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)ii) Balance the object to be weighed with a number of unit weights

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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:

i) Select a unit of height measure

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ii) "Balance" the height of the object to be measured with a number of unitsiii) 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 measurement of height. Personality doesn't count, age doesn't count. Only

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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

A

B

C

D

E

FFigure A2-1compared. Rather than Sue, John, and Mary, these representations could be called AB , CD, and E F , with the pair of letters naming the endpoints of the segment. If the 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 coincide, is called congruence. In Figure A2-1, one 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

4

called the length of5

or the distance from A to B for each segment6

AB7

. The process which assigns such a number to a segment 8

AB9

, usually called linear measurement, mimics exactly the measuring processes we've already encountered.

a) Select a unit segment

10

b) Cover the segment 11

AB

12

with segments congruent to the unit segment laid end to end, and with no overlapc) Count the segments

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As an example, look at the segments 14

XY and UV

15

in Figure A2-2.16

If we take

U

V

X

Y

S

T

Fig. A2-2

17

UV

18

to be our unit segment, we see that 19

XS, ST, and TY

20

are segments congruent to 21

UV

22

which exactly cover 23

XY24

. Counting these segments we find that the length of 25

XY

26

is three units. It is customary in many texts to use have a "shorthand 27

notation" for the length of a segment 28

XY29

, 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 se

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 segments could cover the segment XY exactly, without overlap. This makes the segment XY have a precise whole number length. In actual measurement situations, the unit segment and the segment to be measured are often both specified in advance and we must do the best we can with this measuring process and the whole numbers. The result will often be that lengths can only 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 0,1,2,... as we move by unit steps from the beginning. To measure a segment XY , we simply lay the

0 1 2 3 4 5ruler along the segment with the 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.

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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 Ba) 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.

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.

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<>

><

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 unitii) Cover the region with units without overlapiii) 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

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A

B

C

D

Figure A3-1of 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 the area of the quadrilateral ABCD is 4 units.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 rectangle as unit , giving an area of 8 units.

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Figure A3-2

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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 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

35

Figure A3-3

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most accurate statement we can make about the area A of the lake is that 16A35 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.

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There is an interesting difference between length measure and area measure related to the congruence property. If, for some unit of length,

38

AB

39

and 40

CD

41

both have length 3 units, then 42

AB

43

44

CD45

. This can easily be demonstrated by building a model of 46

AB

47

and one of 48

CD

49

from unit segments and comparing them. However if R1 and R2 are rectangular re gions, 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 thRelated to this strange behavior of area is another important property of

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A

B C

DE

Figure A3-4

51

area measure, sometimes called the dissection property of area. This property says that if we take one region, cut it up and reassemble the pieces into 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 cut apart to remove the small striped triangle, and 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:

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.

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53

3. Sketch three non congruent regions all with area 6.

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4. A segment can be thought of as a region in a plane which has zero width. What should the area of a segment

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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 three-dimensional 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 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

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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 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

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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

58

Fig. A5-1

59

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 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

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vertex at point A; one formed of segments 61

AB

62

and63

AC64

, one formed of segments 65

AD

66

and 67

AE68

, etc. (How many different angles, by this intuitive 69

A

B

C

D

E

Fig. A5-2

70

definition, are there in the Fig. A5-2 which have vertex A?) This fact might be insignificant to a 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:

an angle is a geometric object formed by two rays (the sides of the angle) with a common endpoint

(the vertex of the angle)

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where by a ray 72

AB→

73

we understand informally the indefinite extension of 74

AB

75

from A to B and beyond. With this understanding, we see that 76

AB→

77

= 78

AD→

79

, and 80

AC→

81

=82

AE→

83

. Now whether we say that the angle at A in Fig. A5-2 consists of 84

rays 85

AB→

86

and 87

AC→

88

, or of 89

AD→

90

and 91

AE→

92

, we have described exactly the same object. The price we pay for this precision is that the angles occurring in nature mus

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

93

by the rays 94

AB→

95

and 96

AC→

97

we often substitute the shorter symb ol BAC. We apply to angles the same intuitive description of congruence we have used for other geometric figures: two angles

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 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 angleb) Cover the angle to be measured with angles congruent to the unit angle, laid side to side with the same vertex and with no overlapc) Count the angles

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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

99

A

B C

X

Y Z

1 unit

4 units

Figure A5-3

100

angle in this system if ABC is the unit). We denote the measure of 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 full name of a concept can lead to confusion and frustration. It is often 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

101

a unit angle falls exactly on 102

YX→

103

. 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, w e mean intuitively the region of the plane lying b

104

P

Q

RInterior

Exterior

Figure A5-4Interior

105

Observe that the angle PQR actually divides the plane into three distinct parts, the exterior of the angle, the interior of the angle, and the angle itself, which is neither in its interior nor its exterior. There are actually two

