Download answers for activities, just for fun and connection questions

Answers: Activities and Connections
Activity Set 1.1
ANSWERS FOR ACTIVITIES, JUST FOR FUN AND
CONNECTION QUESTIONS
Activity Set 1.1
1. a. Here are some of the many ways they differ:
(1) The 7th figure has one more triangle than the 6th figure.
(2) The 6th has the same number of triangles pointing up as pointing down.
(3) The 6th figure forms a parallelogram, the 7th figure a trapezoid.
(4) The length of the base in the 7th figure is longer than the base in the 6th figure.
b. The 15th figure would contain 15 green triangles in a row. It would have the shape of a
trapezoid with 8 triangles pointing up and 7 triangles pointing down.
2. a. New figures are created by adding a shape onto the right side of the previous figure so
that the height of all the figures stays the same and there are no gaps. Odd numbered figures
add a green triangle alternating between pointing up and pointing down. Even numbered
figures add a blue rhombus alternating between slanting left and slanting right.
b. The 10th figure will have a blue rhombus on the right end because all even numbered
figures add a blue rhombus to the right end.
c. 13 triangles and 12 rhombuses. Each even-numbered figure has the same number of
triangles as rhombuses, and a rhombus on the right end. The 24th figure has 12 triangles and
12 rhombuses and a triangle attached to the right end produces the 25th figure.
d. For n an odd number; nth figure will have (n+1) ÷ 2 triangles and (n-1) ÷ 2 rhombuses.
3. a. New figures are created in this sequence by adding a shape onto the right side of the
previous figure. The 1st figure and every 3rd figure after that adds a yellow hexagon. The 2nd
figure and every 3rd figure after that adds a white rhombus. The 3rd figure and every 3rd figure
after that adds an orange square.
b. A rhombus. The pattern repeats after 3 steps so every 3rd figure (figures 3, 6, 9, etc.) has a
square on the right. The 15th figure has a square on the right, the 16th a hexagon on the right,
and the 17th a rhombus.
c. There are 7 hexagons, 7 rhombuses, and 6 squares in the 20th figure. Multiples of 3 have
the same number of each shape. The 18th figure has 6 of each shape. A hexagon and rhombus
is added on for the 19th and 20th figures.
d. 19 of each shape. Multiples of 3 have the same number of each shape.
4. a. The 10th figure has 4 hexagons and 6 triangles. The 15th figure has 5 hexagons and 10
triangles.
Activity Set 1.1
Answers: Activities and Connections
b. If the figure number is 1 more than a multiple of 3 determine the number of hexagons and
triangles for the previous figure (which is a multiple of 3) and then add another hexagon. If
the figure number is 1 less than a multiple of 3 determine the number of hexagons and
triangles for the following figure (which is a multiple of 3) and then subtract 1 triangle.
5.
2nd
1st
5th
a. Here are some of the ways they differ:
(1) The odd numbered figures have a trapezoid at each end and the even numbered figures
have triangles.
(2) The bottom side is the longest side in the odd numbered figures.
b.
1st
2nd
5th
In any even numbered figure the number of hexagons is the figure number divided by 2. In
any odd numbered figure the number of hexagons is the figure number plus 1 divided by 2.
6. Here are 3 ways. There are other possibilities.
Sequence I: Add one square to the top of each column in the previous figure.
Sequence II: Form 3 columns with 4 squares in each column and make the 4th column 1
square more than the preceding 4th column.
Sequence III: Form a square array whose dimensions are each one more than the figure number and remove 1 square from the upper right corner.
Answers: Activities and Connections
Connections 1.1
7. Sequence I: 10 hexagons, 10 triangles, 9 trapezoids, and 9 rhombuses; with a triangle on the
right end. The 36th figure looks like 9 copies of the 4th figure so it contains 9 of each shape.
To get the 38th figure we need 1 more hexagon and 1 more triangle.
Sequence II: 10 squares, 19 triangles, and 9 rhombuses; with a triangle on the right end. The
36th figure looks like 9 copies of the 4th figure so it contains 9 squares, 18 triangles, and 9
rhombuses. To get the 38th figure we need 1 more square and 1 more triangle.
Sequence III: 16 rhombuses, 15 triangles, and 7 squares; with a triangle on the right end. The
35th figure looks like 7 copies of the 5th figure so it contains 14 rhombuses, 14 triangles, and
7 squares. To get the 38th figure we need 2 more rhombuses and 1 more triangle.
Connections Questions 1.1
1. a. One hint might be to lay the existing portion of the sequence down one block at a time;
encouraging the student at each step to help you decide which block—a square or a triangle
to put down next.
b. One hint might be that the sequence is supposed to continue on and on; not to end.
Another hint might be to ask what the next block should be if the sequences continues.
2. Answers will vary but should include a diagram of the sequence, a set of corresponding
questions and a list of the elementary student’s responses.
3. Answers will vary but should include a diagram of two sequences with the same first four
figures and different figures after that and the results of challenging another student to find
the two sequences.
4. Entering Principles and Standards for School Mathematics into the search engine at nctm.org
will help the students find: http://www.nctm.org/standards. From there they should be able to
find the “Overview of Principles and Standards for School Mathematics” page. The
Principles and Standards for School Mathematics has four major components.
First, the Principles for school mathematics reflect basic perspectives on which educators
should base decisions that affect school mathematics.
The first five Standards present goals in the mathematical CONTENT areas of number
and operations, algebra, geometry, measurement, and data analysis and probability.
The second five Standards describe goals for the PROCESSES of problem solving,
reasoning and proof, connections, communication, and representation.
Finally, the document discusses the issues related to putting the Principles into action and
outlines the roles played by various groups and communities in realizing the vision of
Principles and Standards.
http://www.nctm.org/standards/overview.htm
5. Algebra Standard, Expectation: Understand patterns, relations, and functions. Studying
patterns leads to algebraic thinking and connections.
Activity Set 1.2
Answers: Activities and Connections
Activity Set 1.2
1. a. The 10th odd number (that is the number of tiles in the 10th figure) is 19.
a. There are many ways to describe it. Here is one way: The 20th figure looks like the letter
L made with tiles. Twenty tiles are used to make the vertical part and nineteen additional tiles
are needed to complete the horizontal part. Thus, there is a total of 20 + 19 = 39 tiles, and the
20th odd number is 39.
b. Here is one way to describe it: The 50th figure will have a horizontal row of 50 tiles with
a vertical column of 49 tiles on its left end. It will contain a total of 49 + 50 = 99 tiles.
c. Here are a few methods.
(1) Add the figure number plus the figure number minus 1: T = n + (n - 1)
(2) Multiply the figure number by 2 and subtract 1: T = 2n - 1
(3) Square the figure number and subtract the figure number minus 1 squared:
T = n2 - (n - 1)2
2. The 4th figure contains 22 tiles.
a. Here is one way: Make a horizontal row of 8 tiles. Affix 1 additional tile at each end of
the row of eight. Put a column of 8 tiles above each end tile. Then add a column of 8 tiles
below each end piece.
b. 42 tiles
c. 77 tiles
d. The 50th figure will look like an H with 101 tiles on each side and 50 tiles between the
sides. There will be a total of 252 tiles
e.
(1)
(2)
(3)
Here are a few methods.
T = 5n + 2
T = 4n + (n + 2)
T = (2n + 1)(2) + n
3. a. The 4th term in the sequence is 13 and the 10th term is 37. Given any figure number,
multiply it by 4 and then subtract 3 from that result. (T = 4n – 3)
b. The 4th term in the sequence is 20 and the 10th term is 44. Given any figure number,
multiply it by 4 then add 4 to that result. (T = 4n + 4)
Answers: Activities and Connections
Activity Set & Just for Fun 1.2
c. The 4th term in the sequence is 25 and the 10th term is 181. Given any figure number,
square that number and to that result add the square of the number which is one less than the
given figure number. T = [n2 + (n – 1) 2]
4. The 4th term is 20 and the 5th term is 30.
a. Here is one way: The 10th rectangle in the pattern has 11 rows and each row has 10 tiles.
Half the tiles are red and half are black. The rectangle has dimensions 11 units by 10 units,
for a total of 110 tiles.
b. The total number of tiles in the rectangle is 50 times 51 which equals 2550. There are an
equal number of red and blue tiles, so 2550 divided by 2 gives 1275 tiles of each color.
5. a. The 5th stair-step has 15 tiles and the 10th stair-step has 55 tiles.
b. The 50th stair-step has 1275 tiles.
c. The 100th stair-step
d. (100 × 101) ÷ 2 = 5050
e. The sum of consecutive whole numbers from 1 to any specified number (let’s call it the
last number) can be represented by a stair-step of tiles. This number of tiles in the stair-step is
one half of a rectangle with dimensions last number by last number + 1. So,
1 + 2 + 3 + 4 + … + (last number) = (last number) × (last number + 1) ÷ 2
6. Here is one method for doing this. The sum of consecutive even numbers from 2 to any
specified number (let’s call it the last number) can be represented by a stair-step of tile. The
number of tile in the stair-step is one half of a rectangle with dimensions (last number
+ 2) by (last number ÷ 2). So,
2 + 4 + 6 + 8 + … + (last number) = (last number + 2) × (last number ÷ 2) ÷ 2
Just for Fun 1.2
The first Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …
If the sequence begins with, “loves me” the last response will be “loves me” for daisies with
an odd number of petals.
Connections Questions 1.2
1. a. One hint might be to notice that the columns of shaded tiles represent 2, 4, 6, 8, etc. and
ask what the whole rectangle would form. Once the student sees the rectangle is double the
sum of the even numbers you can help them understand that half of the area of the rectangle
will be the sum they seek.
Connections 1.2
Answers: Activities and Connections
2. a. This should be a very simple color tile pattern and the corresponding simple number
pattern such as 1, 2, 3, … or 1, 3, 5,… or 2, 4, 6, …
b. This should be a simple number pattern and the corresponding tile pattern.
c. You might expect them to have difficulties connecting the idea of number with the tiles.
It might help to just count out loud for a while, then count out loud while placing the
corresponding number of tiles on the table to help them see that the tiles can also represent
numbers. Finally you can work on a sequence of numbers represented by a sequence of tiles.
3. a.
1
Sum: 1
1+3
1+3+5
1+3+5+7
4
9
16
1+3+5+7+9
25
b.
This represents 2 × (1 + 3 + 5 + 7);
since the dimensions of the rectangle are 4 × 8, this sum is ½ × 4 × 8 = 16.
c. To sum 1 + 3 + … + 191 note the height of the rectangle will be 192. To find the width
you need the number of odd numbers in the sum.
Note that 1 = 2(1) – 1, 3 = 2(2) – 1, 5 = 2(3) – 1, 7 = 2(4) – 1; in a sequence of n consecutive
odd numbers, the last number is 2(n) – 1, If 191 = 2 n – 1, then n = 96. The dimensions of the
rectangle are 96 × 192, therefore 1 + 3 + … + 191 = ½ × 96 × 192 = 9216.
4. a. Answers will vary but should include the first three figures of a tile sequence, (a) the
number of tiles in the 10th figure and a verbal description of how to form the figure, (b) the
number of tiles in the nth figure and an explanation of how to find this number.
5. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
6. At the Pre-K-2 level, children should be able to recognize, describe, and extend patterns this
pattern, of shapes, and translate the shapes into numbers.
At the 3 - 5 level, children should be able to describe, extend, and make generalizations about
the tile pattern and the corresponding numeric pattern, they should be able to verbally
describe the general nth figure.
At the 6 - 8 level, students should be able to represent, analyze, and generalize the color tile
sequence and show the number pattern with a table and a symbolic rule that the recognize is
a non-linear function.
Answers: Activities and Connections
Activity Set 1.3
Activity Set 1.3
1. a.
b. The consecutive numbers are 13, 14, 15, and 16. The sum of 58 is represented by 4
variable pieces and 6 unit pieces. So 4 variable pieces represent 52, (58 – 6), and each
variable piece represents 13, (52 ÷ 4).
2. a.
b. The perimeter of the triangle can be represented with 5 variable pieces and 6 units. The
perimeter is 66, so the 5 variable pieces represent 60 units. Therefore, each single variable
piece represents 12 units. The sides of the triangle are 12 units, 24 units and 30 units.
3. One variable piece represents the number of dimes. Because there are three times as many
nickels as there are dimes, 3 variable pieces represent the number of nickels. The total
number of coins is represented by 4 variable pieces.
a. When dimes are exchanged for nickels the number of coins that were dimes is doubled.
The single variable piece representing the number of dimes must be replaced by 2 variable
pieces representing the number of nickels.
b. The total of $2.25 is represented by 5 variable pieces. Since 2.25 ÷ 5 = .45, one variable
piece represents 45 cents or 9 coins (nickels).
c. Since one variable piece represents 9 coins, there are 3 × 9 = 27 nickels and 1 × 9 = 9
dimes.
4. short piece of rope
long piece of rope
end to end
End to end the length is 75 m; 2 variable pieces represent 68 m (75 m – 7 m); 1 variable
piece represents 34 m. The short piece of rope is 34 m and the long piece is 41 m (34 m + 7
m).
Activity Set 1.3
Answers: Activities and Connections
5. number of girls
number of boys
altogether
The total number of students, 76, is represented by 4 variable pieces. So each variable piece
represents 19 students (76 ÷ 4) and there are 19 girls and 57 boys.
6. number of dimes Greg has
number of nickels Andrea has
Suppose Greg exchanges his dimes for nickels
number of nickels Greg has
number of nickels Andrea has
Andrea has 80 cents more than Greg. So 2 variable pieces represent 80 cents—which is the
amount of money that Greg has.
7. width of rectangle
length of rectangle
length plus width
Half the perimeter (length plus width) is 65 feet. So 4 variable pieces total 60 feet (65 feet – 5
feet) and 1 variable piece represents 15 feet. The width of the rectangle is 15 feet and the
length is 50 feet.
8. number of men
number of women
double number of men
increase women by 9
Suppose each variable piece represents one-fifth of the students. Then 2 variable pieces
represent the total number of men and 3 pieces the number of women. The equality given by
doubling the number of men and increasing the women by 9, shows that each variable piece
represents 9 people. The original class has 45 people, 18 men and 27 women.
Answers: Activities and Connections
Connections 1.3
9. 1st number
2nd number
3rd number
sum of numbers
The sum of the numbers is 43 and we can see that 4 variable pieces represent 36 (43 – 7). Then
one variable piece represents 9 (36 ÷ 4) and the numbers are 14, 9, and 20.
Connections Questions 1.3
1. One hint might be to ask questions such as “What is 3 more than 4? What is 3 more than 5?
What is 3 more than 6?” and then discuss how you might want to ask this question in general
as “What is 3 more than an unknown number? and When is it equal to 9? or When is it equal
to 12?” Then you might show how to use algebra pieces to represent all of these scenarios.
2. Answers will vary but should include a simple problem that can be solved with algebra
pieces and mental arithmetic as well as questions or hints posed and an elementary student’s
responses.
odd number
3.
odd number + 2
odd number + 4
odd number + 6
This is four consecutive odd numbers; there sum is 64, so 64 – 12 = 52, each piece is 52/4 = 13.
The numbers are 13, 15, 17 and 19.
l = 2w
4. Rectangle:
w
w
l = 2w
w
w
w
w
w
w
Let width = 1 algebra piece
If the perimeter is 54 units, then 6w = 54, w = 8 units. The length is 16 units and the width is 8
units.
5. Answers will vary but there are several lessons that involve growing patterns or looking at
sequences of shapes. Answers should include a summary of the lesson and a connection to
Standards Expectations.
6. The Expectation: “Use mathematical models to represent and understand quantitative
relationships; model problem situations with objects…” is met by using algebra pieces
because it allows students to precisely accomplish the task of using an object to model a
problem situation. This helps them use a concrete model to understand the underlying
algebraic relationship.
Activity Set 2.1
Answers: Activities and Connections
Activity Set 2.1
1. The grid can be completed in several ways. Here is one way.
SYS
SBS
SRS
SRH
SRT
SRC
SYC
SYT
SYH
SBH
SBT
SBC
2. a. By asking yes or no questions that refer to a single attribute a player can identify any piece
in at most 6 guesses. For example: Is it large? Is it red? Is it blue? Is it triangular? Is it
square? Is it circular?
b. A sequence of questions like the following will reduce the number of pieces to two: Is it
large? Does it have opposite parallel sides? Is it blue? Is it red? Then a fifth question can be
asked to identify the piece from the two remaining pieces.
3. It is not possible to place all 24 attribute pieces on the grid.
4. a. SRT, LRT
b. LRC, SRT, SBT, LRS, LRH, SRC, SRS, SRH, LRT, SYT, LYT, LBT
5. a. LRC, LYC, SRC, SYC
b. LBC, SBC, LRC, SRC, LYC, SYC, LRS, SRS, LRT, SRT, LRH, SRH, LYT, SYT, LYS,
SYS, LYH, SYH
c. NOT CIRCULAR AND BLUE pieces
6. a. SBC, SBT, SBH, SBS, SYC, SYT, SYH, SYS
b. LARGE RED pieces; or, LARGE AND RED pieces
c. LYT, SYT, LBT, LBS, LBC, LBH, SBT, SBS, SBC, SBH, LRT, LRH, LRS, LRC, SRT,
SRH, SRS, SRC
d. The entire set of attribute pieces.
