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Mathematics GLE Resource
Materials: Definitions and Examples
for Grades K - 8
Geometry and Measurement
January 2005
New Hampshire Department of Education
Rhode Island Department of Education
Vermont Department of Education
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Overview
The purpose of these resource materials is to provide K – 8 educators with explanations
and examples that facilitate understanding of the mathematics in the Grade Level
Expectations (GLEs)*. These resource materials are organized by content strand and by
GLE “stems.” The definitions and explanations within each section are not alphabetized,
but are organized to parallel the introduction of new concepts and skills within each GLE
“stem” across grades. * Grade Level Expectations are called Grade Expectations (GEs) in Vermont.
Stem
Grade 3
M(G&M)–3– 6 Demonstrates conceptual understanding
of perimeter of polygons, and the area of rectangles on
grids using a variety of models or manipulatives. Expresses
all measures using appropriate units.
Grade 4
M(G&M)–4–6 Demonstrates conceptual understanding of
perimeter of polygons, and the area of rectangles, polygons,
and irregular shapes on grids using a variety of models,
manipulatives, or formulas. Expresses all measures using
appropriate units.
To facilitate access, each definition is coded (e.g., G&M – 1-32). There are multiple
table of contents: 1) An overall table of contents on page 2 contains all the terms or
phrases defined for the content strand in alphabetical order; and 2) A table of contents for
each section in alphabetical order. In addition, if there is a word or phrase that you are
unclear about within a definition, check the overall table of contents to determine if the
word or phrase is defined in another location in this document.
The materials contained in this document focus on the Geometry and Measurement
strand and only address the NECAP GLEs that are common to all three states. The
materials are divided into 3 sections as indicated in the following table.
Section #
Grade Level Expectation “Stem”
NECAP GLE Stem Codes
(Vermont codes)
X represents grades K -8
Section 1
Section 2
Uses properties and attributes… to identify,
describe, classify, or distinguish among…
two-dimensional figures.
Applies theorems or relationships…
Uses properties and attributes… to identify,
describe, classify, or distinguish among…
three-dimensional figures.
Applies the concepts of congruency…
NECAP: M(G&M) – X – 1
Page
Numbers
5-32
(Vermont GE MX: 9)
NECAP: M(G&M) – X – 2
(Vermont GE MX: 10)
NECAP: M(G&M) – X – 3
(Vermont GE MX: 11)
NECAP: M( G&M ) – X – 4
33-51
(Vermont GE MX: 12)
Applies the concepts of similarity…
Section 3
Demonstrates understanding of perimeter,
area, volume, and surface area…
Measures and uses units of measures
appropriately and consistently…
NECAP: M(G&M) – X – 5
(Vermont GE MX: 13)
NECAP: M( G&M ) – X – 6
52-61
(Vermont GE MX: 14)
NECAP: M(G&M) – X – 7
(Vermont GE MX: 15)
Note: Examples are provided throughout this document to illustrate definitions or phrases
used in the mathematics GLEs. However, the kinds of questions students might be asked
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in instruction or on the NECAP assessment about the mathematics being illustrated are
NOT necessarily limited to the specific examples given. It is the intent, over time, to add
release NECAP items to expand the set of examples.
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Geometry and Measurement: Table of Contents
Definition
Angle relationships formed by two or more lines cut by a transversal
Angles
Attributes and properties
Composing and decomposing shapes
Congruent
Demonstrates conceptual understanding of perimeter and area of polygons
and irregular figures on grids
Demonstrates conceptual understanding of perimeter and area using models
and manipulatives to surround and cover polygons
Demonstrates conceptual understanding of perimeter, area, volume, or
surface area using models and manipulatives
Demonstrates conceptual understanding of perimeter, area, volume, or
surface area by solving problems
Demonstrates conceptual understanding of the relationships of circle
measures by solving related problems
Describes the proportional effect on linear dimensions of polygons or
circles when scaling up or down while preserving the angles of polygons
Makes conversions within and across systems
Matching congruent figures using reflections, translations, or rotations
Measurable attribute
Measures and uses units of measure appropriately and consistently
Non-measurable attribute
Parallel lines
Perpendicular
Polygons
Pythagorean theorem
Quadrilaterals
Reflection
Rotation
Similar figures
Solves problems involving scaling up or down and their impact on angle
measure, linear dimensions, and areas of polygons and circles when the
linear dimensions are multiplied by a constant factor
Solving problems on a coordinate plane using reflections, translations, or
rotations
Sum of the measures of interior angles of polygons
Three-dimensional shapes
Translation
Triangle
Triangle inequality
Uses composition and decomposition
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Section
Page
Number Number
1
24
1
21
1
6
2
40
2
34
3
54
Definition
Number
G&M -11
G&M -10
G&M -1
G&M -21
G&M -16
G&M -27
3
53
G&M -26
3
56
G&M -28
3
57
G&M -29
3
59
G&M -30
2
47
G&M -24
3
2
1
3
1
1
1
1
1
1
2
2
2
2
61
39
20
60
20
19
18
9
28
12
36
37
44
49
G&M -32
G&M -20
G&M -9
G&M -31
G&M -8
G&M -7
G&M -6
G&M -2
G&M -14
G&M -4
G&M -17
G&M -19
G&M -23
G&M -25
2
41
G&M -22
1
1
2
1
1
1
27
29
37
11
26
16
G&M -13
G&M -15
G&M -18
G&M -3
G&M -12
G&M -5
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Section 1: Uses properties and attributes of two and three dimensional
figures; and applies theorems and relationships
NECAP M(G&M) – X – 1, M(G&M) – X – 2, M(G&M) – X – 3
Vermont: GE MX: 9, MX: 10 & MX: 11
Definition
Angle relationships formed by two or more lines
cut by a transversal
Angles
Attributes and properties
Measurable attribute
Non-measurable attribute
Parallel lines
Perpendicular
Polygons
Pythagorean theorem
Quadrilaterals
Sum of the measures of interior angles of
polygons
Three-dimensional shapes
Triangle
Triangle inequality
Uses composition and decomposition
5
Page
Number
24
Definition
Number
G&M -11
21
6
20
20
19
18
9
28
12
27
G&M -10
G&M -1
G&M -9
G&M -8
G&M -7
G&M -6
G&M -2
G&M -14
G&M -4
G&M -13
29
11
26
16
G&M -15
G&M -3
G&M -12
G&M -5
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G&M – 1 Attributes and Properties: Attributes and properties
refer to the characteristics of an object. Attributes consider
characteristics like shape, size, color, shading, and so on that are not
necessarily unique to a geometric shape or class of geometric shapes.
