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Module 1: Topic A (slides 14-20)
Lesson 1: Construct an Equilateral Triangle
Opening Exercise
Joe and Marty are in the park playing catch. Tony joins them, and the boys want to stand so that the distance between
any two of them is the same. Where do they stand?
How do they figure this out precisely? What tool or tools could they use?
Fill in the blanks below as each term is discussed:
1.
The _______ between points π΄ and π΅ is the set consisting of π΄, π΅, and all points
on the line β‘π΄π΅ between π΄ and π΅.
2.
A segment from the center of a circle to a point on the circle.
3.
Given a point πΆ in the plane and a number π > 0, the _______ with center πΆ and
radius π is the set of all points in the plane that are distance π from the point πΆ.
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Example 1: Sitting Cats
You will need a compass and a straightedge.
Margie has three cats. She has heard that cats in a room position themselves at equal distances from one another and
wants to test that theory. Margie notices that Simon, her tabby cat, is in the center of her bed (at S), while JoJo, her
Siamese, is lying on her desk chair (at J). If the theory is true, where will she find Mack, her calico cat? Use the scale
drawing of Margie’s room shown below, together with (only) a compass and straightedge. Place an M where Mack will
be if the theory is true.
Bookcase
Recliner
Rug
Table
J
Small rug
Desk
Bed
S
Chair
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Example 2: Euclid, Proposition 1
Let’s see how Euclid approached this problem. Look at this venerable Greek’s very first proposition and compare his
steps with yours.
In this margin, compare your steps with Euclid’s.
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Geometry Assumptions
In geometry, as in most fields, there are specific facts and definitions that we assume to be true. In any logical system, it
helps to identify these assumptions as early as possible, since the correctness of any proof we offer hinges upon the
truth of our assumptions. For example, in Proposition 1, when Euclid said, “Let π΄π΅ be the given finite straight line,” he
assumed that, given any two distinct points there is exactly one line that contains them. Of course, that assumes we have
two points! Best if we assume there are points in the plane as well: Every plane contains at least three non-collinear
points.
Euclid continued on to show that the measures of each of the three sides of his triangle were equal. It makes sense to
discuss the measure of a segment in terms of distance. To every pair of points π΄ and π΅ there corresponds a real number
πππ π‘(π΄, π΅) ≥ 0, called the distance from π΄ to π΅. Since the distance from π΄ to π΅ is equal to the distance from π΅ to π΄, we
can interchange π΄ and π΅: πππ π‘(π΄, π΅) = πππ π‘(π΅, π΄). Also, π΄ and π΅ coincide if and only if πππ π‘(π΄, π΅) = 0.
Using distance, we can also assume that every line has a coordinate system, which just means that we can think of any
line in the plane as a number line. Here’s how: given a line πΏ, pick a point π΄ on πΏ to be “0” and find the two points π΅ and
πΆ such that πππ π‘(π΄, π΅) = πππ π‘(π΄, πΆ) = 1. Label one of these points to be 1 (say point π΅), which means the other point πΆ
corresponds to -1. Every other point on the line then corresponds to a real number determined by the (positive or
negative) distance between 0 and the point. In particular, if after placing a coordinate system on a line, if a point π
corresponds to the number π, and a point π corresponds to the number π , then the distance from π
to π is πππ π‘(π
, π) =
|π − π |.
History of Geometry: Examine the site http://geomhistory.com/home.html to see how geometry developed over time.
Relevant Vocabulary
Geometric Construction: A geometric construction is a set of instructions for drawing points, lines, circles and figures in
the plane.
The two most basic types of instructions are:
1.
Given any two points π΄ and π΅, a ruler can be used to draw the line π΄π΅ or segment π΄π΅.
2.
Given any two points πΆ and π΅, use a compass to draw the circle that has center at πΆ that passes through π΅
Μ
Μ
Μ
Μ
.)
(Abbreviation: Draw circle: center πΆ, radius πΆπ΅
Constructions also include steps in which the points where lines or circles intersect are selected and labeled.
(Abbreviation: Mark the point of intersection of the lines π΄π΅ and ππ by π, etc.)
Figure: A (2-dimensional) figure is a set of points in a plane.
Usually the term figure refers to certain common shapes like triangle, square, rectangle, etc. But the definition is broad
enough to include any set of points, so a triangle with a line segment sticking out of it is also a figure.
Equilateral Triangle: An equilateral triangle is a triangle with all sides of equal length.
Collinear: Three or more points are collinear if there is a line containing all of the points; otherwise, the points are noncollinear.
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Length of a Segment: The length of the segment π΄π΅ is the distance from π΄ to π΅, and is denoted π΄π΅. Thus, π΄π΅ =
πππ π‘(π΄, π΅).
In this course, you will have to write about distances between points and lengths of segments in many if not most
homework problems. Instead of writing πππ π‘(π΄, π΅) all of the time, which is rather long and clunky notation, we will
instead use the much simpler notation π΄π΅ for both distance and length of segments.
Here are the standard notations for segments, lines, rays, distances, and lengths:
ο§
A ray with vertex π΄ that contains the point π΅:
π΄π΅ or “ray π΄π΅”
ο§
A line that contains points π΄ and π΅:
β‘π΄π΅ or “line π΄π΅”
ο§
A segment with endpoints π΄ and π΅:
π΄π΅ or “segment π΄π΅”
ο§
The length of segment π΄π΅:
π΄π΅
ο§
The distance from π΄ to π΅:
πππ π‘(π΄, π΅)
Coordinate System on a Line: Given a line π, a coordinate system on π is a correspondence between the points on the
line and the real numbers such that (i) to every point on π there corresponds exactly one real number, (ii) to every real
number there corresponds exactly one point of π, and (iii) the distance between two distinct points on π is equal to the
absolute value of the difference of the corresponding numbers.