7. b.
(1) (H∩NY):
SRH, SBH, LRH, LBH
(2) (H∩NY) ∩ L:
LBH, LRH
(3) (NY∩L) ∩ H:
SYH, LYH, LRH, LBH, SRH, SBH, LRS, LRT, LRC,
LBS, LBT, LBC
Answers: Activities and Connections
Connections 2.1
(4) (H∩L) ∩ H:
SYH, SRH, SBH, LYH, LRH, LBH
(5) H ∩ (NY∩L):
SBH, SRH, LBH, LRH, LYH
(6) (H∩NY) ∩ L:
SRH, SBH, LRH, LBH, LYH, LYC, LYT, LYS, LRS,
LRT, LRC, LBS, LBT, LBC
Connections Questions 2.1: Selected Answers
1. Discuss with the students that union means everything in Set A, in Set B or in both Set A and
Set B. Therefore figure b is correct.
3. a. not (A ∪ B) = all of the pieces that are neither blue nor large. All of the small red pieces
and all of the small yellow pieces. This is illustrated by figure 1.
not A ∩ not B is all of the pieces that are not blue and also not large; again, all of the small
red pieces and all of the small yellow pieces. This is illustrated by figure 2 where not A is
marked by the vertical lines, not B is marked by the horizontal lines and not A ∩ not B is
marked by the crossing vertical and horizontal lines.
Figure 1
Figure 2
b. not (A ∩ B) = all of the pieces that are not both blue and large. All of the red pieces, all
of the yellow pieces and all of the small blue pieces. This is illustrated by figure 3.
not A ∪ not B is all of the pieces that are not blue or not large; again, all of the red pieces,
all of the yellow pieces and all of the small blue pieces. This is illustrated by figure 4 where
not A is marked by the vertical lines, not B is marked by the horizontal lines and not A ∪ not
B is marked by any space with a vertical or horizontal line.
Figure 3
Figure 4
c. The diagrams in part a illustrate that not (A ∪ B) = not A ∩ not B is true in general. The
diagrams in part b illustrate that not (A ∩ B) = not A ∪ not B is true in general
Activity Set 2.2
Answers: Activities and Connections
Activity Set 2.2
1. a.
4
1
−2
3
b.
c.
2. a.
2
3
d.
−3
4
e.
2
4
b.
c.
d.
e.
All of the line segments with the same slope are parallel.
3. a.
b.
A
c.
C
B
d.
A
e.
C
A
B
B
B
C
B
A
C
C
A
AB
−
1
2
2
2
−
2
1
−
2
2
2
1
BC
−
1
2
1
1
−
2
1
−
1
1
2
1
AC
−
2
4
3
3
−
4
2
−
3
3
4
2
The line segments are on the same line; the slope of AB = the slope of BC = the slope of AC .
2
3
b. The point of intersection is (0, 3) and the y coordinate of this point is the value of b in y =
mx + b.
4. a. The slope is the coefficient of x; it is −
c. If you know the slope m and the y-intercept b, the equation of the line is y = mx + b.
5.
a. y =
1
x+2
3
e. y = -2x + 6
b. y =
4
x
3
f. y = x - 1
c. y = −
g. y =
1
x+2
2
1
x+1
2
d. y = 2
h. y = -x + 4
Answers: Activities and Connections
1
x+3
4
1
y= x+1
4
a. y =
6. 1
2
b. y = -4x + 8
y=
1
x+1
4
Activity Set 2.2
3
x+3
2
3
y= − x+7
2
c. y = −
d. y = −
1
x+2
2
y = 2x - 2
If two lines are parallel; their slopes (m) are the same, there is no relationship between their yintercepts.
If two lines are perpendicular; their slopes (m) are negative reciprocals of each other, there is no
relationship between their y-intercepts.
7.
1
2
3
4
Common
features
2
x+2
3
2
y= x+1
3
2
y= x
3
2
2
y= x3
3
a. y =
Parallel
Same slope
8. a. 2 x 2 geoboard: 2 lines
b. y = x + 1
y=
1
3
x+
2
2
y = -2x + 4
y = -x + 3
Through (1,2)
Different slope
and y-intercepts
1
x+1
2
1
y= − x+2
2
1
y= − x+3
2
1
y= − x+4
2
c. y = −
Parallel
Same slope
d. y = 4x
y = 2x
y=x
y=
1
x
2
Through (0,0)
Different slope
3 x 3 geoboard: 4 lines
4 x 4 geoboard: 8 lines
b. 6 x 6 geoboard
0 1 2 3 4 5 1 2 3 1 2
0
5
, , , , , , , , , ,
and the reciprocals of all but and : 20 lines.
5 5 5 5 5 5 4 4 4 3 3
5
5
Connections 2.2
Answers: Activities and Connections
Connections Questions 2.2: Selected Answers
4. a. All lines with y-intercept of 3 are of the form y = mx + 3, where m is any number.
b. All lines with slope 3 are of the form y = 3x + b, where b is any number.
c. All lines with y-intercept 3 and slope 3 are of the form y = 3x + 3.
5. Answers will vary; here are some possible categories:
Parallel
1
y= x+3&
3
1
y= x–3
3
1
y=- x+3&
3
1
y=- x–3
3
Perpendicular
1
y= x+3&
3
1
y = -3x +
3
1
y=- x+3&
3
1
y = 3x 3
Same y-intercept
1
y= x+3&
3
1
y=- x+3
3
1
y= x–3&
3
1
y=- x–3
3
Positive Slope
1
y= x+3&
3
1
y = 3x +
3
1
y= x–3&
3
1
y = 3x 3
Negative Slope
1
y=- x+3&
3
1
y = -3x +
3
1
y=- x–3&
3
1
y = -3x 3
6. a. Standards references to plotting points
Standard
Geometry Standard: Specify
locations and describe spatial
relationships using coordinate
geometry and other
representational systems
Algebra Standard: Understand
patterns, relations, and functions
Level Expectation
3-5 Make and use coordinate systems to specify
locations and to describe paths; find the distance
between points along horizontal and vertical
lines of a coordinate system
6-8
Represent, analyze, and generalize a variety of
patterns with tables, graphs, words, and, when
possible, symbolic rules; identify functions as
linear or nonlinear and contrast their properties
from tables, graphs, or equations
b. Slope is under the Algebra Standard as finding slope involves finding patterns. It is under
this grade level since the idea of slope is not formally used until these grades.
Answers: Activities and Connections
Activity Set & Connections 2.3
Activity Set 2.3
1. Monty
(shortest)
Veneta
Edith
Marc
Jane
Sally
Uri
Nicky
Paul
(tallest)
2. Juan, Anna, Derek, and McKenna.
3. Robinson
Math prize
4. Smith
Pilot
Smith
English prize
Brown
French prize
Jones
Co-Pilot
Jones
Logic prize
Robinson
Engineer
5.
House
Yellow
Blue
Red
Ivory
Green
Nationality
Norwegian
Ukrainian
English
Spaniard
Japanese
Car Make
Chevrolet
Plymouth
Mercedes
Honda
Ford
Beverage
Water
Tea
Milk
Orange Juice
Coffee
Pet
Fox
Horse
Snails
Dog
Skunk
Connections Questions 2.3: Selected Answers
Go shopping
1. Describe how to use a Venn Diagram to see how Do Chores leads to
Going Shopping. Help them see that the X could be in the position of
Go Shopping without doing their chores. Answers may vary
X
Chores
4. Answer will vary but for any cooperative logic problem, the following
processes may be described:
Problem Solving Process Standard: Apply and adapt a variety of appropriate strategies to
solve problems.
We used a variety of strategies to solve the logic problems.
Reasoning and Proof Process Standard: Make and investigate mathematical conjectures
We made conjectures about the order or grouping of the solutions and then adjusted them
using logical thinking.
Connections 2.3
Answers: Activities and Connections
Communication Process Standard: Organize and consolidate their mathematical thinking
through communication; communicate their mathematical thinking coherently and clearly to
peers, teachers, and others; analyze and evaluate the mathematical thinking and strategies of
others
The cooperative nature of the activities required an ability to communicate our ideas and
to understand the ideas of others.
Connections & Representation Process Standard: Probably not directly used
5.
Students
Two important elements of reasoning for students in the early grades are pattern-recognition
and classification skills. Students should create and describe patterns and make conjectures.
By the end of second grade, students also should use properties to reason about numbers.
Teachers
Teachers should maintain an environment that respects, nurtures, and encourages students so
that they do not give up their belief that the world, including mathematics, is supposed to
make sense.
Teachers should guide students to use examples and counterexamples to test whether their
generalizations are appropriate.
Teachers should create learning environments that help students recognize that all
mathematics can and should be understood and that they are expected to understand it.
Teachers should prompt students to make and investigate mathematical conjectures by asking
questions that encourage them to build on what they already know.
Answers: Activities and Connections
Activity Set 3.1
Activity Set 3.1
1.
Long-flats Flats
Longs
Unit
a.
1
0
0
2
b.
0
2
1
1
c.
0
0
4
4
d. Five base pieces of any one type can always be traded for 1 piece of another type so there
need not be more than 4 base pieces of any one type.
2. a. 0 long-flats
1 flat
0 longs
3 units
b. 0 long-flats
1 flat
1 long
1 unit
c. 1 long-flat
0 flats
0 longs
1 unit
d. 1 long-flat
3 flats
0 longs
0 units
e. 38 unit squares
f. 453 unit squares
g. 624 unit squares
3
b. 111five c. 1001 five d. 1300 five e. 123 five f. 3303 five g. 4444 five
4. a. 329 unit squares
b. 142 unit squares
c. 254 unit squares
5. a. Base three
Flat
Long Unit
Activity Set 3.1
Answers: Activities and Connections
b. Base ten
Flat
Long Unit
6. a. 122three
total number of unit squares is 17
b. 425 seven
total number of unit squares is
215
c. 157 ten
total number of unit squares is 157
7. a. 402five
b. 123 nine
c. 102 ten
Answers: Activities and Connections
Just for Fun 3.1
Just for Fun 3.1
Mind Reading Cards
1. If a person is 22 years old, their age will appear on Cards 16, 4, and 2. The binary numbers
whose sum is 22 are 16, 4, and 2.
2. 27 = 1 + 2 + 8 + 16, cards 1, 2, 8, and 16
3. 11
4. A number k is on a card if and only if the number of the card is a term in the sum of the
binary numbers that equals k. So the sum of the numbers of the cards on which a person’s
age appears is the person’s age.
5. Since 44 = 32 + 8 + 4, the number 44 must be written on cards 4, 8, and 32.
6. Card 32: has numbers 32 through 63
Card 16: has additional numbers 48 through 63
Card 8: has additional numbers 40 through 47; and 56 through 63
Card 4: has additional numbers 36-39, 44-47, 52-55, and 60-63
Card 2: has additional numbers 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63
Card 1: has additional numbers 33, 35, 37, 39, … 59, 61, 63
The greatest age on the six card system is 63.
Game of Nim
1. Here are 3 arrangements for which you can win if it’s your opponents turn to play.
2. The 1-2-3 arrangement is an even situation because there are two 2s and two 1s.
3. Crossing out a stick in the second row leaves an even situation.
Connections 3.1
Answers: Activities and Connections
4. It is not possible in one turn to change an even situation into another even situation.
Therefore, if you leave your opponent with an even situation there will be an odd situation on
your turn. To change an even situation into an even situation you would need to cross out 2
binary groups of the same type. But, if you have 2 binary group of the same type in a given
row this forms the next larger binary group.
5. The winning strategy is to leave an even situation for each of your opponent’s turns. This is a
winning strategy for any number of rows and columns.
Connections Questions 3.1: Selected Answers
3. Digits in base b are 0, 1, 2, …, b – 1 because when you get to b objects you group to the next
place value.
4. a. A Flat-Flat is b flats × b flats.
b flats
b flats
b.
Powers of 7
70 71 72 73
74
75
1 7 49 343 2401 16807
30000 ÷ 16807 = 1.8 (to 1 decimal place)
30000 - 16807 = 13193
13193 ÷ 2401 = 5.5 (to 1 decimal place)
13193 – 5 × 2401 = 1188
1188 ÷ 343 = 3.5 (to 1 decimal place)
1188 – 3 × 343 = 159
159 ÷ 49 = 3.2 (to 1 decimal place)
159 – 3 × 49 = 12
12 – 7 = 5
Answers: Activities and Connections
Connections 3.1
30000 = 153315seven
5 flat-flats (not to scale)
3 long-flats 3 flats 1 long
5 units
1 long-flat-flat (not to scale)
5. a. 21 in Base Two: 10101two
b. 11011two 1 unit, 1 long, 0 flats, 1 long-flat and 1 flat-flat
16
+
8
+
2 +
1
= 27
6. Answers will vary, students may refer to:
Problem Solving: Build new mathematical knowledge through problem solving; Apply and
adapt a variety of appropriate strategies to solve problems; Monitor and reflect on the process
of mathematical problem solving
Communication: Analyze and evaluate the mathematical thinking and strategies of others
Connections: Recognize and use connections among mathematical ideas
Representation: Create and use representations to organize, record, and communicate
mathematical ideas; Select, apply, and translate among mathematical representations to solve
problems; Use representations to model and interpret physical, social, and mathematical
phenomena
Activity Set 3.2
Answers: Activities and Connections
Activity Set 3.2
2. a. Two longs and 2 units
b. Since the maximum amount you can win in a roll is 2 longs and 2 units, it will take at
least 11 turns to win the game. (If you win the maximum amount on each turn, then in 2 turns
you will be 1 unit short of a flat. So in 10 turns you will be 1 long short of a long-flat, and it
will take one more turn to win.)
c. The least you can win in one roll is 2 units. If you rolled 2 units on every turn it would
take 63 turns to obtain a long-flat.
d. Comparing 4 flats, 2 longs and 1 unit to a long flat, you will see that you are 2 longs and
4 units short of a long-flat. You cannot win on the next turn because the maximum amount
you can win on 1 turn is 2 longs and 2 units.
e. It will take at least 3 more turns to reach 1000five. Obtaining the maximum amount on
each turn, 2 longs and 2 units, you would still be 1 long and 3 units short after 2 turns. But
you could reach 1000five by winning 1 long and 3 units on the next turn.
3. a. 122 five
b. 1101 five
c. 1021 five
d. 1110 five
4. 1010 three
b. 3121 four
Regroup
Answers: Activities and Connections
Activity Set 3.2
c. 433six
Regroup
6. a. The game can be won in 11 turns. (The maximum amount that can be discarded on a turn
is 2 longs and 2 units. After 10 turns of maximum discards you would still have 1 long
remaining. So you could discard everything in 11 turns.)
b. Comparing the base pieces for 313 five and 242 five the difference is 2 longs and 1 unit. So
the player must have rolled a five on one die and a six on the other.
7. Method I: The student sketched 232 five. In step 1, 1 flat was removed. In step 2, 1 flat was
traded for 5 longs and then 4 longs were removed. In step 3, 1 long was traded for 5 units and
then 3 units were removed, leaving 3 longs and 4 units.
Method II: The student sketched the 2 collections: 232 five and 143 five. In step 1, common
pieces in each collection were matched and removed: 1 flat, 3 longs and 2 units. In step 2, the
remaining mat from the first collection was traded for 5 longs and l long was matched and
removed. In step 3, 1 long from the first collection was traded for 5 units and 1 unit was
matched and removed. This leaves 3 longs and 4 units of the larger collection.
Method III: In step 1, the 2 collections were sketched so that they line up vertically. In step 2,
pieces were added to the smaller collection until it matched the larger collection.
8. Method I: One way to think of this is as Take Away. The student took away 1 flat, 4 longs
and 3 units.
Method II: One way to think of this is as Comparison. At each step the student compared the
values in the minuend and the subtrahend.
Method III: One way to think of this is as The Missing Addend. The base-ten number pieces
outlined by the dotted lines are the missing addend.
Just for Fun & Connections 3.2
Answers: Activities and Connections
Just for Fun 3.2
1. If you cross off 16, then no matter what your opponent does, you will be able to cross off
numbers through 20 on your next turn and win the game. To ensure that you are able to cross
off 16, you must be the player to cross off 12. Similarly, you can be sure of crossing off 12 if
you cross off 8, and you can be sure of crossing off 8 if you cross off 4. The only way you
can be sure of crossing off 4 is if your opponent begins the game.
2. If your opponent begins the game you make sure that after your first turn 5 toothpicks have
been removed; 10 toothpicks after your second move; 15 after you third; and 20 after your
fourth move. Then your opponent will be forced to pick up the last toothpick.