Properties refer to characteristics that are true for a geometric shape in a
particular class of geometric shapes (e.g., quadrilaterals).
Note: The word
“properties” is used in
such a way as to include
those properties that are
inherent in the definition
of a geometric shape (e.g.,
a rectangle has four right
angles).
Example 1.1 – Sorts by an attribute:
Sort these rectangles by one attribute.
Answer: This set of rectangles can be sorted (or classified) by a number of different attributes:
shading, orientation, size (small, big), relationship between length and width (long/skinny,
wide/short).
(Definition G&M – 1 continued on following page)
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Example 1.2 – Properties and Attributes:
Even though these rectangles have different attributes (shaded, unshaded, long,
skinny) they are all rectangles because they satisfy the properties of rectangles.
Some properties of rectangles: Rectangles are quadrilaterals (four-sided polygons)
in which:
1) the opposite sides are congruent (same length);
2) the opposite sides are parallel;
3) the angles are all right angles (90°);
4) the diagonals intersect at their midpoints…
AB reads – “line segment AB”
AB ≅ CD reads – “line segment AB is
congruent to line segment CD”
AB ≅ CD and BC ≅ DA
∠A, ∠B, ∠C , and ∠D are all right angles.
AB CD (line AB is parallel to line CD)
BC DA (line BC is parallel to line DA)
Similar tick marks (e.g., ) on sides of figures
indicate that the sides are congruent.
The small square at the vertex of a figure
indicates that the angle is a right angle (90°).
(Definition G&M – 1 continued on following page)
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Example 1.3 – Uses property of rectangles to distinguish between figures:
Alan says that if a figure has four sides, it must be a rectangle. Gina does not agree.
Which of the following figures shows that Gina is correct?
A)
B)
C)
D)
Answer: D: Figure D is a quadrilateral in which the opposite sides are congruent and parallel, but
the angles are not right angles.
Source: NAEP – 2003 Grade 4
Example 1.4 – Distinguish between polygons using properties:
Laura was asked to choose 1 of 3 shapes N, P, and Q that is different from the other 2.
Laura chose shape N. Explain how shape N is different from shapes P and Q.
N
P
Q
Source: NAEP Grade 4 – 1996
Sample Answers: N has four sides, (or vertices or angles), but P and Q each have 3; all the angles
of N are equal (all right angles), but this is not so with P and Q.
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G&M – 2 Polygons: A polygon is a two dimensional closed figure consisting of line
segments connected from endpoint to endpoint. (e.g., Quadrilateral ABCD is formed by
line segments AB, BC , CD , and DA .)
Side
Vertex: A vertex of a
polygon is the point
where two edges (sides)
intersect.
Polygons are named according to the number of
sides. A general way to name a polygon with n
sides is to call it a n-gon (e.g., A 7-gon is a
polygon with 7 sides).
Number of Sides
3
4
5
6
7
8
And so on…
Name
Triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
(Definition G&M – 2 continued on following page)
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Example 2.1 – Sorts polygons using properties:
Consider these polygons.
D
C
E
A
B
I
F
L
K
H
J
G
This is one way that the polygons can be sorted or classified based upon the number
of sides and parallelism.
At least one set of opposite
sides are parallel
No parallel lines
3 sided
L
G
4 sided
B
J
H
I
More than four
sides
10
F
E
A
D
C
K
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G&M – 3 Triangle: A triangle is a polygon with three sides. Triangles are classified
according the length of their sides and the measures of their angles. Table 3.1 contains
examples of different types of triangles.
Table 3.1 – Triangles:
Triangle
Equilateral
Side or Angle Relationships
An equilateral triangle is a triangle that has
three congruent sides.
Example
AB ≅ BC ≅ CA
Scalene
A scalene triangle is a triangle that has no
congruent sides.
Isosceles
An isosceles triangle is a triangle that has
at least two congruent sides.
Acute
An acute triangle is a triangle that has 3
acute angles that measure more than 0°and
less than 90°.
Right
A right triangle is a triangle with 1 right
angle (an angle that measures 90°).
∠BAC is a right triangle.
Obtuse
An obtuse triangle is a triangle with 1
obtuse angle (an angle that measures more
than 90° and less than 180°).
Equiangular
An equiangular triangle is a triangle in
which all 3 angles are congruent.
∠BCA is an obtuse angle.
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G&M – 4 Quadrilaterals: Quadrilaterals are polygons with four sides. Quadrilaterals
can be classified by their side lengths, and the measures of their angles. Table 4.1
contains examples of different types of quadrilaterals.
Table 4.1 – Quadrilaterals:
Quadrilaterals
Parallelogram
Description
A parallelogram is a quadrilateral in
which both pairs of opposite sides are
parallel. An additional property of a
parallelogram is that both pairs of
opposite sides are congruent.
AB ≅ CD and BC ≅ DA
AB CD and BC AD
Rhombus
A rhombus is a parallelogram with four
congruent sides. Both pairs of opposite
sides are parallel.
AB ≅ BC ≅ CD ≅ DA
AB CD and BC AD
Rectangle
A rectangle is a parallelogram that has
four right angles. Since rectangles are
parallelograms both pairs of opposite
sides are congruent.
Examples
Note: Angled tic marks indicate parallel
lines. Congruency is shown with similar
sets of tic marks.
∠A, ∠B, ∠C , and ∠D are all right angles.
AB CD and BC AD
AB ≅ CD and BC ≅ DA
A square is a parallelogram with four
congruent sides and four congruent
angles. Therefore, a square is a
rectangle and a rhombus.
Square
∠A, ∠B, ∠C , and ∠D are all right angles.
Trapezoid
AB ≅ BC ≅ CD ≅ DA
A trapezoid is a quadrilateral that has
exactly one pair of parallel sides.
(Note: A trapezoid is sometimes defined in a
more inclusive manner – a quadrilateral with at
least one pair of parallel sides (allowing for
parallelograms to be trapezoids); however, for
the NECAP assessment, we will use the more
exclusive definition given above.)
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Example 4.1 – Uses properties to identify, describe, classify, and distinguish among
quadrilaterals:
Investigating Quadrilaterals
Quadrilateral is the “family” name given to closed four-sided shapes.
A rectangle is a special type of quadrilateral because it has four right angles and two
pairs of parallel sides.
Rectangles also have two lines of symmetry.
This quadrilateral has no right angles, no parallel lines, and no
lines of symmetry.
