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Supplement A
Measurement with whole numbers
A1: Weight and height in nonstandard units
The counting concept is the basis for the processes we call measurement.
Let's look first at the idea of weight. According to Webster's New
Collegiate Dictionary, "weight" is "the force with which a body is attracted
toward the earth or a celestial body by gravitation". If we think of it this
way, it is quite easy to see how we could check if two children were of the
same weight--take them to a balanced seesaw on the playground, have each
mount a side, and check to see if the seesaw remains balanced. If so, they
are of the same weight, if not the heavier child will be on the side that ends
up on the ground. In a sense, the seesaw here plays the role of the balance
scale, used to compare the weights of two objects.
There are several situations in which we might like to have more
information than simple balancing can give. One occurs when we want to
compare the weights of two children who are never in the same place, so
they can never get on the same seesaw. A second occurs when we want not
only to know which child weighs more, but how much more the child
weighs. If we know how to count, we can handle both of these situations in
the following manner. We take one child and a large number of identical
heavy textbooks to the seesaw, seat the child on one side, and begin piling
books on the other seat until the book pile balances the child. We count,
say, 23 books and say that the child has the weight of 23 books. We repeat
the procedure with the second child at a different time and/or place (but
with books of the same weight as those used originally) and find that the
child has a weight of 21 books. It is now easy to conclude from the number
fact
21 < 23 that the second child weighs less that the first, and the difference is
two books. If you have ever seen a balance scale used in a market, you will
recognize this process--the salesperson puts the produce on one side of the
balance, and ounce weights on the other until the two sides balance. When
they balance, the cost is determined by the number of weights used.
We might describe our weighing process as follows:
i) Select a unit of weight (a textbook perhaps)
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ii) Balance the object to be weighed with a number of unit weights
iii) Count the units used
It is easy to visualize a large number of classroom activities that could be
carried out to illustrate the weighing process, using only a simple balance
scale. How many paper clips does a pencil weigh? How many pencils does
a notebook weigh? How many marbles does a glass of water weigh. Of
course, if we hear that a glass of water weighs 20 marbles, that doesn't really
give much information about the glass of water unless we know how big the
marbles are. It is to reduce this sort of uncertainty that certain standard
units of weight, such as ounces, pounds, grams and kilograms have been
agreed upon for common usage. Wherever we travel, a gram means the
same thing it did at home. In fact, we could carry a bunch of one gram
weights around with us if we wished. When someone said a glass of water
weighed 100 grams we would know exactly how heavy it was. Once a
standard unit of weight has been agreed upon it is a simple matter, with the
use of available technology, to replace the slow and tedious balance scale
with an instrument that provides an immediate numerical display of the
number of units which would be necessary to balance any object set upon it.
A point to ponder: If we weighed a glass of water on a balance scale by
balancing it with a set of gram weights, we still wouldn't really know
exactly how much the water weighed. Why not? How could we correct the
measuring process to get the exact weight of the water?
Let's think about another attribute of children we might want to compare-height. If we have two children together, it is again a simple matter to
compare their heights. We stand them on a level surface, back-to-back, and
look to see which head rises above the other. If neither does, the two are of
the same height, otherwise we would say that the one rising above is the
taller (has greater height). As with weight we have difficulties with this
process in comparing separated children and in describing how much taller
one child is. Again our textbooks, if thick enough and numerous enough,
can be used to advantage. We simply stack books beside the child until the
top of the top book is level with the top of the child's head. If 25 books are
used, the child is 25 books tall. Repeating this with the second child, and
books of the same thickness, we can compare the heights by comparing the
numbers!
Notice the similarity of our process for measuring height:
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i) Select a unit of height measure
ii) "Balance" the height of the object to be measured with a number
of units
iii) Count the units
If our classroom has concrete block walls, we could measure each child's
height in "concrete block" units by asking each to stand against the wall
and counting blocks until we came to the top of the child's head. In this
case, we would probably observe a drawback to our measuring technique.
In all likelihood, the top of the child's head would not align exactly with the
top of a block, so the number of blocks we count would only give an
estimate of a child's height, rather than the exact height (of course, a similar
problem might also have in our weighing process). In this case, we might
find two children who we would say are 25 books tall, yet who were not
exactly the same height when standing back to back. This sort of
inaccuracy in measurement is one of the motivating factors for introducing
new number systems--fractions, irrational numbers.
As with weight, it is easy to see that information about height can be most
accurately and effectively be communicated with the help of a system of
standard units of height measure--feet, yards, meters, etc. Again, as in the
case of weighing, once we agree upon a common unit, we can simplify the
measurement process by constructing measuring instruments--rulers or tape
measures for instance--which can make the measuring process much easier
that stacking books or other objects and counting.
A2: Measuring segments
Much of the vocabulary of geometry was originally introduced to facilitate
discussions of measurement. This is not surprising, since the origin of the
word geometry is in the Greek for "measurement of the earth". To see how
geometric concepts arise, think about the discussion of height measurement
in A1. In comparing the height of two children, we ignore most attributes of
the children, for instance it matters not at all whether one is fat, the other
thin. All that matters is that they stand upright on level ground, and that we
know where the highest point on their head is. We could as well have each
child stand against a narrow tape pasted vertically on the wall, and mark the
tape at a level even with the top of the child's head. The higher the mark,
the taller the child. The tape stretching from the floor to the top of the
child's head provides a representation of all information that is essential to
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measurement of height. Personality doesn't count, age doesn't count. Only
the marks on the tape. The tape in this illustration, made as thin as we wish,
becomes the “ideal” object representing length and is called a line segment
(it is thought of as having length but no width), or just a segment.
Informally, we think of a segment as a straight path in space which joins
two points, called its endpoints.
From a point of view of height, the line segments pictured in Figure A2-1
might represent three children whose heights are to be
compared. Rather than Sue, John, and Mary, these
E
representations could be called AB , CD, and E F , with the
pair of letters naming the endpoints of the segment. If the
C
A
segment AB were placed on CD , it would fit precisely, so
these segments represent children of the same height, a
similar comparison of segments shows that EF represents
a taller child.
The idea that one segment can be placed to fit exactly
on top of the other so as to make the endpoints
B
D
F coincide, is called congruence. In Figure A2-1, one
Figure A2-1
would say that AB is congruent to CD , but AB is not
congruent to EF . A compact notation for the statement " AB is congruent
to CD " is AB  CD . In simpler language, a child would probably say that
AB is the same length as CD , rather than using the more formal
terminology “congruent”. Notice that the idea of same length can be
understood before the idea of a numerical length is introduced.
Of course, since segments are used to model height, and heights can be
given numerical values as in A1, there should be a numerical measurement
called the length of AB or the distance from A to B for each segment AB.
The process which assigns such a number to a segment AB, usually called
linear measurement, mimics exactly the measuring processes we've
already encountered.
a) Select a unit segment
b) Cover the segment AB with segments congruent to the unit
segment laid end to end, and with no overlap
c) Count the segments
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As an example, look at the segments XYand UV in Figure A2-2.
If we take UV to be our unit segment, we see that XS, ST , and TY are
segments congruent to UV which exactly cover XY. Counting these
segments we find that the length of XY is three units. It is customary in
many texts to use have a "shorthand notation" for the length of a segment
XY, namely XY (the bar above the segment name is simply omitted). The
linear measuring process can be carried out in the schoolroom or the
schoolyard, using a variety of nonstandard units. For instance, one might
determine the length of the classroom in floor tiles (the tiles are the unit
“segment”) ,the length of the sidewalk in concrete squares or the width of
the playground in student shoes. (Do you see a possible problem here if
many different students helped measure?)
A child with enough measuring experiences would likely discover that, for a
fixed unit, any two congruent segments have the same measure. This is a
fundamental property relating congruence and measure. We will call it the
Congruence property:
If the segments AB and XY are congruent then AB=XY
For our example A2-2, we made a very careful selection of our segment and
of the unit, to assure that a whole number of unit
X
U
segments could cover the segment XY exactly, without
overlap. This makes the segment XY have a precise
S
whole number length. In actual measurement situations, V
the unit segment and the segment to be measured are
often both specified in advance and we must do the best
T
we can with this measuring process and the whole
numbers. The result will often be that lengths can only
Y
Fig. A2-2
be specified approximately, to the nearest unit.
In practice one seldom, if ever, measures by actually
covering one segment with unit segments and counting. One constructs or
acquires a measuring device (a ruler). This is a physical model of a
segment--a stick or some other easily portable straight, narrow object--on
which are marked endpoints of a sequence of unit length segments, laid end
to end, beginning at one end of the object. In order, we label these marks
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0,1,2,... as we move by unit steps from the beginning. To measure a
segment XY, we simply lay the
ruler along the
segment with the
0
1
2
3
4
5
point labeled 0
lined up with the
endpoint X. The number on the ruler that lines up with the endpoint Y must
give the length of the segment. If we wish to use standard units, centimeters
or inches for example, we of course do not need to build out own ruler.
Commercially prepared versions of rulers and measuring tapes in these units
are readily available.
A ruler not only provides a tool for measuring, it also provides a useful
visual model of the set of whole numbers which displays them in their
natural order. If the ruler is positioned horizontally, with the point labeled 0
at the left, inequality of whole numbers is given a concrete meaning--a
number m is larger than a number n if m labels a point lying to the
right of the point labeled n. When a ruler is used in this manner to
demonstrate the order properties of the whole numbers, it is often referred to
as a number line.
Exercises
1. i) Taking the segment AB as a unit, make a paper ruler of length 8
units and measure the segments a), b), c) and d) (indicate whether these
measurements are exact or are to the nearest unit)
A
B
a)
b)
c)
d)
ii) Without making a ruler, one could use a compass to mark off
segments congruent to AB on each of the given segments. Do this and
count segments to check the answer you got in i).
2. On a number line, what is the length of the segment between the points
labeled 17 and 21? Sketch this part of the number line and count.
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3. Johnny's ruler broke so that both the 0 and the 1 were lost. He threw it
away, saying that it wouldn't work anymore. Was he right? Explain.
4. Are either of the following statements true? Explain your response.
a) If AB  XY then AB=XY.
b) If AB=XY then AB  XY.
5. Determine which of the following pairs of segments are congruent
without measuring. Explain your method.
a)
b)
c)
d)
>
<
<
>
A3. Measurement of area
Covering and counting, the procedure that allows us give a numerical value
to the length of a segment, also describes a different type of measurement
problem, that of describing the size of a flat region--a plot of land, the floor
of a room, a wall, the surface of a lake. In geometric terminology each of
these regions is a model of a portion of a plane. They are only portions, as
a plane is considered to extend indefinitely, and these regions are enclosed.
The outer edges of the regions we consider are examples of plane curves
(curves in a plane). We can best think of a curve as a figure in a plane that
could be drawn with a pencil without ever lifting the point from the paper
and without ever retracing any part of the figure. Circles, squares,
rectangles, and triangles are examples of curves that are familiar to children.
Beginning as one might in an elementary classroomwe will not try, at this
stage of our discussion, to describe these curves by their attributes, since
there is some evidence that young children do not classify shapes by
attributes but rather understand them as a whole. We will, for now, think of
squares, triangles, etc. by simply picturing their shapes. How should we
attach a number (called the area) to the size of a region? Let's try to mimic
our approach to lengths. We should
i) Select a unit
ii) Cover the region with units without overlap
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iii) Count the units
If we attempt to carry this process out, we are confronted by an initial
decision that did not arise in length measurement. We have a large number
of possible choices for the shape of the unit of area. For instance, to measure
the area of the quadrilateral ABCD below, we might choose as our unit
A
of area one of the four triangles in the ABCD.
If we trace any one of the triangles, lay the
tracing on top of another, and compare, we will
find that it fits exactly. We say that all four
triangles are congruent to the unit triangle. So
B the area of the quadrilateral ABCD is 4 units.
D
On the other hand, if we were to look at the area
of the rectangle in Figure A3-2, we could
choose the small triangle as unit, giving an area
of 16 units; or we could choose the small
C
rectangle as unit , giving an area of 8 units.
Figure A3-1
The conventional choice for a unit of area is a
square with sides of length one unit (in whatever length
measurement system we decide to work). So if we are
working in inches (for instance we might be interested in
the size of a book cover), the unit of area will be a square
one inch on a side (a square inch). If we want to measure
a room for carpeting, we would probably use a square of
side length one yard (a square yard), while for a
Figure A3-2
measurement in a scientific laboratory experiment we
might use a square of side length one centimeter (a square centimeter).
The problem of covering a given region precisely with units is even more
difficult than it is for segments. For example, Figure A3-3 represents a small
lake covered by 35 unit squares. So we know the area is less than or equal
to 35 units. On the other hand, if we look at the unit squares which are
completely contained within the lake, we see that there are only 16. So the
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Figure A3-3
most accurate statement we can make about
the area A of the lake is that 16A35 units.
A smaller unit would make the
approximation better, but as long as we are
restricted to counting only complete squares,
we will have difficulty getting a really good
approximation.
Just as with length, it is clear that two
regions which are congruent (in the sense that one would fit precisely on top
of the other) can be covered with exactly the same number of unit squares.
This is the
Congruence property for area:
Two congruent regions have the same area.
There is an interesting difference between length measure and area measure
related to the congruence property. If, for some unit of length, AB and CD
both have length 3 units, then AB  CD . This can easily be demonstrated
by building a model of AB and one of CD from unit segments and
comparing them. However if R1 and R2 are rectangular regions, both with
area 12 square units, it need not be true that R1 and R2 are congruent. Try
to draw two such rectangles which are not congruent! Of course, there must
also be circular regions of area 12 which are not congruent to any
rectangular region. In fact, we can easily imagine that there are infinitely
many different shaped regions, all of area 12.
Related to this strange behavior of area is another important property of
area measure, sometimes called the dissection
A
property of area. This property says that if we take
one region, cut it up and reassemble the pieces into
E
D
another region (without overlapping pieces), the new
region and the original will have the same area. For
instance, the triangle DABC in Figure A3-4 can be
B
C cut apart to remove the small striped triangle, and
Figure A3-4
reassembled to give the rectangle EBCD, so DABC
and rectangle EBCD have the same area. If the small square is the unit, this
shows that DABC has area 2 units. Tangrams can be used in the classroom
to provide hands-on illustrations of this property (see Lab 27)
Exercises:
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1. Draw a rectangle which is 3 cm wide and 5 cm long. Show how this
rectangle can be covered exactly by unit squares one cm on each side.
Count the units to find the area.
2. If the small square drawn here is the unit, what is the area of the triangle
drawn. Explain your reasoning.
3. Sketch three non congruent regions all with area 6.
4. A segment can be thought of as a region in a plane which has zero width.
What should the area of a segment AB be, measured in square inches.
Explain your reasoning. (Do not resort to some known formula for
area!)
A4. Measurement of volume
Let's just discuss briefly here how the ideas of measurement of length and
area can be copied to attach a numerical value to the volume of a threedimensional object such as a box (a rectangular solid to be precise), a
cylinder, a ball (sphere), or even some irregularly shaped object such as a
rock. To copy the measuring processes we've used before, we should first
choose a unit of volume, then "cover" the object we are measuring by units,
and finally count the units. A natural choice for a unit would be three
dimensional analog of the segment (one dimensional) and the square (two
dimensional), the cube which is one unit long on each side. We'll call that
the unit cube and speak of volume measured in cubic units--one cubic
inch, three cubic centimeters, and so on is we are using standard units.
One big problem we immediately come across in trying to measure volume
is that of "covering" a solid object with unit cubes. As a first example, we
could think about a rectangular solid in the form of a box (closed of course).
To handle this, we could open the box, and place as many unit cubes as
possible inside the box. If we count the cubes, we should have an
approximation of the volume of the box. Of course, it is unlikely that the
cubes will fill the box exactly. There will almost certainly be some space
left over, but it will be shaped wrong for another cube to fit in. In this case
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we have a lower estimate for the volume of the box. If our unit of length
was taken smaller, our unit cubes would also be smaller, so we should be
able to leave less empty space, thus getting a more accurate
measurement.What might we do if the box couldn't be opened, or if our
object of interest was of a shape that couldn't be filled easily with cubes?
One clever approach has to do with displacement, the idea that if we were
to take a solid object (a baseball for instance) and submerge it in a pan of
water which is filled to the top, the volume water which spills over should
be the same as the volume of the ball (since the ball has filled in the space
the water previously filled). Now if we could capture the water in a
rectangular pan, we would have essentially reshaped the volume of the ball
into a box, and we know how to estimate the volume of a box. In practice,
one can easily acquire graduated cylinders which display the numerical
volume (in standard units of cubic inches or cubic centimeters) of any
amount of water that is poured into the cylinder. So the volume of the
baseball could be determined by pouring the displaced water into a
graduated cylinder. A rough version of a graduated cylinder displaying any
desired (small) nonstandard cubic unit can be made from a cylindrical
drinking glass, with suitable 1 cubic unit gradations marked with a magic
marker.
The procedure of measurement by displacement works well with irregularly
shaped objects like rocks, as well as more regular shapes. A sponge could
give some problems! How could you overcome them without destroying
the sponge? The idea that the box of displaced water has the same volume
as the object that displaced it is an extreme illustration of the dissection
property of volume, which says that if a three dimensional object is cut up
and the pieces recombined to form another object, the new and old objects
will have the same volume.
One of the strong points of the metric system of measurement is that there is
a relationship not only between units of length, area, and volume, but also
of weight. In particular, a cubic centimeter of water weights exactly one
gram! In a classroom provided with a good balance scale and graduated
cylinders, students could experiment and discover this for themselves.
Exercises:
1. Use unit cubes to build a rectangular solid which is 3 units tall, 3 units
wide and 4 units long. Count the cubes. What is the volume of a 3 by 3
by 4 box? Can exactly the same cubes be used to build another
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rectangular solid? If so, what can you say about the volumes of the two
solids? What property of volume is at work here?
2. A square plane region can be considered an object in three dimensional
space, so it should have a volume. What should it be? Explain.
3. A very small rectangular container with vertical walls which is 5 cm long
and 2 cm wide is filled to a depth of 2 cm with water. What is the
weight of the water?
A5. Angles and their measure
The concept of angle seems a bit more subtle and remote than that of
segment or plane region. To be sure, there are many angles to be found in
the everyday world of a child, but they are always part of something else (a
block, a star, a room, a roof, etc.) For a child is at a stage of development in
which all objects are seen holistically, and not as the sum of their parts, the
concept of an angle as an independent geometric object might well be
missed. How would we describe an angle once we began to see objects as
the composite of their parts? Let’s consider a puzzle piece in the shape of a
triangle, Fig. A5-1. A child trying to fit this piece into a square hole on the
puzzle board might well observe that it fails to fit
because the sides of the triangle are longer than the
sides of the square. But even if the sides were the
same length, the triangle would still fail to fit
perfectly because the “corners” of the triangle are a
different size than the corners of the square. The
Fig. A5-1
corner, of course, is a representation of an angle.
How do we describe this angle concept? This puzzle piece representation
suggests that an angle is just a figure consisting of two segments which have
a common endpoint, the vertex of the angle. This must be the picture of an
angle that a child has, but it suffers from a minor drawback which is
illustrated in Fig. A5-2. There are, by our definition, several angles with
vertex at point A; one formed of segments AB and AC , one formed of
segments AD and AE , etc. (How many different angles, by this intuitive
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definition, are there in the Fig. A5-2 which have
vertex A?) This fact might be insignificant to a
E child, but the understanding of angles necessary for
meaningful applications is that there is really just one
angle here with vertex A. How can we resolve this
conflict? The mathematically accepted solution is to
define:
A
C
B
D
Fig. A5-2
an angle is a geometric object formed by two rays
(the sides of the angle) with a common endpoint
(the vertex of the angle)