3. You win the game if you leave a total of 1 for your opponent. (Your opponent must subtract
1.) You can be assured of leaving a total of 1 if you can leave a total of 7 on the preceding
turn. (can you see why?) And, you can leave 7 if you leave 13, 13 if you leave 19, 19 if you
leave 25, and so forth. Strategy: If the game starts on a number which is 1 more than a
multiple of 6, let your opponent begin. Otherwise you begin and keep reducing the number to
1 more than a multiple of 6.
Connections Questions 3.2: Selected Answers
2. Error in 23 + 18 = 311 and 34 + 18 = 412; 23 + 18 = 30 + 11, 34 + 18 = 40 + 12. The student
is not adding the partial sums; they are adding the 1s place digits and placing the sum of the
10s place digits in front of that. The student does not understand the place value associated
with adding the 10s place digits. You can use base-ten number pieces to emphasize how to
add the 10s place sum and the 1s place sum correctly; the student can see adding unit and
long pieces.
3. a. 431five + 233 five = 1214 five
431
+ 233
4 units……………………………………….4
6 longs = 1 long + 1 flat…………………….10
6 flats + 1 flat = 1 long-flat + 2 flats………1200
1214 five
+
Regroup
Answers: Activities and Connections
b. 1982 + 2189
1982
+ 2189
11 units = 1 unit + 1 long……………………….1
16 longs + 1 long = 7 longs + 1 flat…………….70
10 flats + 1 flat = 1 long-flat + 1 flat……………100
3 long-flats + 1 long-flat = 4 long-flats………….4000
4171
c. 431five - 233 five = 143 five
In base five
300 + (20 + 50) + (1 + 5)
- 200 30 3
3 units…………………….3
4 longs……….…………………….40
1 flat……………………………...………..100
143 five
-
d. 2912 + 1189
2000 + 800 + (0 + 100) + (2 + 10)
- 1000 – 100 80 9
3 units…………………….3
2 longs……….……………………...20
7 flats……………………………...…………700
1 long-flat……………………………...………..1000
1723
Connections 3.2
Activity Set 3.3
Answers: Activities and Connections
Activity Set 3.3
1. a. 84
b. 161
7
6
23
14
6 × 14 = 84
7 × 23 = 161
2. Any two-digit number can be represented by, at most, 9 longs and 9 units. Multiplication of a
one-digit number by a two-digit number can be done concretely by laying out the specified
number (1 to 9) of copies of longs and flats for the two digit number. The product of the
numbers can then be determined by regrouping the base pieces. Since 10 units can be
regrouped as 1 long, 10 longs as 1 flat, and so on, it is not necessary to count beyond 10.
3. a. 169
b. 483
c. 204
13
17
21
13
12
23
4. Products can be determined by counting the numbers of flats, longs, and units in rectangular
arrays. Use the numbers being multiplied to outline a rectangle. Then fill in the rectangle
with your base pieces. Regroup the pieces to determine the product. The following example
shows this procedure for 21 × 23.
So, 21 × 23 is represented by 4 flats, 8 longs, and 3 units, and 21 × 23 = 483.
Answers: Activities and Connections
Activity Set 3.3
5. a.
b.
6.
Sketch & Product
a.
12 x 23
4 Partial Products
23
× 12
6
40
30
200
276
(2 × 3)
(2 × 20)
(10 × 3)
(10 × 20)
2 Partial Products
23
× 12
46 (2 × 23)
230 (10 × 23)
276
b.
22 × 43
22
× 43
6
60
80
800
946
(3 × 2)
(3 × 20)
(40 × 2)
(40 × 20)
22
× 43
60 (3 × 22)
800 (40 × 20)
946
Activity Set & Just for Fun 3.3
Answers: Activities and Connections
Sketch & Product
4 Partial Products
2 Partial Products
c.
45
× 45
25
200
200
1600
2025
(5 × 5)
(5 × 40)
(40 × 5)
(40 × 40)
45
× 45
225 (5 × 45)
1800 (40 × 45)
2025
45 × 45
7. The first two products in the four partial products add to the first product in the two partial
products and the second two products in the four partial products add to the second product
in the two partial products.
Just for Fun 3.3
Answers: Activities and Connections
Connections 3.3 & Activity Set 3.4
Connections Questions 3.3: Selected Answers
3. 13five × 4five = 4 sets of 1 long and 3 units; minimal collection is 1 flat, 1 long and 2 units =
112five
Group
Regroup
4. 13five × 12five in a rectangular array; minimal collection is 2 flats, 1 long and 1 unit = 211five
1.2five
Regroup
1.3five
Activity Set 3.4
1. a.
24 ÷ 6 = 4 (4 groups of 6)
b.
.
132 ÷ 12 = 11 (11 in each of 12 equal groups)
Activity Set 3.4
Answers: Activities and Connections
2. a.
57 ÷ 4
(14 in each of 4 groups with 1 remaining)
b.
114 ÷ 12
(9 groups of 12 with 6 remaining)
3. a.
b.
21
24
4
13
c.
22
21
Answers: Activities and Connections
4. a. 221 ÷ 17 = 13
b. 529 ÷ 23 = 23
c. 397 ÷ 34 = 11 with remainder 23
Activity Set 3.4
Activity Set 3.4
5.
6. a.
b.
c.
Answers: Activities and Connections
Answers: Activities and Connections
Just for Fun & Connections 3.4
Just for Fun 3.4
Arabian Nights Mystery
Any 3-digit number abc times 1001 is equal to the 6-digit number abcabc.
The product, 7 × 11 × 13 = 1001. Therefore, dividing by 11, 7, and 13, as directed, is equivalent
to dividing by 1001, and the final result is the original 3-digit number.
Magic Formula
Let N be any number. The following steps show that the final result will always be 3.
1.
2.
3.
4.
5.
6.
7.
8.
Select N
Add 221
Multiply by 2652
Subtract 1326
Divide by 663
Subtract 870
Divide by 4
Subtract N
N
N + 221
2652N + 586092
2652N + 584766
4N + 882
4N + 12
N+3
3
Connections 3.4: Selected Answers
2. Adam is using the measurement method; for 12 ÷ 3, you measure off shares of size 3 and
then determine the number of shares (4). Carlie is using the sharing method (12 chips in 3
piles results in 4 chips in each pile) without noticing the size of the shares or the number of
shares that Adam is forming. Help Carlie by having her form both collections to see the
difference.
4. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
5. Division can been seen as the inverse of multiplication using base ten number pieces by
reading a rectangular array in two ways:
5
20
4
Multiplication
4 × 5 = 20
5
20
4
Division
20 ÷ 5 = length of other edge = 4
Activity Set 4.1
Answers: Activities and Connections
Activity Set 4.1
1. The even numbers are represented by two-high rectangular arrays of tiles; the odd numbers
by two-high rectangular arrays with an extra tile left over.
2. a. The 7th even number is 14.
b. The 125th even number has two rows of 125 tiles. The 125th even number is 250.
c. The nth even number will have two rows of n tiles. The nth even number is 2n.
3. a. The 5th odd number is 9. The 8th odd number is 15.
b. The 15th odd number has a horizontal row of 15 tiles on the bottom and a row of 14 tiles
on the top. The 15th odd number is 29.
c. The 50th odd number is 99; the 100th odd number is 199; the nth odd number is 2n –1.
d. 79 is in the 40th position and 117 is in the 59th position.
4. a. Sum is even.
b. Sum is even.
c. Sum is odd.
d. Sum is odd if exactly one of the numbers is odd; even if exactly one is even.
Answers: Activities and Connections
Activity Set 4.1
e. The difference is even.
f. The sum is odd.
5. a. The sum of the first five consecutive odd numbers is 5 × 5 = 25.
b. The L-shapes representing the first 10 consecutive odd numbers form a 10 by 10 tile, so
the sum of the first 10 consecutive odd numbers is 100.
c. There are 40 odd numbers from 1 to 79. There sum is 40 × 40 = 1600.
6. Only one rectangle is possible for the numbers 1, 2, 3, 5, 7 and 11. The numbers 2, 3, 5, 7
and 11 have exactly two factors and the number 1 has only one factor. Two rectangles are
possible for the numbers 4, 6, 8, 9 and 10. The numbers 6, 8 and 10 have four factors and 4
and 9 have three factors because one of their rectangles is a square. Three rectangles which
are not squares are possible for 12, so 12 has 6 factors.
Activity Set 4.1
Answers: Activities and Connections
7.
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Factors
Number of factors
1
1
1, 2
2
1, 3
2
1, 2, 4
3
1, 5
2
1, 2, 3, 6
4
1, 7
2
1, 2, 4, 8
4
1, 3, 9
3
1, 2, 5, 10
4
1, 11
2
1, 2, 3, 4, 6, 12
6
1, 13
2
1, 2, 7, 14
4
1, 3, 5, 15
4
1, 2, 4, 8, 16
5
1, 17
2
1, 2, 3, 6, 9, 18
6
1, 19
2
1, 2, 4, 5, 10, 20
6
8. a. Prime numbers have exactly 2 factors; composite numbers have more than 2 factors.
b. 1 is the only nonzero whole number with exactly 1 factor.
c. No. Any number greater than 1 will have at least one rectangle with one dimension of 1
and the other dimension equal to the given number.
d. The square numbers on the chart are 1, 4 and 9.
e. Numbers with an odd number of factors have a square among their rectangles. That is,
they are square numbers.
Answers: Activities and Connections
Activity Set & Connections 4.1
9. a. 5 is an odd number, so a number with 5 factors must be a square number. The numbers 16
and 81 have 5 factors. There are infinitely many such numbers.
b. 30 has 4 rectangles: 1 by 30, 2 by 15, 3 by 10 and 5 by 6. 30 has 8 factors: 1, 2, 3, 5, 6,
10, 15 and 30.
64 has 4 rectangles: 1 by 64, 2 by 32, 4 by 16 and 8 by 8. 64 has 7 factors: 1, 2, 4, 8, 16, 32
and 64.
10. The numbers 6, 8, 10, 14, 15, 21, 22, 26, 27 and 33 are a few of the numbers that have
exactly 4 factors. These numbers are the product of two different primes or the cube of a
prime.
Just for Fun 4.1
1. Eleven is the next prime number. Eleven multiplied by any number less than 11 will already
have been crossed off. Therefore, 11 × 11 = 121 is the first multiple of 11 that will be crossed
off. However, the chart only goes up to 120.
2. Multiples of 3 have a blue dot and multiples of 5 have a green dot. So, look for numbers
which have both a blue and a green dot. These will be multiples of 15.
3. Keith used 11 different colors. He placed dots on multiples of 2, 3, 5, 7, 11, 13, 17, 19, 23,
29, and 31. The next prime number is 37. Since 372 = 1369 Keith doesn’t need to find
multiples of 37 or any larger number (see part a).
Connections Questions 4.1: Selected Answers
2. One way is a visual method. The consecutive odd numbers from 1 to 219 can be visualized as
1 3
5
…
219
n 1 2
3
?
You can help the student see the pattern in a variety of ways; for example, the nth odd number
as n + (n – 1). Thus if 210 = 2n - 1, n = 110. The sum of 1 + 3 + … + 219 = 1102.
3. From this perspective; the nth odd number looks like:
with n squares in the bottom row and n – 1 squares in the top row for a total of 2n – 1 square.
Activity Set 4.2
Answers: Activities and Connections
4. Numbers have an odd number of factors if they are square numbers. The square numbers less
than 100 are: 1, 4, 9, 16, 25, 36, 49, 64 and 81. In each case the square root factor forms a
square array and adds only 1 factor to the list of factors for the number.
Activity Set 4.2
1. a. The rods can also be cut into common lengths of 1, 2, 3, 9 and 18 units.
b. 1, 2, 3, 6, 9, 18
c. GCF(36,54) = 18
2. a. GCF(36,24) = 12
b. GCF(18,25) = 1
c. GCF(42,28) = 14
3. a. GCF(20,12) = 4
b. GCF(8,12) = 4
Answers: Activities and Connections
Activity Set 4.2
4. a. GCF(16,24) = 8
b. GCF(14,28) = 14
c. GCF(9,12) = 3
5. GCF(40, 24) = 8, GCF(42,28) = 14, GCF(21,12) = 3
6. a. GCF(144,27) = GCF(117,27) = GCF(90,27) = GCF(63,27) = GCF(36,27)= GCF(9,27) =
GCF(9,18) = GCF(9,9) = 9
b. GCF(280,168) = GCF(112,168) = GCF(112,56) = GCF(56,56) = 56
c. GCF(714,420) = GCF(294,420) = GCF(294,126) = GCF(168,126) = GCF(42,126) =
GCF(42,84) = GCF(42,42) = 42
d. GCF(306,187) = GCF(119,187) = GCF(119,68) = GCF(51,68) = GCF(51,17) = GCF(34,
17) = GCF(17, 17) = 17
7.
Activity Set & Just for Fun 4.2
Answers: Activities and Connections
8. a. The product of A and B is equal to the product of GCF(A,B) and LCM(A,B).
A B GCF(A,B) LCM(A,B) A × B
(1) 6 15
3
30
90
(2) 8 12
4
24
96
(3) 14 21
7
42
294
(4) 5 7
1
35
35
(5) 8 10
2
40
80
b. The LCM of two numbers is equal to their product divided by the GCF of the numbers.
9. a. GCF(9,15) = 3, 9 × 15 = 135, 135 ÷ 3 = 45, and so LCM(9,15) = 45.
b. GCF(8, 18) = 2, 8 × 18 = 144, 144 ÷ 2 = 72, and so LCM(8, 18) = 72
c. GCF(140, 350) = 70, (140 × 350) ÷ 70 = 700, and so LCM(140, 350) = 700.
d. GCF(135,42) = 3, (135 × 42) ÷ 3 = 1890, and so LCM(135,42) = 1890.
Just for Fun 4.2
1. Yes
2. Star (n,s) is congruent to (has the same size and shape) star (n,r).
Answers: Activities and Connections
Just for Fun 4.2
3. Star (5,1) has one path; star (10,4) has 2 paths; star (12,4) has 4 paths; and star (6,3) has 3
paths.
4. n and s must be relatively prime, that is, GCF(n,s) = 1. These stars have 1 continuous path:
(1) star (15,2) = star (15,13); (2) star (15,4) = star (15,11); and (3) star (15,7) = star (15,8).
Star (15,2)
Star (15,4)
Star (15,7)
5. Star (9,2) has 1 path and requires 2 orbits. Star (7,3) has 1 path and requires 3 orbits. Star
(14,4) has 2 paths and each path requires 2 orbits.
Star (9,2)
Star (7,3)
Star (14,4)
6. a. The least common multiple of 15 and 6 is 30. Therefore, 30 points will be needed before
the steps of 6 bring us back to the starting point. Thus, the 15 points will be used twice which
means there will be 2 orbits: 30 ÷ 15 = 2.
b. The number of orbits for star (n,s) is LCM (n,s) divided by n.
7. a. The greatest common factor of 52 and 15 is 1.
b. 15, because LCM(52,15) = 780 and 780 ÷ 52 = 15 (see 6b)
Connections 4.2
Answers: Activities and Connections
Connections Questions 4.2: Selected Answers
1. Help Devon understand that Relatively Prime means that two numbers, relative to each other,
have no common prime factors.
3.
Prime Factorizations
(a, b) GCF(a,b) LCM(a,b)
a
b
GCF(a,b)
LCM(a,b)
4
24
(8, 12)
2×2×2
2×2×3
2×2
2×2×2×3
4
120
(20, 24)
2×2×5 2×2×2×3
2×2
2×2×2×3×5
1
300
1
(12, 25)
2×2×3
5×5
2×2×3×5×5
15
60
(15, 60)
3×5
2×2×3×5
3×5
2×2×3×5
For
a = p1n1 × p2n2 × … × pknk and
b = p1m1 × p2m2× … × pkmk where pi is prime and ni ≥ 0 and mi ≥ 0 for 1 ≤ i ≤ k
a. GCF(a,b) = p1min(n1, m1) × p2 min(n2, m2) × … × pk min(nk, mk). You can find the GCF(a,b) by
using each prime factor the least number of times it occurs in a or in b.
b. LCM(a,b) = p1max(n1, m1) × p2 max (n2, m2) × … × pk max (nk, mk). You can find the LCM(a,b) by
using each prime factor the greatest number of times it occurs in a or in b.
4. GCF(a,b) × LCM(a,b) = a × b. This relationship does not hold for three numbers; i.e.
GCF(a,b,c) × LCM(a,b,c) ≠ a × b × c.
This is easy to see by counterexample:
GCF(4,6,12) × LCM(4,6,12) = 2 × 12 = 24, but 4 × 6 × 12 = 288.
5. Standards references to factors and multiples
Level Expectation
Standard
Number and Operation Standard:
3 - 5 Describe classes of numbers according
to characteristics such as the nature of
Understand numbers, ways of representing
numbers, relationships among numbers,
their factors.
and number systems
6-8 Use factors, multiples, prime
factorization, and relatively prime
numbers to solve problems;
Answers: Activities and Connections
Activity Set 5.1
Activity Set 5.1
2.
