Activity 10. 1. a. Trapezoid, area units; hexagon, area 6 units b. The square has an area of slightly more than units and white parallelogram has an area of slightly more than 1 unit.. a. Trapezoid, area 1/ unit; blue parallelogram, area 1/ unit; triangle, area 1/6 unit b. Square, area about /8 unit; white parallelogram, area about 1/5 unit. a. Section 1 has the area of the square; sections and each have an area which is about 1/ the area of a square, and section has an area which is about / the area of the square. b. Trapezoid, area about 1 1/ units; blue parallelogram, area about 8/9 units; triangle, area about /9 units c. A triangle and a square exactly cover white parallelograms and a triangle. So the square and the white parallelograms have the same area. Math Activity 10. 1. a. units b. Figures whose perimeters are odd numbers cannot be obtained. c. No d. Each tile has a perimeter of units. Whenever two or more tiles are joined on a complete edge the perimeter will change by an even number.. a. 1 b. c. The 50th figure will have 6 tiles. With one tile as "center" three arms will have 9 tiles each and the fourth arm will have 98 tiles. d. Area: 6 Perimeter: 9 e. Expressions equivalent to the following: Area: 5n - ; Perimeter: 10n - 6 f. The perimeter will always be an even number. One way to determine this is to note that each tile except the four on the ends of the four arms contribute two edges to the perimeter and each of the four tiles on the ends of the arms contribute edges. Another method of reasoning is explained in 1. d.. Area: n Perimeter: 8n
Math Activity 11.1 1. c. The area of the tile is one-half the area of the large right triangle. The total area of the two small semicircles is π times the area of the tile. The total area of the two semicircles is equal to the area of the large semicircle.. a. When the two tiles are in the positions shown at the right, Figures 1 and can be traced out by point B by moving tile ABCD as described below: Figure 1: slide up, slide left, rotate point B 180 about point D, slide down, slide right, rotate point B 180 about point D Figure : slide up, rotate point B 90 about point D; slide left, rotate point B 90 about point D; slide down; rotate point B 90 about point D; slide right; rotate point B 90 about point D b. If the area of the tile is 1 then the area of both figures is 5 + π. Figure 1 Figure
Math Activity 11. 1. Triangle 1 5 6 7 8 9 10 11 Center of Rotation A,G A A,F F F,G G E,M B,S G G. Triangle 1 5 6 7 8 9 10 11 Line of Reflection AG AE AF FB GF MG SG BE. Triangles, 6, 8, 10. Any of the following triangles:,,, 6, 7, 9, 10 5. a. b. Triangle 1 5 6 7 8 9 10 11 Center of Rotation E,A M E E K,B F F,J R,C Triangle 1 5 6 7 8 9 10 11 Center of Rotation AE BF EF BE KF c. To 7: reflect about BF onto square, rotate about G To 8: reflect about JB onto square 5, rotate about K To 10: reflect about JB onto square 5, rotate about M To 11: reflect about KF onto square 10, rotate about T Other ways are possible. Math Activity 11. 1. a. to 1 b. The two equal acute angles of the larger trapezoid are the acute angles of the smaller trapezoid. The two equal obtuse angles of both figures have a measure of 10 degrees. c. The area of the larger trapezoid is times the area of the smaller. d. There are different ways to construct enlargements of each figure. The area of the enlargement is times the area of the pattern block.. The area of each enlargement is 9 times the area of the pattern block.. One method of building an enlargement of a figure by a scale factor of k is to first build an enlargement of each pattern block by a scale factor of k and then build an enlarged figure which is similar to the original figure by using the enlargements of the pattern blocks. If the scale factor is k, then the area of the enlargement is k times the area of the smaller figure.