where by a ray AB we understand informally the indefinite extension of AB


from A to B and beyond. With this understanding, we see that AB = AD ,


and AC = AE . Now whether we say that the angle at A in Fig. A5-2 consists




of rays AB and AC , or of AD and AE , we have described exactly the same
object. The price we pay for this precision is that the angles occurring in
nature must have their sides extended to form an angle as defined
geometrically.
We again come up against this difference between definition and perception
when we try to describe a triangle in terms of component parts--three sides
(which are segments) and three angles (which are not really part of the
triangle as a geometric figure). Rather than always saying “the angle


formed by the rays AB and AC we often substitute the shorter symbol
BAC. We apply to angles the same intuitive description of congruence
we have used for other geometric figures: two angles are congruent if a
tracing of one would fit precisely on top of the other.
As we look around our classroom for examples of angles, we are struck by
the fact that one particular type of angle predominates. The corner of the
room, the corner between ceiling and wall, the corner of the desk, the
corners of our books. Such angles are so common as to deserve a name of
their own. For small children, that name might be square corners. For us,
as we begin to teach and learn correct geometric terminology, the name is
right angle.
Let’s think a bit about what it takes to define a right angle, that is, to give an
unambiguous description of what is and what is not a right angle. The
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easiest way, and the way we presumably learned the right angle concept, is
to provide a picture of a right angle and say, an angle is a right angle if and
only if it is congruent to this angle. But suppose we want to do without a
picture? Most people would probably agree that a right angle is an angle
with a measure of 90 degrees. This is true, but it is only helpful in
describing a right angle to someone who has never seen one, if that person
knows exactly what one degree looks like and is willing to draw 90 angles
of this size. Another possibility would be to say that it is the angle formed
by perpendicular rays. But what shall we say “perpendicular” means?
Usually we define perpendicular rays to be rays with the same endpoint
which form a right angle. We cannot use both of these definitions at the
same time without creating circular reasoning, in which case neither is
useful. Challenge: define a right angle without using angle measure or
the word perpendicular.
Just as with segments or regions, it would be convenient to be able to assign
to each angle a whole number measure, which would somehow describe
how big the angle is. Since we have developed in other situations a
measuring process that was useful, we might begin our investigation of
angle measure by trying to mimic what has been done before, and seeing
what if anything needs to be changed. The process we propose is:
a) Select a unit angle
b) Cover the angle to be measured with angles congruent to the unit
angle, laid side to side with the same vertex and with no overlap
c) Count the angles
Fig. A5-3 demonstrates how this process would show that the measure of
XYZ is approximately (but slightly less than) four units if we take ABC
as the unit angle. This process does indeed describe a usable method of
assigning numbers to angles (Try to determine the measure of a right
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A
B
C
1 unit
Y
Figure A5-3
angle in this system if ABC is the
unit). We denote the measure of
X
XYZ by the symbol m(XYZ). It is
worth thinking about the fact that for
many students, young or old, the
frequent use of symbols in place of the
Z full name of a concept can lead to
confusion and frustration. It is often
4 units
wise to keep the use, and even
introduction of, symbols to a minimum
until it is clear that the students have a
firm grasp of the
concept at hand. The worst case scenario is that in which a student
mistakes memorization of symbols for the learning of concepts.
There are some subtle problems with our description of the measuring
process for angles. These deal step b). Remember that an angle consists
only of two rays. Then notice that in Fig. A5-3 it is not actually true that we
have "covered" the angle XYZ with copies of the unit angle, since no ray

of a unit angle falls exactly on YX. In fact, this could happen only if the
measure of the angle was exactly four units. What we really need here is to
be able to talk about covering the interior of XYZ with copies of ABC
and its interior, where by interior of an angle, we mean intuitively the
region of the plane lying between the sides on the angle. The interior of
an angle is shaded in Fig. A5-4.
Q
Observe that the angle PQR actually
Exterior
divides the plane into three distinct
parts, the exterior of the angle, the
P
Interior
interior of the angle, and the angle
R
itself, which is neither in its interior nor
its exterior. There are actually two
different ways we could try to measure
Interior
Figure A5-4

PQR beginning at QP and using a

given unit angle. We would start at QP

and lay off unit angles until we reached QR , moving either through the
interior or through the exterior. Certainly this would result in two different
measurements for the same angle, an ambiguous situation we would prefer
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to avoid at the time angle measure is first introduced. To clear up this
ambiguity, we should replace step b) of the measurement process by:
b') Cover the interior of the angle to be measured with angles
congruent to the unit angle and their interiors, laid side to side (with the
same vertex), and with no overlap
At a later date in the educational process, of course, one needs to be able to
measure through the exterior of the angle as well as the interior. This will
be accomplished without ambiguity by assigning to that measurement a
negative value (once negative numbers have been introduced.)
It should be clear that with this angle measurement process, once the unit
angle is selected, congruent angles have the same measure. Of course, as
with segments or regions, selection of a different unit can make a substantial
change in the numerical measure of an angle. There are two standard
angle measurement systems which are in use, degree measure and radian
measure. In the elementary grades, the degree system is by far the most
common, but in science and engineering applications radian measure
dominates. In either case, the process is as we have described above, the
difference between the systems is only in the selection of the unit. In degree
measure, a one degree angle is defined by the property that 360 angles
congruent it, laid side by side with a common vertex and with no overlap,
would exactly cover the entire plane. In other words, after 360 one degree
steps around a point A, we would end right back where we started. It is
believed that this selection of a unit angle dates to the Babylonians, who
visualized the sun moving around the earth at a rate of one degree per day.
So if one year were 360 days, at the end of one year the sun would be back
where it started one year earlier.
The radian measure system is in many ways a more logical system than
degree measure, since it is tied to the system of segment measure we have
already studied. To find the measure of an angle ABC in radians, we
choose a unit segment, then take a circle of radius one unit with center B


(See Fig. A5-5.) We now measure the distance from BA to BC along the arc
of the circle lying in theinterior of the angle (as if the arc were a segment).
The resulting length is the radian measure of the angle. While this method
17
has the advantage of making use of
something we already know, namely
A
segment measure, it has some serious
C
drawbacks also. One is that it is difficult,
1
if not impossible, to accurately measure the
length of an arc of a circle with a ruler. A
B
second is that commonly occurring angles,
such as right angles, not only do not have
exact whole number measures as they do in
Figure A5-5
degree measure, they have in fact irrational
number measures (  for a right angle). Not a good situation, particularly
2
with young children who have no idea what an irrational number might be,
but who are comfortable with whole numbers. A third difficulty, if we were
trying to introduce measure to young children, is that the idea of radian
measure depends on an understanding of center and radius of a circle, not
likely if the child sees the circle holistically.
One strange aspect of angle measure must be emphasized. As opposed to
segment or area measure, measures of angles cannot be arbitrarily large. If
we think about measuring segments, we know that if we start with a
segment QP of any given length, we could always lay off one additional

unit segment along the ray QP and get a segment one unit longer than QP .
If we continued doing this, we would end up segments one million units
long, or more. There are segments which are as long as we wish to have
them. What happens if we try the same thing with angles. Let’s use degree
measure, beginning with the right angle XYZ in Figure A5-6.
If we now lay off one degree angles starting at


YZ and moving toward YW we will get larger
Z
W
T
Y
Y
X
and larger angles, just as we did with segments,
until we have “added” 89 unit angles, at which
time we will have an angle of measure 179
degrees. One more unit angle will get us to 180

degrees, which we see is formed of the ray YZ
and its opposite ray. This angle is called a
straight angle. What happens next? When we add one more unit angle, we
get XYT, which has its interior shaded in the diagram. Since we have
agreed to measure angles through the interior, we see that this angle has
measure 179 degrees again. Reasoning in this way, we see that there cannot
Figure A5-6
18
be any angle with measure greater than 180 degrees (or p radians in the
radian measurement system). Probably you have studied enough
mathematics to know that, as one moves more deeply into the subject, one
must change one’s view of angles so that in fact angles too can have
arbitrarily large measures. From the standpoint of the elementary school
curriculum, however, 180 degrees is the most that is possible.
Just as a ruler makes the process of measuring lengths of segments easier
than the process of laying off unit segments, the measuring device called a
protractor is used to make measurement of angles easier. A protractor is a
semicircular piece of material, usually metal or plastic, on which
consecutive 1 degree angles are marked and labeled: 0,1,2,3,...,179,180
(though in reality, only every tenth angle is usually labeled), with 0 labeling
the straight edge of the protractor. To measure an angle, one need only lay
the protractor on the angle being measured, the center falling on the vertex,
the edge labeled 0 precisely on one side of the angle and the protractor
positioned so that the other side of the angle is covered by the protractor.
The number on the protractor angle which falls on this other side is the
measure of the angle in question.
Exercises:
1.
If D is the unit angle, what is the largest whole
number which could arise as the measure of an angle?
D