Total Number
of Tiles
37
42
20
17
20
24
3. a.
–
b.
+
Number of
Red Tiles
8
26
6
6
12
13
Number of
Black Tiles
29
16
14
11
8
11
Net Value
+
21
10
+
8
+
5
–
4
–
2
–
3 + +2 = –1
7 + –3 = +4
–
4 + –3 = –7
–
8 + +2 = –6
c. 50 + –37 = +13; –34 + –25 = –59; –132 + +70 = –62
4. a. Add the number of red tile which represent the collections and make the answer negative.
b. If there are more red tiles than black tiles, subtract the number of black tiles from the
number of red tiles and make the number of remaining red tiles negative. If the number of
black tiles is greater than or equal to the number of red tiles, subtract the number of red tiles
from the number of black tiles and make the number of remaining black tiles positive.
5. a. –4 – +3 = –7
b.
+
5 – –2 = +7
Activity Set 5.1
c.
–
Answers: Activities and Connections
3 – -5 = +2
6. Notice that in each of the preceding examples, taking out a given number of tiles of one color
is equivalent to putting in the same number of tiles of the opposite color. For example in 5b,
2 red tiles and 2 black tiles were put into the collection and then 2 red were withdrawn from
the collection. The same result could be accomplished by simply adding 2 black tiles.
7. a. +3 × –2 = –6; three times—put in 2 red tile
b.
–
2 × –4 = +8; two times—take away 4 red tile
c.
–
3 × +2 = –6; three times—take away 2 black tile
8. You need to take out 4 red tile 3 times so there must be at least 12 red tiles in the initial
collection. For the initial collection to have a net value of zero, there must also be as many
black tiles as red tile in the collection. After taking out 3 groups of 4 red tile the net value of
the collection is +12.
Answers: Activities and Connections
Activity Set & Connections 5.1
b. –12 ÷ –3 = 4
9. a. –12 ÷ 3 = –4
c. In the sharing approach –3 would represents the number of groups, which is not possible.
In the measuring approach, groups of 3 red tiles cannot be formed from 12 black tiles.
10. a.
–
12 ÷ 3 = –4 because 3 × –4 = –12 and –12 ÷ –3 = 4 because –3 × 4 = –12
b. 12 ÷–3 = –4 because –3 × –4 =12
c. i) 15 ÷ 3 = 5
Share or measure
ii) 15 ÷ –3 = –5
–
3 × –5 =15
iii) –15 ÷ 3 = –5
Share
iv) –15 ÷ –3 = 5
Measure
Connections Questions 5.1: Selected Answers
2. To subtract a negative you must have the same number of black and red pairs (net value 0) as
the number of red tiles in the subtrahend. Thus, to remove the subtrahend; first add in the
needed number of black and red pairs and then take out the red tiles. This is the same as just
adding in the black tiles.
4. a. Starting with no collection, six times—put in 2 black tiles; 6 × 2 = 12
b. Starting with a collection with net value 0, six times—take out 2 black tiles; -6 × 2 = -12
c. Starting with a collection with net value 0, six times—take out 2 red tiles; -6 × -2 = 12
Connections 5.1 & Activity Set 5.2
Answers: Activities and Connections
d. Share 6 red tiles in two groups, the size of each share is -3. -6 ÷ 2 = -3
The size of each group is -3
e. Measure off 2 red tiles from a group of six red tiles; there are 3 groups. -6 ÷ -2 = 3
There are 3 groups
5. Standards references to integers
Standard
Number and Operation Standard:
Understand numbers, ways of
representing numbers, relationships
among numbers, and number systems
Level Expectation
6-8 Develop meaning for integers and represent
and compare quantities with them.
Number and Operation Standard:
Understand meanings of operations
and how they relate to one another
6-8
Understand the meaning and effects of
arithmetic operations with fractions,
decimals, and integers;
Number and Operation Standard:
Compute fluently and make reasonable
estimates
6-8
Develop and analyze algorithms for
computing with fractions, decimals, and
integers and develop fluency in their use;
Activity Set 5.2
1. a.
Fractions
0
4
1
2
2
3
6
6
1
4
Number of bars with
equal fractions
5
4
3
5
2
Answers: Activities and Connections
Activity Set 5.2
b.
Twelfths
Sixths
Fourths
Thirds
Halves
2. a.
b.
0
12
0
6
0
4
0
3
0
2
1
12
2
12
1
6
3
12
4
12
2
6
1
4
5
12
6
12
3
6
2
4
7
12
1
3
8
12
4
6
9
12
3
4
2
3
1
2
10
12
5
6
11
12
12
12
6
6
4
4
3
3
2
2
1
1 1
1
5
1
7
2 3
5 11
< <
< <
<
<
< < < <
12
6 4
3 12
2 12
3 4
6 12
1
2
3
4
5
6
7
8
9
10 11
<
<
<
<
<
<
<
<
<
<
12 12 12 12 12 12 12 12 12 12 12
3. a. Here are 10 possible equalities; there are others.
1
2 1
3 1
6 2
1 3
1 4
1 8
4 8
2 9
3
= ,
= ,
=
,
= ,
= ,
= ,
= ,
= ,
=
2
4 2
6 2 12 12
6 12
4 12
3 12
6 12
3 12
4
b. Here are 10 possible inequalities; there are others.
1
2 1
3 1
4 1
5 1
2 1
3 1
1 1
1 1
4 1
3
< ,
< ,
< ,
< , < , < ,
< ,
< ,
< ,
<
2
3 2
4 2
6 2
6 3 4 3 4 4
2 4
3 4
6 4
6
1
1
>
. A bar cut into 3 equal parts produces larger pieces than a bar cut into 10
3
10
equal parts.
4. a.
2
3
< ; A bar cut into 3 equal parts produces larger pieces than a bar cut into 4
3
4
2
3
parts. In each case one part is missing, for , a part larger than what is missing for .
3
4
b.
c.
5
1
2
1
5
2
is less than and
is greater than so
< .
12
2
3
2
12
3
Activity Set 5.2
d.
Answers: Activities and Connections
7
5 7
5
= ; and are both whole bars
7
5 7
5
5. a. 4
2
, nearest whole number 5.
3
b.
1
, nearest whole number 0
8
c.
10
nearest whole number 1
12
d. 2
5
nearest whole number 3
7
6. a.
4
8
=
9 18
3
6
=
10
20
1
2
=
5 10
6 18
=
7
21
6 15
=
8 24
1
3
=
6 18
0
0
=
3 12
3 12
=
4 16
1
4
=
2
8
b.
c.
d. There will be 3a shaded parts out of a total of 3b parts.
3a
The fraction for this bar is
3b
Answers: Activities and Connections
7. a.
2
5 × 2 10
=
=
3
5 × 3 15
b.
Activity Set 5.2
2 17 × 2 34
=
=
3 17 × 3 51
c. It multiplies both the numerator and the denominator by 17.
1
7
=
2 14
5 10
=
7 14
8. a.
1
8
=
3 24
3
9
=
8
24
b.
4
12
=
10
30
4
20
=
6
30
c.
3 21 5
25
=
; =
5 35 7
35
b.
5
7
c.
1
7
2
6
1
2
=
and =
, so >
3 21
7
21
3 7
d.
7
70
8
72
8
7
=
and
=
so
>
9
90
10
90
10
9
e.
4 12
6 12
4
6
=
and =
, so
=
6 18
9 18
6
9
f.
9. a.
10. a. 3 ÷ 2 =
3
2
b. 2 ÷ 3 =
2
3
c. 4 ÷ 3 =
4
3
11 121
8
120
11
8
=
and
=
, so
>
15 165
11 165
15 11
Connections 5.2 & Activity Set 5.3
Answers: Activities and Connections
Connections Questions 5.2: Selected Answers
1
1
and
with Fraction Bars; emphasizing the whole bars are each
6
4
1
1
the same size, that the bar is divided into six parts and the
bar is divided into four
6
4
parts. Ask which parts are smaller or bigger.
1. Help Jeremy model
2. The student is comparing two different sized pizzas. They need to understand that to
compare fractional parts; they must compare fractional parts of the same whole. Have
1
of this
them draw a single size pizza and divide it into several fractions. Emphasize
2
2
pizza is less than of this pizza.
3
1
1
and , sort out Fraction Bars with smaller parts
3
4
1
1
1
1
than in the
or
bars; compare until you find bars that are between
and .
3
3
4
4
Students should have several example pictures sketched.
3. One way: To find fractions between
4. One way (a) is to divide each of the six parts into two parts; two of 12 parts are shaded.
1
2
1
2
bars directly and see that lines up with
A 2nd way (b) is to compare the and
6
12
6
12
(a)
(b)
6. Answers may include a discussion about determining the relative values (equalities and
inequalities) of fractions using the Fraction Bars and then transferring these relationships
to the number line.
Activity Set 5.3
1. a.
1
2 10
+
=
3 4 12
Answers: Activities and Connections
b.
5 1
3
- =
6 3 6
c.
1 8
2
+
=1
3 9
9
d.
3
5
4
=
4 12 12
e.
3 1
5
+
=
8
4
8
f.
4
4
2
+
=1
10
5
10
Activity Set 5.3
3.
5 10
1
3
=
and
=
6 12
4 12
d.
4. a.
13
1
=1
12
12
1
5
2
+ =1
2
6
6
e.
19
4
=1
15
15
3
9
2 10
=
and
=
5 15
3 15
f.
13
18
g.
7
12
2
4
3
9
=
and =
9 18
6 18
h.
1
15
i.
5
18
Activity Set 5.3
b.
c.
Answers: Activities and Connections
1
3
1
+
=1
3 4
12
3
3
3
=
4
6 12
d.
1
2
3
=
2
7 14
5. a.
1
1 1
× =
2
3 6
1
3
3
× =
2
5 10
1
6
6
× =
2
8 16
b.
1 1
1
×
=
3 4 12
2
1
2
× =
3
6 18
2
4
8
×
=
3
5 15
c.
1
1
1
×
=
4
2
8
3 1
3
× =
4
5 20
3
2
6
×
=
4
3 12
3
Fraction Bar. As you split each shaded part of the bar into 3 equal parts, double
4
1
shade 2 of the 3 parts. Since each split part represents
of the bar, the 6 double shaded
12
6
parts are
of the whole bar.
12
6. Draw a
7. a. 1
f.
1
2
b. 5
7
1
1
÷ =3
8
4
2
c. 3
1
2
d. 1
1
2
e. 2
Answers: Activities and Connections
Activity Set 5.3
3
1
3
1
÷
≈ 5. There are 3 thirds in 1 whole bar and about 2 thirds in
bar. So
can be
4
3
4
3
measured off about 5 times.
8. a. 1
b.
5 1 1 1
1
4
1
– ≈ .
is close to , so the difference is about = .
8 6 2 6
8
8
2
4
1
1
4
1
+
≈ 1 Looking at the Fraction Bar it can be seen that
+
is 1 whole plus some
5
2
3
5
2
1
more. This additional part looks like about of a bar.
3
c.
2
5
1
5
×
≈ . First, of a Fraction Bar is shaded in and then each of these 5 parts is divided
3
6
2
6
2
into 3 equal parts with 2 parts shaded in darker to show . Some of these parts are moved as the
3
1
diagram shows to fill in as many sixths as possible. Approximately of the bar is shaded.
2
d.
9. a. Here are two ways.
(a)
1 1
5
+
=
6 4 12
3×
3
3
=
12
4
2
0 1
+ =
4
3 2
3×
1
3
=
4
4
Connections 5.3
(b)
Answers: Activities and Connections
2 1
0
–
=
4 2
3
2
2
÷
=2
3
6
5
3
1
=
12 12
6
2
2
÷
=2
3
6
Connections Questions 5.3: Selected Answers
2
8
9
72
3
×
=
×
=
, the student is finding a common denominator
4
3
12
12 12
correctly but then multiplying numerators and denominators. Working with the student with
1
1
1
1
1
×
=
illustrated by taking
of
easier examples such as
2
3
6
2
3
might help the students see that you do not multiplying the denominators.
1. In the example:
2. One way to introduce division of fractions is as “how many times does the second number
1
(divisor) go into the first number (dividend)?” or “if we measure off pieces how many are
6
3
there in ?” Illustrate with Fraction Bars.
4
3. First model
3
4
Second divide each part into 6 parts (for a total of 24 parts)
1
3
of
by taking 1 of 6 parts from each shaded region, or a total of 3 out of 24
6
4
1 3
3
=
.
parts. ×
24
6 4
Third; take
4. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
5. Grade 6 - 8 Number and Operations Standard Expectation: Select appropriate methods and
tools for computing with fractions … from among mental computation, estimation,
calculators or computers, and paper and pencil, depending on the situation, and apply the
selected methods.
Answers will vary but may include discussions about using Fraction Bars to achieve these
goals as illustrated in the activity set.
Answers: Activities and Connections
Activity Set 6.1
Activity Set 6.1
1.
a.
b.
c.
.4; Four tenths
.05; Five hundredths
035; Thirty five thousandths
d.
e.
f.
.7; Seven tenths
.80; Eighty hundredths .325; Three hundred twenty five thousandths
2. a. .30; thirty hundredths
b. .330; three hundred thirty thousandths
c. .84; eighty-four hundredths
d. .9; nine tenths
e. 13; thirteen hundredths
f. .09; nine hundredths
g. .008; eight thousandths
h. .090; ninety thousandths
3. a. 5 sets of three cards, 9 sets of two cards, 15 individual cards
b. Answers will vary
c. 3 parts out of 10 is equal to 30 parts out of 100; .3 = .30
d. 42 parts out of 100 is equal to 420 parts out of 1000; .42 = .420
e. 470 parts out of 1000 is equal to 47 parts out of 100; .470 = .47
f. 1 part out of 10 is equal to 10 parts out of 100; .1 = .10
g. 1 part out of 100 is equal to 10 parts out of 1000; .01 = .010
h. 1 part out of 1000 is equal to 10 parts out of 10,000; .001 = .0010
Activity Set 6.1
Answers: Activities and Connections
4.
a.
b.
c.
8 tenths 3 hundredths
2 thousandths
0 tenths 5 hundredths
5 thousandths
7 tenths 0 hundredths
5 thousandths
d. Every collection of 10 like parts (thousandths, hundredths or tenths) can be regrouped to
form 1 larger part and is recorded in the adjacent column, to the left.
5. a. Each part would be 1⁄10,000 of the unit square. The decimal is written as .0001 and named
1 ten-thousandth.
b. Each part would be 1⁄100,000 of the unit square. The decimal is written as .00001 and
named 1 hundred-thousandth.
6. a. .7 > .43 because the Decimal Square for .7 has 7 full tenths shaded while the Decimal
Square for .43 has less than 5 tenths shaded.
b. .042 > .04 because the Decimal Squares for both .042 and .04 have 0 tenths and 4
hundredths shaded but the Decimal Square for .042 has 2 extra thousandths shaded.
c. .3 > .285 because the Decimal Square for .3 has 3 tenths shaded but the Decimal Square
for .285 has less than 3 tenths shaded.
d. Answers will vary
e. Answers will vary
f. Answers will vary
7. The decimals .07, .08, and .075 each represent a Decimal Square with less than a tenth
shaded. Now .08 can be eliminated as it has 8 full hundredths shaded and .07 and .075 have
less than 8 hundredths shaded. Both .07 and .075 have 7 hundredths shaded, but .075 has 5
additional thousandths shaded. Therefore, .07 is the smallest.
Answers: Activities and Connections
Activity Set & Connections 6.1
8.
a. .3
b. .33
.333
c. .2
.17
.167
Connections Questions 6.1: Selected Answers
1. Work with the students and the Decimal Square model. After they are comfortable with
thinking of the decimals as parts of a whole, have them model both .15 and .125 with
Decimal Squares to see that the shaded region in the .15 Decimal Square is larger than the
shaded region in the .125 Decimal Square.
4. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
5.
b.
a.
7 of 10 parts are shaded, .7 =
7
10
15 of 100 parts are shaded, .15 =
15
10
Connections 6.1 & Activity Set 6.2
Answers: Activities and Connections
c.
d.
250 of 1000 parts are shaded, .250 =
250
100
125 of 1000 parts are shaded, .125 =
125
100
6. Some Standards and Expectations that are addressed by the activities in this section:
Number and Operations Standard
Understand numbers, ways of representing numbers, relationships among numbers, and number
systems
Grade 3 – 5 Expectations
• understand the place-value structure of the base-ten number system and be able to
represent and compare whole numbers and decimals;
• recognize equivalent representations for the same number and generate them by
decomposing and composing numbers;
• develop understanding of fractions as parts of unit wholes, as parts of a collection, as
locations on number lines, and as divisions of whole numbers
• recognize and generate equivalent forms of commonly used fractions, decimals, and
percents;
Grade 6 – 8 Expectation
• compare and order fractions, decimals, and percents efficiently and find their
approximate locations on a number line;
Activity Set 6.2
1. a. .3 + .4 = .7
d. .8 + .42 = 1.22
b. .38 + .35 = .73
c. .225 + .355 = .580
e. .7 + .470 = 1.17
f. .8 + .42 = 1.22
2. a. .4 + .29 = .69. The total shaded amount shows 6 tenths and 9 hundredths.
b. .179 + .37 = .549. The total shaded amount shows 4 tenths, 14 hundredths and 9
thousandths. By regrouping the hundredths an additional tenth is formed and 4
hundredths remain.