106

different ways we could try to measure PQR beginning at 107

QP→

108

and using a given unit angle. We would start at 109

QP→

110

and lay off unit angles until we reached 111

QR→

112

, 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 to avoid at the time angle measure is f rst introduced. To cle

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

113

(See Fig. A5-5.) We now measure the distance from 114

BA→

115

to 116

BC→

117

along the arc of the circle lying in theinterior of the angle (as if the arc were a segme

118

1

A

B

C

Figure A5-5

119

has the advantage of making use of something we already know, namely segment measure, it has some serious drawbacks also. One is that it is difficult, if not impossible, to accurately measure the length of an arc of a circle with a ruler. A second is that commonly occurring angles, such as right angles, not only do not have exact whole number measures as they do

120

in degree measure, they have in fact irrational number measures (121

π2

122

for a right angle). Not a good situation, particularly 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 childr en, i s that the idea of radian measure depends on an understanding of center a

One strange aspect of angle measure must be emphasized. As opposed to segment or area measure, measures of angles cannot be arbitrarily large. If

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we think about measuring segments, we know that if we start with a segment 124

QP

125

of any given length, we could always lay off one additional unit segment 126

along the ray 127

QP→

128

and get a segment one unit longer than 129

QP130

. 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

131

XY

Z

W

Figure A5-6T

Y

132

If we now lay off one degree angles starting at 133

YZ→

134

and moving toward 135

YW→

136

we will get larger and larger angles, just as we did with segments, until we have “added” 89 unit angles, at which time we will have an angle o measure 179 degrees. One more unit angle will get us to 180 degrees,

137

which we see is formed of the ray 138

YZ→

139

and its opposite ray. This angle is called a s traight 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 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 elem

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.

140

D

141

If D is the unit angle, what is the largest whole number which could arise as the measure of an angle?

142

2. If 143

BK→

144

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

145

AB

D

C

146

right angle, and that m(ABD)=1. What is m(DBC)? (The drawing may not be to scale)

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.

147

AB

C148

i) ii)149

150

iii) iv)151

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 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

152

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, we thought that the general relationship was that n(ST)=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

153

is on the segment 154

AC155

.

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:

156

AB+BC=AC provided that B is on 157

AC158

.

159

A natural extension of this discussion would be to ask what happens if the segments

160

AB

161

and 162

BC

163

have a common endpoint (B), but do not lie on the same line (see Fig. B1-2). In thi s case, a bit of experimentation sh

164

A B

CFigure B1-2

165

for more, namely a formula that would give us the length AC if we knew AB and BC. In fact, there is 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 one of length 4, end to end on the number line. The resulting segment, beginning at 0 and

166

1 2 30 4 5 6

2 units 4 units

Figure B1-32+4=6

167

ending at 6, is clearly of length 6 so we have here a picture of the arithmetic fact 2+4=6. Of course, every whole number addition fact 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?

168

4

2

C

B A

D

5.

169

170

Suppose that in the quadrilateral (four sided figure) ABCD, we know that the sides

171

AB172

, 173

BC174

, and 175

CD

176

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.

177

A X Y B

178

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

179

A

B

C

D

Figure B2-1

m( ABC)=2m( CBD)=3

m( ABD)=2+3

180

ABD which share a common ray but for which it is not true that m(CBA)+m(ABD)= m(CBD). This problem could be taken care of by specifying that the two angles be adjacent angles: 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

181

AB

C

D Figure B2-2

m( ABC)=70

m( CBD)=140

182

straight angle (Fig. B2-2). In this case, ABC and CBD are adjacent angles, but the sum of their measures is greater than 180 degrees, so it cannot equal m(ABD). What exactly should be the statement of the 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 B2-3.

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 side-by-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.

183

A

B

CA BCFigure B2-3

184

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 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?

185

2. The quadrilateral EFGH is divided into two triangles by the diagonal 186

EG187

.188

E

F

GH

189

a) The sum of the measures of the angles of the quadrilateral, i.e. FEH, GFE, HGF, and EHG is equal to the sum of the measures of the angles of EFG and EHG. Explain carefully why this is so.b) In degree measure, what is the sum of the measures of the angles of this quadrilateral.

190

X Y Z

W

191

c) With a protractor, measure the angles of the quadrilateral XYZW. Do you get the result predicted in b)? If not, why not?

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.

192

193

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.

5. Suppose that the outer quadrilateral shown here is a square, and that the four right triangles are congruent.

194

A

195

a) Determine the degree measure of the angle labeled A in the diagram.

b) What does this tell you about the inner quadrilateral?

6. Find the degree measure of the angle marked ?

196

197

20

6525

?