2. If BK bisects (divides into two congruent pieces) ABC, and if
m(ABC)=6 units, what is m(ABK)?
3. Suppose that in some measurement system a straight angle has measure 8
units. Use paper folding to make a unit angle for this system. Explain
why this must be a unit angle.
4. Suppose that the measure of a straight angle is 10 units, that ABC is a
19
right angle, and that m(ABD)=1. What is
m(DBC)? (The drawing may not be to scale)
C
D
B
A
5. Suppose that 10 is the measure of a straight angle, that m(ABC)=7 and
that m(CBD)=5. If the interiors of ABC and CBD do not overlap,
what is m(ABD)? If the interiors do overlap, what is m(ABD)?
6. Using ABC as a unit, find the measure of each of the other angles to the
nearest unit.
C
B
iii)
i)
ii)
A
iv)
Supplement B
Arithmetic and Geometry in Measurement
Since counting is required in a wide variety of measuring situations--weight,
length, area, angles--and since addition and multiplication of whole
numbers were introduced to make counting more efficient, we should expect
that these operations will be useful in solving many measuring problems,
and that the geometric aspects of these problems will provide models of
addition and multiplication in real world terms. The starting point is the
important “addition principle for sets”: If S and T are sets having no
20
elements in common (i.e. ST=), then n(ST)=n(S)+n(T). Here, of
course, n(S) means the number of elements in S.
B1: Measuring segments
Suppose we want to apply this idea to measurement of distances. From a
map we read that the distance from Cincinnati to Columbus is 110 miles,
while the distance from Columbus to Cleveland is 140 miles. This means
that the segment (road?) joining Cincinnati to Columbus is made up of 110
unit (1 mile) segments laid end to end with no overlap, while the segment
from Columbus to Cleveland is made up of 140 unit segments laid end to
end. If we think of the segments used to get from Cincinnati to Columbus
as a set S of 110 segments, and the Columbus-Cleveland segments as a set T
of 140 segments, the addition principle above tells us that the total
collection of mile-long segments used to get from Cincinnati to Cleveland
should be n(S»T)=110+140=250 miles, which is exactly what the map says.
This and similar examples suggest a
Conjecture :
For any three points A, B, and C, the distance from point A to point
B, plus the distance from point B to point C is equal to the distance
from point A to point C.
Too often, an elementary text leaves the topic of addition in measurement at
this (incomplete) stage, reinforcing this “addition principle” with a number
of similar exercises. In fact, this is just part of the story (as was the case for
the addition principle for sets). The Cincinnati-Columbus-Cleveland
example was chosen very carefully to make a point. Suppose the cities
chosen had been Toledo, Columbus, and Cleveland. Then the distances
would have been 135 miles from Toledo to Columbus, 140 miles from
Columbus to Cleveland, yet only 110 miles from Toledo to Cleveland. Our
Conjecture seems a bit off base here. Certainly 135+140 is not 110! What
has gone wrong? If we look at a map, we see immediately that there is a big
difference in how the three cities are positioned relative to each other in the
latter case as compared to the former. While the shortest route from
Cincinnati to Cleveland may well pass through Columbus, there is certainly
a quicker way from Toledo to Cleveland than traveling via Columbus. Yet
the conjectured principle measures a route that goes through Columbus! It
appears that we need to be a bit more careful in describing the conditions
under which the principle for distance really works. This is no different than
what happened when we first thought about addition and counting. At first,
21
we thought that the general relationship was that n(ST)=n(S)+n(T) for any
sets S and T. Later, looking at Venn diagrams, we realized that this
statement would be true only if the sets S and T had no elements in
common. What should the conditions be to assure that the addition
principle for distance will be true? This is a good subject for a classroom
discussion, but let’s bypass the discussion here and observe that the critical
distinction between our two examples is that Columbus is on the segment
joining Cincinnati and Cleveland, while it is not on the segment joining
Toledo and Cleveland. This is exactly what is needed to make precise the
addition principle of distance:
The distance from point A to point B, plus the distance from point B to
point C is equal to the distance from point A to point C provided that B
is on the segment AC .
Since distance and length are just different views of the same measurement,
we could also have phrased this somewhat more compactly (using the
shorthand notation for length of a segment) as:
AB+BC=AC provided that B is on AC .
A natural extension of this discussion would be to ask what happens if the
segments AB and BC have a common endpoint (B), but do not lie on the
same line (see Fig. B1-2). In this case, a bit of experimentation shows that
the triangle inequality: AB+BC>AC will always be true. We might hope
for more, namely a formula that
A
B
would give us the length AC if we
knew AB and BC. In fact, there is
Figure B1-2
C such a formula provided we also
know the measure of the angle
ABC but it is well beyond the scope of the elementary school curriculum.
It is known as the “law of cosines” and is usually first encountered in a
course on trigonometry. In §B4 we will investigate a special case of the law
of cosines, the theorem of Pythagoras, which is particularly useful in
indirect measurement problems.
A number line provides an excellent tool to model both the addition
process and the addition principle for length. For example, consider the
number line sketched in Fig. B1-1. It pictures a segment of length 2 and
22
one of length 4, end to end on the number line. The resulting segment,
beginning at 0 and
ending at 6, is clearly
2 units
4 units
of length 6 so we have
2
6 here a picture of the
0
3
4
5
1
arithmetic fact 2+4=6.
2+4=6
Of course, every whole
number addition fact
Figure B1-3
could be represented
geometrically on the number line in a similar manner and subtraction,
particularly in its missing addend form, can be similarly modeled.
Exercises:
1. Suppose that the distance from city A to city B is 140 miles, and the
distance from city B to city C is 125 miles. What is the least possible
distance from city A to city C? What is the greatest possible distance?
Draw diagrams to illustrate these two situations.
2. Two pipes, one 26 centimeters long, the other 18 centimeters long, are
joined to form a longer, straight pipe. To make the connection secure, there
is a 4 cm. overlap between the two pipes. How long is the combined pipe?
3. Suppose you wish to enclose a lawn with a fence. The measurements of
the sides of the lawn are 42 ft., 36 ft., 81 ft., and 78 ft., each measured to the
nearest foot. What can you say about the actual perimeter (distance
around) of the lawn? How much fence should you buy?
4. What is the perimeter of your classroom, measured to the nearest
kilometer?
5. B
4
A
2
D
C
Suppose that in the quadrilateral (four sided
figure) ABCD, we know that the sides AB, BC ,
and CD are congruent, and that the lengths of two
sides are as indicated in the diagram. What is the
perimeter of the quadrilateral.
6. (An extension of the addition property of length) Suppose that A,B,X
and Y are situated as in the diagram.
23
A
X
Y
B
a) Explain, using the additive property for length, why AB= AY+XB-XY.
b) We have learned a similar property for counting sets. What is it?
B-2 Additivity of angle measure and the (tri)angle sum formula
Since the process of angle measurement is analogous to the process of
length measurement, we might expect a similar addition principle for angle
measurement. Roughly speaking (see Figure B2-1), it should say that the
sum of the measures of two angles sharing a common ray is equal to the
measure of the angle formed by their non common rays. As with segments,
we have to be careful to spell out exactly how these angles are positioned,
since Fig. B2-1, looked at another way, shows two angles CBA and
ABD which share a
ABC)=2
m(
common ray but for
A
which it is not true that
m( CBD)=3
m(CBA)+m(ABD)=
C
m(CBD). This
m( ABD)=2+3
problem could be taken
care of by specifying
B
D
that the two angles be
adjacent angles:
Figure B2-1
angles with the same
vertex and one common side, but with interiors which do not overlap.
Another difficulty arises because of our agreement that no angle should
have measure greater than that of a
m( ABC)=70
straight angle (Fig. B2C
2). In this case, ABC
m( CBD)=140
and CBD are adjacent
angles, but the sum of
their measures is greater
than 180 degrees, so it
A
B
cannot equal m(ABD).
Figure
B2-2
What exactly should be
D
the statement of the
24
addition principle? Here is one version which will deals with the problems
we have noted.
The addition property of angle measure:
If C is in the interior of ABD, then
m(ABC)+m(CBD)=m(ABD).
Draw a picture of the situation described in this statement to see that it
demonstrates the situation as you understand it.
The addition property of angle measure leads easily to the important
Angle sum formula: The sum of the measures of the angles of any
triangle is equal to the measure of a straight angle.
In particular, this says that in the degree measurement system, the sum of
the measures of the angles of a triangle is 180 degrees, while in the radian
measurement system the sum is p radians. If we were discovering it for the
first time, as an elementary school student might, this could be a rather
surprising fact. There are a vast number of very different looking triangles,
with widely varying angles and side lengths. How can we convince
ourselves (or our students) that the sum of the angle measures is always the
same, no matter how the triangle looks? One idea is pictured in Figure B23.
Take a triangle ABC, cut it out, and tear off the corners (the angles of the
triangle). Rearrange the pieces as indicated so the triangle angles are sideby-side with no overlap and a common vertex. Notice that the outermost
sides of angles A and C form a straight angle, that is, they lie on a line.
B
A
C
C
B
A
Figure B2-3
Now the additive property for angle measure says that
m(A)+m(B)+m(C) must be the measure of that straight angle, so the
sum of the measures of the angles of ABC is just the measure of a straight
angle. To make this argument convincing it may be helpful to have class
members do the "cut and tear" exercise with different sizes and shapes of
25
triangles, until it becomes clear that the size and shape are irrelevant in
arriving at the result.
Exercises:
1. If the unit for an angle measurement system were chosen so that the
measure of straight angle was 6, what would the measure of a right angle
be? What would the sum of the measures of the angles of a triangle be?
2. The quadrilateral EFGH is divided into two triangles by the diagonal
EG .
a) The sum of the measures of the angles of the
F
quadrilateral, i.e. FEH, GFE, HGF, and EHG
is equal to the sum of the measures of the angles
E
of EFG and EHG. Explain carefully why this is
G so.
b) In degree measure, what is the sum of the
H
measures of the angles of this quadrilateral.
c) With a protractor, measure the angles of the
quadrilateral XYZW. Do you get the result
Z
X
predicted in b)? If not, why not?
Y
W
Note: The argument in part a) and b) depends on the fact that the
quadrilateral is convex, that is--for any two points in the interior of the
quadrilateral, the segment joining them is also completely in the interior of
the quadrilateral.
3. a) Use an argument similar to that of #2 a) to decide what the sum of the
angles of the convex pentagon pictured here must be.
b). Suppose that a convex pentagon ABCDE has all its
angles congruent. In degree measure, what will the
measure of each angle be?
4. Do Exercise 3 for a convex hexagon with all angles congruent.
26
5. Suppose that the outer quadrilateral shown here is a square, and that the
four right triangles are congruent.
a) Determine the degree measure of the angle
labeled A in the diagram.
A
b) What does this tell you about the inner
quadrilateral?
6. Find the degree measure of the angle marked ?
20
65
25
?
7. Suppose we were to define a right angle (remember the challenge
question in A5?) as: an angle such that four angles congruent to it, laid
side by side with a common vertex, exactly cover a plane. Explain why
the measure of a right angle, in degree measure, is 90 degrees.
B3: Indirect measurement
The measurement processes described in Supplement A deals with
“accessible” situations--we have in front us some object and we develop a
method to attach a number to the size of that object. Part of the "magic" of
geometry is that it also helps us find measures of certain angles or segments
which are beyond our reach. As a first example, let's consider the
following problem which was posed to a group of students attending a
summer mathematics program at OSU:
27
"Can you determine the width of the Olentangy
River without ever leaving the east bank?" (Not
River
allowed were such strategies as shooting an arrow
C
with a string attached across the river, retrieving the
arrow and string, and measuring the string; or
swimming across with a rope in tow.) After some
thought, the students proposed the following
A
B
solution. (See Fig. B3-1). A small bush (B) was
sighted on the west bank of the river, and a student
(A) took a position directly across from the bush. A
Figure B3-1
second student walked along the river bank on a
path making a right angle with the ray leading from the student to the bush.
As she walked, she continually checked the angle ACB. When that angle
was 45 she stopped and a third student measured the distance AC. The
group then asserted that this was actually the distance AB as well, so they
had measured the width of the river!
Why does this method work? The first important fact is that the sum of the
measures of the angles of any triangle is 180 . Since we know that the
measure of a right angle is 90 and that of ACB is 45 , the measure of
the angle ABC must be 45 . (Here is the magic! The students have
measured ABC even though they cannot reach its vertex!). Since they
have equal measures, ACB and ABC are congruent. One student in the
group remembered from high school geometry that a triangle with two
angles congruent has the sides opposite those angles also congruent (we'll
discuss this point in § C.4.) So AC is congruent to AB and the congruence
poperty for segment measure says that the length AC is equal to AB. Pretty
clever!
Exercises:
1. Suppose that, due to a bend in the river, the students above found it
impossible to walk at a right angle to AB, and
C
instead walked in a direction so that
River
m( BAC)=120 . To use this same
technique, the walker should stop when
A
ABC has what measure?
B
28
2. Suppose you were at the Great Pyramid in Egypt, and wanted to know
what the angle at the top of one of the four faces was. Could you get the
answer without climbing the pyramid? How?
B4: Additivity of area and Pythagoras' Theorem
Fig. B4-1 illustrates that for regions which can be covered precisely by unit
squares, area measure (like length and angle measure) also has an addition
property, that is:
If the region R is made up of two regions S and T which do not overlap,
then A(R)=A(S)+A(T).
S
T
A(S)=6
A(T)=4
A(R)=A(S)+A(T)
R
unit
Figure B4-1
To be absolutely correct, we would
have to admit that the regions S
and T in Fig. B4-1 do overlap, but
they only have a segment in
common and this is of no real
importance here because we have
seen earlier a segments contributes
zero to a region's area.
If we insist that the addition property be true for any regions that do not
overlap, we can discover formulas for the area of a number of types of
regions that are not rectangular (triangular or circular, for instance.) To take
a simple example, let’s look at the triangle ABC in Fig. B4-2. If the small
squares are unit squares, we clearly cannot cover the triangle exactly with
C units, so we might expect to have to estimate its area.
However, from the addition property we know that the area of
ABC is the area of the black region plus the area of the
A
B small shaded triangle. Since the small shaded triangle and the
white triangle are congruent, they have the same area. That
Fig. B4-2
means that the area of ABC is exactly the same as the area
of the black area plus the area of the white triangle. But these
two areas together are covered exactly by two small squares, so their areas
total 2 units. Now we know that ABC also has area exactly 2 units.
29
Before we move on to
a discussion of
S
A(T)=9
formulas for areas of
T
non rectangular
A(R)=A(S)+A(T ) -A(S  T)
regions, we should
= 18-4=14
think a bit about what
Figure B4-3
happens to the
addition principle for area if there is an overlap of the two regions?
Looking at simple cases such as that in Fig. B4-3 we can see that our
formulas for counting overlapping sets or for measuring overlapping
segments should work here as well. That is:
A(S)=9
If R=ST then A(R)=A(S)+A(T)-A(ST).
T
Fig. B4-4
An interesting consequence
of the addition property of
area is Pythagoras’
Theorem (apparently first
discovered by the Chinese
about 2000 BC, but taking
its name from the Greek
Pythagoras who lived in the
6th century BC). The
theorem relates the areas of
squares which are associated
with a right triangle, i.e. a
triangle in which one angle
is a right angle.
Look at Fig. B4-4. The right triangle T is surrounded by three squares, one
having sides the length of the hypotenuse of T, the others having sides the
lengths of the two legs of T. Just looking at this picture, we have no idea
that there is a special relationship between the areas of these three triangles.
However, if we trace the two squares on the legs of T (including the dotted
lines in the larger square), cut out the two squares, and then cut the larger
square along the dotted line we will end up with five pieces. Try this!
Assemble the five pieces to see if they fit precisely on top of the square on
the hypotenuse. AHA! You have just illustrated Pythagoras’ Theorem:
The area of the square on the hypotenuse of a right triangle is equal to
the sums of the areas of the squares on the sides.
30
It is hard to imagine how anyone could discover such a
result, but it is believed that it was discovered by Chinese who were
investigating tessellations. Lab 9 includes an activity which indicates how
this might have occurred. Supplement C also contains a proof of
Pythagoras' Theorem based on properties of similar triangles. It has been a
pastime of amateur mathematicians for centuries to search for new proofs of
Pythagoras' Theorem, and many have been found. A whole book has been
published which contains nothing but proofs of this famous theorem. We
will find that this theorem is very useful in indirect measurement problems.
Exercises:
1. On the geoboard picture, assuming that the shortest vertical or horizontal
distance between pegs is the unit of length.
a) Find a square with area 4 and a rectangle
which is not a square that also has area 4. Do
these regions have the same perimeter?
b) Use the fact that 5=12+22 to find a square
of area 5 square units.
c) Find a square of area 13 square units
having its vertices at pegs.
d) Explain why there cannot be a square of
area12 square units with its vertices at pegs.
2. Without using any formulas for computing areas, find the areas of these
geoboard figures, assuming that the smallest squares with corners at pegs
are unit squares. Explain how you got your answer.
.
31
(a)
(b)
.B5: The Rectangle Area formula
The procedure in § A3 allows us to measure the area of a rectangular region
by covering it with unit squares and counting. In § B4 we saw how to use
addition to find the area of a large region if it can be broken up into smaller
regions which have known areas. So far, we have not discussed the
possibility that the area of a rectangular region might be computed if we
knew only the measurements of its sides, though it would surely be nice if
we could do that, since measuring segments with a ruler is a much easier
process than cutting out unit squares and trying to cover a region exactly
with them.
The idea behind the formula for the area of a rectangle is illustrated (for the
product 34) in Fig. B5-1, (a) and (b). (a) pictures three groups of four
blocks each, representing the repeated sum 4+4+4.Since the product 34 is
defined to be this sum,
we have a model of that
product. If we
rearrange the blocks as
indicated in (b), we still
have a model for the
product 34, but it has
(c)
(b)
(a)
taken on a geometric
significance. The
Figure B5-1
model has become a
rectangle with width 3
units (provided the block has side one unit long) and length 4 units, so the
product 34 counts the number of unit square blocks necessary to cover this
rectangle precisely, with no overlap. That is, 34 gives the area of the 3by-4 rectangle. Of course, we could have used any whole numbers W and L
in place of 3 and 4 in Fig. B5-1, and arrived at the same result, namely:
The area of a rectangle with width W units and length L units is given
by the product WL. (Briefly written, A=WL)
On the one hand, this shows why the well known formula for the area of a
rectangle works. On the other hand it gives a very visual way to model
32
products of whole numbers. This is usually referred to as the area model
for multiplication.
One of the very useful features of the area model is its adaptability for
demonstrating why the various properties of multiplication hold. For
instance, if we compare the rectangles in Fig. B5-1 (b) and (c), we see that
they are congruent (for instance (c) is just (b) rotated 90 ) . Since they are
congruent, they must have the same area, but (b) models the product 3 4,
while (c) models 4  3, so we see that 3  4=4  3. Looking at similar
models with other length sides, we see that m  n=n  m for any whole
numbers m and n. That is, we have illustrated the commutative property
of multiplication.
In a similar way the distributive property of multiplication over addition
can be demonstrated (see Fig. B5-2). Diagram (a) represents two regions of
areas 3  4 and 3  3 respectively. When they are
put together
they give a
rectangle with
width 3 and
length 3+4, so
they give an
3x 4
3 x 3
3 x (4+3)
+
area of
(a)
(b)
3  (3+4)
square units.
Figure B5-2
The addition
property for
area then tells us that 3  (3+4) = 3  4 + 3  3. Again this could be done
with any whole numbers W, L, and M in place of 3, 4 and 3 to illustrate that
the distributive property: W(L+M) = W  L + W  M holds.
33
Before we move on to search for
formulas for the areas of other types
of regions, let’s look again at
Pythagoras’ Theorem in the light of
our new idea relating areas to
C
lengths. Recall that Pythagoras’
Theorem says that for any right
triangle, the area of the square on the
hypotenuse is equal to the areas of
the squares on the two legs. For the
A
B
triangle ABC in Fig. B5-3, the
formula for the area of a rectangle
(which certainly must give the area
of a square) tells us that the area of
the square on the hypotenuse is BC2,
while the areas of the squares on the
legs are AB2 and AC2. Rewriting
Figure B5-3
Pythagoras theorem in these terms
gives the relationship: BC2=AB2 + AC2 between the lengths of the sides
of a right triangle. This is the most familiar form of Pythagoras” Theorem,
and the form in which it is most useful in applications.
To look at just one application, suppose that a storm has loosened the roots
of a tree in the school yard, and you are asked to stabilize the tree by tying a
rope from a crotch in the tree nine feet above the ground, to the base on the
bicycle rack 12 feet away on the level ground. How much rope will you
need? Assuming that the tree is still standing vertically, you can visualize a
right triangle with vertices the base of the tree, the base of the bicycle rack,
and the crotch of the tree. Since the lengths of the legs of the triangle are 9
and 12 feet, the square of the length of the hypotenuse of the triangle will be
92+122=225. A little checking (or a short computation using the square
root key on your calculator) will show that you will need 15 feet of rope. A
little warning is in order here. The lengths of the legs of this triangle were
very carefully selected to make the hypotenuse come out to be a whole
number. A little investigation of various right triangles with whole number
length sides, making use of Pythagoras’ Theorem to compute the length of
the hypotenuse, should convince you that very few of these triangles can
have a whole number hypotenuse. While this makes it difficult to do a great
deal with indirect measurement via Pythagoras’ Theorem in the early
34
grades, it also provides a good incentive to enlarge the number system to
provide enough numbers so that we could measure these lengths exactly.
Exercises:
1. Sketch two other parallelograms on the geoboard grid which have the
same area as the one pictured, but which are not congruent to the
pictured parallelogram.
2. If the lengths of the sides of a square are doubled, what is the effect on
the area?
What is the effect on the perimeter?
What if the lengths of the sides are tripled?
Model this behavior on a geoboard. Try it for triangles as well. Result?
3. Find the perimeter of the parallelograms you have drawn in exercise 1
(write the answer as a sum of a whole number and some square roots).
4. Which of the following triples of whole numbers could represent the
lengths of the sides of some right triangle?
(a) 3, 5, 4 (b) 7, 11, 12 (c) 25 ,7 ,24 (d) 10 ,24 ,26 (e) 40 ,41 ,9
5. A right triangle has sides with whole number lengths, If two of those
lengths are 15 and 17, how long is the third side?
6. Determine the value of x.
35
13
5
x
7. Suppose the pegs of the geoboard were labeled, each with a pair of
whole numbers as indicated. Assuming that the distance between
adjacent
horizontal or adjacent vertical pegs is 1
unit, what is the distance from the peg
labeled (0,3) to the peg labeled (2,0)?
(3,5)
What is the distance between the pegs
(1,4)
labeled (1,1) and (3,5)? Between (1,4) and
(0,3)
(2,2)? Can you find a formula which will
give the distance from (a,b) to (c,d) for any
(0,2)
(2,2)
whole numbers a,b,c, and d?
(0,1)
(0,0)
(1,1)
(1,0)
(2,0)
(3,0)
B6: Area formulas for triangles and circles
Once the relationship between whole number products and rectangle areas
Z
is well understood, it is easy to use the ideas of
§B4 to find the area of a right triangle with
base of length B and height H. In Fig. B6-1
H
we have begunwith such a triangle, XYZ,
and combined it with a congruent triangle to
B
Y
X
form a rectangle with width H and length B.
Figure B6-1
(To see quite concretely that these triangles
are congruent, rotate XYZ around the midpoint of its hypotenuse until Y
lands on original position of Z-- a 180 degree rotation). Since the triangles
are congruent, they have the same area, and since they make up the
rectangle without overlapping, the sum of their areas must equal the area of
the rectangle. From this reasoning we see that 2A(DXYZ) = BH, in
other words: A(XYZ) = (BH) 2. So we have illustrated the well known
36
formula for the area of a triangle, at least for right triangles. What can we
do with other triangles? Once we have decided which side of the triangle is
to be the base, (an arbitrary choice, and we want the area formula to work
no
matter which choice
Q
we make), we have
two possible situations
H
H
that can occur--either
P
S
R the line perpendicular
to the base and
(a)
(b)
passing through the
Figure B6-2
opposite vertex hits
the base of the triangle, or it lies outside the triangle and hits the line which
includes the base, but not the base (segment) itself (see Fig. B6-2). In either
case we
call the segment joining the opposite vertex and the line containing the base
the altitude to the given base, and its length is called the height H of the
triangle. If we look at Fig. B6-2, we see that the triangle we are interested
in is either made up of two right triangles (case (b)) or is gotten by taking
one right triangle away from another (case (a)).
Lets look at situation (b). The length of the base of the triangle is
B=PR=PS+SR (by the addition property for lengths). The addition property
for areas gives
(**) A(PQS)+A(QRS)=A(PRQ).
Since we know how to compute areas of right triangles, we know that
A(PQS)= (QS  PS)∏ 2 and A(QRS)=(QS  SR) 2. Substituting in
(**) gives
A(PRQ)= (QS  PQ) 2 + (QS  SR) 2 = (QS  (PQ + SR)) 2 =
=(QS  PQ) 2 = (B H) 2
where B is the length of base of the triangle and H = QS is the height. So
once again we have demonstrated the correctness of the area formula for
triangles. The argument for case (a) is similar, and is left as an exercise.
There is one inconvenience built into the formula that we have found here
for the area of a triangle . The measurements one would usually expect to
37
be given, or be able to find, for a triangle would be the lengths of the three
sides. If, however, we were to want to know the area of a triangle which
had sides of lengths 12, 14, and 21 centimeters, our area formula wouldn’t
be very useful. If we choose the side of length 14 as base, the formula does
not use the lengths of the other sides, it uses the height of the triangle and
we do not know that! Of course, we could draw a scale model of the
triangle and measure the height, but that would involve constructing a
perpendicular accurately. Not an easy chore. It would be nicer, if there
were a formula that gave area directly from the lengths of the three sides.
Such a formula is known, and can be proved correct just using Euclidean
geometry at the high school level. Though the proof is beyond the scope of
this book, and not really accessible to elementary school students, we give
the formula here for interest’s sake.
Heron’s Formula:
If a triangle has sides of lengths a,b, and c units, then its area is given
by A2 = s(s-a)(s-b)(s-c), where s=(a+b+c) 2.
We see here again evidence that the whole number system is not large
enough to allow us to solve all the problems that it allows us to state. If our
triangle has sides with whole number lengths, it may still not have a whole
number area. Even the simple equilateral triangle with sides of length 1,1,
and 1 unit will have an area which squares to give 3 8. It is clear that no
whole number will do this, since 3 8 does not give a whole number answer.
One can show that even extending to fractions will not provide a number
that precisely measures this area!
While we are about the business of finding formulas for areas, we might as
well take a look at parallelograms--quadrilaterals in which opposite sides
are parallel. (Roughly speaking, we say that two segments are parallel if
the lines formed by extending each of them indefinitely in both directions
never meet). Fig. B5-6 depicts such a quadrilateral BCDE. BP is an
altitude of the parallelogram (to base DC ). The segment BD divides the
parallelogram
38
E
B
D
Fig. B6-3
P
into two congruent triangles (Rotate DBC
180 degrees about the midpoint of BD .) As
we saw for rectangles, this gives use
A(BCDE)= 2 A(DBC). But we know
C
that the area of DBC is given by (DC
 BP) 2 so we get
the area of a parallelogram is equal to the product of the length of its
base by its height.
Finally, we can use our knowledge of areas of rectangles, together with a
little creative visualization, to discover an important relation between the
area of a circle and its perimeter (circumference). Let’s continue to think
of geometric objects as whole objects, and not yet try to give to a circle a
precise definition, though we have a perfectly good picture in our minds.
An
important property of a circle, easily
tested by paper-folding, is that it has
lots of “fold symmetries”. That is, a
circle can be cut out of paper and
folded in such a way as to make its
two halves fit perfectly on top of each
other. The segment along which the
paper is folded to give this symmetry
Figure B6-4
is called a diameter of the circle (see
Fig. B6-4). To estimate the area of a
we sequentially draw diameters
through the circle in such a way as to make each new angle formed
congruent to its predecessor (see Fig. B6-5).
Now we can cut the circular region along these diameters, reassembling the
pieces in the form pictured in (b). Since both figures are made up of the
same number of pieces, and all of the pieces are congruent, the areas of the
two figures should be equal. But what is the region (b). It is , roughly,
39
(b)
(a)
Figure B6-5
a parallelogram with width about D 2 (where D is the diameter of the
circle) and length about C 2 (where C is the circumference of the circle).
So we arrive at the conclusion that
A  ( D 2)  (C 2) = (D C) 4 ( meaning: approximately equal to).
It appears from the picture, that the more diameters one used to divide the
circle, the closer the figure (b) is to a rectangle. This argument makes it
plausible to conjecture that the area of a circle is given by the product of
the diameter and the circumference, divided by 4.
Often this area formula is stated in terms of the radius (= diameter 2) of
the circle. In that case, the area formula reads A= (R C) 2, where R is the
radius of the circle. Lab experiments (wrap a string around a tin can,
measure the string, measure the diameter of the can, divide) show that the
quotient of the circumference of any circle by its diameter is a number
slightly larger that 3. As our number systems grow to include not only
fractions but all real numbers, we will find there an irrational number,
called , for which C=D=2R. When this fact is substituted into the area
formula derived above, we get the well known formula: A=R2.
Exercises:
40
1. Find the areas of the regions pictured, assuming that the lengths
indicated are correct, that angles that look like right angles are right angles,
and that segments that look parallel are parallel.
7
4
12
11
(b)
(a)
2
4
2
7
1
2
5
9
(d)
(c)
2. Suppose that, instead of the 1 by 1 unit square, we had chosen ABC to
be our unit of area measure. What would the areas of the regions a), b) and
c) be?
1
A
1
B
2
1
1
C
3
2
(a)
2
2
(c)
(b)
3. Four walls of a rectangular room 16 feet long, 9 feet wide, and 7 feet
high are to be painted, with the exception of one doorway which is 6 feet by
3 feet and a 4 ft. by 3 ft. rectangular window. How many square feet are to
be painted? If paint is sold only by the gallon and one gallon will cover 125
square feet, how many gallons will have to be purchased?
4. A parallelogram is known to have base of length 7 units and a side of
length 3 units. What is the largest its area could be? What is the smallest
its area could be?
41
5. Suppose that a rectangle is measured to have sides of lengths 13 and 15
centimeters, to the nearest centimeter. What can you say about the area of
the rectangle?
6. A rectangular floor is to be tiled with rectangular tiles which are sold in
the following sizes: .1 ft.  1 ft., 1 ft.  2 ft., 2 ft.  2 ft., 3 ft.  3 ft.. If the
room has dimensions 12 ft. by 14 ft. , and all tiles used must be of the same
size, which sizes of tiles could be used without cutting any tiles? For each
usable size, determine the number of tiles that would be needed.
7. Explain carefully why the triangle area formula works for triangles of the
sort pictured in Fig. B6- 2, (a).
8. Use Heron’s formula to compute the area of a triangle with sides of
lengths 5, 5 and 6 units respectively.
42
Supplement C
Geometry revisited--the deductive approach
C1: Introduction
In Supplements A and B, we discovered a number of properties of geometry
which were useful in solving measurement problems (the angle sum formula
and Pythagoras’ Theorem, for example.) We used several other properties
without even mentioning them because they appeared completely obvious
(for example the fact that two congruent right triangles can be put together
to form a rectangle). These are just a few of a long list on geometric facts
that have been discovered and recorded through the centuries. By the time
of Euclid (about 300 BC) the list had grown so long that an attempt to
organize it was in order, a task Euclid attacked with extraordinary success.
His approach, which with minor variations is still taught in many high
schools today, is very similar to the approach scientists use to put their
observations of physical phenomena into some cohesive logical structure.
For scientists, this involves first identifying among the various “properties
of nature” they have observed through the years, the ones that appear to be
most fundamental and at the same time most “self-evident”. (It is important
to note that these observed “laws” have on occasion turned out not to be
correct at all --new and more accurate experimental techniques at subatomic
scales have shown for example that Newtonian laws of mechanics are not
really correct in all cases). The scientists then attempt to predict on the
basis of these laws (without referring to experiments or new observation)
how various phenomena should behave. These predictions can then be
tested experimentally to see if the behavior really occurs. If not, an attempt
is made to reformulate the basic laws in such a way as to make more
accurate predictions possible. At this stage the circle of development
continues--a never ending process of observe, record, identify basic
principles, predict, observe (to test the predictions), etc.
In this Supplement we follow a part of the historical path of the
development of deductive geometry, particularly as it appears in the
writings of the Greeks, with Euclid the major contributor. We proceed in
the order sketched above in the description of the scientific method.
43
Already we have performed much of the necessary experimentation and
recording of apparent “facts”, so we begin by selecting from the many facts
we “know”, a few that appear particularly important and “obvious”. A
scientist might call these the “laws of geometry”, but we will follow Euclid
and call them axioms or postulates. From these basic assumptions we will
attempt to logically deduce (mathematicians call this process proof) new
facts about geometry which can be checked against our experimental
observations to see if they agree. In many cases, this process will simply
reinforce our belief that properties we have discovered experimentally really
are valid. In some cases, conclusions based only on abstract reasoning may
suggest new, as yet unobserved phenomena that experiments can validate.
C2: Points, lines and planes
Our study begins with a decision as to what particular aspects of our
universe we will include in the domain of geometry? In our earlier
discussions we have identified many objects of interest--segments, triangles,
quadrilaterals, circles, angles-- and we could have identified many more.
Each of these is an object in (3-dimensional) space. We think of space as a
collection (or set) of points, each representing a single position. Euclid
characterized points as having no dimensions--no length, no width, no
height--a description that fits our conception of “point” quite well. A
(straight) line is a set of points we picture as having infinite length (in two
opposite directions), but no measurable width or height, while a plane is
visualized as having infinite length and width, but no measurable height (i.e.
it is viewed as a flat surface). We will consider points, lines, and planes the
fundamental objects of our study.
To be formally correct, we should mention that we haven’t really defined
point, line or plane since the words length, height and width have not been
defined. We have just suggested informally the characteristics we visualize
them as having. Advanced geometry books introduce point, line and plane
as undefined objects, a position that is logically correct, but which goes
beyond the understanding we expect from an elementary (or perhaps even
high school) student. In this supplement we will not attempt to give a
formally correct or complete treatment of Euclidean geometry, but rather try
to convey the essence of the subject without getting overly encumbered by
technicalities (though they are important).
A first example of the sort of simple geometric principles (laws of
geometry) we will take as fundamental is
44
Point-line Axiom:
If P and Q are two different points in space, then there is exactly