Connections 6.1 & Activity Set 6.2
Answers: Activities and Connections
Answers: Activities and Connections
Activity Set 6.2
c. .07 + .605 + .2 = .875. The total shaded amount shows 8 tenths, 7 hundredths, and 5
thousandths.
3. a. .25 – .19 = .06
b. .280 – .175 = .105
c. .75 – .08 = .67
4. a. .5 – .07 = .43. If 3 hundredths and 4 tenths are added to the square for .07, you get the
square for .5
b. .205 – .09 = .115 Add 1 tenth, 1 hundredth and 5 thousandths to .09
c. .435 – .3 = .135 Add 1 tenth, 3 hundredths and 5 thousandths to .3
5. a. 6 × .83 = 4.98 because 6 squares, each with 83 parts of 100 shaded, yield a total of 498
shaded parts. Since there are 100 parts in each unit square, the total shaded amount would be
equal to 4 unit squares and 98 parts shaded out of 100.
b. 4 × .725 = 2.900 because 4 thousandth squares, each with 725 parts shaded, yield a total
of 2900 shaded parts. Since there are 1000 parts in 1 unit square, the total shaded amount
would be equal to 2 whole squares and 900 parts shaded out of 1000.
c. Multiplying by 10 makes each shaded tenth column of a Decimal Square equal to a unit
square. Similarly, multiplying by 10 makes each hundredth square equal to a tenth column
and each thousandth piece equal to a hundredth square.
6. a. .4 × .3 = .12 The original shaded portion represents .3. This portion is split into 10 equal
parts and 4 are taken. The result is 12 hundredths.
Activity Set 6.2
Answers: Activities and Connections
b. .4 × 1.3 = .52 The original shaded portion represents 1.3. This portion is split into 10
equal parts and 4 are taken. The result is 4 tenths and 12 hundredths. Regrouping gives 5
tenths and 2 hundredths.
c. 1.4 × 1.3 = 1.82 The original shaded portion represents 1.3 shaded twice. One full portion
is taken representing 1 × 1.3 . The second portion is split into 10 equal parts and 4 are taken
representing .4 × 1.3. The shaded portion represents 1 unit, 7 tenths, and 12 hundredths or 1
unit, 8 tenths, and 2 hundredths.
Answers: Activities and Connections
Activity Set 6.2
d. 2.4 × 1.3 = 3.12 The original shaded portion represents 1.3 shaded three times. Two full
portions are taken representing 2 × 1.3. The third portion is split into 10 equal parts and 4 are
taken representing .4 × 1.3. So, the product can be thought of as (2 × 1.3) + (.4 × 1.3). The
shaded portion represents 2 units, 10 tenths, and 12 hundredths. Regrouping gives 3 units, 1
tenth, and 2 hundredths.
.
7.
a. .90 ÷ 6 = .15
b. .75 ÷ 3 = .25
c. 200 ÷ 4 = .05
Activity Set 6.2
Answers: Activities and Connections
8.
a. 1 ÷ 10 = .10
b. .1 ÷ 10 = .01
c. 01 ÷ 10 = .001
9. Dividing a unit square by 10 results in a region equivalent to a tenth column. Dividing a tenth
column by 10 results in a region equal to a hundredth square, and dividing a hundredth
square by 10 results in a region equal to a thousandth piece. The decimal 2.87 can be
represented by 2 whole squares and a square with 87 hundredths shaded (8 tenth columns and
7 hundredth squares). When the squares representing 2.87 are divided by 10, the result is 2
tenth columns, 8 hundredth squares, and 7 thousandth pieces.
10.
a. .8 ÷ .2 = 4
b. .70 ÷ .05 = 14
c. .600 ÷ .3 = 2
d. Five tenths can be measured off or subtracted from each unit twice. Since there are 250
units, .5 can be subtracted 500 times.
11. a. Here are two solutions
(1)
.3 + .35 = .650
.75 – .400 = .350
4 × .2 = .8
.45 ÷ .15 = 3
(2)
.400 + .350 = .75
.650 – .3 = .35
4 × .2 = .8
.45 ÷ .15 = 3
b. Answers will vary
Answers: Activities and Connections
Connections 6.2
Connections Questions 6.2: Selected Answers
3. a .2 × . 5 = .1
.5
.2
b. 1.2 × 1.5 = 1.8
1.5
1.2
c. 0.75 ÷ 0.15 = 5; divide 75 parts into 15 groups, there are 5 parts in each group. (or
measure off groups of size .15; there are 5 such groups)
Connections 6.2 & Activity Set 6.3
Answers: Activities and Connections
4.
Since the parts in 0.75 and 0.15 are the same size, one way to think of 0.75 ÷ 0.15 in the Decimal
Square model is as “how many times does 15 parts go into 75 parts?” This is the same question
as 75 ÷ 15 in the measurement setting.
Similarly, since the parts in 0.8 and 0.2 are the same size, one way to think of 0.8 ÷ 0.2 in the
Decimal Square model is as “how many times does 2 parts go into 8 parts?” This is the same
question as 8 ÷ 2 in the measurement setting.
Thinking of .7 as .70, .70 and .05 have the same size parts and the analogy still holds.
In general with the Decimal Square model using tenths, hundredths and thousandths, you can use
the measurement setting (of whole numbers) in this way. You can extend the Decimal Square
model to see this always works by imagining more and more subdivisions of the parts of the
Decimal Squares.
Activity Set 6.3
1. a. 30%
b. 13%
c. 103%
d. 130%
e. 300%
2.
a.
c.
e.
b.
d.
f.
3. a. The value is 3. One small square has value 3 because 300 ÷ 100 =3.
b. The value is 99. Each strip has a value 30 and each small square has a value 3.
c. The value is 288. Each small square has value 3 and 300 – 12 = 288.
Answers: Activities and Connections
Activity Set 6.3
d. The value is 330. The 10 by 10 grid has a value of 300 and the strip has a value of 30.
e. The value is 600. Each 10 by 10 grid has a value of 300.
4. a. The value is 45. Each strip (10%) has value 15 and three strips (30%) have value 45.
b. The value is 112.5. Each strip (10%) has value 15 so all the strips (70%) have value 105.
The one-half strip (5%) has value 7.5.
c. The value is 10.5. Every pair of small squares has value 3 and 3 + 3 + 3 + 1.5 = 10.5.
d. The value is 166.5. The 10 by 10 grid (100%) has value 150. The strip (10%) has value
15 and the small square (1%) has value 1.5.
e. The value is 315. Two 10 by 10 grids (200%) have value 300. The strip (10%) has value
15.
5. a. 45 is 10% of 450. Each strip has value 45 and one strip is 10%.
b. 108 is 24% of 450. Each strip has value 45 so 2 strips has value 90. Every pair of small
squares has value 9 so two pairs (4 squares) is 18. 90 + 18 = 108.
c. 441 is 98% of 450. 450 – 441 = 9. Since two small squares have value 9, take away 2
small squares from the 10 by 10 grid leaving 98.
d. 585 is 130% of 450. One 10 by 10 grid has value 450. Each strip has value 45 so 3 strips
have value 135. 450 + 135 = 585.
e. 909 is 202% of 450. Two 10 by 10 grids have value 900. Two small squares have value
9.
6. a. The value of the 10 by 10 grid is 400. Since 6% has value 24, 1% has value 4 and the
value of the 10 by 10 grid is 400.
b. The value of the 10 by 10 grid is 75. Since 40% has value 30, 20% has value 15 and the
value of the 10 by 10 grid is 15 × 5 = 75.
c. The value of the grid is 240. Since 55% has value 132, 1 % has value 2.4 (132 ÷ 55) and
100% has value 240.
Activity Set 6.3
Answers: Activities and Connections
d. The value of the grid is 144. There are 5 sets of 25 in 125. Since 125% has value 180,
25% has value 36 (180 ÷ 5). There are 4 sets of 25 in the 10 by 10 grid, so 36 × 4 = 144. [Or,
the value of the grid is 144. Since 125% has value 180, 1% has value 1.44 (180 ÷ 125) and
100% has value 144.
7. a. Some observations: There are 208 women employees; 35% of the employees are men;
there are 112 men employees.
b. Some observations: 35% of the students were absent; 65% of the students were present.
c. Some observations: The original price of the sweater was $180; I paid 70% of that for the
sweater; the 30% discount came to $54.
Answers: Activities and Connections
Activity Set 6.3
d. Some observations: The charity benefit raised $2450 more than its goal; the charity
benefit raised $9450 this year; 10% of the charity benefit’s goal is $700 and 1% is $70.
e. Some observations: There are 150 schools in the county; 93 (or 62%) of the schools have
a student to teacher ratio that is at or below recommendations for accreditation.
f. Some observations: The price of one share of stock in July was $10.06; the stock had
increased by $22.44 from July to December (this is 223% of $10.06); the price of the stock in
December is 323% of the price of the stock in July. (Note, these figures are rounded to the
nearest penny.)
Activity Set 6.3
Answers: Activities and Connections
1
= 50% = .5
2
split each part of the bar into 50 equal parts
8. a.
3
= 30% = .3
10
split each part of the bar into 10 equal parts
2
5 = 40% = .4
5
split each part of the bar into 20 equal parts
b.
13
= 65% = .65
20
split each part of the bar into 5 equal parts
c.
d.
1
≈ 33% = .33
3
split each part of the bar into 33 equal parts
5
≈ 85% = .85
6
split each part of the bar into 17 equal parts
9. a.
b.
2
≈ 22% = .22
9
split each part of the bar into 11 equal parts
c.
Connections Questions 6.3: Selected Answers
1. Help the students see the following: 30 squares has value $10.99, $10.99/30 ≈ $.3663, the
original cost is 100 × $.3663 ≈ $36.63
These 30 squares represent 30% of the original cost and have value $10.99
Answers: Activities and Connections
Connections 6.3
2. This is not true, here is a counterexample:
10%
10% increased by 50% 15% decreased by 50% = 7.5%
=15%
In general, algebraically: X >> 1.5X >> .5(1.5X) = .75X and .75X ≠ X
3.
a. Doubles; if an amount doubles, there are twice as many small squares, thinking of each
small square as 1% of the whole, twice as many small squares is twice the percent. Thinking
of the original value as 100%, if an amount doubles, it increases by 100% (1 whole square).
b. Triples; if an amount triples, there are three times as many small squares, thinking of
each small square as 1% of the whole, three times as many small squares is three times the
percent. Thinking of the original value as 100%, if an amount triples, it increases by 200% (2
whole squares).
4. a. 3 squares = $45, each square = $15, 100 squares = $15 × 100 = $150
b. 45 squares = $3, each square = $3/45 = $.067, 100 squares = 100% = $3/45 × 100 =
$6.67. Another way: 100/45 = 2.2, $3 × 2.2 = $6.6 (approximately)
c. They are very different questions. Part a is given the percent of the part and the part
(value), what is the whole (value of 100%) whereas part b is given a different part (percent)
and the value of the part, what is the whole (value of 100%)
Activity Set 6.4
Answers: Activities and Connections
Activity Set 6.4
1. a. 7 square units
b. 7 square units
e. 4 square units
f. 2 square units
c. 2 square units
1
g. 3 square units
2
d. 6 square units
b. 4 square units
c. 9 square units
d. 16 square units
f. 8 square units
g. 10 square units
h. 2 square units
i. 6 square units
2. a. 1 square unit
e. 2 square units
3. Area of square:
1
2
4
5
8
9
10
16
Length of side:
1
2
2
5
8
3
10
4
4. a. 13
b.5
c. 17
5. a. One four function calculator gives 2.2360679.
b. 4.9999996 using the calculator value from part a, some calculators will round to 5. The 8
digit number in part a is an approximation of 5 .
c. It is more concise and represents an exact number instead of an approximation.
6.
Square A
Square B
Square C
a. Figure 1
9
16
25
b. Figure 2
9
4
13
c. Figure 3
10
40
50
d. Figure 4
45
20
65
Answers: Activities and Connections
Activity Set & Just for Fun 6.4
7. a. The area of the square on the hypotenuse is equal to the sum of the areas of the squares on
the legs of a right triangle.
b.
Total area of squares
on two legs
Area of square on
hypotenuse
Length of hypotenuse
(1) Figure 1
16 + 16 = 32
32
32
(2) Figure 2
1 + 25 = 26
26
26
(3) Figure 3
9 + 16 = 25
25
5
(4) Figure 4
9 + 49 = 58
58
58
Just for Fun 6.4
The middle rectangle is closest to a golden rectangle.
To 6 decimal places,
1+ 5
equals 1.618034.
2
If the golden rectangle is folded in half the resulting rectangle is not a golden rectangle.
The ratio of the 11th Fibonacci number (89) to the 10th (55) is 1.618181818…
Connections Questions 6.4: Selected Answers
3. There are 12 line segments of different slope that can be formed on a 5 pin by 5 pin geoboard
(see Activity Set 2.2). To find the lengths; use the Pythagorean Theorem; there are only 7
line segments of distinct length on a 5 pin by 5 pin geoboard.
5
32
13
20
10
4
17
4. See Just for Fun 6.4 Answers
5. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
Connections 6.4
Answers: Activities and Connections
6. A Standard and Expectation that addresses working with roots:
Geometry Standard
Understand meanings of operations and how they relate to one another
Grade 6 – 8 Expectation
• understand … squaring and finding square roots to simplify computations and solve
problems
Students may also choose some of the following, there are many possibilities:
Algebra Standard
Understand patterns, relations, and functions
Grade 3 – 5 Expectation
• describe, extend, and make generalizations about geometric and numeric patterns;
• represent and analyze patterns and functions, using words, tables and graphs.
Algebra Standard
Represent and analyze mathematical situations and structures using algebraic symbols
Grade 3 – 5 Expectation
• represent the idea of a variable as an unknown quantity using a letter or a symbol;
• express mathematical relationships using equations.
Answers: Activities and Connections
Activity Set 7.1
Activity Set 7.1
1. A typical distribution might look like
Central Angle
Eye color # Percentage
(degrees)
Brown
10
30.3
109.1
Blue
5
15.2
54.5
Green
7
21.2
76.4
Hazel
8
24.2
87.3
Gray
3
9.1
32.7
Total
33
100.0
360
a. A typical bar graph might look like
b. A typical Pie Graph might look like
Green
21.2%
Blue
15.2%
Brown
30.3%
Hazel
24.2%
Gray
9.1%
c. Answers will vary
d. Answers will vary
e. Key features of a bar graph: Emphasizes relative frequency of data, minimum and
maximum data values show, data values easily show. Key features of pie graph: Emphasizes
relative proportion of data, does not show number of data points in each category, can see
larger and smaller data slices easily.
f. Data that can be categorized into distinct categories, usually with a fairly small data
spread. Typical answers might include; voting results for a population, favorite type of pizza
topping or similar “favorite” polling result, hair color for a population.
g. Data that can be categorized into distinct categories, larger data spreads or data given as
percentages can easily be shown. Typical answers might include; ingredient proportions,
percent of population spreads (in favor of, against, etc.)
Activity Set 7.1
Answers: Activities and Connections
2. a. A typical line plot might look like:
Number of Writing Tools in Math Class
Number of students
10
5
X
X
1
X
X
X
X
X
2
X
X
X
X
X
X
3
X
X
X
X
4
X
X
X
X
X
X
X
X
X
X X X
X X
X X X X X X
5 6 7 8 9 10
Number of writing tools
X
X
X
X
X
11
X
X
12
X
13
X
X
14
X
X
X
15+
b. Answers will vary
c. Key features of a line plot: Shows numerical data that falls into distinct numerical
categories clearly. Each data value is displayed. Relative sizes of data points easy to see.
Minimum and maximum data values are easy to see.
d. Numerical data that can be categorized into distinct numerical categories. Typical
answers should start with “number of ….”
3. a. 26
b. The highest temperature is 74°, the lowest temperature is 20°.
c.
Interval 15-24 25-34 35-44 45-54 55-64 65-74
Frequency
1
5
14
6
17
7
d.
Answers: Activities and Connections
Activity Set 7.1
e. The most common range of temperatures is 55° - 64° F.
f. Answers will vary
g. Stem and leaf emphasizes value of individual data points, categorizes data in groups 2029, 30-39, etc., shows repeat data points. Histogram emphasizes proportion of data points
in categories, categorizes data in groups 15-24, 25-34, etc., and does not show repeated
data points.
h. Numerical data, perhaps with many repeat data values
i. Numerical data that falls into distinct categories.