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Topic B: Unknown Angles (slides 21-24)
Lesson 8, Exercise 2:
Lesson 9, Discussion: What can be established about x, y, and z?
Μ
Μ
Μ
Μ
β₯ Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
β₯ πΈπΉ
Μ
Μ
Μ
Μ
. Prove that π∠π΄π΅πΆ = π∠π·πΈπΉ.
Lesson 10, Problem Set 1: π΄π΅
π·πΈ and π΅πΆ
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Topic C: Transformations/Rigid Motions (slides 25-34)
Lesson 14, Exploratory Challenge:
β³ π΄π΅πΆ is reflected across π·πΈ and maps onto β³ π΄′π΅′πΆ′.
Use your compass and straightedge to construct the
perpendicular bisector of each of the segments connecting π΄
to π΄′, π΅ to π΅′, and πΆ to πΆ′. What do you notice about these
perpendicular bisectors?
Reflections and the Perpendicular Bisector:
Lesson 14, Example 1:
Construct the segment that represents the line of reflection
for quadrilateral π΄π΅πΆπ· and its image π΄′π΅′πΆ′π·′.
What is true about each point on π΄π΅πΆπ· and its corresponding
point on π΄′π΅′πΆ′π·′ with respect to the line of reflection?
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Definition of Reflection:
For a line π in the plane, a reflection across π is the transformation ππ of the plane defined as follows:
For any point π· on the line π, ππ (π·) = π·, and
2. For any point π· not on π, ππ (π·) is the point πΈ so that π is the perpendicular
bisector of the segment π·πΈ.
1.
Lesson 14, Exit Ticket:
1. Construct the line of reflection for the figures.
2. Reflect the given figure across the line of reflection
provided.
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Topic D: Congruence (slides 36-40)
Lesson 22, Discussion
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Example from Mid-Module Assessment
Μ
Μ
Μ
Μ
Given in the figure below, line π is the perpendicular bisector of π΄π΅
Μ
Μ
Μ
Μ
.
and of πΆπ·
Μ
Μ
Μ
Μ
≅ π΅π·
Μ
Μ
Μ
Μ
using rigid motions.
a. Show π΄πΆ
b. Show ∠π΄πΆπ· ≅ ∠π΅π·πΆ.
c.
Μ
Μ
Μ
Μ
β₯ πΆπ·
Μ
Μ
Μ
Μ
.
Show π΄π΅
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Module 2 Topic A: Scale Drawings (slides 46-52)
For π > 0, a dilation with center π and scale factor π is a transformation π·π,π of the
plane defined as follows:
1. For the center π, π·π,π (π) = π, and
2. For any other point π, π·π,π (π) is the point π on
the ray ππ so that |ππ′| = π β |ππ|.
Lesson 2, Example 2
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Lesson 3, Example 2
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Topic B: Dilations (slides 53-54)
Lesson 7, Example 3
Topic C: Similarity and Dilations (slides 55-64)
Lesson 12, Example 1
Z
Z'
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Lesson 15, Discussion
Lesson 15, Exercise 9
x
3.14
4
12
y
16.5
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Lesson 15, Problem Set 4
Μ
Μ
Μ
Μ
, Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
, and Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
, prove that the triangles are similar.
If Μ
Μ
Μ
Μ
π¨πͺ||π¬π«
π¨π©||π¬π
πͺπ©||π«π
Lesson 17, Discussion
Theorem (SAS). Given two triangles β³ π΄π΅πΆ and β³ π΄’π΅’πΆ’ so that
π΄′π΅′
π΄π΅
=
π΄′πΆ ′
π΄πΆ
and π∠π΄ = π∠π΄′ then the triangles are
similar, β³ π΄π΅πΆ ~ β³ π΄′ π΅′ πΆ ′.
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Lesson 17, Discussion
Theorem (SSS). Given two triangles β³ ABC and β³ A’B’C’ so that
A′ B′
AB
=
B′ C′
BC
=
A′ C′
AC
then the triangles are similar, β³
ABC ~ β³ A′ B ′ C ′ .
Topic D: Applying Similarity to Right Triangles (slides 65-73)
Lesson 21, Example 1
How can we show that β³ π΄π΅πΆ~ β³ π΅π·πΆ~ β³ π΄π·π΅?
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Lesson 21, Example 2
Lesson 22, Discussion
How can the ratios in right triangles (e.g. shorter leg:hypotenuse) be used to prove the Pythagorean Theorem?
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Which right triangle ratios can be used to express a in terms of other variables?
Topic E: Trigonometry (slides 74-81)
Lesson 27, Opening Exercise
B
β
a
c
What conclusions can be drawn based on each of the ratios?
α
A
b
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Lesson 27, Example 2
Imagine that the hypotenuse has a fixed length of π = 1 in β³ π΄π΅πΆ.
ο·
What happens to the value of the sine ratio and cosine ratio as π
approaches 0Λ?
ο·
What happens to the value of the sine ratio and cosine ratio as π
approaches 90Λ?
Lesson 27, Example 2
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Lesson 28, Exercise 6
Lesson 29, Example 2
From an angle of depression of 40Λ, John watches his friend approach his building while standing on the rooftop. The
rooftop is 16 meters from the ground, and John’s eye level is at about 1.8 meters from the rooftop. What is the
distance between John’s friend and the building?
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