one line PQ which contains them both.
P
Figure C2-1
This first axiom, which describes the most important
and obvious relationship between points and lines, is
surely is supported by all of our experimental evidence
Q and observations. Without specifically saying so, the
statement nicely reflects our picture of lines as straight
paths, since if curved paths were also allowed to be
called lines, we could easily find many lines through
both P and Q (See Fig. C2-1).
Perhaps not so obvious, but easy to visualize after a bit of experimentation,
is the basic relationship between points and planes:
Point-plane axiom:
If P,Q, and R are three points which do not all lie on some line, then
they are contained in exactly one plane
To make this statement a bit more understandable, you might want to look at
some simple examples (for instance, suppose P, Q, and R are three corners
of the blackboard, and the blackboard represents the plane containing them).
Then think about why the conclusion would not be true if all three points
were on the same line.
The relationship between points, lines and planes is governed by
Line-Plane Axiom:

If P and Q are two points in a plane, then the line PQ lies entirely
within the plane.
In combination with our view of lines as straight, this axiom reflects the

flatness of a plane, since if it were curved, the line PQ would also have to
curve to fit into the plane.
As we identify the important relationships we find between the fundamental
objects of geometry, we should remember that we often refer to one point
45
lying between two others. This is a relationship between three distinct
points, which makes sense only for points lying on the same line. It is
pictured in Fig. C2-2. A is between P and Q, while B is not. Since the
concept “between” cannot be described in terms of our few fundamental
objects (points, lines, and planes) only, it
P
A
Q B
enters our vocabulary as a new fundamental
Figure C2-2
idea.
Exercises:
1. Assuming that Axiom 1 is true, explain why the following statement
must be also true: If L and M are distinct lines, then L and M have at most
one point in common.
C3: New objects from old
From the fundamental objects of geometry (points, lines and planes), we
need to be able to build all of the objects we consider part of the subject of
geometry. Probably the most important new object we want to have in our
vocabulary is a segment. This we describe as follows:

The segment PQ is the set of all points on the line PQ which are
between P and Q, together with the (end)points P and Q.
Notice that we have followed the rules of the deductive geometry game
here. We have mentioned in the definition of segment only concepts already
introduced--point, line, and between.
Of course, once the concept of segment is available to us, we can describe
the concept of triangle:
If A,B, and C are points which do not all lie on the same line, then the
triangle ABC is the union of the segments AB , BC and CA .
The points A, B, and C are called the vertices of the triangle. Again, as
required by the rules of the game, this definition depends only on concepts
previously introduced (we assume that the vocabulary of set theory--union,
intersection, subset, etc.--is available to us in advance of our study of
geometry). We could now proceed to give definitions of quadrilateral,
pentagon, hexagon, polygon, using only the terms at hand. We leave these
as exercises, noting that they are more difficult that the definition of a
46
triangle since one must worry about “sides” that cross each other. A more
subtle problem that occurs in defining these figures is caused by the fact that
if we try to begin as with a triangle, namely picking four points X,Y,Z, and
W for vertices of a quadrilateral, we have no assurance that they will lie in a
plane, even though our idea of a quadrilateral is a figure lying completely in
some plane (Axiom 2 gets us out of this possible predicament in the case of
triangles). For more complicated figures one must stipulate as part of the
definition that the vertices lie in a plane.
The name “triangle” for the three sided figure defined above is rather
interesting. It suggests that perhaps we are looking at the figure differently
than the person who settled on this name. The way we defined it, one would
expect the name to be “trilateral” or “trigon”, indicating a figure with three
sides. Of course, if we look at a picture of a triangle, and think about it as a
composite of parts, we see that it could also be seen as built of angles (or at
least parts of angles). To fit this discussion into our deductive geometry
scheme, we need to define the term “angle” before we can legitimately use
it. We know from our descriptive work in Supplement A what we mean by
the term, so we need only see whether we can fit it into our scheme without
introducing any new fundamental terms. We first need to define ray:

The ray PQ (with endpoint P) is the union of the set PQ with the set of

points X on PQ such that Q is between P and X
There’s no need for any new terms in this definition, so it is admissible.
You might want to draw a picture of the set of points described here and see
if it looks like you think a ray should look. On your picture you will see a


second ray which is part of the line PQ , called the opposite ray of PQ . It

can be described as the point P together with all points of PQ which are not

contained in PQ . Now that we have defined rays, angles are simple:
An angle is the union of two rays with a common endpoint.
Notice that this definition does not use fundamental terms, but it does use
only terms that have already been defined using fundamental terms. What
we are trying to do is make a geometry dictionary which begins with the
fundamental terms and contains no circular definitions. That is, we don’t
47
want to define concept X in terms of concept Y and then turn around and
define concept Y in terms of concept X.
A square would seem to be a logical next choice for definition. Here we
run into some difficulties, despite the fact that a square seen holistically is a
simple and easily identifiable object. The problem is that what we see when
we look at a square, is that the sides all appear to be identical segments,
differing only by their position in space. We have nothing in our
“dictionary” so far that could be used to describe this phenomenon, so we
must add another fundamental term, congruence (of segments and of
angles), describing a relationship between these figures. Intuitively, we
think of congruent figures as figures of the same size and shape, but
conceivably positioned differently in space.
Euclid, and many geometers after him, used the word equal for this
relationship. In modern-day mathematics this usage creates problems, since
in the terminology of sets, two segments should be considered to be equal
only if they consist of exactly the same points. This is more than we mean
to say, so we need a new word to describe the relationship we want, hence
congruent. There is a standard shorthand notation we will use for the
statement “is congruent to”. We will write AB  CD to mean that the
segment AB is congruent to the segment CD , and similarly A  B.
We could now try to define a square as a quadrilateral having all four sides
congruent. Try to draw a few quadrilaterals with four congruent sides. Are
they all squares? Hopefully not, or you haven’t been creative enough. A
quadrilateral with all four sides congruent is called a rhombus. So a square
is a rhombus, but a very special kind of rhombus--a rhombus with all four
angles congruent!
A square is a quadrilateral with all four sides congruent and all four
angles congruent.
We are now pretty well supplied with fundamental concepts of our
geometry, let’s recap them: point, line, plane, between, congruence.
Hopefully, every geometric object we will be interested in can be defined
starting only with these and with the vocabulary of sets. Let’s try another-parallelogram. We decided in our descriptive phase that this should be a
quadrilateral with opposite pairs of sides parallel. But what does
parallel mean?
48
Lines L and M are parallel if they have no point in common (in set
terms L « M={ }).
With this agreement, we have a usable definition of parallelogram which
reflects an essential idea behind parallelism. (You might prefer a definition
saying something about lines which are everywhere equidistant, a
reasonable reflection of our experimental observations. We have chosen the
other definition in part because we do not yet have a way to define
“equidistant”.)
Finally, we should look at the concept of a circle. This is difficult, since a
picture of a circle does not suggest any definitive characteristics (“round” is
not part of our dictionary, so it cannot be used in a definition). If we think
about making a circle with a compass, however, a good definition is
suggested:
A circle with center O and radius AB is the set of all points X in a plane
containing O for which OX  AB .
In the compass construction, the center of the circle is the point where the
compass point is set, the radius AB is the segment between the compass
point and the point of the pencil. The difficulty with this definition, from
the point of a young child, is that a circle in nature is just a round object. It
no center marked, and without a center, it is hard to visualize a radius.
Notice that we have chosen here to define a circle in terms of congruence,
rather that using the more common statement that a circle is the set of all
points in a plane equidistant from a given point O. Again we have done so
because “equidistant” is not yet in our dictionary, and we wish to keep the
list of fundamental terms as short as possible.
Exercises:
49
1. Give a careful definition of a quadrilateral, taking care that it exclude a
figure like
B
A
C
D
Do the same for a pentagon.
2. A supplement of an angle A is the angle formed by one side of A
and the ray opposite to the other side.
a) Sketch an angle and its supplement. How many supplements does an
angle have?
b) Give a careful definition of a right angle based on the relationship
between the angle and its supplement.
3. This diagram illustrates a strange fact about segments--there is a
one-to-one correspondence between the points on the
P'
shorter segment XY and the longer segment AB! To
A
B
each point P of XY one associates a point P' of AB as
X
P

Y
follows: P' is the point of intersection of the ray OP
and the segment AB.
O
Explain how to make similar one-to-one correspondence between the points
of the small circle and those of the large circle in (a). Do the same for the
points of the semicircle and the points of the entire line in (b).
50
(a)
(b)
C4: More “Laws of Geometry”
Once we have our basic definitions in place, we can get down to the real
business of deductive geometry--identifying a few basic principles from
which we hope to deduce all of the facts that we have earlier discovered via
hands-on exploration. We have noted three of these (Axioms 1, 2 and 3 of
§C2) which deal with relationships between points, lines, and planes.
(There are several other, very technical axioms which one would need to
include if this were an exhaustive course in deductive geometry. Many of
these were introduced by the German mathematician David Hilbert in the
late 19th century, when he undertook to bring the axiomatic setting
developed by Euclid up to the logical standards required of a mathematical
theory in modern times. We will pass over these, and concentrate on
principles which stand out as important and self-evident to an interested
person who is not trained in sophisticated mathematics.) Given the long list
of known geometrical facts, one could make the choice of basic principles
from this list in many ways. We will choose principles which make it easy
to deduce some of the basic measurement principles--the angle sum in a
triangle, Pythagoras’ theorem, and the AAA-principle for similar triangles-in a relatively short and painless manner. Our axiom set is thus rather
different from that found in Euclid.
We have mentioned before that the principles chosen as fundamental should
be “self-evident”, and they should be useful. Beyond the three axioms
relating points, lines and planes stated in §C2, we take as a new principle
The Vertical Angle Axiom:
If two lines intersect at a point P, then each pair of vertical angles
formed at P consists of congruent angles
51
Here vertical angles are any pair of angles such that the sides of one are
the opposite rays of the sides of the other. Fig. C4-1 displays vertical
angles a and b. Notice that there is
another pair of vertical angles in that
diagram. Axiom 4 clearly satisfies our
b
a
requirement of being self-evident. If we
rotate the diagram 180 about the point of
intersection of the two lines, it appears
Figure C4-1
obvious that a will come to rest precisely
atop b, and conversely, so the angles
satisfy our intuitive meaning of congruent. The usefulness of this axiom will
not be apparent until we discuss the exterior angle theorem.
More generally useful, but not as “self-evident”, is an axiom related to
congruence of triangles. Its usefulness is tied to the fact that triangles play a
very critical role as building blocks for regions in the plane. We saw in our
discussion of area formulas and angle measure that rectangles,
parallelograms, and even pentagons and hexagons, can be broken down into
triangular regions, and properties of the triangles can be used to derive
similar properties for these regions. Since Axiom 5 will refer to
congruence of triangles, we need to first define this concept in terms of
words already in our vocabulary, if possible. We want to be able to describe
formally what t means to say informally that when one triangle is placed
upon another it
will fit precisely. Let’s try to
Z
C
describe it for the triangles
ABD and XYZ pictured in
Fig. C4-2. In this case, the
obvious requirement is that
A
B X
Y after moving ABC and
setting it on XYZ, the
Figure C4-2
vertices A, B, and C fall
exactly on the vertices of XYZ. Of course, A may not fall on X, or B on Y
when they fit, but they will match up in some order. Mathematically
speaking, this gives a one-to-one correspondence between the vertex set
{A,B,C} and the vertex set {X,Y,Z} (there are several such correspondences
but in Fig. C4-2 only AY, BX, CZ makes the triangles fit exactly).
What else do we see when we fit the triangles together? AB is lying
precisely on top of YX, AC on top of YZ , and BC on top of XZ. That is to
52
say, the pairs of corresponding sides (for the given correspondence) of the
two triangles are congruent. In the same way we see that the pairs of
corresponding angles A and Y, B and X, C and Z are congruent.
On the other hand, if we could have matched up the vertices so that all these
corresponding parts were congruent, it is clear (at least after a bit of
experimentation) that the triangle ABC can be fit precisely on XYZ.
Summarizing these observations, we define:
ABC is congruent to XYZ (ABC  XYZ for short) if and only if
there is a one-to-one correspondence between the sets of vertices of
the two triangles such that the pairs of corresponding sides are
congruent and the pairs of corresponding angles are congruent.
The one-to-one correspondence between vertex sets for which each pair of
corresponding parts is congruent is called a congruence between the two
triangles.
Warning: It is customary in many geometry texts to write ABC  QXYZ
to mean not only that the triangles are congruent, but that the congruence is
AX, BQY, CQZ . We will not use that convention here,since we wish to
emphasize importance of identifying the one-to-one correspondence each
time there is question of whether two triangles are congruent.
So far, we have only showed that we can increase our dictionary of
geometric terminology to include a definition of congruence of triangles,
using only terms already in our dictionary (as long as we include our
fundamental terms in the dictionary). An important observation here is that
if two triangles are congruent, then six pairs of triangle parts, three pairs
of sides and three pairs of angles, are congruent. The reason that this is
important, is that our experiments in Lab 16 for instance, have suggested
that only knowing three pairs are congruent is enough to force congruence:
Side-Angle-Side Congruence Axiom:
ABC is congruent to QXYZ if there is a one-to-one
correspondence between the sets of vertices of the two triangles such
that two pairs of corresponding sides are congruent and the pair of
angles formed by these sides (included angles) are congruent.
This axiom, which we denote by SAS, is one of the fundamental principles
for Euclidean geometry. It becomes “self-evident” if we work through
numerous exercises such as those in Lab 16. To gauge its usefulness, let’s
53
look at how this might be applied to a problem. Suppose we are to measure
the distance between two trees on opposite sides of Mirror Lake, and we are
not allowed to enter the water (See Fig. C4-3). We could move away from
the lake until we reached a point S where we could see both trees with no
water
tree
S
?
tree
Y
X
?
Z
Figure C4-3
between us and the trees (of course, if the shore is irregular enough this
might be impossible, but let’s assume this works). Notice that we need to
know the length of one side of a triangle to solve our problem. Now we run
strings from the position (S) to each tree and mark off the distance to the
tree on the string. We also copy the angle between the two strings on a
large sheet of paper so we can duplicate it. We move off to some level
piece of land, and there use our string and angle to reproduce a triangle
XYZ with two sides and an included angle congruent to those formed at
the lake. Since SAS says that this triangle must be congruent to the original,
the side XZ must be congruent to the segment between the trees. So, if we
measure the length of this side, which we can do directly since no water
now interferes, we have in fact found the distance across the lake.
Of course, this is a fairly simplistic application of SAS to an indirect
measurement problem. The applications of SAS in the development of
geometry are much more far reaching. An important example is pictured in
Fig. C4-4. In that diagram, ACE is called an exterior angle of ABC.
54
A
G
A
F
B
C
E
B
C
E
Figure C4-4
The Exterior Angle Theorem (which seems intuitively clear from the
picture) says:
An exterior angle is greater than either of the remote interior angles
(ABC or BAC).
That this is called a theorem, rather than an axiom or a postulate, is no
accident. The word “theorem” reflects the fact that one need not decide the
truth or falsehood of the statement on the basis of experiments or
observations, but that its truth can be deduced logically as a consequence of
the axioms (provided that the axioms themselves are true). To verify this
theorem, one need not look at any empirical evidence, one need only apply
deductive reasoning to principles already accepted. Let’s work through that
reasoning.
To show that BAC is smaller than ACE it is enough to show that an
angle congruent to BAC lies entirely within the interior of ACE. To do
this, we draw some extra points and segments in the diagram, helping us
envision what is happening. We label by F the midpoint of AC , that is