4. Answers will vary
5. Answers will vary
6. a. The trend line for this data has a negative slope. It indicates that the taller the father the
shorter the mother. Or, the shorter the father, the taller the mother.
b. The trend line for this data has zero slope. Short fathers are paired with short and tall
mothers and the same case is true for the taller fathers. There seems to be no apparent trend.
c. Answers will vary for the data in activity 4. 6a. a negative relationship;
6b. no relationship.
d. Examples of possible negative relationships:
(1) Scores on a test of math anxiety and scores on a math examination.
(2) Horsepower of gasoline engines in automobiles and trucks and corresponding miles
per gallon of gasoline.
(3) Hours of sleep deprivation and ability to solve problems.
(4) Air temperature and altitude above sea level.
Activity Set & Just for Fun 7.1
Answers: Activities and Connections
Examples of possible positive relationships:
(1) The greater the distance below sea level, the greater the water pressure.
(2) The greater a person's height, the greater the shoe size.
(3) The age of a Douglas fir tree and the length of its circumference.
7. a. The trend line indicates a negative relationship. Based on this trend line, the average
7.5 year-old would hop 50 feet between 7.6 and 7.8 seconds.
b. This trend line indicates a positive relationship. Using this line we would expect a tree of
diameter 5.5 inches to be about 26 years old and a tree 40 years old to have a diameter of 8
inches.
Just for Fun 7.1
1. There is negative relationship. As the number of blue increases, the number of brown
decreases. (A correlation of about -.5)
2. Based on a scatter plot trend line, if you had 14 blue you could expect 3 to 5 brown. If you
had 10 brown you could expect 3 to 5 blue.
3. There is a negative relationship between yellow and green. (A correlation of about -.4)
Answers: Activities and Connections
Connections 7.1
Connections Questions 7.1: Selected Answers
1. One approach would be to ask a series of questions to determine the student’s knowledge
about the given scatter plot and the relationship of the data points to the trend line. Then ask
where a data point for a monthly income of $4000 might be located if the dining frequency
was not known but you want to make a best guess based on the known data.
2. a. This would be an opportunity for the students to list the types of graphs they know and
compare each type to their data eliminating inappropriate graphs as they proceed.
b. Answers will vary; bar graph is probably best
3. See Just for Fun 7.1 Answers
5. a. Both give visual comparisons of relative frequencies of different categories by heights (of
bars or number lists); Stem and leaf categories are intervals and bar graphs are distinct
categories. Both allow one to reconstruct the data for each category.
b. Bar graph bars are separated and histogram bars are adjacent because bar graphs
represent distinct categories and histograms represent intervals. Both use bars to represent
frequencies.
c. Stem and leaf graphs, like histograms, show data over intervals. All data can be retrieved
from a stem and leaf but not from a histogram.
6. Standards and Expectations that address working with graphs:
Data Analysis and Probability Standard
Formulate questions that can be addressed with data and collect, organize, and display relevant
data to answer them
Pre-K – 2 Expectations
• represent data using concrete objects, pictures, and graphs.
Grade 3 – 5 Expectations
• represent data using tables and graphs such as line plots, bar graphs, and line graphs;
Grade 6 – 8 Expectations
• select, create, and use appropriate graphical representations of data, including histograms,
box plots, and scatterplots.
Activity Set 7.2
Answers: Activities and Connections
Activity Set 7.2
1. a. Some possible observations:
• Short and long time intervals alternate.
• The shortest time interval is 51 minutes and the longest is 98 minutes.
• There is a difference of 47 minutes between the shortest and longest time intervals.
• The table does not show how long an eruption lasts, just the time interval between
eruptions.
• Among the 15 time intervals there are four pairs of equal time intervals.
b.
Stem
5
6
7
8
9
Leaf
12779
2
23
7788
448
c. Both plots show that the time intervals between eruptions cluster at the low and high end
of the time line. The line plot shows more clearly that the long times cluster around 90
minutes and the shorter times around 57 minutes. The line plot shows the gaps between the
times more clearly. Neither plot shows the alternating pattern of the time intervals.
d. Mean: 74.6 minutes
Median: 73 minutes
e. 7 shortest time intervals
mean: 60.4 minutes
median: 58 minutes
mode: 57 minutes
Modes: 57, 87, 88, 94 minutes
7 longest time intervals
mean: 90.8 minutes
median: 88 minutes
modes: 87, 88, and 94 minutes
Because the short and long time intervals between eruptions cluster around different
times, these averages are more representative of the short and long times, respectively.
f. If the last time interval was short (at the lower half of the line plot) then the next interval
will probably be long (between 85 and 95 minutes). If the last time interval was long (at the
upper half of the line plot) then the next interval will probably be short (between 55 and 65
minutes).
Answers: Activities and Connections
Activity Set & Connections 7.2
2.
Starting
Duration
Time
11:22 am 1 min 50 sec
8:30 am 4 min 26 sec
12:40 pm 3 min 46 sec
Predicted Interval
Predicted Time of
Before Next Eruption
Next Eruption
between 55 and 65 min 12:17 pm to 12:27 pm
between 80 and 90 min 9:50 am to 10:00 am
between 75 and 85 min 1:55 pm to 2:05 pm
3. Answers will vary
4. a. Answers will vary
b. $287,000 is the average for the 6 greatest incomes; $120,667 is the average for the 6
smallest incomes.
c. Answers will vary
d. $264,167 is the greatest average and $139,667 is the least average. Both are closer to the
mean of $195,611 than the greatest average and least average from part b.
5. a. In one sample of number pairs from the table of random digits eight consecutive pairs had
to be picked to obtain five representing type O blood: 23 98 34 97 76 29 28 10.
b. Answers will vary
c. Answers will vary
Connections Questions 7.2: Selected Answers
1. One way, is to ask the students to think of each amount in pennies. If they were to share the
money evenly (amongst all 10 students), how would they do it and how much would each
person get?
4. a. A stratified sample of six should have 3 City, 1 Luxury and 2 Country
b. Answers will vary but median of all is $195,000
c. $25,000
5. a. The mean is a helpful average when there are no extreme values in the data – like the
average height of 10 year old girls in a school.
Activity Set 7.3
Answers: Activities and Connections
b. The median is useful when there are extreme values in the data set – like the number of
deaths from earthquakes in the last century.
c. The mode is important when data values are repeated – most common dress size, bike tire
size or hat size.
Activity Set 7.3
1.
Penny
H
H
T
H
T
T
H
T
Dime
Nickel
2 heads and 2 tails
H
T
T
H
H
H
T
T
H
T
T
H
4 heads or 4 tails
H
H
T
T
Quarter
Penny
T
T
T
H
H
H
H
H
H
T
Dime
Nickel
3 heads and 1 tail
H
H
H
T
T
H
H
H
T
T
T
H
1 head and 3 tails
T
T
T
H
H
T
T
T
H
T
Quarter
T
H
H
H
H
T
T
T
c. At least 3 girls means 3, 4, 5, or 6 girls. Count the number of outcomes in the chart that
have 3, 4, 5 or 6 heads and divide that number by 64.
2. It appears close to a normal distribution.
b. The pulse rate distribution comes close to satisfying the property because 68.3% of the
pulse rates are within plus or minus 1 standard deviation of the mean and 95% are within
plus or minus 2 standard deviations of the mean.
Answers: Activities and Connections
3. a. letter
percent
e
16.3%
s
7.2%
c
2.6%
Activity Set 7.3
w
1.7%
k
6%
b. Here’s one: This is not such a difficult task that you ask of us.
c. The following assignment of code symbols requires 43 time units, as compared to 47
time units for Morse code. Note: It is also possible to use • • — for s and — • • for h.
4. One hundred eleven, or 67%, of the populations begin with digits 1, 2, 3, or 4.
First Digit
Total
1
41
2
34
3
19
4
17
5
12
6
11
7
13
8
11
9
8
a. Yes, over 50% of the populations begin with digits 1, 2 or 3.
b. Some possible conjectures based on this set of tallies:
(1) In general, the frequency of occurrence of leading digits decreases from a high for the
digit 1 to a low for the digit 9.
(2) The total of leading digits 1, 2, and 3 will be greater than the total of leading digits 4,
5, 6, 7, 8, and 9.
(3) The leading digit 3 occurs less than half as frequently as the leading digit 1.
c.
Interval
Number of Numbers Win or Lose
7,092 to 9,999
2,908
lose
10,000 to 49,999
40,000
win
50,000 to 99,999
50,000
lose
100,000 to 499,999
400,000
win
500,000 to 999,999
500,000
lose
1,000,000 to 2,784,075
1,784,076
win
d. There are 2,776, 984 numbers from 7,092 to 2,784,075 (the range of the populations).
Totals from the chart show that 2,224,076 numbers begin with digits 1, 2, 3, and 4, while
only 552,908 numbers begin with digits 5, 6, 7, 8, and 9. If the populations of the
metropolitan areas are randomly distributed you can expect a greater number of them to start
with digits 1, 2, 3, and 4.
Just for Fun & Connections 7.3
Answers: Activities and Connections
Just for Fun 7.3
I came, I saw, I conquered. Here is a frequency distribution for the letters which occur in the
message.
The letter M occurs most frequently so it is reasonable to guess that this is the enciphered letter
for E. If M represents E, then I represents A and Q represents I. This agrees with one of the
patterns for the distribution of letters in large samples. Also notice that the 5th and 6th letters
beyond M did not occur in the message. This also agrees with the fact that the 5th and 6th letters
beyond E are the low frequency letters J and K. It appears that the message has been coded by
replacing each letter by the letter that is 8 letters beyond. The message can be decoded by
reversing that process. The result is:
Galileo showed men of science that weighing and measuring are worthwhile. Newton
convinced a large proportion of them that weighing and measuring are the only
investigations that are worthwhile
Connections Questions 7.3: Selected Answers
3. a. Bar graph; somewhat bell-shaped, therefore somewhat normally distributed.
b. 47%, 90%, and 100% within ±1, ±2, and ±3 deviations, respectively.
c. Close but there should be a few more sums in the middle to be closer to a true normal
distribution..
Answers: Activities and Connections
Activity Set 8.1
fActivity Set 8.1
1. Here are the results of this experiment for one type of tack.
Trials
1
2
3
4
5
6
7
8
9
10
Points Up
4
4
9
6
6
8
5
4
4
4
a. The tacks landed with points up 54 times. This is not a good bet to accept because the
experiment suggests that more than half the time this type of tack will land with its point up.
b.
54
or .54
100
c. .54 × 300 = 162
d.
46
or .46 This probability plus .54 equals 1.
100
2. a. Here are the results of 24 spins of a penny: 5 inside and 19 on a line.
b.
5
≈.21
24
c. Points A and C
d. Subdivide the square into smaller squares whose sides equal the radius of a penny. The
centers of winning pennies must lie within the 4 center squares. The ratio of the winning
4
1
=
= .25.
region to the total region is
4
16
e. No. There is only a 25% chance of winning (theoretically).
Activity Set 8.1
Answers: Activities and Connections
3. . Here are the results of 24 spins of a penny: 11 inside and 13 on a line. The experimental
11
probability for this experiment is
≈.46.
24
b. Subdivide the square into smaller squares whose sides measure the
radius of a penny. By shading the region in which the center of a
winning penny must fall we can see that the ratio of shaded area to the
16
4
whole square is
=
≈.44.
36 9
c. This game is not fair because you have a less than 50% chance of
winning.
4. Here are the results of 24 draws: 17 had the same letter facing down and 7 had a different
letter facing down. For this experiment the experimental probability that the letter facing up
7
≈ .29.
will be different from the letter facing down is
24
a. 2
b.
2
1
=
6
3
c. A1
A2
d.
A2
A1
A
B
1
3
e. No. If an A is facing up, part c shows there is a
Similarly, if the letter facing up is B, there is a
1
chance the letter facing down is B.
3
1
chance the letter facing down is A.
3
5. For comparison: The theoretical probability that a student will wear the same sweater more
than once in a 5-day week is .6976.
The chance of wearing a different sweater each day is 1 ×
and 1 - .3024 = .6976.
9
8
7
6
×
×
×
= .3024
10
10
10
10
Answers: Activities and Connections
Just for Fun & Connections 8.1
Just for Fun 8.1
Race track Game: The three most likely tracks to produce a winner are tracks 6, 7, and 8 with
5 6
5
,
, and
, respectively.
probabilities of
36 36
36
Variation: On the track for sums of three dice, the tracks most likely to produce a winner are
tracks 9, 10, 11, and 12 with probabilities of 11.5%, 12.5%, 12.5% and 11.5%, respectively.
Connections Questions 8.1: Selected Answers
1. The game is fair when the grid is between a 3.25 and 3.5 penny grid (about 3.41). It is fair
when the area of the inner square equals the difference in area between the inner square and
the grid square. The proof of this follows. Many students will probably approach this more
intuitively.
x 2 = 4r 2 + 4rx
x 2 − 4rx − 4r 2 = 0
r2
4r ± 16r 2 + 16r 2
2
x = 2r + 2r 2 = ( 2 + 2 ) r ≈ 3.41r
rx
x=
rx
r2
x2
r2
rx
rx
r2
2. Here are two possibilities: Put objects numbered 1 to 6 in a container so students can draw,
replace and draw again; or, use a able of random numbers
3. For the 6 card set there are 12 equally likely outcomes and the probability of a different letter
6
1
down is
or . The probability that the letter facing up will be different than the letter
12
2
facing down increases as the number of cards increases.
A1 A2 B1 B2 C1 C2 A A B B C C
UP
A2 A1 B2 B1 C2 C1 B C A C A B DOWN
4. a. Possible outcomes: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36
1
; the numbers 2, 3, 5, 8,
36
1
2
10, 15, 18, 20, 24 and 30 each have a probability of
= ; the number 4 has a
36
9
3
4
probability of
; the numbers 6 and 12 each have a probability of
.
36
36
b. The numbers 1, 9, 16, 25, and 36 each have a probability of
Activity Set 8.2
Answers: Activities and Connections
Activity Set 8.2
1. a. Here are the results of one set of 40 draws: 21 were 2 tiles of different colors and 19
were two tiles of the same color. Therefore, for this trial the experimental probability of
21
drawing 2 tile of different colors is
= .525 and the experimental probability of drawing
40
19
2 tile of the same color is
= .475.
40
b. Probability of 2 of different colors:
c. red-green:
9 3
6 2
= . Probability of 2 of same color:
= .
15 5
15 5
1
3
3
1
3
3
1
2
2
× =
green-red: × =
green-green: × =
.
2
5 10
2
5 10
2
5 10
d. From the probability tree we see that the probability of drawing first a red and then a
3
3
, and the probability of drawing first a green and then a red is
. The
green is
10
10
3
3
3
probability of drawing 2 tiles of different colors is
+
= . The same as the
10
10
5
result in part b.
2. b. There are 16 total outcomes and 6 of them have exactly 2 heads and 2 tails.
c.
6
3
or
16
8
3. a. Here are the results of doing this experiment 30 times:
Total
2 red tiles 14
1 or more green tiles 16
The experimental probability of drawing 2 red tiles for this experiment is
b. The theoretical
42
probability is
90
≈.47
14
7
or
.
30
15
Answers: Activities and Connections
Activity Set 8.2
4. a. Here are the results of doing this experiment 30 times:
Total
2 red tiles 12
1 or more green tiles 18
The experimental probability of drawing 2 red tiles is
12
2
or .
30
5
b. The theoretical probability
7
7
is
×
=.49
10 10
5. a.
1
≈ .03
36
b.
35
≈.97
36
c.
35 35
35 2
×
= ( ) ≈.95
36 36
36
d. (
35 24
) ≈.51
36
e. Approximately 1 – .51 = .49
6. a.
1
3
b. Suppose you chose a door with junk behind it (but you didn’t know it). The host would
then open another door which has junk behind it. Using the switch strategy the door
you switch to would contain the prize. Using the switch strategy the theoretical
2
probability of winning the prize is
because if you randomly pick a junk door, you
3
win.
c. If your first spin selects junk, the host would open a second junk door. If your second
spin selects switch you would then win the prize.
First spin
Junk
junk
prize
Prize
Second spin
switch
stick
switch
stick
Prize
yes
no
no
yes
Activity Set & Just for Fun 8.2
Answers: Activities and Connections
d. The probability of winning with this
1
strategy is .
2
e. The first door is picked at random. If you stick with that choice the probability of
1
2
winning the prize is ; if you switch the probability of getting the prize is ; and, if
3
3
you make your switch or stick decision by a random process the chance of winning is
1
. So the best strategy is to always switch.
2
Just for Fun 8.2
Here are the results of 21 rolls of Die 2 and Die 4.
Die 2 5 1 5 5 1 1 1 5 1 5 5 1 1 1 5 1 1 5 5 5 5
Die 4 2 2 2 6 6 6 6 6 2 2 2 6 2 6 2 2 2 6 6 2 6
2.
24
2
=
36
3
3.
4
1
4
1
× =
=
6
2 12
3
4.
2
1
=
6
3
5.
1 1
2
+ =
3 3 3
Die 4
1. The experimental probability of Die 4
winning over Die 2 for this experiment is
15
5
= ≈.7.