198

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:

199

AB

C

Figure B3-1

River

200

"Can you determine the width of the Olentangy River without ever leaving the east bank?" (Not allowed were such strategies as shooting an arrow 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 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 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

201

discuss this point in § C.4.) So 202

AC

203

is congruent to 204

AB

205

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

206

B

A

CRiver

207

impossible to walk at a right angle to 208

AB209

, and instead walked in a direction so that m( BAC)=120∞. To use this same technique, the walker should stop when ABC has what measure?

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).

210

R

ST A(S)=6

A(T)=4A(R)=A(S)+A(T)

Figure B4-1unit

211

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

212

A B

C

Fig. B4-2

213

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 small shaded triangle. Since the small shaded triangle and the white triangle are congruent, they have the same area. That 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.

214

ST

A(S)=9A(T)=9

A(R)=A(S)+A(T= 18-4=14Figure B4-3

-A(S ∩ T))

215

Before we move on to a discussion of formulas for areas of non rectangular regions, we should think a bit about what 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:

If R=ST then A(R)=A(S)+A(T)-A(ST).

216

T

Fig. B4-4

217

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.

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.

218

219

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.

220

221

222

. 223

224

(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 34) 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 34 is

225

(a) (b) (c)

Figure B5-1

226

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 34, but it has taken on a geometric significance. The model has become a rectangle with width 3 units (provided the block has side one unit long) and length 4 units, so the product 34 counts the number of unit square blocks necessary to cover this rectangle precisely, with no overlap. That is, 34 gives the area of the 3-by-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 WL. (Briefly written, A=WL)

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 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

227

(a) (b)

Figure B5-2

3 x 4 + 3 x 3 3 x (4+3)

228

put together they give a rectangle with width 3 and length 3+4, so they give an area of 3 (3+4) square units. 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.

229

A B

C

Figure B5-3

230

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 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 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 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 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

231

232

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.

233

5

x

13

234

7. Suppose the pegs of the geoboard were labeled, each with a pair of whole numbers as indicated. Assuming that the distance between adjacent

235

(0,0) (1,0) (2,0) (3,0)

(0,1) (1,1)

(0,2) (2,2)

(0,3)

(1,4)

(3,5)

236

horizontal or adjacent vertical pegs is 1 unit, what is the distance from the peg labeled (0,3) to the peg labeled (2,0)? What is the distance between the pegs labeled (1,1) and (3,5)? Between (1,4) and (2,2)? Can you find a formula which will give the distance from (a,b) to (c,d) for any whole numbers a,b,c, and d?

B6: Area formulas for triangles and circles

Once the relationship between whole number products and rectangle areas

237

H

BFigure B6-1

X Y

Z238

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 we have begunwith such a triangle, XYZ, and combined it with a congruent triangle to form a rectangle with width H and length B. (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 2A(DXYZ) = BH, in other words: A(XYZ) = (BH)∏2. So we have illustrated the well known 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

239

(a) (b)P

Q

RS

Figure B6-2

H H

240

matter which choice we make), we have two possible situations that can occur--either the line perpendicular to the base and passing through the 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 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

241

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

242

never meet). Fig. B5-6 depicts such a quadrilateral BCDE. 243

BP

244

is an altitude of the parallelogram (to base 245

DC246

). The segment 247

BD

248

divides the parallelogram 249

E B

CD

Fig. B6-3

P

250

into two congruent triangles (Rotate DBC 180 degrees about the midpoint of

251

BD252

.) As we saw for rectangles, this gives use A(BCDE)= 2 A(DBC). But we know 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

253

Figure B6-4

254

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 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,

255

Figure B6-5

(b)

(a)

256

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 π, for which C=πD=2πR. When this fact is substituted into the area formula derived above, we get the well known formula: A=πR2.

Exercises:

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.

257

7

12

(a)11

4

(b)

21

2

5

(c)

29

74

(d)

258

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?

259

260

A

B C(a)

11

11 2

3

(b)(c)

22

2

261

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?

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.

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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

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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. 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

264

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

Point-line Axiom:

265

If P and Q are two different points in space, then there is exactly one line

266

PQ↔

267

which contains them both.268

P

Q

Figure C2-1

269

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 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:

270

If P and Q are two points in a plane, then the line 271

PQ↔

272

lies entirely within the plane.273

In combination with our view of lines as straight, this axiom reflects the flatness of a plane, since if it were curved, the line

274

PQ↔

275

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 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

276

P A Q BFigure C2-2

277

objects (points, lines, and planes) only, it enters our vocabulary as a new fundamental 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:

278

The segment 279

PQ

280

is the set of all points on the line 281

PQ↔

282

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:

283

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

284

AB285

, 286

BC

287

and 288

CA289

.