AF . FC On the ray BF we label by G the point such that F is the midpoint
of AG . The theorem will be proved to be true if we can show that BAF
(=BAF) is congruent to FCG. How can we do this? We consider the
correspondence BG, AC, FF between vertex sets of BAF and FCG.
The vertical angle axiom tells us that the corresponding angles QBFA and
QGFC are congruent, and our selection of F as midpoint tells us that both
BF  FG and AF Q FC . Since the hypotheses of SAS are fulfilled, we can
apply it to conclude that the given correspondence is a congruence between
BAF and FCG. Once we know this, the definition of congruent triangles
tells us that the corresponding angles BAF and FCG are congruent,
which is what we set out to show.
55
The above argument demonstrates how one can discover new facts about
geometry, not only by physical experimentation but by pure logical
experimentation, once one has identified the fundamental principles of the
system. Of course, it would now be a good test of the axioms we have
selected to experiment with a number of triangles and see whether the
theorem really does hold experimentally.
Another result we can deduce from SAS and reasoning is:
The Angle-Side-Angle Theorem (ASA):
DABC is congruent to DXYZ if and only if there is a one-to-one
correspondence between the sets of vertices of the two triangles such
that two pairs of corresponding angles are congruent and the pair
of sides shared by these angles (the included sides) are congruent.
We include this theorem not only because it will prove useful a bit later (and
of course is something that experimentation similar to Lab 16 would lead us
to suspect was true), but because the reasoning that we use in verifying it is
rather different than the “standard” reasoning we are often used to.
In applying the most common reasoning process, direct reasoning, one
begins with one or more statements which are presumed to be true, and by a
sequence of arguments of the sort “if this statement is true, then that
statement must be true” concludes that some additional statement is also
true. The proof of the exterior angle theorem followed this reasoning
pattern. It began with the assumption that the vertical angle axiom was true,
that the SAS axiom was true, and that it was true that certain segments were
congruent. The truth of the vertical angle theorem led to the truth of the
congruence of a pair of angles; the truth of the congruence of these angles
and the two pairs of sides, together with the truth of SAS, led to the truth of
the congruence of the triangles; the truth of the congruence of the triangles
led to the truth of the congruence of the targeted pair of angles.
In some situations, this type of reasoning does not fit the information one
has access to, so alternative (but equally valid) types of reasoning--indirect
reasoning-- must be used instead. One common form of indirect reasoning
is called reasoning by cases. Roughly speaking this works as follows. We
want to prove that Statement A is true, but we do not know how to do this
by a direct proof. However, we are convinced that either Statement A,
Statement B or Statement C is true (3 cases). If we can prove that both
Statement B and Statement C are false, we are left with the conclusion that
56
Statement A must be true. This type argument can be applied to any number
of cases.
Let’s apply this technique to the ASA theorem (see Fig. C4-5).
Suppose that the hypotheses of the ASA theorem, namely AX,
Z
B Y and AB  XY , are
D
satisfied. We want to look at
C
the cases possible in
comparing AC to XZ . One of
a) AC  XZ
X
Y
b) AC is longer than XZ
A
B
c) AC is shorter than XZ
is surely true.
Figure C4-5
We want to show case a) is
true, since then the hypotheses of SAS are satisfied so ABCXYZ, which
is what we are trying to show.
We will do this by showing that neither b) nor c) can possible be true. To
show that c) cannot be true, we again use a proof by cases, noting that there
are two possible cases: c) is true or c) is not true. We’ll show that if c) were
true, then B would not be congruent to Y, which is absurd since we
started out with the information that B Y. We have to conclude that this
case is impossible, so the remaining case-- c) is not true-- must be valid. A
similar argument would show that b) is also not true.
Now let’s convince ourselves that if c) is true, then B is not be congruent

to Y. If c) is true then XZ is longer than AC , so there is a point D on AC
with XZ AD . If this is true, then SAS tells us that ABDXYZ, and then
the definition of congruence tells us that the corresponding angles DBA
and Y are congruent. Since DBA is larger than B, B and Y are not
congruent.
After all this work, the proof of ASA is complete. It is most certainly a
complicated proof, but a clever one.
Remember the problem of measuring the distance across the river in §B3.
57
To make it work, we needed to know that a triangle
with two congruent angles, has congruent sides
opposite those angles. If we look at a few pictures,
we can easily believe that this statement is true.
Without looking at pictures, we could also convince
ourselves that it is true, basing our deduction only on
the “self-evident” axioms we have written down, and
B the theorems we have proved using those axioms.
A
Let’s consider the situation pictured in Fig. C4-6,
Figure C4-6
and suppose that we know that AB. Now
consider ABC, and the correspondence AB, BA, CC of its vertex set
with itself. We know that AB, BA and AB BA (in fact they are
equal), so by ASA this correspondence is actually a congruence of ABC
with itself. So the corresponding sides AC and BC must be congruent,
which is what we set out do show.
C
Exercises:
1. Using the SAS axiom as a starting point, give a convincing argument that
a triangle with two congruent sides (an isosceles triangle) has
congruent angles opposite those sides.
2. Suppose that the correspondence AX, BY, CZ is a congruence
between ABC and QXYZ, and that AC  BC . What other
correspondences between the vertex sets must also be congruences?
Which correspondences would be congruences if the triangles were
equilateral, that is, all three sides of ABC were congruent?
3. IS, JT, KR is a correspondence (but not necessarily a congruence)
between the vertex sets of triangles IJK and RST. List the pairs of
corresponding sides (for example IK  SR ). List the pairs of corresponding
angles.
4. The triangles below are not accurately drawn, but you are to assume that
they are accurately labeled. Which of the pairs of triangles are congruent?
For each pair, explain how you know they are or are not congruent, or note
that not enough information is given to come to a conclusion. For each
congruent pair, give the relevant correspondence between vertex sets.
58
Z
3
C
E
40
A
2
57
(b)
X 40
(a) 3
W
2
U
4
B
4
57
57
Y
A
V
C
2
S
I
Z
40
(d)
(c)
55
R
85
X
85
Y
7
4
3
40
3
B
7
K
4
55
T
C
A
5. Find x and y.
C
4
x
8
Y
63
A
9
4
B
76
y
63
X
9
Z
C5: Parallel lines
One of the enduring mysteries in geometry, from the time of Euclid until the
19th century, was concerned with the role of the concept of parallel in the
setting deductive geometry. Already by the time of Euclid, it was known
that if we accept his axioms as true, then we must logically accept the fact
that there are parallel lines. One can deduce this from the following
theorem (see Fig. C5-1).
Alternate Interior Angle Theorem (part 1):
J
59
If L and M are distinct lines in a plane, with transversal T, and if a
pair of alternate interior angles formed are congruent, then L and
M are parallel.
Before trying to deduce this from our axioms and previous theorems, we
should make sure the terminology is clear. By a transversal to two lines L
and M, we mean a line which intersects them both.
By alternate interior angles formed
by a transversal, we mean angles
A
L

such as  and  in Fig. C5-1. It is, of
course, possible to give a formal

definition of these angles in terms of
B
M L, M, T and their points of
intersection. We leave that as an
exercise. There are two pairs of
T
alternate interior angles in Fig. C5-1,
only one of which has been labeled.
Figure C5-1
Note also that there are transversals
(and accordingly pairs of alternate
interior angles) for any pair of
M
distinct lines in a plane (see figure
C5-2), not just for parallel lines. The

distinct feature of parallel lines is
L that the pairs of alternate interior

angles are congruent.
We will use indirect reasoning again
in deducing that the theorem stated
T
above is true. So we begin with the
fact that the angles a and b are
Figure C5-2
congruent, and reason by cases: a) L
and M are parallel, or b) L and M are
not parallel. We want to rule out case b) by showing that it leads to an
absurd conclusion. So suppose that the lines L and M are not parallel, that
is, they meet at some point P. Let’s suppose that P is on the right side of T
in Fig. C5-1, so b is an exterior angle of the triangle ABP, and a is a
remote interior angle of that triangle. (Sketch this picture to better
understand the proof). Now we know from the exterior angle theorem that
b is greater than a. But this is absurd, since
60
P

M

L
we started out knowing that they
were congruent. Reasoning by
cases, we conclude that b) is false,
so a) must be true. So the lines
must be parallel.
This alternate interior angle
theorem gives us a way to
construct parallel lines provided
T
we know how to copy an angle
(See Fig. C5-3). If we begin with
Figure C5-3
a line L and a point P not on L, we
draw any line T through P which intersects L. At P we copy the angle a
formed by T and L as in Fig. C5-3, , with one side a ray lying on T. By our
alternate interior angle theorem the remaining side of this angle lies on a
line M through P parallel to L. This argument gives us the
Existence of Parallels Theorem:
Given any line L and any point P not on L, there is a line M through
P which is parallel to L.
The alternate interior angle theorem also provides a practical test for
parallelism: To determine whether lines L and M are parallel, draw a
transversal and compare the alternate interior angles formed. They are
congruent if, and only if, the lines are parallel. This is particularly useful
because the definition of parallel, given the fact that lines extend
indefinitely, is one that could never be checked directly. (How far along the
lines would you have to go before you concluded that they would never
meet?).
The real controversy related to parallel lines which followed the publication
of Euclid’s Elements dealt with the converse of the theorem we have stated
above. Euclid took as an axiom a version of the following:
The Parallel Axiom:
Given a point P not on a line L, there is exactly one line through P
parallel to L.
That is, once we use our theorem to find a parallel through P, there are no
other parallels possible. As axioms should, this one seems self-evident, and
in fact no one doubted that it should be true. The question was--should this
61
really be called an axiom, or should we be able to prove it as a theorem on
the basis of the other axioms we have already selected. Many people
attempted to prove this through the centuries, until 19th century
mathematicians succeeded in showing that it could not be proved from the
other axioms--Euclid had guessed right. We will take the Parallel Axiom as
our final fundamental principle of geometry (it joins the Point-Line Axiom,
the Point-Plane Axiom, the Line-Plane Axiom, the Vertical Angle Axiom,
and SAS). As a first consequence of this axiom, we can deduce that the
converse of the alternate interior angle theorem (part 1) is true.
Alternate interior angle theorem (part 2):
If two parallel lines L and M are cut by a transversal T, then the
alternate interior angles formed are congruent.
Before we look at the reasoning behind this claim, you should make sure
that you understand that this says something different than part 1 of the
theorem. The words are nearly the same, and this leads many first time
readers to assume that these are the same statement. Look carefully at the
two statements, and compare the statements to the following two: i) When it
rains, the Clippers don’t plan: ii) When the Clippers don’t play, it rains.
The Part 1 and Part 2 of the Alternate Interior Angle Theorem have exactly
the same relationship as these two statements about the Clippers. They are
converses of each other, and the truth of one does not automatically imply
the truth of the other.
So how do we “prove” part 2? Let’s suppose that L, M, and T are the lines
pictured in Fig. C5-1, and that A is the point where L and T intersect. Now
forgetting L, we construct as in the proof of the Existence of Parallels
Theorem, an angle at A congruent to the angle –b formed by M and T. One
side of this angle lies on a line L’ which is parallel to M and which forms
congruent alternate interior angles by the manner in which it was formed.
Now we have both L and L’ being lines parallel to M and passing through
the point A. The Parallel Axiom tells us that these are the same line, so
L=L’. We conclude that since L’ gives us congruent alternate interior
angles, so must L, which is what we set out to show.
Let’s recall that one of the rules of the deductive geometry game is that one
periodically checks the predictions of the theory against observations from
experiments. We can do that now, returning to an old question about the
62
angles of a triangle (see Fig. C5-4). We begin with a triangle ABC, and

take a line L through C parallel to AB (which exists because of the
Existence of Parallels Theorem).
C
L

 and  are alternate interior
 



A
B
Figure C5-4
angles (for transversal AC ), so
they are congruent by Theorem 2.
For the same reason,  is
congruent to  (use transversal

BC ). Looking now at the angles
formed at the point C, we see that our deductive geometry predicts that
congruent copies of the three angles of the triangle, laid side by side,
form exactly a straight angle--something we observed experimentally by
cutting out triangles and tearing off their angles.
During a recent workshop, teachers who were asked to determine a method
to measure the distance across Mirror Lake (see §C4),
generally
X
chose the
tree
method
pictured in
Fig. C5-5.
?
They
Y attempted to
move from
tree
the two trees
along parallel
Figure C5-5
paths, and
thus to form a parallelogram with side XY. They then measured this side
and asserted that they had found the distance between the trees! Why would
this work? The required fact, which seems to be true if we experiment with
various parallelograms, is: opposite sides of a parallelogram are
congruent.
D
C This is again something we can attempt to
deduce directly in our geometry, without
experimentation (see Fig. C5-6). Here
ABCD is a parallelogram, so AD is parallel
to BC and BD is
A
B
a transversal to the lines containing these two
Figure C5-6
63
segments. The alternative interior angle theorem tells us that
ADBDBC. Now using the parallel sides DC and AB, with transversal
BD we see that the alternate interior angles ABD and CDB are
congruent. Combining these facts with the fact that BD is congruent to DB ,
we can use ASA to conclude that AC, BD, DB gives a congruence
between ABD and BCD. Since corresponding sides of congruent
triangles are congruent, we see that AB DC and AD  BC . Which is what
we set out to show. In fact, we actually showed more, since if we look at
corresponding angles we will discover also that opposite angles of a
parallelogram are congruent.
Exercises:
1. Is it possible to find a pair of lines which are not parallel, yet which do
not have any points in common? Explain.
2. Explain carefully why the measure of each angle of an equilateral
triangle must measure 60 .
3. Which of the following pairs of lines do you know are parallel (assuming
all angle measures are in degrees?
(a)
(b)
45
45
(c)
140
40
75
(d )
70
70
75
4. Assuming that the labeling (in degrees) is correct, are there any parallel
lines in the diagram below? Any right angles?
64
25
65
65
20
4. The diagram below is not drawn to scale. Assume that L is parallel to M
and that m(a)=75 , m(b)=138 and m(c)=35 .
c
L
a
Find the measures of the other
marked angles.
u
e
d
b
M
5. Suppose that DABC is isosceles.
a) What are the measures of the other angles if m(a)=92 ?
b) What are the measures of the other angles if m(a)=38 ?
7. Suppose that AB is a diameter of a circle with center P, and that K is a
point on the circumference of the circle.
K
a) Explain why PAK  PKA
b) Explain why PBK  PKB
A
P
B
c) Explain why 2m(PKA)+2m(PKB)=180
d) Conclude that BKA is a right angle (this
result is often stated as:
an angle inscribed
in a semicircle is always a right angle)
8. Look back at the proof of the Exterior Angle Theorem in section C4.
There we showed that ACG  A. Explain why it then follows that
GCE  B. Then explain why m(A)+m(B)=m(ACE) (in other
65
words the measure of an exterior angle of a triangle is equal to the
sum of the measures of the remote interior angles).
9. Find the area of this isosceles triangle. (Hint: you may want to sketch in
a segment from the top vertex to the midpoint of the base). Explain
carefully why your technique for computing the area works.
5
5
6
10. In ABC m(A)=70 . In XYZ m(Y)=115Q. Is it possible that the
triangles are congruent?
11. Suppose you are standing on the side of a deep stream, and a tree stands
on the other side. If you sight from point A (on your side of the stream)
to the top of the tree, the line of sight makes an angle of 30 with the
(level) ground. If you walk directly away from the tree 15 feet from
point A, the line of sight makes angle of 15 with the ground. Describe
how you could use this data to determine the height of the tree. What is
it?
12. In the diagram below, the pair of angles a and b are called
corresponding angles formed by the transversal T to the lines L and M.
Using the Vertical Angle
Axiom and the Alternate

L
Interior Angle Theorem, explain
why L is parallel to M if and

M only if a and b are congruent.
T
C6: Similar triangles and more indirect measurement
66
One of the most useful measurement tools based on geometry is the theory
of similar triangles. Intuitively, similar triangles are of the same shape
but not necessarily of the same size. In particular, congruent triangles are
always similar, but similar triangles aren’t always congruent. Perhaps the
simplest example of the latter are two equilateral triangles, one with sides of
length 2, one with sides of length 3. If we try to describe similar triangles in
terms of relationships among their parts, as we have done with congruent
triangles, it is clear that we cannot specify congruence of pairs of sides, but
examination of a number of pairs of triangles for which there is a one-toone correspondence between vertex sets such that all pairs of
corresponding angles are congruent shows that this condition, which we
will call the Angle-Angle-Angle Similarity Criterion (AAA for short),
produces triangles which we would agree of the same shape. It is natural to
ask what, if any, relationship
Z
there is between lengths of sides.
(a)
C
We hope, of course, that some
useful relationship will exist,
since that could be the basis of
useful applications of similar
A
B
X
Y triangles.
Z
Lets look at a simple example.
Suppose that for triangles ABC
W
V
and XYZ as pictured in Fig.
C6-1, a) we know that AX,
BY, and CZ (that is, the
X
U
Y
AAA similarity criterion holds)
Figure C6-1
and suppose also that we know
that XZ=2AC, that is, we know that one side of XYZ is twice as long as
the corresponding side of ABC. In Fig. C6-1 b), we have copied XYZ
and labeled the midpoint of XZ by W, so we have ZW  XW AC . Now we
(b)


draw a line parallel to YZ through W and label the point where it meets XY

by U; and we draw a second line through W, this time parallel to XY and

meeting YZ at the point V. Now we can apply practically everything we
know about parallel lines and congruence of segments.
First, we have that UY  WV since UYVW is a parallelogram. Then we look
at the two triangles XUW and WVZ with correspondence XW, UV,
67
WZ. We know WXUZWV. Why? Similarly we know that
XWUWZV. Why? Since we started with the fact that ZW  XW, we
have exactly the right facts to apply ASA to see that our correspondence is a
congruence. Since other corresponding parts must also be congruent, we
get that XU WV . Combined with UY  WV , which we already knew, we
have XU UY , which means that U is the midpoint of XY and XY=2XU.
Now we started the problem knowing that AX, BY, and CZ,
and we just saw that XWUWZV= Z. So we know that CXWU and
AX, Plus we have that AC=XW so AC  XW. This is enough to show
(using ASA again) that AX, CW, BU is a congruence between ABC
and QXWU, and in turn that AB XU so AB=XU. What we really showed
here is that our process succeeded in copying ABC inside XWU with A
landing on X.
All together, we have seen that XY=2AB, that is, side XY is twice as long as
side AB. If we had made a similar argument, copying ABC inside XWU
with C landing on Z we would have showed that YZ=2BC.
This formally rather complicated (but visually fairly simple) argument
verifies one special case of the important
AAA Similarity Theorem:
If there is a one-to-one correspondence between the sets of vertices
of two triangles such that all three pairs of corresponding angles
are congruent, then the ratios of the lengths of the pairs of
corresponding sides are equal.
In this particular case we showed that if one ration is 2:1, then all three
ratios are 2:1. In Fig. C6-2 we sketch a diagram that could be used to show
the same result if the ratio of XZ to AC were 3:2, rather that 2:1. You
should be able to show that the three small triangles are all congruent, and
thus
68
Z

conclude that the three parallel lines YZ , BC

C


and the line bisecting AC divide the side XY
into three congruent parts, two of which make

