Download NC–PIMS Geometry 6-12 - MELT-Institute

NC–PIMS Geometry 6-12
Day 4



circles: tangents, inscribed angle theorem and consequences
curves of constant width
Pythagorean theorem and similar right triangles
Block 1: Short Sighted
D4.B1.A1
Work in small groups. Strategize methods to solve the problem and determine salient
concepts and needed information. After sketching out and discussing solution strategies,
attempt to model and solve the problem using Geometer’s Sketchpad.
Problem
Joe is 5’8” tall. He is standing at the shore and looking out over the ocean at the
horizon. He then stands on a rock 18” high. How much farther can he see to the
horizon than before?”
Instructor Notes
This problem allows students to investigate concepts by first deciding how to
model the situation. The instructor should not intervene too quickly and should
allow groups of students to deeply contemplate their problem-solving techniques.
Students will typically create a circle with radii. However, it is recommended that
the instructor create the circle by using the perpendicular bisectors of two sides of
a triangle. This will allow the circle to be flattened out to better model the Earth.
The sketch which students should reproduce should be similar to that below.
Note that the points Eye 1 and Eye 2 should be independent and one should not
move if the other does.
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NC-PIMS Geometry 6-12
1
To create this diagram, students will need to be able to create a tangent to a circle
from a given point outside the circle. However, this skill should not be
immediately shown to the students. The instructor should allow students to make
incorrect tangent lines which become secants and do not intersect the radii at right
angles.
To create a tangent to a circle from a point outside the circle, find the midpoint of
the line segment from the center of the circle to the point outside the circle (Point
C on AB). From this midpoint, create a circle to point B. The intersection of
these circles (Point D) is the point of tangency. See below.
In order to complete the problem, certain information must be known. It is
necessary for the instructor to provide facts 1-3 after a few minutes of
investigation. However, points 4-6 should be delayed and students allowed to
hypothesize whether the angle at the point of tangency is a right angle.
1.
2.
3.
4.
5.
6.
Day 4
Average radius of the Earth is 3960 miles.
5280 ft = 1 mile.
12 inches = 1 foot.
The triangles formed at the points of tangency are right triangles.
The Pythagorean Theorem
Dimensional Analysis.
NC-PIMS Geometry 6-12
2
This means that the radius of the Earth is 250,905,600 inches. Some students will
choose 68 inches as Joe’s eye height and others will choose a smaller number,
noting that the eyes are not on the top of his head. Let us say that the height
chosen by the students is x inches. Then the length of the tangent from Joe’s eyes
to the horizon is l  (250905600  x) 2  2509056002 . (Note, l is approximately
184,725 inches or 15,394 ft., depending on x)
After finding this length, students must redo the problem with the new height,
(250,905,600 + x + 18) inches.
At this point, students have found the length of the two straight lines of sight.
Ask them to compare their findings with the arc length. To do this, they will need
to find the central angles formed using an inverse trigonometric function.
l
l
Since tan  
,   tan 1
.
250905600
250905600
Note that θ will be very small.
Using θ, students should be able to find the arc length along the Earth.
Students should then be asked why the arc length is so close to the sight line
length. At this point, the circle created from the perpendicular bisectors can be
significantly flattened out and students can see that these measures indeed should
be very close.
Much of the value of this problem is in modeling geometric scenarios and
realizing the number of theorems which can be used to solve a real world problem
Questions which can accompany the student’s experience are:
Meta-Problem Solving. Consider the following questions:
How did you solve the problem?
Could you have solved it another way?
What information did you need? What mathematical concepts did
you need to know/use?
Could your students have answered this question? If not, how
could you have modified the question for them to be able to
come to a solution?
If you measured the line of sight as straight from the eyes to the
horizon, what is the difference in length if you measured
along the surface of the Earth (along an arc) instead?
Consider the additional questions:
If Joe stood in a lighthouse 50’ above the ground, how far to the
horizon could he see?