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 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:

290

The ray 291

PQ→

292

(with endpoint P) is the union of the set 293

PQ

294

with the set of points X on 295

PQ↔

296

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

297

second ray which is part of the line 298

PQ↔

299

, called the opposite ray of 300

PQ→

301

. It can be described as the point P together with all points of 302

PQ↔

303

which are not contained in 304

PQ→

305

. 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 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

306

statement “is congruent to”. We will write 307

AB

308

309

CD

310

to mean that the segment311

AB

312

is congruent to the segment 313

CD314

, 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?

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:

315

A circle with center O and radius 316

AB

317

is the set of all points X in a plane containing O for which 318

OX

319

320

AB321

.322

In the compass construction, the center of the circle is the point where the compass point is set, the radius

323

AB

324

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:

325

1. Give a careful definition of a quadrilateral, taking care that it exclude a figure like

326

A B

CD

327

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

328

A B

X YP

P'

O

329

one-to-one correspondence between the points on the shorter segment 330

XY

331

and the longer segment 332

AB333

! To each point P of 334

XY

335

one associates a point P' of 336

AB

337

as follows: P' is the point of intersection of the ray 338

OP→

339

and the segment 340

AB341

.

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).

342

(a) (b)

343

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

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

344

a b

Figure C4-1

345

angles a and b. Notice that there is another pair of vertical angles in that diagram. Axiom 4 clearly satisfies our requirement of being self-evident. If we rotate the diagram 180∞ about the point of intersection of the two lines, it appears 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

346

A B

C

X Y

Z

Figure C4-2

347

will fit precisely. Let’s try to describe it for the triangles ABD and XYZ pictured in Fig. C4-2. In this case, the obvious requirement is that after moving ABC and setting it on XYZ, the 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 AY, BX, CZ makes the triangles fit exactly). What else do

348

we see when we fit the triangles together? 349

AB

350

is lying precisely on top of 351

YX352

, 353

AC

354

on top of 355

YZ356

, and 357

BC

358

on top of 359

XZ360

. That is to say, the pairs of corresponding sides (for the given correspondence) of the two triangles are congruent. In the same way we see t hat 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 ex rimentation) that the triangle ABC can be

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 AX, 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 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

361

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

362

tree

tree

?

Figure C4-3

X

Z

Y

?

S

363

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

364

SAS says that this triangle must be congruent to the original, the side 365

XZ

366

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

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.

367

A

B C

F

E

G

Figure C4-4

A

B C E

368

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

369

envision what is happening. We label by F the midpoint of 370

AC371

, that is 372

AF373

.374

FC

375

On the ray 376

BF→

377

we label by G the point such that F is the midpoint of 378

AG379

. The theorem wi ll be proved to be true if we can show that BAF (=BAF) is congr uent to FCG. How can we do this? We c onsid r the correspondence BG, AC, FF between vertex sets of BAF and FCG. The vertical angle axiom tells us that the corresponding angles QBFA and

380

QGFC are congruent, and our selection of F as midpoint tells us that both 381

BF382

383

FG

384

and 385

AF386

Q387

FC388

. Since the hypotheses of SAS are fulfilled, we c a n apply i t to conclude that the given correspondence is a congruence between BAF and FCG. Once we know this, the defi nition of congruent triangles tells us that the corre 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

389

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 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 AX,

390

A B

C

X Y

ZD

Figure C4-5

391

B Y and 392

AB393

394

XY395

, are satisfied. We want to look at the cases possible in comparing 396

AC

397

to 398

XZ399

. One of a) 400

AC

401

402

XZ403

b) 404

AC

405

is longer than 406

XZ407

c) 408

AC

409

is shorter than 410

XZ

411

is surely true.

We want to show case a) is true, since then the hypotheses of SAS are satisfied so ABCXYZ, 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.

412

Now let’s convince ourselves that if c) is true, then B is not be congruent to Y. If c) is true then

413

XZ

414

is longer than 415

AC416

, so there is a point D on417

AC→

418

with 419

XZ420

421

AD422

. If this is true, then SAS tells us that ABDXYZ, and then the defini t i n of congruence tells us that the correspo nding angle s DBA and Y are con

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.

423

A B

C

Figure C4-6

424

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 the theorems we have proved using those axioms. Let’s consider the situation pictured in Fig. C4-6, and suppose that we know that AB. Now consider ABC, and the correspondence AB, BA, CC of its vertex set

425

with itself. We know that AB, BA and 426

AB427

428

BA

429

(in fact they are equal), so by ASA this correspondence is actually a 430

congruence of ABC with itself. So the corresponding sides 431

AC

432

and 433

BC

434

must be congruent, which is what we set out do show.

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.

435

2. Suppose that the correspondence AX, BY, CZ is a congruence between ABC and QXYZ, and that

436

AC

437

438

BC439

. What other correspondences between the vertex sets must also be congruences? Which correspondences would be congruenc es if the triangles

3. IS, JT, KR is a correspondence (but not necessarily a congruence) between the vertex sets of triangles IJK and RST. List the pairs of

440

corresponding sides (for example 441

IK

442

443

SR444

). 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.