21 7
2
2
2
2
6
6
5
2,5
2,5
2,5
2,5
6,5
6,5
5
2,5
2,5
2,5
2,5
6,5
6,5
Die 2
5
1
2,5 2,1
2,5 2,1
2,5 2,1
2,5 2,1
6,5 6,1
6,5 6,1
1
2,1
2,1
2,1
2,1
6,1
6,1
1
2,1
2,1
2,1
2,1
6,1
6,1
Answers: Activities and Connections
Connections 8.2
Connections Questions 8.2: Selected Answers
1. One way: Have the students number the red tiles R1, R2, R3, and the green tiles G1, G2, G3.
Then list all possible pairs they can draw grabbing two at a time. Make another list of all
possible pairs when they draw one, don’t replace, and draw again.
2. One way is to make a tree diagram. Another way would be to make a table of all possible
combinations that can come up on two different colored dice.
12
); this is the same
240
12
12 1
for drawing two 2s, two 3s and 2 4s. Therefore, P(a pair) = 4 ×
=
= .
240
60 5
3. a. The partial tree diagram shows the probability of drawing two 1s (
1st card
3
15
2nd card
Outcome
1
1,1
2
Probability
4
3
×
16 15
3
4
16
4
1
2
b. The pairs of cards that sum to more than 5 are (4,4), (4,3), (4,2) and (3,3). The probability of
4
4
3
3
×
, the probability of drawing a (3,3) is
×
, the probability of
drawing a (4,4) is
15
15
16
16
4
4
4
4
drawing a (4,3) is
×
and the probability of drawing a (4,2) is
×
Thus the
15
15
16
16
12
16
7
probability of drawing a sum more than 5 is (2 ×
) + (2 ×
) =
. Therefore the
240
240
30
7
21
probability of drawing a sum of at most 5 is 1 =
30 30
4. The probabilities for rolling doubles with a pair of 4-sided, 6-sided, and 8-sided dice are
1
1
, and respectively.
6
8
1
,
4
Activity Set 9.1
Answers: Activities and Connections
Activity Set 9.1
1. There are 20 different geoboard triangles with the given base. The triangles are different in
the sense that the third vertex can be at 20 different pins on the geoboard.
b. All triangles that have their third vertex on the same row parallel to the base have the
same area.
c. There are 12 noncongruent triangles that can be formed on the given base.
d.
e. It is not possible to draw a triangle that belongs to the regions marked 0 in part d. A right
triangle must be either isosceles or scalene, but cannot be both isosceles and scalene. No
triangle can be scalene and isosceles.
2. There are 14 geoboard segments of different length.
3. Here are seven ways. There are many other ways.
Answers: Activities and Connections
Activity Set 9.1
4.
Quadrilaterals
Square
Rectangle, but not square
Parallelogram, but not rectangle
Trapezoid
Convex
Concave
Figures
a, b, c
d
h, i
e, f, g
a, b, c, d, e, f, g, h, i, j, k, l
m, n, o, p
5. Here is one possible solution for each polygon.
convex octagon
concave octagon
concave dodecagon
convex nonagon
165°
195°
105°
6. a. 75°; 90°; 210°; 120°
b.
45°
7. a. 45°; 30°; 45°; 30°; 45°
b. Inscribed angles that intercept arcs of equal length in the same or congruent circles have
the same measure.
c. Inscribed angles whose sides intersect the ends of a diameter are right angles.
Activity Set & Just for Fun 9.1
Answers: Activities and Connections
8. Angle ABC measures 30° and angle AOC measures 60°.
b.
inscribed angle 37.5° inscribed angle 67.5° inscribed angle 75.0° inscribed angle 22.5°
central angle 75.0°
central angle 135.0° central angle 150.0°
central angle 45.0°
c. The measure of an inscribed angle is half the measure of the central angle that intercepts
the same arc.
Just for Fun 9.1
1. a.
b.
Number of
pieces used
1
2
3
4
5
6
7
Triangle
Square
Rectangle
(not a square)
Parallelogram
(not a rectangle)
Answers: Activities and Connections
Connections 9.
Connections Questions 9.1: Selected Answers
1. One way: Cut out paper patterns of each “half” and compare.
2. Perhaps students need to look at many examples of squares and rhombuses and discuss the
attributes of each class of quadrilateral to see that all squares are rhombuses but not all
rhombuses are squares. Also, when sorting geometric figures note the orientation in the plane
doesn’t change the category the figure is in.
3. Congruent pairs: Any pair Tx, Ty where x + y = 24 (and x, y ≠ 0, 1, 22)
Obtuse triangles: T2 to T10, and T14 to T22 (note T11 and T13 are right triangles)
Acute triangles: T12
Scalene triangles: All except T12, T21, and T3
Isosceles triangles: T12, T21, and T3
Equilateral triangles: none
Right triangles: T11 and T13
4. When the vertices of the polygons are on geoboard pins, there are 6 regular polygons, the
number of side must be a factor of 24: 3, 4, 6, 8, 12 and 24.
Activity Set 9.2
1. a. Here is one other way to tessellate with the red trapezoid. There are others.
Activity Set 9.2
Answers: Activities and Connections
b. All the other pattern blocks have more than one tessellation pattern. Here are two each for
the square and the blue rhombus. Some have more.
Square:
Blue rhombus:
2. Angle 1 measures 150°. A straight line has an angle measure of 180°. 180° – 30° (the
measure of the small angle of the tan rhombus) = 150°.
Angle 2 measures 60°. When 2 acute angles of the tan rhombus are put together at a point the
sum of their measures is the same as one of the angles of the green triangle. The acute angle
of the tan rhombus measures 30°.
Angle 3 measures 120°. Two angles of the triangle (angle 2) have the same measure as angle
3.
Angle 4 measures 60°. It is the same measure as angle 2.
Angle 5 measures 120°. It is the same as angle 3. Another way to figure this is that three
angle 5’s make a complete circle (360°), so each one measures 360° ÷ 3 = 120°.
Angle 6 measures 120°. It is the same measure as angle 5. Angle 7 measures 60°. It is the
same measure as angle 2.
3.
a.
Answers: Activities and Connections
Activity Set 9.2
b. This figure will not tessellate. There is an angle of 210°.To tessellate other angles must
combine with 210° to equal 360° (otherwise there would be a gap or overlap). Therefore,
there needs to be a combination of angles which adds up to 150° (360° – 210°). Because the
other angles are 90° and 120° this can’t be done.
4.
a. The triangle, square and hexagon will tessellate. The pentagon, heptagon and octagon
will not tessellate.
b. The number of degrees in each angle of a regular polygon must be a factor of 360.
c. Each interior angle of a regular polygon with more than 6 sides is greater than 120° and
less than 180°. The measures of such angles are not factors of 360.
5. a.
Activity Set & Just for Fun 9.2
Answers: Activities and Connections
b.
c. Here are two examples of pattern block tessellations that are neither regular nor semi-regular.
Just for Fun 9.2
The winning strategy on a 3 by 3 or 5 by 5 grid is to occupy the center hexagon on the first
move. On a 4 by 4 grid, the winning first move is any of those hexagons marked below.
Answers: Activities and Connections
Connections 9.2
Connections Questions 9.2: Selected Answers
2. One way: For each of the figures compare the side lengths using a ruler, or simply making
marks on a piece of paper for one side and compare it to the other. The corner of a sheet of
paper can also be used to check that the angles of the “square” are right angles.
4. Here is one way
5. a.
c. Every vertex of the tessellation is surrounded by the four angles of the original concave
quadrilateral.
6.
Standards addressed by studying tessellations:
Geometry Standard
Level
Pre-K-2
Apply transformations and use
symmetry to analyze
mathematical situations.
3-5
6-8
Expectation
Recognize and apply slides, flips, and turns;
Recognize and create shapes that have
symmetry
Predict and describe the results of sliding,
flipping, and turning; Describe a motion or
series of motions that show two shapes
congruent
Describe sizes, positions, orientations of shapes
under informal transformations such as flips,
turns, slides and scaling
Activity Set 9.3
Answers: Activities and Connections
Activity Set 9.3
1.
Polyhedron
# of faces
Type of face
4
equilateral triangle
Tetrahedron
6
square
Cube
8
equilateral triangle
Octahedron
12
regular pentagon
Dodecahedron
20
equilateral triangle
Icosahedron
2. a. The tetrahedron has 4 triangular faces and a rubber band would hold 2 edges of 2 triangles
together. The 4 triangular faces have 4 × 3 = 12 edges total, that are bound together in pairs,
so there are 12 ÷ 2 = 6 rubber bands needed. The cube requires 12 bands; the octahedron
requires 12 bands; the dodecahedron requires 30 bands; and the icosahedron requires 30
bands.
b. 2 colors
3. a.
Polyhedron
Vertices (V) Faces (F) Edges (E)
4
4
6
Tetrahedron
8
6
12
Cube
6
8
12
Octahedron
20
12
30
Dodecahedron
12
20
30
Icosahedron
b. The number of vertices plus the number of faces is two more than the number of edges.
Or, V + F = E + 2
c.
(1) (2) (3) (4)
Vertices 5 10 12 10
Faces
5
7
8
7
Edges
8
15
18
15
4. Patterns b, c and e are nets for a cube.
5. Figure b - Cutting off 6 corners of the octahedron produces 6 squares. The 8 hexagons are the
remaining parts of the 8 faces of the octahedron. Figure b is called a truncated octahedron.
Answers: Activities and Connections
Activity Set & Just for Fun 9.3
Figure e - Cutting off the 12 corners of the icosahedron produces 12 pentagons. The 20
hexagons are the remaining parts of the 20 faces of the icosahedron. Figure e is called a
truncated icosahedron.
Figure f - Cutting off the 8 corners of the cube produces 8 triangles. The 6 octagons are the
remaining parts of the 6 faces of the cube. Figure f is called a truncated cube.
Figure g - Cutting off the 12 corners of an icosahedron midway down the edges produces 12
pentagons and 20 triangles. Or, by cutting off the corners of a dodecagon midway down the
edges produces 20 triangles and 12 pentagons. This figure is called an icosidodecahedron.
Figure h - Cutting off 20 corners of the dodecahedron produces 20 triangles. The 12
decagons are the remaining parts of the 12 faces of the dodecahedron. Figure h is called a
truncated dodecahedron.
Figure j - Cutting off the 8 corners of the cube produces 8 triangles. If these cuts are made
such that the vertices of the triangles are at the midpoints of the edges of the cube then the
squares are the remaining parts of the 6 faces of the cube. This polyhedron can also be
obtained from an octahedron. Cutting off the 6 corners produces 6 squares. If these cuts are
made such that the vertices of the squares are at the midpoints of the edges of the octahedron
then the 8 triangles are the remaining parts of the 8 faces of the octahedron. This figure is
called a cube octahedron.
Just for Fun 9.3
Circle the first 5 for Cube A, 4 for Cube B, 5 for Cube C, and 6 for Cube D.
Connections Questions 9.3: Selected Answers
1. Your students could use three-dimensional models and count the vertices and edges. Or, this
could be an opportunity to see the relationship between the number of faces and the vertices
(gumdrops) and edges (sticks) of the solid. For example, the tetrahedron has four equilateral
triangle faces. The four triangles have a total of 12 edges and 12 vertices. To make the
tetrahedron three triangle vertices make one vertex in the tetrahedron (12 ÷ 3 = 4 gumdrops) and
two triangle edges make one edge in the tetrahedron (12 ÷ 2 = 6 sticks). A similar process can be
used for the other solids.
3. Here is one net for the four cube figure – where B marks the bottom of the cubes in the figure
Connections 9.3 & Activity Set 9.4
Answers: Activities and Connections
4. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
5. Here is one possibility but there are other choices as well.
Geometry Standard
Analyze characteristics and
properties of two- and threedimensional geometric shapes
and develop mathematical
arguments about geometrical
relationships
Level
Expectation
Recognize, name, build, compare, and sort twoPre-K-2
and three-dimensional shapes (activities 1, 2)
Identify, compare, and analyze attributes of twoand three-dimensional shapes and develop
3-5
vocabulary to describe those attributes
(activities 1, 2, 3 and 4)
Precisely describe, classify, and understand
relationships among types of two- and three6-8
dimensional objects using their defining
properties (activity 5)
Activity Set 9.4
1. a.
b. 180° for both figures
c. Here are some possible results. There are other ways.
Answers: Activities and Connections
Activity Set 9.4
2. a. The completed design is symmetric about two perpendicular lines and has rotational
symmetries of 180° and 360°.
b.
The completed design does not have any line symmetries.
c.
3. a. A square is formed by making the cut such that AC = BC.
b. Here are some of the figures that can be obtained: rectangle, square, rhombus, convex
hexagon, concave hexagon, and concave decagon.
4. a. Angle ABC has a measure of 135°, AD = DC and AB = BC.
Octagon:
Activity Set 9.4
Answers: Activities and Connections
b. Angle EFG has a measure of 120°, HF = HG = FG and EF = 1⁄2 of FG.
Hexagon:
c. If AC = CB and RC ≠ CS, there will be a rhombus inside a square. If AC ≠ CB and RC =
CS there will be a square inside a rhombus.
5. a. One possibility, a 12-sided figure with 2 lines of symmetry.
d. One possibility, a 12-sided figure with 2 lines of symmetry.
e. One possibility, 2 similar 12-sided figures with one rotated 90° to the other and both with
2 lines of symmetry.
f. The eight-pointed wind rose if the cuts are made to form a parallelogram as shown on the
figure.
g. A regular 16-sided polygon can be obtained by cutting a 157.5° angle as shown here,
with AB = AC = AD. Note: Some of the lines of symmetry are not crease lines.
6. a. The result could be decagons (10-sided polygons) or pentagons (5-sided polygons)
depending upon the relative lengths and ZN and ZY.
h. A cut perpendicular to ZY or ZN produces a regular pentagon. When ZN = ZY the cut
NY results in a regular decagon.
Answers: Activities and Connections
Connections 9.4
Connections Questions 9.4: Selected Answers
1. It is true that any line through the center of square divides the square into two congruent
regions; perhaps this is what she is thinking. One idea is to have the student fold and cut
squares to show you her strategy and discover which folds result in lines of symmetry.
2. Rydell is not correct; have him work with non-symmetrical quadrilaterals to see this.
3. There are five symmetries, one rotational and four reflective.
4. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
Activity Set & Connections 10.1
Answers: Activities and Connections
Activity Set 10.1
2. Width 20 cm, length 25 cm.
3. a. 156 mm
b. 19 mm
d. Determine the number of pages that have a thickness of 10 mm and then divide 10 by that
number of pages. This will give you the thickness of one page as a fraction of a millimeter.
6. a. 10 meters is 1000 centimeters. Divide 1000 by the number of paces needed to walk 10
meters to obtain the length of a single pace in centimeters.
b. Multiply the number of paces needed to walk 10 meters by 100.
c. Multiply the time, in seconds, needed to walk 10 meters by 100, then divide by 60
seconds/minute.
7. a. A quart contains approximately 946 cm3. A liter contains slightly more than a quart.
b. Approximately 10.9 cm
Connections 10.1: Selected Answers
2. One idea would be to start with metric ruler and tapes that are marked only in centimeters (no
millimeter marks) to measure various lengths.
3. a. quarter-of-an-inch—millimeter; inch—centimeter; foot—meters; mile—kilometers
b. The units in the metric system are related by powers of ten, like our base ten numeration
system.
4. Freezing to boiling ranges from 0 to 100 on the Celsius scale and from 32° to 212° (or 180°)
1
on the Fahrenheit scale. Twenty is one-fifth of the Celsius scale. Because of 180 is 36, the
5
corresponding number on the Fahrenheit scale is 32° + 36° = 68°.
0C
32°
20C
68°
100C
212°
Answers: Activities and Connections
Activity Set 10.2
Activity Set 10.2
1. a. 9 square units
b. 6 square units
c. 14 square units
2. a. 3 square units
b. 4 square units
c. 1 square unit
3.
Height from Vertex A: 1 unit
4 units
Base BC length:
Area:
2 units2
Height from Vertex E: 2 units
3 units
Base FG length:
Area:
3 units2
Height from side ZW : 2 unit
3 units
Base ZW length:
Area:
6 units2
Height from side RS :
2 units
Base RS length:
2 2 units
Area:
4 units2
4.
a.
b.
Area:
8 units2
Perimeter: 4 + 2 17 units ≈ 12.25 units
Area:
8 units2
Perimeter: 4 + 2 20 units ≈ 12.94 units
c.
d.
Area:
8 units2
Perimeter: 10 units
Area:
8 units2
Perimeter: 4 + 2 32 units ≈ 15.32 units
Activity Set & Just for Fun 10.2
Answers: Activities and Connections
e. The area of a parallelogram is the product of the length of a side and the height from that
side.
f. The area is not directly related to the perimeter of the parallelogram. In the above
examples, the height and base of each parallelogram are the same, therefore the area is the
same, but the perimeter is clearly different.
1
square units. Two triangles which have a side of equal
2
length and an altitude (to that side) of equal length, have equal areas.