Part I: Use the grid on the next page. In each box sketch a quadrilateral that
satisfies both properties for the box.
A sample has been provided in the box 0, 0. This quadrilateral has no right angles
and no parallel sides. Fill in the remaining boxes. If there is no quadrilateral that
satisfies both properties, explain why.
Part II: Build another grid and use it to investigate a different pair of
characteristics by sketching a quadrilateral that satisfies both properties for the box.
Use any two of the characteristics below.
•
•
•
•
Number of right angles
Number of parallel sides
Number of lines of symmetry
Number of pairs of equal sides
(Definition G&M – 4 continued on following page)
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Part 1
0
Number of Pairs of Parallel Sides
1
0
1
2
3
4
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Sample answer: Part I
0
Number of Pairs of Parallel Sides
1
2
0
Impossible: If 1 pair of
sides is parallel, then at
least two angles must be
right angles.
1
Impossible: If 2 pair of
sides are parallel and 1
angle is a right angle, then
all 4 angles must be right
angles.
2
3
4
Impossible: If 2 pair of
sides are parallel and 1
angle is a right angle, then
all 4 angles must be right
angles.
Impossible: The sum of the measures of three right angles is equal to 3 · 90° = 270°.
Since the sum of the measures of the angles in a quadrilateral is 360°, the remaining
angle measure is 360° – 270° = 90°. Therefore, if a quadrilateral has 3 right angles,
then the fourth angle must also be a right angle.
Impossible: If 4 angles are
right angles, then both
pairs of sides would be
parallel.
Impossible: If 4 angles are
right angles, then both
pairs of sides would be
parallel.
Part II: Answers will vary with properties used.
Source: Grade Eight Guide to Diversity of Content and Problem
Solving Tasks, VISMT(1995). (Task adapted from a Balanced
Assessment Task)
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G&M – 5 Uses composition or decomposition to sort or classify polygons:
To use composition or decomposition to sort or classify polygons means to analyze a set
of polygons and determine if they can be decomposed into other polygons, or if polygons
can be combined to form other polygons.
Example 5.1 – Use decomposition to sort or classify:
Which polygons pictured below can be decomposed into exactly two triangles using
exactly 1 line segment. Show one way that each polygon that you circled can be
decomposed into 2 triangles.
Answer: All but the octagon can be decomposed into two triangles. These are two ways that each
of the polygons can be decomposed into 2 triangles.
Example 5.2 – Composing/decomposing shapes to create a figure:
Refer to the following information.
1
2
3
Triangles 1, 2, and 3 shown above can be arranged with no overlap to form either of
the following figures.
(Definition G&M – 5 continued on following page)
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3
2
1
3
2
1
Draw lines on the figure below to show how triangles 1, 2, and 3 can be rearranged
without overlap to form this rectangle.
Answer:
Source: NAEP 2003 Released Item – Grade
8
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G&M – 6 Perpendicular: Two rays, line segments, etc., are said to be perpendicular
uuur
uuur
if they intersect to form a right angle. (e.g., AB and BC intersect at point B to form a
uuur
uuur
right angle. Therefore, AB ⊥ BC .)
The symbol, ⊥ , is used to
denote perpendicular.
uuur
uuur
AB ⊥ BC
Perpendicularity is one property that can help distinguish between, or be used to identify,
describe or classify polygons.
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GM – 7 Parallel lines: Two or more lines (rays, segments) that are the same distance
apart, have no points in common, and therefore, will never meet are said to be parallel
lines.
A
The symbol, , is used
to denote parallelism.
B
suur suur
AB CD
C
D
Parallelism is one property that can help distinguish between, or be used to identify,
describe or classify polygons.
Example 7.1 – Describe parallelograms using properties:
In the figure above, WXYZ is a parallelogram. Which of the
following is NOT necessarily true?
A) Side WX is parallel to side ZY.
B) Side XY is parallel to side WZ.
C) The measures of angles W and Y are equal.
D) The lengths of sides WX and ZY are equal.
E) The lengths of sides WX and XY are equal.
Source: NAEP Grade 8 2003
Answer: E
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G&M – 8 Non-measurable attribute: Non-measurable attributes are attributes that,
in general, are typically not described by a measurement (e.g., the texture of material, a
type of material (wood vs. metal), or the shape of an object).
G&M – 9 Measurable attribute: Measurable attributes are attributes that could be
described by a measurement (e.g., weight, length, height).
Example 9.1 – Sorts or classifies objects by two non-measurable attributes:
Sort these shapes by two attributes.
Answer: Two non-measurable attributes of the shapes include shading/color and type of shape.
Triangles
Circles
Octagons
Spotted
Gray
(Definition G&M – 9 continued on following page)
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Example 9.2 – Sorts or classifies objects by two attributes – one non-measurable
and one measurable attribute:
Sort these shapes by two attributes
Answer: One way to sort these shapes is by coloring (non-measurable) and size (measurable).
Small
Large
Spotted
Gray
G&M – 10 Angles: An angle is formed by two rays that intersect. The point at which
they intersect is called the vertex of the angle. Example 10.1 contains descriptions and
examples of some angles.
Vertex of
the angle
(Definition G&M – 10 continued on following page)
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Table 10.1 – Angles:
Angles
Acute
Definitions
An angle with a measure greater than 0°
and less than 90°
Obtuse
An angle with a measure greater than 90°
and less than 180°
Right
An angle with a measure of 90°
Straight
An angle with a measure of 180°
Adjacent
Two angles that have a common side and a
common vertex that do not overlap
Examples
e.g., ∠1 and ∠2 in the example are one set
of adjacent angles.
Complementary Two angles whose measures total 90°
e.g.,
∠1 and ∠2 are complementary angles
2
1
∠AED and ∠CEB are complementary
angles.
(Definition G&M – 10 continued on following page)
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Angles
Supplementary
Vertical
23
Definitions
Two angles whose measures total 180°
e.g.,
∠DEC and ∠FEG are supplementary
angles.
∠1 and ∠2 in the example are one set of
supplementary angles in this example.
Examples
Non-adjacent angles formed by two lines
(rays, line segments) intersecting
∠1 and ∠3 are vertical angles.
∠2 and ∠4 are vertical angles.
Note: Vertical angles are congruent.
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G&M – 11 Angle relationships formed by two non-parallel lines, or
parallel lines intersected by a transversal: When two non-parallel lines or two
parallel lines are intersected by a transversal a number of angle relationships are formed.