445

A B

CX

Y

Z

3

44

3

40

40(a) (b)

A C

E

U

V

W

2

2

57 2 57

57

446

447

(c)

X Y

Z

R

S

T

85 40

40

857

7(d)

A

B

C

I

JK3 4

55

3

455

448

5. Find x and y.

449

4

9

8

A B

C

63

x

X

Y

Z

4

963

76 y

450

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):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.

451

α

β

L

M

TFigure C5-1

A

B

452

By alternate interior angles formed by a transversal, we mean angles such as α and β in Fig. C5-1. It is, of course, possible to give a formal definition of these angles in terms of L, M, T and their points of intersection. We leave that as an exercise. There are two pairs of alternate interior angles in Fig. C5-1, only one of which has been labeled.Note also that there are transversals (and accordingly pairs of alternate

453

L

M

α

β

T

Figure C5-2

454

interior angles) for any pair of distinct lines in a plane (see figure C5-2), not just for parallel lines. The distinct feature of parallel lines is that the pairs of alternate interior angles are congruent.

We will use indirect reasoning again in deducing that the theorem stated above is true. So we begin with the fact that the angles a and b are 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

455

L

T

α

α M

Figure C5-3

P

456

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 we know how to copy an angle (See Fig. C5-3). If we begin with 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 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,

457

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 angles of a triangle (see Fig. C5-4). We begin with a triangle ABC, and

458

take a line L through C parallel to 459

AB↔

460

(which exists because of the 461

A B

C

α β

g e σL

Figure C5-4

462

Existence of Parallels Theorem). α and σ are alternate interior angles (for transversal

463

AC↔

464

), so they are congruent by Theorem 2. For the same reason, β is 465

congruent to e (use transversal 466

BC↔

467

). Looking now at the angles for med at the point C, we see that our deductive geometry predicts that congruent copies of the three ang les of the triangle, laid side by side, form exactly a straight angle--something we observed

During a recent workshop, teachers who were asked to determine a method to measure the distance across Mirror Lake (see §C4),

468

tree

tree

?

Figure C5-5

X

Y

469

generally chose the method pictured in Fig. C5-5. They attempted to move from the two trees along parallel paths, and thus to form a parallelogram

470

with side 471

XY472

. 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 o be true if we experiment wit h

473

A B

CD

Figure C5-6

474

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

475

AD

476

is parallel to 477

BC

478

and 479

BD

480

isa transversal to the lines containing these two segments. The alternative interior angle theorem tells us that ADBDBC. Now using the parallel

481

sides 482

DC

483

and 484

AB485

, with transversal 486

BD

487

we see that the alternate interior angles ABD and CDB are congruent. 488

Combining these facts with the fact that 489

BD

490

is congruent to 491

DB492

, we can use ASA to conclude that AC, BD, DB gives a congruence between ABD and BCD. Since corresponding sides of congruent

493

triangles are congruent, we see that 494

AB495

496

DC

497

and 498

AD499

500

BC501

. Which is what we set out to show. In fact, we actually showed more, since if we look at corresponding angles we w

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?

502

(a) 45

45

(b) 140

40

503

504

(c) 75

75

70

70(d)

505

4. Assuming that the labeling (in degrees) is correct, are there any parallel lines in the diagram below? Any right angles?

506

2065

6525

507

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∞.

508

L

M

ac

ub

ed

509

Find the measures of the other marked angles.

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∞?

510

7. Suppose that 511

AB

512

is a diameter of a circle with center P, and that K is a point on the circumference of the circle.

513

A P B

K514

a) Explain why PAK PKA

b) Explain why PBK PKB

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 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.

515

5

6

5

516

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.

517

T

L

M

α

β

518

Using the Vertical Angle Axiom and the Alternate Interior Angle Theorem, explain why L is parallel to M if and only if a and b are congruent.

C6: Similar triangles and more indirect measurement

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-to-one 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

519

A B

C

X Y

Z

(a)

(b)

X Y

Z

W

UFigure C6-1

V

520

ask what, if any, relationship there is between lengths of sides. We hope, of course, that some useful relationship will exist, since that could be the basis of useful applications of similar triangles.

Lets look at a simple example. Suppose that for triangles ABC and XYZ as pictured in Fig. C6-1, a) we know that AX, BY, and CZ (that is, the AAA similarity criterion holds) 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

521

labeled the midpoint of 522

XZ

523

by W, so we have 524

ZW525

526

XW527

528

AC529

. Now we draw a line parallel to 530

YZ↔

531

through W and label the point where it meets 532

XY↔

533

by U; and we draw a second line through W, this time parallel to 534

XY↔

535

and meeting 536

YZ↔

537

at the point V. Now we can apply practically everything we know about parallel lines and congruence of segments.