5. All of the triangles have area 4
6. a. Parallelogram, 6 square units; triangle, 3 square units
b. Parallelogram, 6 square units; triangle, 3 square units
c. Parallelogram, 9 square units; triangle, 4
1
square units
2
d. Parallelogram, 3 square units; triangle, 1
1
square units
2
7. The area of a parallelogram is equal to the length of a side multiplied by the height from that
side. A triangle with base of length b units and height of length h units has half the area of a
parallelogram with base length b and a height of length h from that side.
Just for Fun 10.2
1. Here are some additional solutions.
2. Here are two of the many solutions.
Answers: Activities and Connections
Connections 10.2
3. Here is one solution to the grid with the missing corners.
4. Here are two ways that player B can win. The squares with an X in the figure below show
two possible final moves that player B can make.
5. Here are five pieces that work.
Connections 10.2: Selected Answers
1. Since there are “no pins on the inside,” this means you can only form the following shapes
and combinations of the following shapes:
Right Triangles of height 1 unit
Squares on 4 pins
4−2 2
Area = 1 u2=
u
2
Triangles of height 1 unit: Let B = the number of pins on the boundary.
The length of the base of the triangle = # pins on the base - 1, therefore:
B – 2 can be thought of (B – 1) – 1 which is the length of base take away the 1 pin that is the
top vertex for the height, i.e., the length of the base.
Additionally the height is 1 unit so
B−2 2 1
u =
× base length × height u2 = area of the triangle.
2
2
Connections 10.2
Answers: Activities and Connections
Combinations
Since there are no pins on the inside, you can see by inspection that the above argument works
by decomposing combination shapes into 4 pin squares and triangles of height 1 unit.
To help the student; form figures with 3 to 6 pins on the boundary with no pins inside to review
the student’s conjecture with them. Then repeat with the same number of pins on the boundary
but 1 pin inside the polygon to see if there is a pattern before answering the student.
3. a. 8 units2 b. 5.5 units2 c. 5.5 units2 d. 5 units2 e. 1 units2 f. 4.5 units2
4. a.
1
square units is the least and 8 square units is the greatest.
2
.5 u2
8 u2
b.
.5 x
2 u2
4 u2
5 u2
1
u2
16 - 4 - 3 - 2 = 7
10 x 10 = 5
3
6 u2
7 u2
8 u2
u2
5. The Grades 3 – 5 and 6 – 8 Expectations are listed here:
Measurement Standard
Understand measurable
attributes of objects and the
units, systems, and processes
of measurement
Level
Expectation
Understand how to measure using nonstandard
Pre-K-2
and standard units
Understand the need for measuring with
standard units and become familiar with
3-5
standard units in the customary and metric
systems
Understand both metric and customary systems
6-8
of measurement
Answers: Activities and Connections
Activity Set 10.3
Activity Set 10.3
1. The volume of the crown was greater than the volume of the gold. King Heron was cheated.
2.
Hexagonal prism
Hexagonal pyramid
Cylinder
Area of base
9.1 cm2
9.1 cm2
12.6 cm2
Height (altitude)
5.0 cm
5.0 cm
3.0 cm
Volume
45.5 cm3
15.2 cm3
37.7 cm3
Surface area
75.2 cm2
38.7 cm2
62.8 cm2
The volume of the pyramid is
1
the volume of the prism because both have the same altitude
3
and base.
The altitude of the pyramid can be computed by using the slant height of the pyramid, 5.2
cm, and the distance from the edge of the base to the center of the base, 1.6 cm.
The surface area of the side of the cylinder is computed by multiplying the height by the
distance around the cylinder, which is just the circumference of the circle (2πr). (Note: the
surface areas of the prism, pyramid, and cylinder include the surface areas of their bases.)
3. Cylinder A: 1030 cm3 Cylinder B: 1342 cm3. The radius of the base of the cylinder A is
21.5⁄2π. So the volume of cylinder A is 28 × (21.5⁄2π)2 ×π = 1030 cm3.
The radius of the base of cylinder B is 28⁄2π. So the volume of cylinder B is 21.5 × (28⁄2π)2
× π = 1342 cm3.
4. The radius of the base is approximately 8.92 cm and the volume is approximately 2684 cm3.
This cylinder has twice the volume of cylinder B in activity 3.
In general, if one cylinder B has height h and radius r and a second cylinder C has half the
h
height ( ) and twice the radius (2r), cylinder C will have twice the volume of cylinder B.
2
h
h
Volume of C = π × (2r) 2 × = π× 4r 2 × = 2(π × r 2 × h)
Volume of B = π × r2 × h
2
2
Activity Set 10.3
Answers: Activities and Connections
5. b.
1
2
3
4
5
6
Central angle of disc
45°
90°
135°
225°
270°
315°
Radius of base (cm)
1.25
2.50
3.75
6.25
7.50
8.75
Base area (cm2)
4.91
19.63
44.16
Height of cone (cm)
9.92
9.68
9.27
122.66 176.63 240.41
7.81
6.61
4.84
Volume of cone (cm 3) 16.24 63.34 136.45 319.32 389.17 387.86
c. For central angles of less than 45°, the smaller the angle is, the smaller the volume is. For
central angles of more than 315°, the larger the angle is, the smaller the volume is. The
maximum volume of approximately 403 cm3 occurs with a central angle of 294°.
1
the volume of the whole cone. A cone with a base of
8
radius r and a height of h has a volume of 1⁄3πr2h. The “half cone” has a base with radius r⁄2
r
h
1
1
1
and a height of h⁄2 . Its volume is × π( )2( ) which is equal to ×πr2h × . Therefore,
3
2
2
3
8
1
the volume of the “half cone” is the volume of the original cone.
8
6. c. The volume of the “half cone” is
7. a. 7235 cm3
c. Approximately 24 cm in diameter. The diameter is twice the radius r and r can be found
4
from the equation: 7235 = ×π×r3
3
d. Approximately 203.6 cm3
e. A can with a radius of 4 cm and height of 11 cm has a volume of approximately 553 cm3.
Answers: Activities and Connections
Activity Set & Connections 10.3
1
the space is wasted. The volume of the cube is 253 cm3 and the volume
2
253
4
25 3 3
of the sphere is ×π ( ) cm which is approximately
cm3 if we approximate π by 3.
3
2
2
8. a. Approximately
1
of the space is wasted. Suppose the can and the balls have diameter 2r. Then the radius
3
4
of the balls and the can is r and the height of the can is 6r. Volume of 3 balls = 3 × × π × r3
3
= 4πr3. Volume of can = π × r2 × 6r = 6πr3.
b.
4
×π(3.15) 3 cm3 ≈ 393 cm3. The volume of the can
3
1
is 20 × π(3.5)2 cm3 ≈ 769 cm3. Therefore, approximately
of the space is wasted.
2
c. The volume of the three balls is 3 ×
Just for Fun 10.3
The figure with 13 cubes cannot be formed because no combination of 4s and one 3 total 13.
Connections 10.3: Selected Answers
3. The volume of the larger cubical box must be 81 cubic inches (3 × 27 cubic inches) so each
edge must measure 4.3 inches, to the nearest tenth of an inch ( 3 81 ≈ 4.3).
4. a. 6 × 4 inches × 4 inches = 96 square inches
b. The clay cube has volume (4 inches)3 = 64 cubic inches and surface area 96 square
inches.
A clay sphere of volume 64 cubic inches has radius about 2.48 inches.
48
4
64 = πr3, so r = 3
≈ 2.48
3
π
A clay sphere of volume 64 cubic inches has surface area about 77.4 square inches
2
⎛ 48 ⎞
⎟ ≈ 77.4
4πr = 4π × ⎜⎜ 3
⎟
π
⎝
⎠
2
The clay cube has a greater surface area with the same volume.
Connections 10.3
Answers: Activities and Connections
6. Here are some of the Expectations that address volume, there are others.
Measurement Standard
Understand measurable
attributes of objects and the
units, systems, and processes
of measurement
Level
Expectation
Recognize the attributes of length, volume,
Pre-K-2
weight, area, and time
Understand such attributes as length, area,
weight, volume, and size of angle and select the
3-5
appropriate type of unit for measuring each
attribute
Understand, select, and use units of appropriate
6-8
size and type to measure angles, perimeter, area,
surface area, and volume
Answers: Activities and Connections
Activity Set 11.1
Activity Set 11.1
1. a. The points form a line which is perpendicular to segment KL at the midpoint of segment
KL.
b. The points form 2 perpendicular lines that pass through the intersection of lines m and n
and bisect the angles formed by the intersection of m and n.
m
n
2. a. The points form a circle with P as center and radius equal to the distance from P to the
midpoint of the chords.
b. The points form a circle whose center is the midpoint of PR and whose diameter is PR.
Activity Set 11.1
Answers: Activities and Connections
c. The points form a circle whose center is the midpoint of segment PS and whose diameter
is PS.
3. a. A square, centered inside the original square, whose sides are 3.0 cm shorter than the
original.
b. The figure consists of 4 congruent segments located outside the original square and
parallel to the sides of the original square and 4 quarter arcs of a circle of radius 1.5 cm,
centered at the vertices of the original square.
4. A circle with center P on ray AB (where BP is
1
2
AB) and radius of the circle is AB.
3
3
Answers: Activities and Connections
Activity Set & Just for Fun11.1
5. One fourth of a circle whose radius is 2.5 cm and whose center is the intersection of the
perpendicular lines.
6. a. A semicircle with center at the midpoint of PQ and diameter PQ. (A circle if you lift the
card and do the same thing on the other side of segment PQ.)
Q
P
b. The major arc of a circle. (Three-fourths the circumference of a circle with chord RS.)
R
S
c. A minor arc of a circle. (An arc shorter than a semi-circle.)
R
S
Just for Fun 11.1
When the front ends of the chords travel 3 times as fast the resulting curve has 2 cusps.
Connections 11.1, Activity Set 11.2
Answers: Activities and Connections
Connections 11.1: Selected Answers
1. Here is one way: Cresedeo could put a marker (say a yellow marker) on a table and then use
a piece of string to arrange many other markers around the yellow marker a strings length
away. The yellow marker can be named the center, the string the radius, and the many
markers placed around form the circle. Diameter, chord, and arc can be illustrated by placing
colored yarn on Cresedeo’s circle. For a whole class, Cresedeo could be the center and the
other students the markers forming the circle.
4. A line parallel to the given line at a distance equal to half the distance from the point to the
line.
P
5. Solutions to the online Math Investigations may be found in the Conceptual Approach IRM
and also downloaded from the Instructor’s side of the Online Learning Center (Activity
Approach).
6.
Geometry Standard
Specify locations and describe
spatial relationships using
coordinate geometry and other
representational systems.
Level
Expectation
Describe, name, and interpret relative positions
Pre-K-2
in space and apply ideas about relative position
Describe location and movement using common
3-5
language and geometric vocabulary.
Make and use coordinate systems to specify
6-8
locations and to describe paths.
Activity Set 11.2
1. The new figure will tessellate because the exact shape that has been adjoined to one edge of
the rectangle was deleted from the opposite side of the rectangle.
2. There are six rotational symmetries about point P: 60°, 120°, 180°, 240°, 300° and 360°.
3. a. The top row of the fish tessellation is mapped onto the second row by a glide reflection.
The top row is mapped onto the third row by a translation.
Answers: Activities and Connections
Just for Fun 11.2
Just for Fun 11.2
Puzzle 1
Now fold along the adjoining edges of each pair of squares to form a cube.
Puzzle 2
Puzzle 3
Fold a sheet of paper in half once and cut on all the dotted lines shown. This will produce a hole
large enough to pass over your body. Experiment with even more cuts.
Connections 11.2, Activity Set 11.3
Answers: Activities and Connections
Connections 11.2: Selected Answers
3. Alterations to half of a side can be rotated about the midpoint of that same side (similar to the
midpoint rotation of side BC in activity 2, figures c and d, in this section.)
5. Templates for each of the three tessellation shaping techniques can be constructed using the
cut and tape process.
6. Escher-type drawings and Escher’s artwork have become very popular in recent times. These
drawings are done in art classes and there are software programs that enable the user to make
Escher-type drawings. Mathematicians and math teachers have been drawn to this art
because of the mathematical elements, as we have experienced in this section. On this basis
we can make the argument that the Connections Process Standard applies through the
expectation, “recognize and apply mathematics in contexts outside of mathematics.”
Activity Set 11.3
1. a. Both triangles contain the angle at E, and the angles at A and A′ are right angles. When
two angles of one triangle are congruent to two angles of another the triangles are similar.
b. Answers will vary.
c. Answers will vary; here is an example of the set up. Assume EA = 56 cm, your thumb is
9 cm down the ruler and your estimate of the height of the person is 150 cm. Then EA′ can
EA' 150
be determined by using the proportion:
=
.
56
9
2. a. Triangle E1E2A is similar to triangle TRA because angle E1AE2 and angle RAT are
congruent and opposite the side connecting congruent angles in the two isosceles triangles;
triangle E1E2A and triangle TRA.
b.
AT
AE2
75
=
=
= 10 , thus, AT = 900 meters
90 E1 E2 7.5
Answers: Activities and Connections
Activity Set 11.3
c. Answers will vary.
d. When you are estimating large distances, the distance from eye to finger is relatively
small and ignoring it will not significantly affect the estimate. However, when you are
estimating small distances, which may be only a few times greater than the distance from eye
to finger, ignoring the distance from the eye to the finger will give a larger error. In the latter
case, the distance should be added to the estimated distance.
3. a. Both triangles contain the angle at point E and right angles. Therefore the corresponding
angles of the triangles have the same degree measure.
b.
ED
3
=
so ED = 20.25 m. Therefore the person is about 20 m from the truck.
27
4
c. 24.3 m
4. a. Angle BVW and angle BAC are both compliments to the congruent angles marked X and
are therefore congruent angles.
V
X
B
X
A
W
C
b. The scale model triangle ETR and the outside triangle ETR are both right triangles with
the same angle RET and are therefore similar triangles.
c. ET in the figure measures 80 mm. So 80 mm represents 60 m or 1 mm represents .75 m.
The height of the tree from T to R in the diagram is 24 mm so the height of the actual tree
between those two points is 24 × .75 = 18 m.
d. The distance from the ground to the observer’s eye plus the answer in part c gives an
approximate tree height of 19.5 m.
e. Answers will vary.
Activity Set 11.3
Answers: Activities and Connections
5. a. Approximately 33 m plus 1.5 m the height of the hypsometer above the ground, for a total
of 34.5 m.
Triangle ABC is similar to triangle A’B’’C as angles A’CB’’ and ACB are congruent.
Triangle A’B’’C is also similar to triangle A’B’C’ as they are right triangles that share angle
A’. Therefore triangle ABC is similar to triangle A’B’’C which is similar to triangle A’B’C’.
We can conclude that triangle ABC is similar to triangle A’B’C’.
B ' C ' ≅ B' ' D are parallel sides of a rectangle formed inside triangle A’B’’C.
A'
B'
C'
C
A
D
B
C
C'
B''
A'
D
25
55
B'
B''
b. The scale on the hypsometer is proportional to the scale of the actual triangle formed
outside by your eye and the top of the building (as marked on the triangle).
c.
25 H
=
, H ≈31.8. Adding 1.5 meters to 31.8 meters, the total height is ≈ 333.3 meters.
55 70
d.
55
D
=
, D ≈ 62.4. The distance is about 62.5 feet.
25 28.5
e. 20 m minus the height of the hypsometer above the ground might be about 18.5 m. Thus
the distance to the tree is approximately 40 m.
6. a. The distance can be computed by first measuring the length of AB and AF in mm. Then,
divide 500 by the length of AB to determine the number of meters represented by each mm.
Use that result to determine the actual length of AF. The actual length of AF is
approximately 643 m.
b. A map can be made by measuring 1 distance and 10 angles. Use a line from the land
marker to the pine tree as a base line and measure the distance between these 2 points. From
the pine tree, measure the 5 angles between the base line and lines to the remaining 5 objects.
From the land mark, measure the 5 angles between the base line and lines to the remaining 5
objects. With these 10 angles a map can be made of the locations of each object. Using the
distance from the land mark to the pine tree a scale can be set up to determine all distances
between any pairs of objects in the field.
Answers: Activities and Connections
Connections 11.3
Connections 11.3: Selected Answers
1. Two polygons are similar if their corresponding angles are
equal and the corresponding sides are proportional. One way to
explore the similarity of squares with a student would be to
draw a square on grid paper as shown below. Then if the sides
XA and XB are increased (or decreased) by the same factor, and
another square is constructed on those sides, the student can
observe that the remaining sides change by the same factor.
C
2. The student is correct. In the figure below, triangle AYX and triangle
ABC are similar (because their angles are congruent) so the sides of the
smaller rectangle are proportional to the sides of the larger rectangle.
X
A
Y
B
3. Note there are many ways to make the “hexagonal” and “trapezoidal” regions here.
4. The side lengths of the new figure are
of the new figure is
3
times the side length of the original figure. The area
2
9
times as large as the original area.
4