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 I don’t get I no respect!
ME by the SEa June 14, 2013 Richard Rupp Professor of Mathematics Del Mar College The Call to Action! If you ask a random person to name three geometric figures, what will they say? Typically it’s circle, square, and triangle. Alas, the bane of the parallelogram – to be forgotten by society at large. But truly, this fate is ill deserved as the parallelogram is the basis of the formulae used to calculate the areas of most geometric figures, including the great three so prominently named earlier. So rise up my brothers and sisters and join the revolution! Let us no longer tolerate the deplorable regime of circle, square, and triangle! Let us utilize The Geometer’s Sketchpad to properly investigate the role of parallelogram in the derivation of geometric area formulae. Let us finally give parallelogram the respect it deserves! Viva la Parallelogram! Humble Beginnings…Creating a Parallelogram So where do we begin our parallelogram odyssey? Why with its most basic properties. At its simplest, a parallelogram is a quadrilateral with two pairs of parallel sides. Let’s construct one now. In The Geometer’s Sketchpad, let’s begin by creating a line segment. Do this with the line segment . Left click and hold, which will create an initial point, and construction tool, which looks like this then drag the mouse to where you want the segment to end and release the mouse button. You will now have a line segment which may look like this. We now need to create a second side of the parallelogram. Move your mouse over one of the points at the end of the parallelogram. The point should grow slightly larger and in the notification area at the bottom of the Geometer’s Sketchpad window you should see the text “…from Point…”. From here, left click, hold, and drag the mouse to where you want the new segment to end and release the mouse button. You will now have a new line segment starting from the point on your first segment that you selected. The diagram may now look like this. We now need to create the parallel sides. To do this we will use the construct a parallel line function within The Geometer’s Sketchpad. Since we no longer need to construct line segments, choose instead the selection arrow tool, which looks like this . This is the primary tool you will use in The Geometer’s Sketchpad and you should think of it as a natural, default setting. The construction functions in The Geometer’s Sketchpad are turned off or “greyed out” by default and only activate when the objects necessary for them to operate are selected. Note, any “highlighted” or “glowing” objects are considered to be selected. To make sure no objects are selected. Left click anywhere in the blank Page 1 canvas. Now, to create a parallel line, we need two objects: (1) a line to which we shall be parallel and (2) a point to anchor the new line. These can be selected in any order. Select the original line you created and the final point from the second segment you created by clicking on both. From the Construct menu above, notice that Parallel Line is now a selectable option. Click on the Parallel Line option. You will get a new line, not a segment, which may appear as follows. When you are viewing the canvas of The Geometer’s Sketchpad, lines do not terminate with arrows as in the picture above, they merely continue off the canvas. If you were to copy and paste the images from The Geometer’s Sketchpad to another program, such as Microsoft Word, then the terminating arrows are rendered. Clear any selected objects by clicking on the blank canvas. Now, create a second parallel line by selecting the second segment you created and the second point from your first line. Then use the construct a parallel line function as before. Your diagram may now look like this. We don’t need the actual parallel lines we constructed, only their point of intersection. To find this point, select only the two parallel lines and use the Intersection function found in the Construct menu. This will create the point of intersection as shown in the following diagram. Page 2 In The Geometer’s Sketchpad, there is an important distinction that must be made when removing objects from the diagram. Deleting an object permanently removes it and any objects created via it, called its “children”, from the sketch. This can have disastrous and unintended consequences. Most of the time, we simply don’t want to have to look at an object, but we still want its “children” around for our use. Such is the case with the parallel lines we constructed, we want the point of intersection, but don’t need the lines themselves. Deleting either parallel line would delete the point of intersection since the point of intersection was dependent on the line’s existence. So, instead of deleting the parallel lines, we want to hide them. To do this, select only the two constructed parallel lines and then select the Hide Parallel Lines function found in the Display menu. Your diagram should now look like this. But you say, “this is not a parallelogram”…and you would be correct. We need to construct the two missing segments. To create one, select only the endpoints of the line segment you wish to create and then use the Segment function in the Construct menu. Repeat this with the endpoints of the second segment you need to create. Sequentially, your diagram should evolve as follows. Congratulations! You have now created a parallelogram with The Geometer’s Sketchpad. Now let’s examine the properties of the parallelogram, which will be essential to the derivation of area formulae later. Page 3 Dissecting the Parallelogram…Discovering its Properties and Area Our life will be substantially easier if we label the vertices of our parallelogram. To do this, select the text tool, which looks like this , and then click on each of the vertices of the parallelogram. As you click on each point, a letter label should appear next to each vertex. When completed, it should look like this. C