538

First, we have that 539

UY540

541

WV

542

since UYVW is a parallelogram. The n we l ook at the two t ri angles XUW and WVZ with corresponden ce XW, UV, WZ. We know WXUZWV. Why? Similarly we know that

543

XWUWZV. Why? Since we started with the fact that 544

ZW545

546

XW547

, 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

548

get that 549

XU550

551

WV552

. Combined with 553

UY554

555

WV556

, which we already knew, we have 557

XU558

559

UY560

, which means that U is the midpoint of 561

XY

562

and XY=2XU.

Now we started the problem knowing that AX, BY, and CZ, and we just saw that XWUWZV= Z. So we know that CXWU and

563

AX, Plus we have that AC=XW so 564

AC565

566

XW567

. This is enough to show (using ASA again) that AX, CW, BU is a 568

congruence between ABC and QXWU, and in turn that 569

AB570

571

XU

572

so AB=XU. What we really showed here is that our process succeeded in copying ABC inside XWU with A landing on X.

573

All together, we have seen that XY=2AB, that is, side 574

XY

575

is twice as long as side 576

AB577

. 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

578

C

Z

A=X B Y

Figure C6-2

579

conclude that the three parallel lines 580

YZ↔

581

, 582

BC↔

583

and the line bisecting 584

AC↔

585

divide the side 586

XY↔

587

into three congruent parts, two of which make up 588

AB↔

589

, 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 AC were m:n, then the ratios of the other pairs of corresponding sides would also be m:n. Note that these pictures are th

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

590

pole

Pmeter

stick

Figure C6-3

h

591

ground level to the top of the pole, another rests a meter stick perpendicular to the ground in the line of sight so the sighter can read the height (d centimeters) on the stick which lines up with the top of the pole. 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 there are clearly two similar triangles in this 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=80620, which gives a solution of h=400 cm.

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

592

A

B C

D

Figure C6-4

593

ABC and BDC satisfy the AA hypotheses, since each has a right angle (and these are certainly congruent), and the triangles share the angle C. So for the correspondence AB, BD, CC the ratios of the lengths of corresponding sides are congruent. In 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 AA, BD, CB 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:

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

594

15 cm

15 cm

595

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.

3. Find the lengths a, b, and c.

596

597

12

123 5

c

ba

598

4. In the diagram below a) Is DABC similar to DDEC? Explain

599

A

BD

C E

600

b) Is 601

AB↔

602

parallel to 603

DE↔

604

? 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, findDE.

5. Two similar triangles have the lengths of corresponding sides in the ratio of 1:3. What is the ratio of their areas?

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.

605

3

36

xi) ii)

35

10 x38 71

71

38

110 3232

110

606

8. In ABC, m(A) = 62∞, m(B) = 54∞. In XYZ, m(x) = 54∞, m(Y) = 64∞. Is ABC similar to XYZ? Explain.

607

9. Suppose that 608

AB

609

is a diameter of a circle, P a point on the circumference of the circle and 610

XŒ611

AB

612

such that 613

XP

614

is 615

P

X BA

616

perpendicular to 617

AB618

. Argue that the triangles AXP, XPB, and APB are similar.

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?

619

Lab 1

Patterns (Triangular Numbers)

1. Look at these triangular arrays of dots

620

(a) (b) (c) (d)

621

Count the number of dots in each triangle. How many dotswould there be in the next (undrawn) triangle?

Fill in the table:

622

n number of dots

1

2

3

4

5

6

7

1

3

6

10

623

Do you see a pattern which relates the number of dots in the nth array to the number n? If so, describe the pattern.

You may have noticed the following pattern:

624

n number of dots

1

2

3

4

5

6

7

1

3

6

10

1=1

3=1+2

6=1+2+3

10=1+2+3+4

625

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.

His reasoning: if the sum is called S, then

S=1 + 2+ 3+ 4+...+ 99+100 but alsoS=100+ 99+ 98+ 97+...+ 2+ 1 so we can add to get

2S=101+101+101+101+...+101+101.

Now we have 2S=101100=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 (54)∏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 ______

626

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?

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?

627

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

628

629

How many 1 1 squares are there?

How many 2 2 squares are there?

How many 3 3 squares are there?

How many squares all together?

b) In this 4 4 square

630

631

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?

632

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?

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.

633

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?______

634

Lab 4

Pascal's Pizza

Equipment: Cardboard pizzas and paper toppings--anchovies, pepperoni, sausage, mushrooms, onions

1. If we use sausage and pepperoni onlya) 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 onlya) 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

635

Ingredients

Number of pizzas possiblewith

0 items 1 item 2 items 3 items 4 items 5 items

1

2

3

4

5

121

636

4. Often the table you filled in above are arranged a bitdifferently, 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.