up AB, so the ratio of XY to AB is again 3:2
as the theorem asserts. It should be clear that
for any whole numbers m and n, this same
picture would generalize to show that if XZ to
Y AC were m:n, then the ratios of the other pairs
A=X
B
of corresponding sides would also be m:n.
Figure C6-2
Note that these pictures are the same as those
we encountered in studying triangular tessellations in Lab 28.
How is this useful? Suppose we were to need to find the height of a
flagpole erected vertically on level ground. Fig. C6-3 shows an approach
that will give the height (without climbing the pole). One person (P) sights
from
ground level to the top of the pole, another
pole
rests a meter stick perpendicular to the
ground in the line of sight so the sighter can
h
read the height (d centimeters) on the stick
which lines up with the top of the pole.
meter
stick
P Another team member measures the distance
from P to the meter stick and from P to the
base of the pole, all in centimeters. Now
Figure
there are clearly two similar triangles in this
C6-3
picture, since both share the angle at P, both
have a right angle, and the third angles in each must be congruent since the
three angles of a triangle together make up a straight angle. The AAA
Theorem tells us that, if h is the height of the pole, the following proportion
is valid:
h:d = distance to pole:distance to meter stick.
Solving this proportion would give the height of the pole in centimeters. Of
course, one could also measure everything in meters. (If only whole
numbers are to be used, what would be the advantages and disadvantages of
using meters rather than centimeters?) For example, if the height read on
the meter stick were 80 cm, and the distances to the pole and meter stick
were respectively 620 cm and 124 cm, we would have to solve the
proportion h:80=620:124. This is equivalent to 124h=80620, which gives
a solution of h=400 cm.
69
In many problems dealing with similar triangles, it is only feasible to show,
as we did in this example, that two pairs of corresponding angles of the two
triangles are congruent. The remaining angle is often positioned where it
cannot be directly measured. Thanks to the Angle Sum Theorem, however,
this is enough to invoke the AAA Theorem. Suppose, for example, two
triangles each had an angle of measure 75 , and each had an angle
measuring 38 . The Angle Sum Theorem then tells us that the third angle
in each triangle has measure 180 -(75 +38 )=67 , so we conclude that all
three pairs of corresponding angles are congruent. Using this reasoning at
the outset, we could have renamed our theorem the AA Similarity
Theorem, with the necessary change in hypotheses, and no change in the
conclusion. This is, of course, a better form of the theorem, since we need
only feed in two pieces of information (rather than three) to get the same
output.
As an interesting application of the AA Similarity Theorem we can give a
short proof of Pythagoras’ Theorem, without using areas as we did in our
first discussion of that result. We begin with the right triangle ABC in
Fig. C6-4, and draw in an altitude through vertex B to the opposite side.
Now
A
ABC and BDC satisfy the AA hypotheses,
D
since each has a right angle (and these are
certainly congruent), and the triangles share
the angle C. So for the correspondence
AB, BD, CC the ratios of the lengths of
B
C corresponding sides are congruent. In
Figure C6-4
particular BC:DC=AC:BC or (BC)2 =
=AC  DC. Comparing the triangles ABC and BDA, which share the
angle A, we see similarly that AA, BD, CB satisfies the hypotheses of
the AA Similarity Theorem, so AB:AD=AC:AB or (AB)2= AC  AD.
Adding the underlined equalities, and using the distributive property we get
(AB)2+(BC)2 = (AC  DC)+(AC  AD)= AC  (DC + AD) = AC  AC
= =(AC)2. That is, the square of the length of the hypotenuse is equal to the
sum of the squares of the lengths of the sides!
Exercises:
70
1. A triangle has sides of lengths 2, 3, and 4 units. A second triangle,
similar to the first, has perimeter 18. What are the lengths of the sides
of the second triangle?
2. A child’s stack toy is make up of ten discs of increasing radius, which fit
over a central spindle. Each disc is 3 cm thick. When viewed from the
side, the stack looks like this. Assume the
“edges” lie on a line as shown. Use
similar triangles to find the radius of each
disc.
15 cm
15 cm
3. Find the lengths a, b, and c.
b
12
a
3
12
5
c
4. In the diagram below
a) Is DABC similar to DDEC? Explain
D
B
A
C
E


b) Is AB parallel to DE ? Explain.
c) If m(–D) = 40∞, find the measures
of all the other angles in the diagram.
.d) If CD = 3, AC = 9 and AB = 6, find
DE.
5. Two similar triangles have the lengths of corresponding sides in the ratio
of 1:3. What is the ratio of their areas?
71
5. Is it true that any 2 equilateral triangles are similar? Explain. What about
two isosceles triangles?
7. Suppose the triangles below, while not accurately drawn, are accurately
labeled, with angle measures in degrees. Find x in each pair.
i)
3
ii)
38
x
38
71
6
3
3
10
110
x
32
5
71
110
32
8. In ABC, m(A) = 62 , m(B) = 54 . In XYZ, m(x) = 54 ,
m(Y) = 64 . Is ABC similar to XYZ? Explain.
9. Suppose that AB is a diameter of a circle, P a point on the circumference
of the circle and XΠAB such that XP is
P
perpendicular to AB. Argue that the triangles
AXP, XPB, and APB are similar.
A
X
B
10. A triangle with sides of lengths 3, 5, and 7 is similar to a triangle with
longest side of length 21. How long are the other sides of the second
triangle.
11. A man who is 6 feet tall stands 8 feet from a lamp pole which is 12 feet
tall. If the light from the lamp casts a shadow on the ground, how long
is the man’s shadow?
72
73
Lab 1
Patterns (Triangular Numbers)
1. Look at these triangular arrays of dots
(a)
(b)
(c)
(d)
Count the number of dots in each triangle. How many dots
would there be in the next (undrawn) triangle?
Fill in the table:
74
n
number of dots
1
1
2
3
3
6
4
10
5
6
7
Do you see a pattern which relates the number of dots in the n th array to the
number n? If so, describe the pattern.
You may have noticed the following pattern:
75
n
number of dots
1
1
1=1
2
3
3=1+2
3
6
6=1+2+3
4
10
10=1+2+3+4
5
6
7
Check to see whether this pattern holds in the rest of the table.
Question: What is the 100th triangular number, i.e. the number of dots in the
100th array?
Of course, assuming that our pattern always works, we could use a
calculator to compute the sum of the numbers from 1 to 100. That would
take a long time however, so we might look for a cleverer, quicker way. It
is said that the mathematician Carl Friedrich Gauss was given this problem
by a teacher when he was 10 years old, presumably to keep him quiet while
the teacher did some necessary classroom chores. Gauss almost
immediately gave the answer, however--5050.
76
His reasoning: if the sum is called S, then
S=1 + 2+ 3+ 4+...+ 99+100 but also
S=100+ 99+ 98+ 97+...+ 2+ 1 so we can add to get
2S=101+101+101+101+...+101+101.
Now we have 2S=101100=10,100 so S=10,100 2=5050.
Of course, this same trick can be used to find the number of dots in any
array. For instance, it would tell us that in the fourth array we had
(54) 2=10 dots, which agrees with our count.
Write, in terms of n, the formula for the number of dots in the nth array.
2. A related counting problem:
a) Let X be the set {a,b}. List all two element subsets of X. The
number of two element subsets of X is ______
b)Suppose that X is the set {a,b,c}. List all two element subsets of X.
The number of such subsets is _______
c)Suppose that X is the set {a,b,c,d}. List all two element subsets of X.
The number of such subsets is _______
Do you see a relationship between the numbers in this problem and the
"triangular numbers" of problem 1? Try to explain, in a complete
sentence, what relationship you are observing.
How many two letter subsets do you think could be formed if the entire
English alphabet were available?
77
Lab 2
Patterns
Equipment: cubes, stickers
1. Build a 3  3  3 cube from the small cube you are given. Place a sticker
on each face of a small cube that is on the outside of the large cube.
a) How many small cubes did you use to build the large one?
b) How many small cubes have no sticker on them?
c) How many small cubes have exactly one sticker on them?
d) How many small cubes have exactly two stickers on them?
e) How many small cubes have exactly 3 stickers on them?
f) How many small cubes have at least 4 stickers on them?
What would the answers to these questions be for a 2  2  2 cube? for a
4  4  4 cube? Can you guess formulas that would give the right
answers for an n  n  n cube and any n?
2. a) In this 3  3 square
How many 1  1 squares are there?
How many 2  2 squares are there?
How many 3  3 squares are there?
78
How many squares all together?
b) In this 4  4 square
How many squares all together?
(Make a table as in part a))
c) Can you discover a formula that would give the number of squares to
be found on an n  n square?
d) How many squares would there be on a checkerboard?
79
Lab 3
Comparing without counting
1. A small shepherd boy was responsible for taking to pasture each day 23
sheep. These sheep would graze freely all day, and in the evening the boy
was to round them all up and drive them home. Unfortunately, the boy
could not count, so he had difficulty each day deciding whether he had
found all the sheep before returning home.
a) Can you devise a method (other than teaching the boy to count) that
would make it possible for him to be sure he had them all? If so,
describe it.
b) The sheep's owner came up with a simple scheme. It involved
making, for each sheep, a string necklace that would fit around the
sheep's neck. In the morning, the boy would take the necklaces and
place them all around his own neck. What should he do with them in
the evening?
80
c) The "necklace idea" spread throughout the region and soon all the
young people tending sheep went to the meadows with string necklaces
around their necks. One day, an argument arose between a shepherd and
a shepherdess as to which was tending the largest number of sheep.
Resolving the argument was made difficult since neither could count.
Can you describe a method by which they could resolve their argument
while leaving their sheep grazing undisturbed?
2. Suppose that X is the set {a,b,c} and Y is the set {a,b,c,d}.
a) List all subsets of X.
b) List all subsets of Y that are not subsets of X.
c) Explain how one can tell, without counting, that the number of
subsets in a) is equal to that in b)
d) What can you say about the relationship between the number of
subsets of {a,b,c} and the number of subsets of {a,b,c,d}?
e) If you knew that a set X has 32 subsets, how many subsets do you
think X»{t} would have if t is not an element of X?______ What if t is
an element of X?______
81
Lab 4
Pascal's Pizza
Equipment: Cardboard pizzas and paper toppings--anchovies, pepperoni,
sausage, mushrooms, onions
1. If we use sausage and pepperoni only
a) How many 1-item pizzas can we make? List the possibilities.
b) How many 2-item pizzas can we make? List them.
c) How many 0-item pizzas can we make?
2. Now suppose that we can use sausage, pepperoni and mushroom only
a) How many 1-item pizzas can we make? List the possibilities.
b) How many 2-item pizzas can we make? List them.
c) How many 3-item pizzas can we make?
3. Complete the following chart by listing and counting the possibilities
82
Number of pizzas possible with
Ingredients
0 items 1 item 2 items 3 items 4 items 5 items
1
2
1
2
1
3
4
5
4. Often the table you filled in above are arranged a bit
differently, as below, where the top 1 counts the number of 0-topping
pizzas that can be made if 0 toppings are available. Complete the first
6 rows of this triangle, which is called Pascal's Triangle.
1
1
1
1
1
2
1
3
3
1
Do you see some symmetry in this triangle? Describe it.
5. From your results,
a) How many 2-item pizzas are possible if 5 ingredients are available?
b) How many 3-item pizzas are possible if 5 ingredients are available?
83
c) Explain why these two answers are related.
d) What has this to do with the symmetry you observed above?
Lab 5
Numeration
Equipment: bundling sticks (popsicle sticks will do), rubber bands, Dienes
blocks
1. Bundling Sticks
From the box of bundling sticks, count out 6 piles of 8 sticks each.
These will form your set of sticks.
a) Form as many bundles of 10 sticks each as your stock will allow
Record the result:
Tens
Ones
84
What is the base 10 numeral representing the number of sticks in your
stock?
b)Unbundle your stock and rebundle into bundles of 4 sticks each.
Suppose that you can only count to 4, can you record the result in the
table below? If not, why not?
Fours
Ones
From the bundles, form "superbundles" each containing four
bundles.
Record your result:
Four fours
Fours
Ones
What is the base four
numeral for the number of
sticks in your
stock?_______
c) Carry out the same procedure as in a) and b), now bundling into
bundles of five. What is the base five numeral for the number of sticks
in your stock?_______
d) From your stock, select a subset of sticks containing 221three sticks.
(Do not convert 221three to a base 10 numeral and count. Pretend that
you can count only to three).
2. Dienes blocks (base 10)
85
a) If the small cube represents the number one, then a "long"
represents_______; a "flat" represents__________; and a large cube
represents______________________.
b) If you were working with base 10 bundles as in problem 1, what
kind of a bundle would represent the same number as the large Dienes
block?
c) Assuming that a small cube represents the number one
i) How would you represent 23?
_________________________________________________
ii) How would you represent 2437?
_________________________________________________
iii) Start with 13 flats, 9 longs and 22 small cubes. Make necessary
trades (10 longs for a flat, for instance) until no more that 9 of any
shape is used. You now have ______ large cubes, ______ flats,
______ longs, and _____ small cubes. The base 10 numeral
represented by these blocks is __________.
What base 10 addition problem did you just do?
Lab 6
Addition/Subtraction with Manipulatives
Equipment: Bundling sticks, Dienes blocks
1. a) Represent 27 with bundling sticks. It takes ______ bundles (of ten) and
______ single sticks. Represent 18 with a different collection of
bundling sticks. It takes ______ bundles (of ten) and ______ single
sticks.
b) Combine the two sets of sticks you have selected and bundled (i.e.
take a union of the two sets). After making necessary trades until no
more that 9 of any size bundle remains, you have _______ bundles and
________single sticks.
c) What trades, if any, did you make in this process?
86
d) This process just demonstrated that 27 + 18 = ______.
e) Using the "standard addition algorithm" you know from school,
compute
27
+18
f) Did you have to "carry" to complete e)?
part c)?
How is this related to
2. Solve the addition problem 122four + 31four = _______four using
bundling sticks, trading and counting. (Do not convert to base 10 and
add). Describe the process you used and the trades you made.
3. a) Represent 273 with Dienes blocks. It takes ______ small cubes,
______ longs and _______ flats. Represent 1740 with Dienes blocks.
b) Combine the two sets of Dienes blocks to represent the sum of 273
and 1740 (we are thinking of each block as a set of 1,10,100, or 1000
small cubes here). Make necessary trades until no more that 9 of any
shape is used.What you end up with represents __________ in base 10.
c) You have shown, by counting and trading, that 273+1740 =
_______.
d) Would you like to do this problem using bundling sticks? Why or
why not?
87
4. a)Solve the problem 1272 - 457=______ using Dienes blocks. Describe
how you did it.
b) What trades did you make? How do these trades appear in the
"standard subtraction algorithm" you learned in school?
c) Describe how this same problem could be done with bundling sticks.
Do not actually carry out the process.
Lab 7
Measuring Segments and angles
Equipment: Straight edge and compass
1. Segments.
For this exercise, we will use A ___________ B as our unit segment,

that is, we will say that this segment AB has length 1.
88

a) Using only your compass, find and label a point P on the segment XY

such that XP has length 1.
X ________________________________ Y
b) Again using only your compass, find and label a point Q on the


segment XY such that XQ has length 1

c) What is the length of XY to the nearest whole number? Explain how
you decided. In particular, explain how you can tell that this really is
the nearest whole number.
2. Angle construction
For this exercise, we will use  ABC as out unit angle.
A
The procedure for copying this angle is
much more complicated than that for
copying a segment. Try this procedure,

B
C
starting with ray XY, to make an angle
ZXY which is congruent to ABC (and
so measure of ZXY is 1)
a) With compass point at B, strike an arc hitting both sides of  ABC.
b) Without changing the compass setting, set the point at X and strike an

arc hitting XY.
89
c) Go back to ABC, set the
compass point at one of the two
intersection points of the original
arc, and strike an arc passing
through the other intersection
point.
P
X
Y
d) Without changing the compass
setting, set the point of the
compass at the intersection point

of the arc hitting XY and strike a
new arc intersecting the existing
one.
e) Now draw the ray from X to the intersection point P of the two arcs.

Together with the ray XY this will make the desired angle.
To be sure that you understand this procedure, draw an angle and a ray
with your straightedge on a separate sheets of paper. Copy the angle on
the ray as described. Then set one angle on top of the other, looking
through the paper to see that they fit exactly, i.e. you have made an
exact copy of the first angle.
3. Angle measure
90
a) Beginning with the fact that ABC of exercise 2 has measure 1,