Example 11.1 – Parallel and non-parallel lines cut by a transversal:
Figure A
Figure B
A transversal is a line that
intersects two or more
lines.
Example 11.2 – Some angle relationships formed by parallel and non-parallel
lines being cut by a transversal:
The measures of vertical angles are
equal.
The measures of angles forming a linear
pair total 180°. If you know one angle in
a linear pair, then you can determine the
other angle.
The measures of alternate exterior
angles are equal.
The measures of alternate interior
angles are equal.
The measures of the same-side interior
angles are supplementary (sum to 180°).
The measures of corresponding angles
are equal.
Examples
∠ 1 and ∠ 4
∠ 2 and ∠ 3
∠ 1 and ∠ 2
∠ 5 and ∠ 6
∠ 1 and ∠ 8
∠ 2 and ∠ 7
∠ 3 and ∠ 6
∠ 4 and ∠ 5
∠ 3 and ∠ 5
∠ 4 and ∠ 6
∠ 1 and ∠ 5
∠ 2 and ∠ 6
Applies to Figure
Figure A
Figure B
yes
yes
yes
yes
yes
no
yes
no
yes
no
yes
no
(Definition G&M – 11 continued on following page)
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Example 11.3 – Uses properties of angle relationships formed by two lines cut by
a transversal to solve a problem:
Sam said that he could determine the measure of every angle in Figures A and B
without actually measuring the angles if he knew just one of the angles in each of the
figures.
Is Sam correct? Explain why or why not using an example.
Figure A
Figure B
Answer: Sam is incorrect. You can determine the angles in Figure A if you know the measure of 1
angle, but not figure B. Given the measure of one angle in Figure A all the angles can be
determined. However, because the lines being cut by the transversal are not parallel in Figure B, it
is not possible to determine the measures of the corresponding angles in Figure B.
Figure A: For example, if the measure of ∠ 1 is 120°, then the measure of ∠ 4 is 120° because
the measure of vertical angles is the same. Angles 1 and 2 form a linear pair. Therefore, the
measure of ∠ 2 is 60° (180° – 120° = 60°). Angles 2 and 3 are vertical angles and, therefore, the
measures of their angles are equal. Because alternate interior and alternate exterior angles are
equal when two parallel lines are cut by a transversal, the measure of ∠ 5 is 120°, ∠ 6 is
60°, ∠ 7 is 60°, and ∠ 8 is 120°.
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G&M – 12 Triangle Inequality: The Triangle Inequality states that the sum of the
lengths of any two sides of a triangle is greater than the length of the third side.
Example 12.1 – Solve a problem using the Triangle Inequality:
Can 5 cm, 6 cm, and 12 cm be the sides of a triangle?
Answer: No, according to the Triangle Inequality the sum of the lengths of any two sides must be
greater than the length of the third side.
6 + 12 > 5
5 + 12 > 6
However, 5 + 6 < 12.
Therefore, a triangle can not be constructed using the dimensions 5 cm, 6 cm, and 12 cm.
Example 12.2 – Solve a problem using the Triangle Inequality:
A hotel is 25 miles from Airport A and 40 miles from Airport B. The hotel and
Airports A and B form a triangle. Use this information to determine the range of the
distances that Airport A and Airport B can be apart.
Answer: Since 25 miles + 40 miles = 65 miles. In order for the three segments to form a triangle
the distance between the two airports must be less than 65.
To determine the smallest possible distance between the airports, determine the smallest value that
satisfies the inequality 25 + x > 40, since 40 plus any positive distance will be greater than 25.
25 + x > 40
If x > 15
Therefore, the distance between the two airports is between 15 miles and 65 miles.
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GM – 13 Sum of the measures of the interior angles of polygons: The sum
of the measures of the interior angles in any polygon, S, is equal to the number of sides of
the polygon, n, minus two, multiplied by 180° (S = (n – 2) 180°).
Example 13.1 – Solve a problem using the Sum of the Interior Angles Theorem:
Mr. A’s mathematics class is building a set of benches for a patio that is in the shape
of a regular hexagon. The benches need to be built to meet at the vertices of the
hexagon as shown in the diagram.
Regular polygons
are polygons that are
equilateral and
equiangular.
If ∠a and ∠b are congruent, what is the measure of angle a?
Answer: Since the sum of the measures of the interior angles is 180(6-2) = 720°and each angle
has the same measure (the polygon is regular), the measure of each angle is
o
the benches to meet as shown, the measure of ∠a is 120
27
720o
= 120o . For
6
÷ 2 = 60o .
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G&M – 14 Pythagorean Theorem: The Pythagorean Theorem states – In a right
triangle, the square of the length of the hypotenuse of the triangle is equal to the sum of
the squares of the length of the legs of the right triangle (a2 + b2 = c2, where c is the
length of the hypotenuse in the right triangle, and a and b are the length of the legs of the
right triangle.)
a 2 + b2 = c 2
Example 14.1 – The Pythagorean Theorem modeled with a right triangle with
sides 6 units by 8 units by 10 units:
a2 + b2 = c2
62 + 82 = 102
36 + 64 = 100
(Definition G&M – 14 continued on following page)
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Example 14.2 – Solve a problem using the Pythagorean Theorem:
A store, an airport, and a house form the vertices of a right triangle. The store is at the
vertex with the right angle. The distance from the house to the airport is 20 miles. The
distance from the store to the airport is 12 miles. What is the distance from the house
to the store?
Answer: The distance from the house to the store is 16 miles. The distance from the airport to the
house forms the hypotenuse of the right triangle, and the distance from the store to the airport
forms one leg of the right triangle. Therefore, 122 + b2 = 202; 144 + b2 = 400; b2 = 256; b =
256 or 16 miles.
GM – 15 Three-dimensional shapes: Three dimensional shapes are often described
by the polygon that forms the face(s). The number of edges, faces, and vertices depend on
the type of figure.
(Definition G&M – 14 continued on following page)
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Table 15.1 – Three dimensional figures:
3 – Dimensional
Description
Figure
A prism is a three-dimensional figure
Prism
with two parallel congruent faces,
called bases, and lateral faces in the
shape of parallelograms. Prisms are
named according to the shape of their
bases.
Examples
Oblique Rectangular Prism
Right Hexagonal Prism
Rectangular prism A prism whose bases are rectangles.
Cube
A rectangular prism whose faces are
squares.
Triangular prism
A prism whose bases are triangles
Cylinder
A three-dimensional figure with two
parallel congruent circular faces whose
cross section is circular.
Cone
A cone is like a pyramid with a circular
base.