637

1

3

1 2 1

1

1

1 13

638

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?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 SticksFrom 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

639

Record the result:640

641

Tens Ones

642

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?

643

644

Four Oness

645

From the bundles, form "superbundles" each containing four bundles.

Record your result:

646

OnesFoursFour fours

647

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)

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?_________________________________________________

648

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?

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)? How is this related to part c)?

649

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?

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?

650

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.

651

For this exercise, we will use A ___________ B as our unit segment, that is, we will say that this segment

652

AB↔

653

has length 1.654

a) Using only your compass, find and label a point P on the segment 655

XY↔

656

such that 657

XP↔

658

has length 1.

X ________________________________ Y

659

b) Again using only your compass, find and label a point Q on the segment

660

XY↔

661

such that 662

XQ↔

663

has length 1664

c) What is the length of 665

XY↔

666

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.

667

B

A

C

668

669

The procedure for copying this angle is much more complicated than that for copying a segment. Try this procedure, starting with ray

670

XY→

671

, 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.

672

b) Without changing the compass setting, set the point at X and strike an arc hitting

673

XY→

674

.675

X

P

Y

676

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.

677

d) Without changing the compass setting, set the point of the compass at the intersection point of the arc hitting

678

XY→

679

and strike a new arc intersecting the existing one.680

e) Now draw the ray from X to the intersection point P of the two arcs. Together with the ray

681

XY→

682

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

683

a) Beginning with the fact that ABC of exercise 2 has measure 1, construct an angle RST with measure 2 on the ray

684

ST↔

685

below.686

687

S T

688

b) What is the measure, to the nearest whole number of the angle DEF below. How did you find the measure?

689

690

D

E F

691

Lab 8

Triangular Tessellation's

Equipment: Colored pens or pencils, plain paper, straightedge

692

If we start with the triangle 693

694

we can build a pattern from triangles congruent to it that can be extended695

696

697

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?

2. Choose 3 colors, and color the interiors of the angles in the basic tile with these colors as indicated.

698

699

700

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

701

Fold line

702

the other vertices to meet at the same point. What do you notice? Does this strengthen your belief in your conjecture?

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.

703

704

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'sA 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?

705

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.

706

707

708

709

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

710

711

712

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)

713

714

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:

715

716

and

717

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 _______

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 Prime side Other side2 2 13456

Table continued:Number Prime side Other side

78910111213141516

718

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?____________________________________________ ______________________________________

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

719

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

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?

720

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.

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

721

722

The small square on the geoboard enclosed by a rubber band will be taken to have area 1.

1. Inside each of the regions below which are marked as enclosed by a rubber band, write the area of the region.

723

724

725

726

2. On your geoboard, form rectangles of areas 2, 3, and 6.

Draw sketches of these rectangles on the "geoboard" below.

727

728

729

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.

730

731

732

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.

733

734

735

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).

736

737

738

Lab 13

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.

739

740

741

In the interior of each triangle below, put its area.In one case you have to be a bit extra careful.

742

743

744

On the geoboards below, create triangles with areas of 3, 6,8 and 9 if possible.

745

746

747

748

749

What is the largest possible area for a triangle on your geoboard?

What is the area of the geoboard figure sketched below?

750

751

752

On your geoboard, make a square with area 2 if you can. Then try to get area 3. Is this possible?

Lab 14

Star Polygons

1. A star polygon is a geometric pattern of the form:

753

754

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:

755

(5,1)

756

(5,3)

757

(5,4)

758

What do you notice about these star polygons?_____________________________ Could you have predicted this behavior?

Draw star polygons of the following types:

759

760

(7,1)

761

(7,2)

762

(7,3)

763

(7,4)

764

(7,5)

765

(7,6)

766

What patterns do you see?

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.

767

768

(6,2)

769

( , )

770

2. Sketch the remaining three star polygons with six points:

771

772

(6, )

773

(6, )

774

(6, )

775

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, )?

Sketch all star polygons of type (8, ) and see if your conjecture is correct.

776

777

778

779

780

781

782

783

784

785

786

3. Guess how manydisconnected paths each of these star polygons will have, then sketch the polygon to check your answer:

787

788

(10,1)

789

(10,2)

790

(10,3)

791

(10,4)

792

(10,5)

793

794

795

(9,2)

796

(9,3)

797

(9,4)

798

(9,5)

799

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)?_____________

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(CR)=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(CR)=N(____).

800

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 CR in terms of the loops?________________________________

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(ABC) ... .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(AB)=3. Find N(AB) 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(AB)=1. Find N(AB) 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.

801

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(ST)=0?______ Explain why or why not.

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.

802

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 _______.