construct an angle RST with measure 2 on the ray ST below.
T
S
b) What is the measure, to the nearest whole number of the angle DEF
below. How did you find the measure?
D
E
F
91
Lab 8
Triangular Tessellation's
Equipment: Colored pens or pencils, plain paper, straightedge
If we start with the triangle
we can build a pattern from
triangles congruent to it that can be extended
indefinitely in all directions. Such a pattern is called a (or tiling) by
triangles. (It is required of a tessellation that no two "tiles" overlap, and
that there are no spaces between tiles.
a)Sketch a different tessellation by triangles which is made up of
triangles congruent to the given one.
b) Compare your with that of the other members of your group. How
many different tessellation's do there seem to be? How many would
there be if we had started with an equilateral triangle as our basic tile?
c) Do you think that there are any triangles that could not be fit into a
tessellation's? Why or why not?
92
2. Choose 3 colors, and color the interiors of the angles in the basic tile with
these colors as indicated.
a) In the tessellation beginning the lab, color each angle interior the
same color as the congruent angle in the original tile
b) Look for patterns, in particular look at all the angles meeting at some
fixed point. Does the picture suggest any relationship between the
three angles of the triangle and a straight angle?
c) Test your conjecture. On a separate piece of paper, draw a triangle
using your straightedge. Cut out the triangle. Fold one vertex to the
opposite side, making a fold which is parallel to the side. Then fold
the other vertices to meet at the same
point. What do you notice? Does this
Fold
strengthen your belief in your conjecture?
line
3. If we agree that the measure of a straight angle is 180 degrees, #2 tells us
that the sums of the measures of the angles of a triangle is 180 degrees.
Use this fact to find the sum of the measures of the angles of this
quadrilateral. Hint: Divide the quadrilateral into triangles.
93
4. Draw a pentagon and a hexagon, and try to determine the sum of the
measures of its angles by the same reasoning used in #3. Pentagon
____________ , hexagon ___________
5. Other tessellation's
A regular polygon is one which is convex and with all sides congruent.
For instance, a regular triangle is equilateral and a regular quadrilateral
is a square.
You probably decided above that you could make a tessellation using
equilateral triangles. Can you make one using squares?_____ If so,
sketch it, if not say why not.
Using regular hexagons?______ If so, sketch it, if not say why not.
For which regular polygons is a possible?
94
Lab 9
Area (nonstandard units)
1. a) Turn back to the triangular tessellation in Lab 8. Suppose that we
decided that the unit of area measure was the basic triangular tile, so we
would say that that triangle had area 1. Sketch as many different
parallelograms as you can that appear in the triangular tessellation and
that have area 2.
b) What would you conclude about the Conjecture: Any two
parallelograms with the same area are congruent?
2. a) Explain why these two figures have the same area.
95
d) Perhaps you did the straightforward thing in a) and just counted
triangle unit in each figure, getting the same answer each time. A
different approach would be to notice that you can add two triangles to
each figure and end up with the same new figure, so each has area A - 2,
where A is the area of the new larger figure. Show on the figures in a)
where the two triangles should be added on.
3. These two large squares are congruent, so they have the same area. The
small triangles are congruent right triangles
a) How many triangle areas can be taken away from the left square?
What is left?
b) How many triangle areas can be taken away from the right square?
What is left?
c)
96
The right triangle here is
congruent to those in parts
a) and b) above. What do a)
and b) tell us about the
relationship between the
areas of the three squares in
this picture. Do you
recognize this as a famous
result in mathematics?
Lab 10
Factors and Multiples
Equipment: Cubes
1. In Blockland all houses have rectangular floor plans and are one story
high.
a) If we are given 6 blocks, there are 2 distinctly different ways of
building a 1-story house with these blocks:
and
Using blocks to actually construct models, determine how many
different 1-story houses can be made from
2 blocks________ 3 blocks _______ 4 blocks
5 blocks _______ 7 blocks ________ 8 blocks ________ 12 blocks
24 blocks _______
97
b) Which of the numbers in the previous problem have only one model?
These are the first few prime numbers. Notice that there is also only one
1-block house, but 1 is not called a prime for special reasons. The
numbers that allow more than one model are called composite numbers.
For each of the numbers given above, there is a at least one model which
has a side of prime length. Complete the following table (if you haven't
already made a model for some of these numbers, either make one or
visualize one):
Number
2
3
4
5
6
Table continued:
Number
7
8
9
10
11
12
13
14
15
16
Prime side
Other side
2
Prime side
1
Other side
Compare your answers with your neighbors. If they do not agree, figure
out why.
2. Suppose that an ordinance is passed in Blockland forbidding construction
of any house with a length or width less than 3. Which numbers less
than 20 can then have a model satisfying this
ordinance?____________________________________________
______________________________________
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Do any of these have two "legal" models?_________
What is the smallest number that has more than one "legal"
model.__________ (We might call the numbers with only one "legal"
model "ordinance primes", and those with more than one "legal" model
"ordinance composites"
Is it true that every "ordinance composite" number is a product of
several "ordinance primes"?___________ If not, explain why not.
Lab 11
Greatest Common Factors
Equipment: Cuisenaire Rods
1. Make a train of red rods (red rods placed end-to-end) which has the same
length as a dark green rod. Can you make a train of red rods which has
the same length as a black rod? ______ To express what we have
observed we will say that "red divides (is a factor of) dark green" and
that "red does not divide black". Test whether the following statements
are true or false.
a) red divides orange
b) orange divides red
c) black divides black
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d) light green divides brown
e) light green divides blue
f) purple does not divide orange
g) purple does not divide brown
2. Rather than saying "red is a factor of dark green" we could also say that
"dark green is a multiple of red". Among your ten different rods, which
ones are
a) factors of dark green?
b) factors of brown?
c) factors of blue?
d) multiples of light green
e) multiples of blue
3. Since red is a factor of purple and red is a factor of dark green, we say
that red is a common factor of purple and green. Which rods are
common factors of
a) dark green and brown?
b) dark green and blue?
c) brown and blue?
4. For each part of exercise 3, name the longest rod which is a common
factor. This could be called the greatest common factor (GCF) of the
two rods.
100
5. a) Make a train of dark green rods and another train of purple rods so
that both trains have the same length? Call the shortest such train the
least common multiple (LCM) of green and purple. If the white rod is
taken to have length 1, what is the length of the LCM of dark green and
purple?______ Since dark green has length 6 and purple has length 4,
we might also denote this number as LCM(6,4).
b) Use rods to find LCM(3,7).
c) Use rods to find LCM(6,9).
Lab 12
Quadrilateral Area on a Geoboard
Equipment: Geoboards, rubber bands
The small square on the geoboard enclosed by a
rubber band will be taken to have area 1.
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1.
Inside each of the regions below which are marked as enclosed by a
rubber band, write the area of the region.
2. On your geoboard, form rectangles of areas 2, 3, and 6.
Draw sketches of these rectangles on the "geoboard" below.
What are all possible areas of rectangles you could form on your
geoboard, if the sides of the rectangles were horizontal and vertical?
Can you find a rectangle on the geoboard that has sides that are not
horizontal or vertical? Sketch it on the geoboard below.
102
The parallelogram (dark boundary) on the geoboard above has the same
area as the rectangle (dotted boundary). Explain why this is true.
On your geoboard, make a rectangle of area 6. Using the same base,
make as many parallelograms as you can with the same area.
Did you find a parallelogram that looks like the one below? Explain
carefully how you know that this parallelogram has area 6 (without
using any formula for area).
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Lab 13
104
Triangle Area on a Geoboard
Equipment: Geoboards, rubber bands
Let's suppose we know how to determine the area of parallelograms on the
geoboard. Now suppose we want to find the area of the triangle with the
solid boundary sketched below. After introducing the congruent triangle
with the dotted boundary, we see a parallelogram of area _______.
What is the area of the original triangle?_____
Explain your reasoning.
In the interior of each triangle below, put its area.
In one case you have to be a bit extra careful.
105
On the geoboards below, create triangles with areas of 3, 6,8 and 9 if
possible.
What is the largest possible area for a triangle on your geoboard?
What is the area of the geoboard figure sketched below?
On your geoboard, make a square with area 2 if you can. Then try to get
area 3. Is this possible?
106
Lab 14
Star Polygons
1. A star polygon is a geometric pattern of the form:
In this case we have five points on a circle, with chords
joining every second point. For brevity we will call this a star
polygon of type (5,2). Other examples with five points are:
(5,1)
(5,3)
(5,4)
What do you notice about these star polygons?__________
___________________ Could you have predicted this behavior?
Draw star polygons of the following types:
(7,1)
(7,2)
(7,5)
(7,6)
What patterns do you see?
(7,3)
(7,4)
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Based on your observations, what type star polygon would you expect to
be the same as one of type (6,2)?
Draw star polygons of both types to check your conjecture.
(6,2)
( , )
2. Sketch the remaining three star polygons with six points:
(6, )
(6, )
(6, )
Notice that there is a difference in the types of paths that occur for
polygons of type (6, ) than for (5, ) or (7, ), for instance (6,2) consists of
two disconnected paths. Can you make a conjecture about what
properties of the numbers 5,6, and 7 account for the fact that the paths
for 5 and 7 can be drawn without lifting the pencil from the paper but
those for 6 cannot?
Based on your conjecture, what do you think the situation will be for
star polygons of type (8, )?
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Sketch all star polygons of type (8, ) and see if your conjecture is
correct.
3. Guess how manydisconnected paths each of these star polygons will
have, then sketch the polygon to check your answer:
(10,1)
(10,2)
(10,3)
(10,4)
(9,2)
(9,3)
(9,4)
(9,5)
(10,5)
4. How many disconnected paths would you expect for a star polygon of
type (15,3)_______ ? Why?___________________
_________________. How many would you expect for one of type
(20,12)?_____________
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Source: "Star Polygons" by Albert B. Bennett, Jr. Arithmetic Teacher,
January, 1978.
Lab 15
Counting Unions (Venn Diagrams)
Equipment: Unifix cubes (5 red, 6 other; Poker chips (8 red, 4 other); 3 yarn
loops
1. Place all cubes and chips on your table.
a) Count the cubes, N(cubes)=_______
b) Count the red objects, N(red objects)= _______
c) How many objects are there that are either red or cubes?________
d) If we let C represent the set of all cubes and R the set of all red
objects, we might expect from the definition of addition of whole
numbers that N(CR)=N(C)+N(R). Is this true?______ If not, why
doesn't the addition formula work here?
e) N(C)+N(R) actually tells how many things you counted in a) and b).
Some of these you counted twice so you got too big an answer when
you added. How many of these things did you actually count
twice?________. In set notation, how would we denote the set of all
things that were counted twice (in terms of C and R)?____________.
Complete the following statement
N(C)+N(R)-N(CR)=N(____).
f) Put a yarn loop around the set of cubes, and another around the set of
all red objects. This gives you a Venn Diagram representing the problem
we have been looking at. Describe the set of elements that are inside
both loops.______________ How could you describe CR in terms of
the loops?________________________________
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g) Suppose we try the same exercise with C representing the set of all
chips and R the set of red objects. Does the formula you came up with
in e) work here also?
2. Suppose that A,B, and C are sets. Conjecture a formula (like the one you
got in 1 e))that would give
N(A  B C) using terms like N(A), N(B), N(C), N(ABC) ... .
N(A  B  C)= _____________________________
Check whether your conjecture works by picking sets A, B and C of
chips and cubes, encircling each set by a yarn loop, counting the sets
you have used in your formula, and applying the formula. Compare
your formula and results with those of students at nearby tables.
3. a) Use two yarn loops to encircle sets C and R with N(C)=7, N(B)=5 and
N(AB)=3. Find N(AB) two ways:
by counting ______, by formula e)________. Do the results agree?
b) Use two yarn loops to encircle sets A and B with N(A)=7, N(B)=5
and N(AB)=1. Find N(AB) two ways:
by counting ______, by formula e)________. Do the results agree?
c) Are you convinced that the formula e) is correct for any two
sets?____
Do these exercises prove that it is correct?
d) Suppose that you know only that sets A and B have N(A)=7 and
N(B)=5. What is the largest that N(A«B) could be?_______ What is
the smallest it could be.
e) If S and T are subsets of the original set of chips and cubes, and we
know N(S)=15 and N(T)=12, is it possible that N(ST)=0?______
Explain why or why not.
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Lab 16
Arithmetic properties
Equipment: interlocking cubes
1. a) Using the cubes provided, form five rectangular sheets with sides of
length 4 and 3 (the unit being the length of the side of the original cube).
Combine these sheets to form a rectangular solid of length 5, width 4,
height 3. Explain why the solid has volume 5  (4 3) cubic units. (A
similar rectangular solid would have volume length  (width  height)
by the same sort of argument).
b) Using cubes, form 5 4 columns of length 3, and combine them to
form a rectangular solid exactly the same shape and size as the one in
part a). Explain why this solid has volume (5  4)  3 cubic units.
112
c) Since the two solids are the same size and shape, they must have the
same volume. What property of multiplication does this illustrate?
2. Using the cubes, form a 3 by 7 sheet. This sheet is an area model of the
multiplication ______ . Now rotate the sheet 90 degrees. You now
see a model of the product ______ . Since the same sheet is a model
for two different products, this provides an illustration of what property
of multiplication?
3. a)Using the cubes, form a 3 by 4 sheet and a 3 by 6 sheet. The first sheet
is an area model of the multiplication ______ while the second is a
model of ________. Combine the two sheets to form one rectangular
sheet , without breaking the original sheets apart. What are the
dimensions of this sheet?__________ This sheet is an area model of
____  _____.
b) Since the area of the new sheet must be the sum of the areas of the
two smaller sheets we see that 3  4 + 3  6 = _________.
c) What property of arithmetic does this illustrate?
4. a) Begin with a set of 9 blocks, and model the solution of (9 - 4) -3 using
the “take-away” interpretation of subtraction. Answer______
b) Again using blocks, model 9 - (4 - 3). Answer _______.
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c) What does this activity tell you about the associative property of
subtraction?
Lab 17
Pascal and paths
1. Copy the six row Pascal triangle from Lab 4.
b)For each row of the triangle, sum the entries:
Row 1 __________
Row 2 __________
Row 3 __________
Row 4 __________
Row 5 __________
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Row 6 __________
c) Predict the sum of the entries of the 10th row _______
d) Give a one sentence interpretation of the sum of the entries of the 5th
row in terms of subsets of a 4 element set.
e) How many different topping combinations do you think are possible if
Pizza Shop offers any combination of 5 toppings?
2. John and Sue go from home (H) to school (S) each day by various
routes. The one thing consistent about the
F
S routes is that John's route always passes by
the fudge shop (F) and Sue's route passes by
I the ice cream parlor (I). They always stick to
the streets shown on the map, and always
take the shortest route possible.
H
a) How long is John's route to school?_____
b) How long is Sue's route?______
c) Is there any route that both John and Sue might take?
________ Explain.
d) How many different routes could John take?_______
e) How many different routes could John take?_______
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f) Circle the answers to d) and e) somewhere side-by-side in Pascal's
triangle.
g) Is there any route to school from home that isn't either among John's
or Sue's routes?______ How many different routes from school to home
are possible?_____
h) Find and circle the total number of routes from school to home in
Pascal's triangle (near the two other circled numbers).
3. Repeat exercise 2 for the map:
F
S
What pattern is there among the circled numbers?
I
H
Use this pattern to produce the seventh row of Pascal's
triangle from the sixth row just using arithmetic.
___ ___ ___ ___ ___ ___ ___
Lab 18
Ratios
Equipment: Dice, straightedge, compass, ruler (mm)
1. Preprobability
A. Experiment: Roll two dice and record their sum. Perform this
experiment 24 times. Put your results together with a partner’s, and
summarize using a bar graph how often each outcome occurred.
_____________________________________________________
Sum: 1 2 3
4
5
6
7
8
9
10
11
12
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What is the ratio of the number of 7’s to the total number of
rolls?___:___
What is the ratio of the number of 5’s to the total number of
rolls?___:___
Which do you think is more likely to appear, a sum of 7 or a sum of
5?__________
B. Analysis: Among all possible outcomes of the experiment of rolling two
dice (of different colors), list on scratch paper all those showing a sum
of 2 (1+1). Of 3(1+2, 2+1). Of 4. etc. Record your results in the table
below.
Ratio
Sum # Representations of sum # Rep: # possible sums
2
3
4
5
6
7
8
9
10
11
12
Compare the computed ratio for the sum 7 to the experimental ratio for sum
7 (Part A). Do they appear to be about equal?
_______Do the same for the sum 5.______
2. Using the angle copying procedure sketched in Lab 7, copy the angle
ABC at the point P to get an angle APQ congruent to ABC and Q is