Sphere
A sphere is the set of all points in space
that are a fixed distance from a
common point called the center.
Pyramids
A three-dimensional figure with one
polygonal base and lateral faces the
shape of triangles that meet at a
common vertex, called the apex.
(Definition G&M – 15 continued on following page)
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Example 15.1 – Applies properties to describe three-dimensional shapes:
The figure below is a triangular prism. Which of the following describes the
triangular prism?
A)
B)
C)
D)
All faces are congruent.
Because it has two triangles as bases, it has six edges.
Because it has two triangles as bases, it has six vertices.
All the angles formed by the edges are congruent.
Answer: C
(Definition G&M – 15 continued on following page)
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Example 15.2 – Use properties to describe three-dimensional shapes:
Below are descriptions of some top, front, and side views of three-dimensional
shapes. For each description identify a three-dimensional shape whose views match
the given description, and sketch a view of it that shows its principle features.
a)
b)
c)
d)
e)
f)
Side and front views are triangles. Top view is a circle.
Side and front views are rectangles. Top view is a circle.
Side and front views are triangles. Top view is a square.
Side and front view are rectangles. Top view is a rectangle.
Side, top, and front views are all congruent squares.
Side, top, and front views are all congruent triangles.
Source: Foundations of Success, Achieve, Inc. 2002
Answer:
32
a)
Cone
b) Cylinder
c)
Square pyramid
d) Rectangular prism
e)
Cube
f) Triangular pyramid
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Section 2: Demonstrates Conceptual Understanding of Congruency;
Demonstrates Conceptual Understanding of Similarity
NECAP: M(G&M) – X – 2, M(G&M) – X – 3
Vermont: MX: 12, MX: 13
Definition
Composing and decomposing shapes
Congruent
Describes the proportional effect on linear
dimensions of polygons or circles when scaling
up or down while preserving the angles of
polygons
Matching congruent figures using reflections,
translations, or rotations
Reflection
Rotation
Similar figures
Solves problems involving scaling up or down
and their impact on angle measure, linear
dimensions, and areas of polygons and circles
when the linear dimensions are multiplied by a
constant factor
Solving problems on a coordinate plane using
reflections, translations, or rotations
Translation
33
Page
Number
40
34
47
Definition
Number
G&M -21
G&M -16
G&M -24
39
G&M -20
36
37
44
49
G&M -17
G&M -19
G&M -23
G&M -25
41
G&M -22
37
G&M -18
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G&M – 16 Congruent: Congruence is often defined as figures that have the same
shape (see similar figures G&M – 23) and size. Because congruent figures have the same
shape and size, they
For figures that do not
• differ only in location;
have “sides”, such as
• can be matched by a rotation translation, or reflection
circles, all distances are
(see G&M –17, 18, and 19), or any combinations of
in a 1:1 ratio (e.g., the
these; and
ratio of the radii
between congruent
• are similar figures whose corresponding sides are in a
circles).
1:1 ratio.
(The symbol ≅ is used to denote congruence.)
Example 16.1 – Congruent line segments:
Line segment AB is congruent to line segment CD ( AB ≅ CD ) since they have the
same length (i.e., they only differ in location).
Example 16.2 – Congruent circles:
Circle R is congruent to Circle S since they have radii of equal lengths (i.e., they only
differ in location).
(Definition G&M – 16 continued on following page)
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Example 16.3 – Congruent triangles:
Triangle ABC is congruent to Triangle DEF since they have the same shape and size
(i.e., they only differ in location). The following diagram shows that Triangle DEF
can be obtained by reflecting Triangle ABC over line segment k and then rotating
Triangle DEF 90o counterclockwise about point B’.
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G&M – 17 Reflection (flip): A reflection over a line is an operation that replaces each
point in a figure with a new point by flipping the original figure over the line of
reflection. The original figure is called the pre-image and the resulting figure is called the
image.
Example 17.1 – Reflecting a figure over a line segment:
CE ≅ C ' E
Line of Reflection
…the pre-image point
and the image point lie
the same distance from
the line of reflection…
k
Pre-image
C
E
C'
D
A
36
Image
B'
B
D'
A'
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G&M –18 Translation (slide): A translation is an operation that replaces each point
in a figure with a new point by sliding all points making up the original figure the same
distance in the same direction. The original figure is called the pre-image and the
resulting figure is called the image.
Example 18.1 – Translating a figure:
Image
C'
Pre-image
B'
D'
C
A'
B
D
A
3 units up
6 units right
Figure A’B’C’D’ is the image of Figure ABCD after translating (sliding) each point of
figure ABCD the same distance in the same direction – 6 units to the right and 3 units
up. The gray lines show how the vertices of Figure ABCD were translated.
G&M – 19 Rotation (turn): A rotation is an operation that replaces each point in a
figure with a new point by turning all points making up the original figure about a fixed
point (called the center of rotation) by the same number of degrees in the same direction
(counterclockwise or clockwise). The original figure is called the pre-image and the
resulting figure is called the image.
(Definition G&M – 19 continued on following page)
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Example 19.1 – Rotating a figure about a point not on the figure:
Triangle A’B’C’ is the image of Triangle ABC after a rotation of 90o
counterclockwise about point O (the center of rotation). The gray line indicates how
point A was rotated about point O.
Center of rotation
A
O
90
C
C'
B
B'
A'
Example 19.2 – Rotating a figure about a point on the figure:
Figure A’B’C’D’ is the image of Figure ABCD after a rotation of 90o clockwise about
point C (the center of rotation). The gray line indicates how point A was rotated about
point C.
A'
B'
D
D'
Center of rotation
90
C = C'
A
B
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G&M – 20 Matching congruent figures using reflections, translations, or
rotations: Matching congruent figures using reflections, translations, or rotations
means to identify if figures are congruent by determining if the figures only differ in
location. This may be accomplished through the use of manipulatives and flipping,
sliding, or turning them (applying reflections, translations, or rotations) to see if the
figures only differ in location.
Example 20.1: Name all pairs of congruent figures.
A
C
B
D
G
F
E
H
I
J
L
K
Answer: Figure A and Figure G are congruent; Figure B and Figure I are congruent; Figure C and
Figure F are congruent; Figure D and Figure J are congruent (these pairs of congruent figures can
be verified by flipping, sliding, or turning manipulatives in the shapes of these figures to
determine if they only differ in location)
Example 20.2 – Demonstrates understanding of congruence:
Quinn states that all squares are congruent. Explain whether or not Quinn is correct.