If Joe stood on top of Mt. Katahdin in Maine, how far to the
horizon could he see? (Web Quest for height)
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NC-PIMS Geometry 6-12
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If Joe is at 35,000’ in a jet, how far to the horizon can he see?
Is there a limit to how far to the horizon a person could see? Can
you demonstrate this on GSP?
The following mathematical concepts will be discussed/covered through this
experience:
Problem Solving and Collaboration
Radius of the Earth
Dimensional Analysis & Conversion
Tangents to Circles
Pythagorean Theorem
Inverse Trigonometric Functions
Length of an Arc on a Circle
Modeling
Geometric Constructions using Sketchpad
Conjecturing
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Block 2: Circles
D4.B2.A1
Find the length x.
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D4.B2.A2
Find x in the following figures. Use similar triangles to justify your answers.
2
x
2.25
6.7
10
7
x
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NC-PIMS Geometry 6-12
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D4.B2.A3
Find the distance CP and the radius of the circle. Note that segment CP is perpendicular
to segment AB.
C
3
A
4
P
B
O
6
D
Generalize: Express the square of the radius in terms of x, y, z, and w.
x
z
w
y
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Instructor Notes
Activities 1 and 2 illustrate three circle theorems which are based on similar
triangles and are consequences of the Inscribed Angle Theorem.
The Inscribed Angle Theorem is important and also easy to prove. The key is to
break it into three cases.
Case 1: One chord is a diameter. Note that you will need to remind
students that the measure of a minor arc is defined to be the measure of the
corresponding central angle.
A
x
x
B
O
C
Angles B and A are congruent and angle AOC is an exterior angle, so two
times measure of B is x degrees. Therefore, measure B is half x.
Case 2: Center O is in the interior of angle ABC. Split angle ABC with a
diameter and apply case 1 twice.
Case 3: Center O is not in the interior of angle ABC. Again use case 1
with a diagonal and subtract.
Point out the important special case: an angle inscribed in a semi-circle is a right
angle.
The GSP file “Inscribed Angle.gsp” provides a demonstration for the Inscribed
Angle Theorem.
Day 4
NC-PIMS Geometry 6-12
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Circle Theorems: Angles
C
B
y
x
A
mA =
1
2
x + y 
The key is that marked angle at A is an exterior angle of triangle ABC.
B
x
y
A
D
m A=
C
1
2
x - y 
The key is that angle BDC is an exterior angle of triangle ABD.
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NC-PIMS Geometry 6-12
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D
B
O
x
A
m  BAC =
C
1
2
x
The key is that arc DB measures 180 – x.
Circle Theorems: Lengths
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NC-PIMS Geometry 6-12
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A
C
D
P
AP * PB = CP * PD
B
A
AP * PB = CP * PD
B
C
D
P
A
AP * PB = PC2
B
C
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NC-PIMS Geometry 6-12
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11
The key to the problems in activity D4.B2.A3 is that you can find the distance
from the center to the horizontal chord, then use the Pythagorean theorem to find
the radius.
Block 3: Why Are Manhole Covers Round?
The reason is simple: We do not want the cover to be able to drop through the hole. A circle, because it is
the same width in every direction, cannot be dropped through the opening. Demonstrate why this property
would not hold if the cover were a square, a triangle, or a rectangle. Would it work if the manhole cover
were a regular hexagon? Is it possible to find shapes other than a circle that have the same "width" in every
direction? Figures with this property are called curves of constant width, and their study leads to some
interesting mathematics and to some alternative manhole cover shapes.
Curves of Constant Width
Any closed curve in a plane can be supported by two parallel lines. The parallel lines of
support may be tangent (as in the circle) or not tangent (as in the quadrilateral).
The distance between the parallel lines of support is called the width of the figure for a
given direction. In the circle, the width is the same in every direction (it is equal to the
diameter of the circle). Therefore, the circle is said to have constant width. Clearly, the
quadrilateral does not have constant width (show why).
However, a curve does not have to be a circle to have constant width. We will build what
is called a Reuleaux Triangle. Start with an equilateral triangle with side w. Then from
each vertex, construct the circle through the other two vertices.