on AC .
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P
B
A
C
a) Measure AB in millimeters,_________
b) Measure AC in millimeters,_________
c) The ratio AB:AC = ___:___
d) Measure AP in millimeters,_________
e) Measure AQ in millimeters,_________
f) The ratio AP:AQ = ___:___
g) Are the ratios AB:AC and AP:AQ approximately equal?
h) Notice that the triangles ABC and APQ have congruent pairs of
angles: A  A Why?
 B   P Why?
 C   Q Why?
3. Make two other triangles with angles paired such that each pair is
congruent. Measure the sides and check to see that the ratios of lengths
of corresponding sides are equal.
Lab 19
Building triangles
Equipment: Ruler (mm), protractor, compass, blank paper
118
1. For each set of measurements, make as many different (i.e. non
congruent) triangles DABC as you can.
a) AB = 24 mm, BC=18 mm, m( B)=48°
b) AB = 20 mm, BC=35 mm, m( B)=128°
Do you think that you could make a triangle ABC for any given
lengths AB and BC and measure of  B?
If not, what would you look for in these numbers to see that the triangle
was impossible?
In what cases, if any, could you make at least 2 different triangles with the
same given measurements AB and BC and measure of  B?
c) AB = 15 mm, BC=18 mm, AC=26 mm
d) AB = 22 mm, BC=46 mm, AC=18 mm
Do you think that you could make a triangle ABC for any given
lengths AB ,BC and AC?
If not, what would you look for in these numbers to see that the triangle
was impossible?
In what cases, if any, could you make at least 2 different triangles with the
same given measurements AB, BC and AC?
e) AB = 30 mm, BC=24 mm, m( A)=35°
f) AB = 23 mm, BC=20 mm, m( A)=90°
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Do you think that you could make a triangle ABC for any given
lengths AB ,BC and AC?
If not, what would you look for in these numbers to see that the triangle
was impossible?
In what cases, if any, could you make at least 2 different triangles with the
same given measurements AB, BC and m( A)?
g) AB = 30 mm, BC=24 mm
How many different triangles can you make with this information?
120
Lab 20
Symmetry Motions of a Rectangle
Equipment: Index cards
1. Label the two sides an one of your index cards "front" and "back" so that
when you turn it over like the page of a book the writing is right side up
on both sides
Front
Back
Fig. 1
Fig. 2
2. Place your card "front side up in the rectangle below as shown in Fig. 1.
This will be called the original position.
3. Now turn the card over using a "flip" over the vertical line through the
center of the rectangle. It should now look like Fig. 2. Name this
flipping motion V. Notice that after this motion, the card still fits
precisely on it's original outline. Such a motion is called a symmetry
motion of the rectangle.
121
4. Return the card to the Original Position. Another symmetry motion is the
"half turn", a 180 turn (rotation) about its center. Call this motion T.
The result is shown below
5. Return the card to the Original Position. Suppose we rotate the card 360
about its center. It looks just like it did when we started, so this motion
has no effect on the card. We will call it I, for "identity".
6. Describe a symmetry motion other than V,T, and I.
Call it H.
7. Are there any other symmetry motions of this rectangular card? If so, list
and name them.
8. Return the card to the Original Position. You can think of a type of
"multiplication" of symmetry motions. It consists applying first one
motion, then applying the second to whatever resulted from the first. For
instance, the "product" H V means the motion gotten from first doing
V, then H. Verify by actually carrying out these motions on the card
that H V=T and that T T=I.
9. Complete the multiplication table below:
I
T
V
H
I
T
I
V
H
T
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j) Using this "multiplication table" Check to see whether this operation
is commutative. Is it?_____ If not, give an example of two motions
which do not commute.
k) Check to see whether this operation is associative. Is it?____ If not,
give an example of three motions which do not satisfy the associative
property.
l) Recall what it meant for an element to have an inverse for the
operations of addition and multiplication in the integers. What would it
mean to say that V has an inverse in the set of symmetry motions?
If there is one, what is it?
What symmetry motions of the rectangle, if any, do not have inverses.
123
Lab 21
Symmetry Motions of a Triangle
Equipment: Index cards, scissors, compass, straightedge
1. Using your compass and straightedge, construct an equilateral triangle on
an index card. Cut out the triangle and label it "front" and "back" as you
did with the rectangle in Lab 17.
2. Which of the motions I,H,V,T introduced in Lab 17 are also symmetry
motions of the triangle?
3. Describe, and name with letters of your choice, as many new motions as
you can find which are symmetry motions of the triangle.
4. Make a multiplication table for the "product" of the symmetry motions of
the triangle, as you did for the motions of a rectangle in Lab 17.
124
(If closure fails, i.e. if the
product of two symmetries does
not appear in your list in 3. You
have probably not found the
complete list of symmetries in
3.)
5. Use the table in 4. to determine whether the operation is commutative. Is
it?______ If not, show by example that it is not.
6. Check a number of cases of the associative property. Do you think that
the operation is associative? If not, did you find a counterexample?
What is it?
Interesting (?) question: How many different instances of the associative
property would have to be checked before you know that the operation
is or is not associative?
7. Can you tell by looking at the table whether the symmetry motion I acts
as an identity for this operation?_____ Does it?_____. If not, give an
example where it fails to act as an identity.
8. Can you tell by looking at the table whether every symmetry has an
inverse?______ Do they?_______ If not, give an example.
125
Lab 22
Fraction Arithmetic
Equipment: Cuisenaire rods
1. Representation and addition of fractions
a) What rod has the same length as a red rod together with a purple
rod?_________
b)If we assume that a white rod has length 1, what is the length of a red
rod?______ a purple rod?______a dark green rod?_______ In this
setting, what arithmetic statement is modeled in part a)?_____________
c) If we assume that a brown rod has length 1, what is the length of a red
rod?______ a purple rod?______
a dark green rod?_______. In this setting, what arithmetic statement is
modeled in part a)? _________________
d) If the dark green rod has length 1, which rod represents
1/6?_________ Which represents 1/3?________
126
e) Use rods to model the statement that 8/6=4/3.
f) Use rods to model the difference 4/3 - 1/6. What is the
answer?________ Did you have to trade rods to get the exact answer?
g) For each of the following problems, first choose a rod to have length
1, then use the rods to solve the following:
3/4 + 1/8 = ______
4/5 - 1/2 = _______
Unit used _________ Unit used _______
2. a) Assume that a white rod has length 1. How can you illustrate using
rods that
6 ÷ 3 = 2?
8 ÷ 4 = 2?
9 ÷ 2 = 4 1/2?
b) Assuming that a brown rod has length 1, illustrate and solve the
following problems:
3/4 ÷ 1/8 = _______?
3/8 ÷ 3/4 = _______?
127
c) For each of the following, first choose a rod to have length 1, then use
the rods to solve the problem:
1/2 ÷ 1/3 = ______?
2 1/2 ÷ 1/2 = ______?
Lab 23
Fractions/Decimals
Equipment: Graph paper, Dienes blocks
1. On the graph paper, suppose that the side of each small square is of
length 1/8.
a) With your pen, outline a square on the graph paper that has area
1._______ How many small squares are there in this "unit square"
?_________ What is the area of one small square?___________
128
b) Within the unit square you have outlined, color a rectangle with sides
3/8 and 5/8._________ How many small squares are colored? _______
What is the area of the colored rectangle? _________ What is the
product of 3/8 and 5/8? ___________ Compare the size of the
product to the size of the factors. Which is smallest?_______ Does this
surprise you?_______ Would you expect it to surprise a child who was
just learning to multiply fractions?________ Why or why not?
c) Make another unit square on your graph paper and use it to find the
product of 1/2 and 3/8. Answer = _______
d) Solve the problem 5/8  11/8 by making an area model of this
product on your graph paper as above. Answer_____
What difficulties arise in doing this problem which do not arise in the
earlier examples?
2. On the graph paper below suppose that the square surrounded by
boldface is of area one (a unit square).
a) What is the area of each
small square (as a
decimal)?_______
b) What are the lengths of the
sides of the shaded rectangle
(as decimals)?_______
c) The shaded rectangle has
area (as decimal) ________
d) What is .3  .7 ? _______
129
e) On your graph paper, solve .5  1.2 by drawing a suitable area model
and counting small squares.
.5  1.2 = __________
f) Use area models on your graph paper to solve:
.8  2 = ______
.5  ? = .20, ? = ______
?  .7 = .42, ? = ______
g) What arithmetic operation are you actually doing when you solve the
last two parts of f)? ___________________
3. Representing decimals by Dienes blocks
a) If a flat represents 1 unit of volume, then a long will represent .1;
Why?______________________________
a small cube will represent _______ ; a large cube will represent
_______; three longs represent _____ ; 7 small cubes represent _____ ;
and 2 large cubes, 2 flats, and 4 small cubes together represent
________.
b) If a large cube represents 1, then 3 large cubes, 4 flats, 5 longs and 6
small cubes together represent ____________________________
c) If a large cube represents 1, how would one represent .1?
______________,.04? ______________,.007? ___________,
.302?___________ 3.45? ___________________
130
d) Working with your group, use Dienes blocks to:
i) Decide whether .302 is larger than .32
ii) Add .37 and .21
iii) Add .37 and .25
iv) Add .375 and .086
v) Subtract .42 from .58
vi) Subtract .21 from 1.4
vii) Subtract .36 from .2
viii) Multiply 3 times .008
ix) Divide .09 by 3
Lab 24
More Fractions/Decimals
Equipment: Four function calculator with memory.
1. Suppose you are to use the calculator to add the fractions 23/79 and 47/92
a) Write down a sequence of calculator key strokes that will compute
the numerator and denominator of the sum of these fractions using
whole number arithmetic on the calculator, no decimals.
131
b) The exact answer, as a fraction, is ________ . Convert this fraction to
a decimal on your calculator __________
b) Convert the fraction 23/79 to a decimal on your calculator
__________
Convert the fraction 47/72 to a decimal on your calculator
__________
Add the decimal representations of the two fractions. 23/79 + 47/92 =
_______.
c) Compare your answer in b) with that in a). If they are not exactly the
same, explain why not?
2. a) Use your calculator to convert 371/87 to a decimal.
371/87 = ___________
b) We know that if we work only with whole numbers, we divide 371 by
87 to get a quotient Q and a remainder R (which is less than 87).
Just from the decimal representation in a) can you decide what Q must
be? ______________
Describe a process which, beginning with the decimal representation in
a), will give R.
132
c) On a certain manuscript, one can decipher only the following parts of
an arithmetic statement:
/ 65 =
.476923 .
Can you figure out from this information exactly what the numerator of
the fraction is?
Can you figure out what the remainder is when the numerator of the
fraction is divided by the denominator?
d) Make up a similar problem in which missing information about a
fraction can be found if one knows part of the decimal representation of
the fraction.
Lab 25
Arithmetic on a Broken Calculator
Equipment: Four function calculator
1. Suppose that the  key on your calculator does not work. Solve the
problem 76  14 on the calculator without using that key. 76  14 =
________
133
2. Suppose that the key on the calculator does not work.
Solve the whole number problem 945 45 (i.e. find the quotient and the
remainder) without using that key.
945 45 = ___________
3. Some calculators have a yx key. Without using such a key is it possible
to compute 138?______ How?
If $500 is invested at 6% per year, compounded annually, the principal
will grow to $500 (1.06)5 after five years. Without using the yx key, if
it is available, figure out how much money this will be.
______________
4. Suppose that the key with the number 4 does not work. Devise, if
possible, a method of computing 812 - 245 on your calculator. Explain.
Devise a method of computing 45  273 without using the "4" key.
Explain.
Is it possible to solve every whole number arithmetic problem on a
calculator with a broken "4" key?
5. Ammo target game
In this came you may use only the keys: , , +, -, 8, 5, and =. You may
use the same key more than once in a sequence. Using this list as
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"ammo", try to his the following "targets. In each case write down your
calculator sequence.
Target
a) 10
b) 2
c) 16
d) 1
e) -3
f) -1
g) .5
h) -.25
Calculator Sequence
5+5=
5+5-8=
Lab 26
Indirect Measurement
135
Equipment: Stadia device and hypsometer (description at end of lab), metric
stick, cm ruler, trundle wheel, measuring tape
1. Measurement with a stadia device
a) Measure the length of the stadia device: L = _____mm
b) Distance between crosshairs: D = ______mm
c) Ratio L:D = _______:_______
d) Identify an object (tree, building), the distance to which you wish to
measure.
Have Person A in your group sight (parallel to the ground) through the
stadia device at a meter stick held in a vertical position against the
object by Person B. Person C should determine how many cm on the
meter stick can be viewed by Person A crosshairs as lying between the
crosshairs. Call this distance C. If the distance from Person A to the
object is X, set up a proportion involving L,D,X and C (find similar
triangles
_____ : _____ = ______:______
L
D
Meter
stick
C
X
object
e) Since L,D and C are numbers we know, we can solve this proportion
for X. X = _________ cm.
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f) Measure the same distance from Person A with a trundlewheel or
measuring tape. If this differs from the answer in e), what could be the
sources of error in the measurements?
g) Could you use the stadia device to measure the distance across a
river? What difficulties would this present?
2. Measurement with a hypsometer.
a) Identify a tree, the height of which you wish to measure (without
climbing the tree.)
b) With the hypsometer as close to the ground as possible, sight through
the straw to the top of the tree.
The situation is pictured below.
G
Hypsometer
A C
B
D
F
H
E
A partner should read off the lengths EF and CF from the graph paper
on the hypsometer (the side of a small square is one unit). EF = ______ ;
CF = _______.
c) The triangles AGH and CEF are similar. Why? (You may need to
look at some other triangles also)
d) Set up a proportion relating EF, CF, GH and AH. ____:_____ =
_____:_____ .
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e) Have one group member measure AH, for instance using a trundle
wheel. AH = ______ m.
f) Since AF, CF and EF are known numbers, you can solve the
proportion in d) for GH. GH = ______.
g) Have another group measure the same tree with their hypsometer. If
they do not get exactly the same measurement, describe the possible
sources of errors in this procedure.
STADIA DEVICE: Begin with a cardboard mailing tube with a cover
on one end. Punch a small whole in the center of the cover. At the open
end of the tube, tape two pieces of thread parallel across the opening.
threads
HYPSOMETER:
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Straw
Cardboard
Paper
Fastener
String
Weight
Lab 27
Probability
Equipment: Brown bags containing 20 poker chips each (in each bag the
distribution of colors should be the same, e.g. 3 red chips in each); deck
of 12 index cards numbered 1 to 12; calculator
1. Brown bag experiment (DO NOT LOOK INSIDE YOUR BAG!!!) Draw
a chip from the bag and record its color, then return the chip to the bag
and shake the bag. Do this ten times.
a) How many of each color did you get in your 10 draws?
b) What would you guess to be the colors in the bag? How many of
each do you think there are?
c) Repeat the drawing process, taking 10 more draws. How many of
each color did you get in the 20 draws.
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Now what would you guess to be the number of chips of each color in
the bag?
d) Your instructor will summarize the results for the whole class. Based
on these totals, has your guess about the number of chips of each color
changed?
e) Now look in your bag. Did you guess correctly?
2. Card drawing (Form groups of at least 6 students for this activity.)
a) Each student in the group should, in turn, draw a card from the 12
card deck, record the number on the card, return the card to the deck and
pass the deck to the next group member. Do not share your number with
the group.
b) Guess the probability that two members of your group drew the same
number _________
c) Now compare the results of your draws. Are any two of them the
same? _____ Compare results with the other groups.
3. Theoretical analysis of activity 2.
a) Computing the probability that all draws are different in a 6 member
group. The first person to draw can draw any one of (how many?)
numbers _______ . If the second person gets a different card, how many
possible numbers could he/she draw?_______ The third person, if there
is no duplication could draw any one of _____ numbers. So, assuming
that no two members draw the same number, the six members could
draw ________________________ different sequences of
numbers. Evaluate this product with your calculator.________________
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b) If there were no requirement that the cards drawn were all different,
there would be ________________________ possibilities.
c) The probability that the six members drew different numbers is
______
d) The probability that at least two members drew the same number is
_______
e) Suppose that you wanted to do a similar experiment, with six
members in each group, but that you wanted the probability of two
members drawing the same number to be at least .9 . How many cards
would you put in the deck? _______
Lab 28
Modeling integers
Equipment: Red and black chips
1. Let each black chip represent 1, and each red chip represent -1. (Can you
think of a reason we chose to use red and black chips).
a) What integer is represented by three red chips?____
three red and two black chips?_____
b) How many different ways are there to represent 5 with red and black
chips? Display three different representations of 5.
c) Since 2 red chips, together with 2 black chips represents 0, we can
think of each red chip "canceling" a black chip. If this is done, then 3
red chips and 4 black chips is equivalent to _________ black chips.
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d) Represent the problems below with red chips for negatives, black
chips for positives, and use cancellation to arrive at the correct sum:
2 + 3 = ________
-2 + -5 = ________
-4 + 1 = ________
e) If we wanted to subtract -2 from 4, we want to "take away" two red
chips from our representation of 4. If 4 is represented by four black
chips there are no red chips to take away. How can we represent 4 in a
way that uses 2 red chips? ________black, 2 red. If we take away 2 red
chips, what is left?______ What does this suggest that the answer to 4 (-2) should be?_______ Use this technique to solve:
-4 - (-3) = ______
1 - 3 = _______
-3 - (-5) = ______
2. The number line model.
Until the negative integers are introduced, we actually work with a
"number ray" rather than a "number line". To label points on a number
line, we pick first a point of the line we label 0, and a segment we call
the unit segment. In one direction along our chosen line, beginning at 0,
we lay off segments congruent to the unit segment, labeling their
endpoints 1, 2, 3, ... . Moving in the other direction from 0 we similarly
label endpoints of laid off segments by -1, -2, -3, -4 etc.
2
0
-3
-2
-1
1
This labeled line models directed distances from the point labeled 0.
The point labeled 3, for instance is three units to the right of 0, while
that labeled -2 is 2 units to the left of 0. We have already seen that
whole number addition is modeled on the line by successive movement
along the line (to the right). For instance, we model 2 + 3 as
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2
0
1
3
2
3
4
5
showing that 2 + 3 = 5
a) Use the same idea, thinking of -2 as a movement to the left, draw on
the number line above a model for -2 + 3.
What does this give as a sum of -2 and 3?
b) Sketch a number line and solve -3 + 3 on that line.
-3 + 3 = _______ . Is it true that for every integer a there is an integer b
such that a + b = 0?_______
For a = 6, b = ______
For a = 0, b = ______
For a = -76, b = ______
c) Think about the model of subtraction of whole numbers on the
number line. Describe the process.
d) Try to mimic the whole number process to model on a number line
the subtraction problem (-3) - 2. What answer did you get?_______ Do
the same for 3 - (-2)=______ and for (-1) - (-3) = _______
e) Multiplication is not so easy to model on a number line (remember
that we use an area model for whole numbers). One way to think of a
product m  n is to think of an object starting at 0 and moving on the
number line n units per hour. What do you think it should mean to
move -5 miles per hour?_______________________________
If the object moves 4 miles per hour, after 3 hours it will be at the point
labeled 12. Sketch this situation on a number line. This provides a
model for the product 3  4. Sketch a similar number line model for
3  (-4)=_____
Let's think about modeling (-3)  4 = ______.
143
According to our multiplication process in the number line model, this
should tell where the object is after -3 hours if it is moving 4 units per
hour. What do you think it should mean to say "after -3 hours"
__________________
_______________________. With your interpretation, does it turn out
that (-3)  4 = 4  (-3)? If not, you might want to look for another
interpretation.
The most intriguing product to model is one like (-2)  (-4). We should
want to decide where an object moving -4 units per hour is after -2
hours. Think carefully and then work out the problem on the number
line. (-2)  (-4) = _______
Lab 29
Modeling irrationals
Equipment: Geoboards, Straightedge and compass, ruler, calculator
1. Square roots on the geoboard
Assume the each small square on the geoboard has area 1.
a) Find a square on the geoboard of area 4. What other areas can you
find easily?
b)
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What is the area of the square pictured at the left. Explain how
you know this is so.
c)
What is the area of the square pictured at the left. Explain
how you know this is so.
d) Can you find a square on the geoboard which has area 3?
2. Square roots with straightedge and compass.
For this activity, we need to be able to construct right triangles using
straightedge and compass. The hard part, constructing a right angle, can
be done as follows. Beginning with a point P on a line L, strike arcs
145
Q
S
P
T
with center P and the same radius,
intersecting L at points S and T.
Now, strike arcs with centers S
and T and equal radii. Label a
point where the arcs intersect Q.
Then QPT is a right angle.
L Explain how you know this is a
right angle (look at triangles
QPS and QPT.
a) Construct a right triangle with both legs of length 1 inch. How long
should the hypotenuse be according to Pythagoras?__________
Measure the hypotenuse and check whether Pythagoras was right.
b) Now construct a right triangle with one side congruent to the
hypotenuse from a) and one side of length 1 inch. How long will the
hypotenuse of this triangle be?_______
Measure to see if it is correct.
c) Describe a process by which you could construct a segment with
length n for any whole number n.
3. Square roots on a calculator
146
Assume that your calculator has no square root key (or if it has one,
assume that it is broken).
a) 22 = 4 and 32 = 9. What does this tell you about the relative sizes of
2, 3, and 7? (for example, is 2 > 7? If not, why not?)
b) Square each of the numbers 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9,
3. Compare the squares to 7 to find two consecutive "tenths" between
which 7 lies.
2.___ < 7 < 2.___
c) Use a process like that in b) to find two successive "hundredths"
between which 7 lies.
d) Find the first 5 decimal places of 7
147
Lab 30
Tangrams and Area
Equipment: Tangrams, ruler
The tangrams used in this lab are based on an ancient Chinese puzzle
consisting of seven geometric shapes which can be cut from a square. The
original purpose of the puzzle was to for silhouettes of various objects.
1. Try to arrange the shapes, using all of them, to form a silhouette of a
duck. Sketch your result here.
We will use these puzzle pieces to study ideas of area and perimeter.
2. Using the two small triangles, make (and sketch here) two different
common geometrical shapes.
(1)
(2)
What can you conclude about the area of these different shapes?
3. a) Use all seven pieces without overlap to form (if possible)
i) a square ii) a (non square parallelogram)
iii) a non square trapezoid
iv) a non square rectangle
b) How are the areas of these four figures related?
c) Measure the perimeter of each of these figures:
148
square_________, parallelogram_________, trapezoid_______
rectangle_________
d) Which figure has the largest perimeter?______________
e) Which figure has the smallest perimeter?_____________
f) Try to construct a figure using these seven pieces whose perimeter is
larger than that in d). Sketch it.
g) Try to construct a figure using these seven pieces whose perimeter is
smaller than that in d). Sketch it.
4. Make (if possible) a triangle using the indicated number of pieces. Find
the area of each, assuming that the small triangle has area measure 1.
Number of pieces
1
Sketch
Area
2
3
4
5
6
Compare the perimeters of these triangles. Is there any consistent
relationship between area and perimeter?
149
Lab 31
Geometric Properties via tessellation's
Equipment: Colored pencils
Copied below is the triangular tessellation pattern we saw in Lab 8. In Lab
9 we saw that there was information buried in this picture about the
measures of angles of a triangle. In this Lab we will look for other "hidden"
information.
B
A
C
D
1. Opposite angles of a parallelogram
As you did in Lab 9, color the vertices of each triangle in the tessellation
with three different colors so that congruent angles have the same color.
Look at the angles ABC and CDA in the shaded rectangle ABCD.
Each angle is made up of a pair of smaller angles, identify them.
ABC is made up of ________ and ________
CDA is made up of ________ and ________
Each angle of one pair is congruent to one in the second pair:
 ______   ______
and  ______   ______
What does this say about ABC and CDA?
150
What does this activity suggest about the opposite angles of any
parallelogram?
Check this conjecture by looking at other small parallelograms in the
tessellation which are not congruent to ABCD.
2. If L and M are two parallel lines, and T is a line intersecting both of them
(a transversal) several angles are formed:
The angles shaded here are
examples of a pair of alternate
L
interior
angles.
Shade the other pair of alternate
interior angles formed by these
lines
M and the transversal.
T



In the tessellation above, the lines AB and CD are parallel and CD is a
transversal. ABD one angle in a pair of alternate interior angles for
these lines. What is the other angle of this pair?_______
What do you notice about these two angles?
What does this suggest about alternate interior angles?
Check this conjecture by identifying other pairs of parallel lines and
transversals in the tessellation.
3.
151
Y
X
Z
In this triangular tessellation, two large triangles are outlined in
boldface, XYZ and one other. Label the vertices of the other triangle
X', Y', and Z' so that X X', etc.
Call XY and XY corresponding sides of the triangles. What are the
other pairs of corresponding sides?
Does it appear that XY and XY are congruent?
What does appear to be the relationship between the lengths of these
two sides?
Similarly, compare the other two pairs of corresponding sides. What do
you observe?
Pairs of triangles related as these are, are called similar triangles. Find
some other pairs of similar triangles in the tessellation. Do you note the
same relationship between lengths of corresponding sides?
152
Lab 32
Mirror Lake measurements
Equipment: String, paper, pencil, trundle wheel or measuring tape, ruler,
 1/2  11 transparency with 1/2 in. grid,
8 1/2  1111 transparency with 1/4 in. grid
1. Indirect measurement
The instructor will select two objects (trees, for instance) on opposite
sides of the lake. You are to determine the distance between the two
objects with no tools other than those listed above. You may make use
of any facts about geometry you know. You may not step in the water,
and you may not stretch string across the water between the two objects.
Describe your method, including a statement of any geometry fact you
may have used. What do you think the distance is?___________ What
are the possible sources of error in this measurement?
153
2. At the end of this lab is a scale map of Mirror Lake, drawn to a scale 1
inch = 15 feet. On the map, mark the approximate positions of the two
objects used in #1. Measure the distance between these points on the
map. What appears to be the actual distance between the
objects?_______
3. Again using the map, we want to determine the area of mirror lake.
a) Lay the 1/2 " transparency grid on the map. How many small squares
are required to cover the lake?_______
b) Using a) approximate the area of the lake (on the map) in square
inches ___________
c) Using b) find an approximation for the actual area of the lake in
square feet.________ Do you think this approximation is too large or
too small?___________Why?
d) Repeat the above process using the 1/4 " grid. What approximation
does this give for the actual area of the lake in square feet? _________
Do you think this approximation is better or worse than that in
c)________
Why?
154
Lab 33
Distance, Time and Velocity
The following data was recorded on a recorder in Joe’s auto as he went on a
short jaunt one recent weekend.
155
Mileage
37.8
39.0
40.2
41.8
42.6
43.4
44.2
45.0
46.2
47.0
48.2
49.0
50.2
51.0
52.2
53.0
54.2
55.0
56.2
57.0
58.2
MPH
15
35
35
0
40
55
55
55
55
55
55
35
65
65
65
65
65
65
65
65
65
Time
11AM
11:02
11:06
11:09
11:31
11:33
11:34
11:34
11:36
11:37
11:38
11:39
11:40
11:41
11:42
11:43
11:44
11:45
11:46
11:46
11:48
Mileage
59.0
60.2
61.0
62.2
63.0
63.8
64.6
65.0
66.2
67.0
67.8
68.6
69.0
70.2
71.0
72.2
72.3
MPH
Time
55
11:48
25
12:11
55
12:12
55
12:13
55
12:14
55
12:15
55
12:16
55
12:16
55
12:17
55
12:18
55
12:19
47
12:20
45
12:21
40
12:22
40
12:23
25
12:26
0 12:27
1. Make a graph of this data, plotting velocity on one axis, time on the other
2. Make a graph which plots distance on one axis and time on the other
axis.
156
3. Distance travelled (d) is related to time travelled (t) and speed (r) by the
formula d = r  t.
Use this formula, plus the given mileages at t=53 and t=54.2 to estimate
the speed at t=53. r=______ mph.
Do the same for t=54.2 and t=55. r=______ mph.
Why do you suppose that these numbers differ from the actual
speedometer reading?
If you were not given the speed data, how might you get a better
estimate of the actual mph at t=54.2 using the distance/time data you
have available?
4. Make up a description of Joe's trip that would fit the data given in the
table.