Answer: Quinn is not correct. Here is an example of two squares that are not congruent since
there is no way to flip, slide, or turn Square A so that it coincides with Square B.
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G&M – 21 Composing or Decomposing shapes: Demonstrates conceptual
understanding of congruency as a result of composing or decomposing shapes means to
identify or create congruent figures by putting shapes together with no gaps or overlaps
or by taking shapes apart.
Example 21.1 – Composing/decomposing shapes to create a figure:
Refer to the following information.
1
2
3
Triangles 1, 2, and 3 shown above can be arranged with no overlap to form either of
the following figures.
3
2
1
3
2
1
Draw lines on the figure below to show how triangles 1, 2, and 3 can be rearranged
without overlap to form this rectangle.
Source: NAEP 2003 Released Item – Grade 8
Answer:
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G&M – 22 Solving problems on a coordinate plane involving reflections,
translations, or rotations: Refers to any problems involving reflections,
translations, or rotations that occur on the coordinate plane.
Example 22.1 – Describe how to translate a figure:
Describe how to translate Figure A to obtain Figure B
Answer: Translate each point on Figure A 7 units to the right and 3 units down to obtain Figure B.
(Definition G&M – 22 continued on following page)
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Example 22.2 – Solving problems using translation on a coordinate plane:
The Translation Game
The Translation game is played by moving an object (square) across the coordinate
plane from a start position until it is touching some part of the end post using a series
of translations (slides).
The distance that the object (square) is translated is based upon the values rolled by
using two number cubes. The first three rolls are in the following table. You job is to
determine values for two additional rolls that will result in the object (square)
touching some part of the circle (End).
Show all moves on the board. Show that the points were being translated
the correct distances.
Number Cube Roll
1
2
3
4
5
Move to Right
Move Up
(Number of First Cube)
(Number of Second Cube)
5
3
5
3
4
1
(Definition G&M – 22 continued on following page)
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Sample Answer: In this solution the final two translations are 6 units right and 6 units up, and
then 5 units right and 3 units up.
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G&M – 23 Similar figures: Two figures are similar if one is a
magnification or reduction of the other. The two figures have
corresponding angles that have the same measures, and the lengths of the
corresponding sides are proportional (for figures that do not have “sides”
such as circles, all distances are increased or decreased in the same ratio).
Congruent figures
are similar figures –
they can be thought of
as having a
magnification ratio of
1:1.
We often say that similar figures have the same shape but not necessarily the same size
(if they have the same size, then they are similar figures whose corresponding sides are in
a 1:1 ratio and are therefore congruent).
Example 23.1 – Similar triangles:
D
Sides AB and DE are
corresponding sides in a
2:1 ratio.
10 cm
6 cm
A
5 cm
3 cm
B
4 cm
E
8 cm
F
C
Triangle ABC is similar to Triangle DEF since Triangle DEF is an enlargement of
Triangle ABC by a factor of 2. That is, the measure of angle A is equal to the measure
of angle D, the measure of angle B is equal to the measure of angle E, the measure of
DE EF DF 2
=
=
= .
angle C is equal to the measure of angle F, and
AB BC AC 1
…the lengths of
corresponding
sides are
proportional
(Definition G&M – 23 continued on following page)
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Example 23.2 – Applies concepts of similarity:
Paul states that all rectangles are similar. Explain how you know whether or not Paul
is correct.
Answer: Paul is not correct. Even though the corresponding angles of two rectangles are equal in
o
measure (since all angles in a rectangle measure 90 ), their corresponding sides are not necessarily
in proportion as the following diagram illustrates.
This example shows that
not all rectangles have
the “same shape.” Two
rectangles have the same
shape if and only if they
are similar.
It is not possible to pair vertices in such
a way as to have the ratio of
corresponding sides proportional, since
the ratio of corresponding sides in
rectangle ABCD to rectangle EFGH is
2:1 for two sides and 1:1 for the
remaining sides.
(Definition G&M – 23 continued on following page)
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Example 23.3 – Applies scale on map:
Look at the maps of the United States and Australia.
John looked at the two maps and said, “The distance from Brisbane, Australia to
Carnarvon, Australia is about half the distance it is from Washington, D.C. to San
Francisco, California.”
Is John correct?
Explain your answer.
Carnarvon
Australia
Brisbane
San Francisco
Washington, D.C.
Answer:
Answer: John is incorrect. John compared the size of the maps, not the distances based on the
scales on the maps. The scale on the U.S. map is 1.4 cm = 400 miles. The scale on the Australian
map is 1 cm = 500 miles. The distances between the two sets of cities is about the same (within 130
miles): Brisbane, Australia to Carnarvon, Australia ≈ 2400 miles ( 1 cm = 4.7 cm ; x
500 miles
x
1.4 cm
9 cm ;x
=
400 miles
x
≈ 2350 miles); Washington, D.C. to San Francisco, California ≈ 2600 miles (
≈ 2571 miles).
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G&M – 24 Describes the proportional effect on the linear dimensions of
polygons or circles when scaling up or down while preserving the angles
of polygons (See G&M – 10): To determine how linear measures (e.g., length, width,
perimeter) are impacted by magnifying or reducing a figure.
Example 24.1 – Impact of scaling on perimeter:
Rectangle ABCD is similar to Rectangle EFGH. How many times as great is the
perimeter of Rectangle EFGH compared to the perimeter of Rectangle ABCD? Show
your work or explain how you know.
E
H
A
B
D
C
3 units
12 units
4 units
F
G
Answer: 3; Since Rectangle ABCD is similar to Rectangle EFGH and the ratio of corresponding
sides in Rectangle EFGH to Rectangle ABCD is 3:1 (
EF
=
12
=
3
), each side in Rectangle
4 1
AB
EFGH is three times as great as its corresponding side in Rectangle ABCD. Therefore, the
perimeter of Rectangle EFGH is three times as great as the perimeter of Rectangle ABCD
42
3
( = ).
1
14
(Definition G&M – 24 continued on following page)
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Example 24.2 – Impact of scaling up on circumference:
If the radius of Circle A is doubled, how many times as great is the circumference of
the new circle compared to the circumference of Circle A? Show your work or
explain how you know.
A
Answer: 2; If the circumference of Circle A is 2π r , then the circumference of the new circle
is 2π (2r ) or 4π r . Therefore, the ratio of the circumference of Circle A to the new circle is 2:1.
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G&M – 25 Solves problems involving scaling up or down and their impact
on angle measures, linear dimensions and areas of polygons and circles
when the linear dimensions are multiplied by a constant factor (See G&M
– 9): To determine how linear measures (e.g., length, width, perimeter) and areas are
impacted by magnifying or reducing a figure.