Day 4
NC-PIMS Geometry 6-12
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w
Now consider the path made up of the arcs on each side of the triangle. This curve is
called the Reuleaux Triangle. Explain why any two parallel lines of support are distance
w apart (hence the curve has constant width).
w
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NC-PIMS Geometry 6-12
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D4.B3.A1
Curves of width 2r
This activity sheet asks you to compare the circumferences and areas of a circle and a
Reuleaux Triangle that have the same width.
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NC-PIMS Geometry 6-12
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D4.B3.A2
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D4.B3.A3
A Reuleaux Triangle in a Square
This activity page illustrates what happens when two pairs of support lines are
perpendicular to each other. This property illustrates how turning the Reuleaux Triangle
results in an “almost” square hole.
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Instructor Notes
An internet search on “Reuleaux Triangle” found over 1000 hits. Two of the best
are:
http://mathworld.wolfram.com/ReuleauxTriangle.html
http://whistleralley.com/reuleaux.htm
Block 4: The Pythagorean Theorem
In all of the figures, ABC is a right triangle with right angle C, the legs have length a
and b and the hypotenuse has length c.
B
c
a
C
A
b
This figure below presents the classical Pythagorean argument given in most
mathematics textbooks today. The picture shows two different ways of filling a square
with side a + b. One square uses a square of side a, a square of side b, and 4 of the
original triangles. The other square uses a square of side c and 4 of the original triangles.
Therefore the area of the square of side c (c2) is equal to the sum of areas of the other two
squares (a2 + b2).
b2
c2
c
a
a2
b
a
A = a2 + b2 + 4(
Day 4
ab
)
2
c
a
b
A = c2 + 4(
NC-PIMS Geometry 6-12
a
ab
)
2
17
D4.B4.A1
On cardstock construct a rectangle, draw one of its diagonals and construct the
perpendicular from one corner to the diagonal, as in the diagram below. Label the edges.
Cut out the rectangle and cut along the diagonal. How can you tell that the two triangles
are congruent?
Now cut along the perpendicular labeled h so that you have three triangles. The edges of
the triangles need to be labeled on both the front and back of the cardstock.
b
y
h
a
h
a
c
x
b
Line up the three triangles to show the corresponding angles and ratios of the sides.
Explain the meaning when we say the following three ratios are equivalent (that is, what
equal rational numbers are implied?).
a:b:c = x:h:a = h:y:b
Use the equivalent ratios you found above to complete these expressions:
ab = _______________
h2 = ________________
a2 = _______________
a2 + b2 = _______________
b2 = _______________
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NC-PIMS Geometry 6-12
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Instructor Notes
x
c
b
b
y
a
h
h
a
D4.B4.A2
This figure creates three triangles, all similar to the original, by multiplying each side by
the scale factors a, b, and c. The three new triangles are then assembled to create an
isosceles triangle, whose properties lead to the conclusion that a2 + b2 = c2. Fill in the
details that complete the argument.
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[This idea appeared in The College Mathematics Journal (v 27, n 5, Nov 1996, p 409)]
D4.B4.A3
This figure uses a rotation and a translation to verify that h2 = xy. This fact has an
interesting relationship to the Pythagorean Theorem, since
c2 = (x + y) 2 = x2 + 2xy + y2 = (x2 + xy) + (xy + y2) = (x2 + h2) + (h2 + y2) = a2 + b2.
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In right ABC, where h is the altitude to hypotunuse c,
dividing c into segments x and y, show that h2 = xy
First rotate
CDB about point C by 90 to
C
h
CEF
E
x
F
h
B
x
D
y
A
c
Then slide
CEF along vector FA to
JKA, and extend EF and KJ.
h
C
E
h-x
h
H
y-h
J
K
F
x
h-x
y-h
D
Day 4
y
G
NC-PIMS Geometry 6-12
A
21
Home Work
1. Create a GSP sketch that illustrates the Inscribed Angle Theorem.
2. Create a GSP sketch to verify your answers for D4.B2.A2.
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