803

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 __________

804

Row 5 __________

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

805

H

SF

I

806

routes. The one thing consistent about the routes is that John's route always passes by the fudge shop (F) and Sue's route passes by the ice cream parlor (I). They always stick to the streets shown on the map, and always take the shortest route possible.

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?_______

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:

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H

SF

I

808

What pattern is there among the circled numbers?

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. PreprobabilityA. 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

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

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2 (1+1). Of 3(1+2, 2+1). Of 4. etc. Record your results in the table below.

RatioSum # Representations of sum # Rep: # possible sums23456789101112Compare 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

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on 811

AC→

812

.813

814

A

B

C

P

815

816

a) Measure 817

AB

818

in millimeters,_________819

b) Measure 820

AC

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in millimeters,_________

c) The ratio AB:AC = ___:___

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d) Measure 823

AP

824

in millimeters,_________825

e) Measure 826

AQ

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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

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?

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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°

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?

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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

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Front Back

Fig. 1 Fig. 2

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2. Place your card "front side up in the rectangle below as shown in Fig. 1. This will be called the original position.

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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.

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

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Front

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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:

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I T V H

I

T

V

H

I

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.

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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.

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(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.

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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?________

e) Use rods to model the statement that 8/6=4/3.

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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 = _______?

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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.

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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?___________

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 byboldface is of area one (a unit square).

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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 ? _______

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 _____ ;

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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? ___________________

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

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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.

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 = ___________

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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.

c) On a certain manuscript, one can decipher only the following parts of an arithmetic statement:

856

857

.47692/ 65 = 3

858

.

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 = ________

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?

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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 "ammo", try to his the following "targets. In each case write down your calculator sequence.

Target Calculator Sequence

a) 10 5 + 5 =b) 2 5 + 5 - 8 = c) 16d) 1e) -3f) -1

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g) .5h) -.25

Lab 26

Indirect Measurement

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 = _______:_______

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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 _____ : _____ = ______:______

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863

object

C

Meterstick

L

X

D

864

e) Since L,D and C are numbers we know, we can solve this proportion for X. X = _________ cm.

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.

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A C D

FB

E

G

H

Hypsometer

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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. ____:_____ = _____:_____ .

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.

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threads

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HYPSOMETER:

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Straw

Cardboard

String

Weight

PaperFastener

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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.

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?

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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.________________

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? _______

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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.

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

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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.

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0 1 2-1-2-3

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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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0 1 2 3 4 5

2 3

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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 = ______.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" __________________

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_______________________. 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)

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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

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LPS T

Q

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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.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

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?)

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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

Lab 30

Tangrams and Area

Equipment: Tangrams, ruler

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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:square_________, parallelogram_________, trapezoid_______rectangle_________d) Which figure has the largest perimeter?______________e) Which figure has the smallest perimeter?_____________

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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 Sketch Area1

2

3

4

5

6

Compare the perimeters of these triangles. Is there any consistent relationship between area and perimeter?

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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.

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A

B C

D

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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?

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:

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L

M

T

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The angles shaded here are examples of a pair of alternate interior angles.Shade the other pair of alternate interior angles formed by these lines and the transversal.

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In the tessellation above, the lines 899

AB↔

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and 901

CD↔

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are parallel and 903

CD↔

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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.

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X

Y

Z

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I n this triangular tessellation, two large triangles are outlined in boldface, XYZ and o ne other.

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Call 908

XY

909

and 910

′ X ′ Y

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corresponding sides of the triangles. What are the other pairs of corresponding sides?

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Does it appear that 913

XY

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and 915

′ X ′ Y

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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?

Lab 32

Mirror Lake measurements

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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?

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.

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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?

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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.

Mileage MPH Time 37.8 15 11AM39.0 35 11:0240.2 35 11:0641.8 0 11:0942.6 40 11:3143.4 55 11:3344.2 55 11:3445.0 55 11:3446.2 55 11:3647.0 55 11:3748.2 55 11:3849.0 35 11:3950.2 65 11:4051.0 65 11:4152.2 65 11:4253.0 65 11:4354.2 65 11:4455.0 65 11:4556.2 65 11:4657.0 65 11:4658.2 65 11:48

Mileage MPH Time 59.0 55 11:48 60.2 25 12:11 61.0 55 12:12 62.2 55 12:13 63.0 55 12:14 63.8 55 12:15 64.6 55 12:16 65.0 55 12:16 66.2 55 12:17

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67.0 55 12:18 67.8 55 12:19 68.6 47 12:20 69.0 45 12:21 70.2 40 12:22 71.0 40 12:23 72.2 25 12:26 72.3 0 12:27

1. Make a graph of this data, plotting velocity on one axis, time on the other

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2. Make a graph which plots distance on one axis and time on the other axis.

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924

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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.

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