Example 25.1 – Solving a problem involving scaling up:
The following is a model of a sign being built for a school, and the final sign. Both
the model and the final sign are made out of wood.
Key: Each
is 1 foot x 1 foot.
Model
Final Sign
A) Are the model and the final sign similar figures? Explain your answer.
B) How many times as great is the perimeter of the final sign than the model?
Explain your answer.
C) Is this statement true? The amount of wood (area) needed to build the two signs is
in the same proportion as the length of the corresponding sides of the sign.
Explain your answer.
(Definition G&M – 25 continued on following page)
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Solution:
A) Yes, the model and the final sign are similar figures. The two figures have corresponding
angles that have the same measures and the lengths of the corresponding sides are proportional. In
this example all the angles are right angles.
Height final sign
HeightModel
=
LengthFinal sign 24 2
12 2
= and
=
=
LengthModel
6 1
12 1
B) The perimeter of the final sign is 2 times as great as the perimeter of the model since the ratio
of the corresponding sides is 2:1.
C) No, this statement is incorrect. The area of the final sign is 4 times as great as the area of the
model since the A = lw and both the length and width of the model are doubled.
Example 25.2 – Solves problem that show the impact of scaling up on perimeter:
Trapezoid ABCD is similar to Trapezoid EFGH. How many times as great is the
perimeter of Trapezoid ABCD compared to the perimeter of Trapezoid EFGH? How
many times as great is the area of Trapezoid ABCD compared to the area of
Trapezoid EFGH? Show your work or explain how you know.
F
G
C
B
5 units
A
4 units
D
E
H
8 units
Answer: 2; The perimeter of Trapezoid EFGH is two times as great as the perimeter of Trapezoid
ABCD, since the ratio of corresponding sides of Trapezoid EFGH to Trapezoid ABCD is 2:1.
Therefore, each side in Trapezoid EFGH is twice as large as its corresponding side in Trapezoid
ABCD. The area of Trapezoid EFGH is four times as great as the area of Trapezoid ABCD.
To determine the area of a trapezoid ABCD you would apply the formula
1
2
h (b1 + b2 ) , where h is
the height of the trapezoid, b1 is one base, and b2 is the second base.
Therefore,
If the area of Trapezoid ABCD is
1
2
h (b1 + b2 ) , then the area of Trapezoid EFGH is
1
1
(2h)(2b1 + 2b2) = 4( h(b1 + b2)), which is four times the area of Trapezoid ABCD.
2
2
(Definition G&M – 25 continued on following page)
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Example 25.3 – Impact of scaling up on area:
If the radius of Circle A is doubled, how many times as great is the area of the new
circle compared to the area of Circle A? Show your work or explain how you know.
A
2
Answer: 4; If the area of Circle A is π r , then the area of the new circle
is π (2 r ) = 4π r , which is four times the area of Circle A.
2
2
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Section 3: Demonstrates understanding of perimeter, area, volume, and
surface area…; Measures and uses units of measures appropriately and
consistently…
NECAP: M(G&M) – X – 6, NECAP: M(G&M) – X – 7
Vermont GEs: MX: 14, MX: 15
Definition
Demonstrates conceptual understanding of
perimeter and area of polygons and irregular
figures on grids
Demonstrates conceptual understanding of
perimeter and area using models and
manipulatives to surround and cover polygons
Demonstrates conceptual understanding of
perimeter, area, volume, or surface area using
models and manipulatives
Demonstrates conceptual understanding of
perimeter, area, volume, or surface area by
solving problems
Demonstrates conceptual understanding of the
relationships of circle measures by solving
related problems
Makes conversions within and across systems
Measures and uses units of measure
appropriately and consistently
52
Page
Number
54
Definition
Number
G&M -27
53
G&M -26
56
G&M -28
57
G&M -29
59
G&M -30
61
60
G&M -32
G&M -31
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G&M – 26 Demonstrates conceptual understanding of perimeter and area
using models and manipulatives to surround and cover polygons: See
examples.
Example 26.1 – Demonstrates conceptual understanding of area by covering
polygons using manipulatives:
This is a model of a floor of a room. One tile has been placed on the floor.
How many tiles will cover the floor without spaces or overlap? (Students would be
given a tile to use to solve this problem.)
Answer: 24 tiles
Example 26.2 – Demonstrates conceptual understanding of perimeter by
surrounding figures:
About how many pieces of string 3 cm long can fit around this shape? (The student is
given a 3 cm long piece of string.)
Answer: About 6 ½
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G&M – 27 Demonstrates conceptual understanding of perimeter and area
of polygons and irregular figures on grids:
Example 27.1 – Finds area and perimeter of a polygon on a grid:
What is the area of this rectangle?
Key
represents 1 square unit
Solution: 84 square units
Example 27.2 – Finds area and perimeter of irregular shape on a grid:
Find the area of the pond.
(Definition G&M – 27 continued on following page)
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Answer: One strategy to estimate the area of this irregular shape is to first identify and total the
full square units – 5 square units as pictured in Figure 1 below. Then most elementary students
will start combining partial boxes with each other. Using this strategy, students arrive at an answer
of about 10 square. The solution below, however, models a Greek method involving successive
approximation that examines the upper bounds and lower bounds, and improves this estimation by
partitioning the area into successively smaller parts.
Figure 1: Count the number of
squares lying completely in the region
to obtain a lower estimate. (5 square
units)
Figure 2: Count the number of squares
containing any portion of the region. (20 square
units)
Using this size grid you know that the area of the region is between 5 square units and 20 square
units.
Figure 3: To increase the accuracy of this
estimate, you can break each original grid into
four squares. Count the number of squares
lying completely within the region to obtain the
lower bound. (7 square units)
Figure 4: The number of squares that cover
some portion of the pond is 13 square units.
Using this size grid you know that the area of the region is between 7 square units and 13 square
units.
If you continue to partition the squares into smaller squares, you will narrow the gap between the
upper and lower bound and successively obtain better approximations to the area of the pond.
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G&M- 28 Demonstrates conceptual understanding of perimeter, area,
volume, or surface area using models and manipulatives:
Example 28.1 – Demonstrates understanding of surface area of a cylinder using
a model:
John made two sketches of a net for a cylinder. Which sketch most accurately
represents the net of the cylinder? Explain your answer.
Sketch A
Sketch B
Answer: Although neither sketch is completely accurate, sketch A best represents a net of a
cylinder. Both sketches model the shape of the sides of the can when it is stretched out forming a
rectangle. However, since the length of the rectangle (representing the side of the cylinder) is
equal to the circumference of each circle forming the ends of the can, the ratio of diameter of the
circular region to the length of the rectangular regions should be about 1:3 (C = 3.14d). The length
of the rectangle in sketch A is 15 units, and the diameter of the circle is about 5 units making a 1:3
ratio. The length of the diameter of the circle is about 7.5 units and the length of the rectangle in
sketch B is 15 units,. The ratio of the diameter of the circle to the length of the rectangle is 1:2 for
Sketch B, not a 1:3 ratio as in Sketch A.
(Definition G&M – 28 continued on following page)
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Example 28.2 – Demonstrates understanding of perimeter and area using a
model:
Look at the rectangle on the grid below. Sketch 2 other rectangles that have the same
area as the rectangle on the grid, but a different perimeter.
Key
represents 1 square unit
Answer: A sketch of 1 unit x 20 unit rectangle, and a 2 unit x 10 unit rectangle.
G&M – 29 Demonstrates conceptual understanding of perimeter, area,
volume, or surface area by solving problems:
Example 29.1 – Solves a problem involving volume:
A rectangular tank 30 cm long and 10 cm wide is filled with water to a depth of 6 cm.
When an object was put into the tank it sank to the bottom of the tank and the water
level rose 3 cm. What is the volume of the object?
Answer: The volume of the object is 900 cm3. The volume of the water in the tank before the
object is put into the tank is (30)(10)(6) = 1800 cm3. The volume of the water and object is
(30)(10)(9) = 2700 cm3. The difference between 2700 cm3 and 1800 cm3 represents the volume of
the object.
(Definition G&M – 29 continued on following page)
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Example 29.2 – Solves a problem involving perimeter:
If both the square and the triangle above have the same perimeter, what is
the length of each side of the square?
A) 4
B) 5
C) 6
D) 7
Source: NAEP Grade 4, 1996
Answer: B
Example 29.3 – Solves a problem involving area:
A rectangular carpet is 9 feet long and 6 feet wide. What is the area of the
carpet in square feet?
A) 15
B) 27
C) 30
D) 54
Source: NAEP Grade 4, 1992
Answer: D
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Example 29.4 – Solves a problem involving surface area:
A cylindrical water tank is made out of steel. The height of the tank is 10 feet. The
radius of the bases is 5 feet. How many square feet of steel is used to make the tank?
Answer: To determine the amount of steel needed to make the tank you must determine the
surface area of the cylinder.
Scylinder = 2 π r 2 + π dh
Scylinder = 2 π 52 + π (10)(10)
Scylinder = 471 ft2
G&M – 30 Demonstrates conceptual understanding of the relationships of
circle measures by solving related problems: To solve problems that involve
an understanding of the relationship between circle measures such as radius to diameter,
and diameter to circumference.
Example 30.1 – Solves problems involving the relationship between circle
measures:
Point O is the center of this circle.
If AB = 2 cm, which answer most closely estimates the circumference of the
circle?
A) 4 cm
B) 6 cm
C) 8 cm
D) 10 cm
Answer: B
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G&M – 31 Measures and uses units of measure appropriately and
consistently: There are two aspects to this GLE. One has to do with measurement, and
the other has to do with using appropriate units given a problem situation. Appendix A of
the GLEs contains Benchmark Measurements for each grade.
Example 31.1 – Measures accurately:
What is the length of the toy car pictured below?
5
8
Answer: 2 inches
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G&M – 32 Makes conversions within and across systems: Uses a conversion
factor (12 in. in 1 foot) to make conversions within and across systems. The emphasis is
on the use of the conversion factor.
Example 32.1 – Converts within systems:
This chart shows how Sam used his time after school.
Activity
Snack
Homework
TV
Playing
Time (minutes)
15 minutes
45 minutes
30 minutes
60 minutes
How much time is this in hours?
1
2
Answer: 2 hours
Example 32.2 – Converts within systems:
Rich bought 5.5 lb of apples. There are 16 ounces in 1 pound.
Which of the following choices is the same as 5.5 lb?
A)
B)
C)
D)
85.0 ounces
88.0 ounces
80.5
5 pounds 5 ounces
Answer: B
Example 32.3 – Converts across systems:
Marathons are about 26 miles long. About how many kilometers are marathons? One
mile is equivalent to 1.61 kilometers.
Answer:
61
26 miles 1.61 km
•
= 41.86 km
1
1 mile
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References:
Agrawal, Piyush C. (1993). Mathematics Applications and Connections. Lake Forest,
IL: Macmillan/McGraw-Hill.
Billstein, Rick, and Williamson, Jim (1999). MATH Thematics. Evanston, Illinois:
McDougal Littell.
Borowski, E.J., and Borwein, J.M. (1991). The HarperCollins Dictionary of
Mathematics. New York, NY: HarperCollins Publishers.
Cavanagh, Mary C. (2000). Math To Know. Wilmington, MA: Great Source
Education Group.
Lamon, S.J., Teaching Fractions and Ratios for Understanding, Lawrence Erlbaum
Associates, Mahwah, New Jersey.
Licker, Mark D. (2003). Dictionary of Mathematics. New York, NY: McGraw Hill.
Lilly, Marsha (1999). Math at Hand. Wilmington, MA: Great Source Education
Group.
Mathworld, Wolfram, http://mathworld.wolfram.com/
Massachusetts Department of Education, MCAS, 2003 Release Item
National Center for Health Statistics Growth Chart
http://www.cdc.gov/nchs/data/nhanes/growthcharts/set2/chart%2007.pdf
National Assessment of Educational Progress release items
http://nces.ed.gov/nationsreportcard/itmrls/
Nelson, David (2003). The Penguin Dictionary of Mathematics. London, England:
Penguin Books.
University of Chicago School Mathematics Project (1993). Geometry. Glenview, IL:
ScottForesman
University of Chicago School Mathematics Project (2004). Everyday Mathematics;
Teacher’s Reference Manual. Columbus, OH: McGraw Hill SRA
Usiskin, Zalman (1992). Transition Mathematics. Glenview, IL: Scott, Foresman.
Van De Walle, John A. (2001). Elementary and Middle School Mathematics: Teaching
Developmentally. New York, NY: Addison Wesley Longman, Inc.
Vermont Institute of Science Math and Technology (1995). Grade Eight Guide to
Diversity of Content and Problem Solving Tasks.
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