D
B
A
Now let’s discuss a few properties. First, recall that opposite sides of our parallelogram are parallel, i.e. AB is parallel to CD and AD is parallel to BC . So if we focus on the parallelism of AB to CD , we can think of AD as a transversal with respect to AB and CD . To emphasize this, we could unhide a few lines and extend our segments to have our diagram appear as such. C
D
B
A
Now we want to discuss some angles. To display these best, let’s select the marker tool, which looks like this . (Note: the marker tool is a new feature of version 5 of The Geometer’s Sketchpad.) The marker tool allows us to indicate angles or add hatch marks to objects to indicate congruency. Let’s label the internal angle of the parallelogram at vertex A. To do this, have the marker tool active and click on the vertex and drag into the interior of the parallelogram. When you have finished this, let’s also mark the alternate interior angle that the parallel line axioms tell us would be congruent to the angle we just marked. An extra point may display, but this is ok. Sketchpad needs three points to indicate the angle and that is all that is happening. We can hide the point later. Your diagram will now look like this. Page 4 C
D
B
A
Since we could also think of CD and AB as transversals across the parallel segments AD and BC , we can mark the congruent, corresponding angles that the parallel line axioms state exist. Marking these as indicated, our diagram looks as follows. C
D
B
A
Thus, we note the interior angles at vertices A and C are congruent. Let’s now mark the interior angles at vertices D and B. After we mark these angles, use the selection arrow tool and right click on the newly marked angles. Select properties from the pop‐up menu. In the marker tab, change the stroke count to two. Thusly, we can easily label our two new angles as congruent, since they are both supplements of congruent angles, and yet indicate they are not necessarily congruent to the originally marked angles. Your diagram should now look like this. C
D
B
A
Thusly, we have proven that in parallelograms adjacent angles are supplementary and opposite angles are congruent. Next, let’s discuss congruency of sides and establish an area formula. Let’s begin by 
constructing perpendicular line segments from the points C and D to AB . To do this, first, make sure Page 5 you are using the selection arrow tool and that only the point D and AB are highlighted. Then, from the Construct menu above, use the Perpendicular Line option. You will get a result like this. C
D
B
A
We don’t want the whole new line, just the segment from point D to AB . So, using the method described earlier, construct the point of intersection between our new perpendicular line and AB ; label this point as E; hide the perpendicular line; construct DE ; and then mark the right angle created at vertex E of the triangle ADE . Note, since this new angle is a right angle, The Geometer’s Sketchpad automatically marks it as such. When these steps are completed, your diagram will look like this. 
C
D
B
E
A

Repeat these steps to create the perpendicular segment from C to AB . Your updated diagram will look like this. C
D
B
F
E
A


Now, since CD is parallel to AB and the distance between parallel lines is constant, CF  DE . Thusly, by the angle‐angle‐side (AAS) postulate, ADE  BCF . Then, since corresponding parts of 
Page 6 
congruent triangles are congruent, AD  BC . A similar argument can be used to prove AB  DC . Thusly, we have proven opposite sides of parallelograms are congruent. Looking at our diagram a little differently, we can consider the polygon EDCFB. Note, this is a quadrilateral and not a pentagon because the two adjacent angles at point B are supplementary. Let’s hide a few angle markings and mark a few new ones. Let’s hide the double marked angle at point D, the marked angles at point B, and the single marked angle at point C. Then mark the interior angles of quadrilateral EDCF at vertices C, D, and E. C
D
F
B
E
A
Note, the newly marked angle at point E is supplementary to a right angle and is thusly a right angle. Furthermore, the newly marked angle at point D is an alternate interior angle to the first right angle at point E and is thereby a right angle. We can see the newly marked angle at point C is a right angle via either a combination of the two methods just used or the fact that the sum of the interior angles of a quadrilateral is 360 . Hence, quadrilateral EDCF is a rectangle and, by definition, its area can be calculated via the formula Area   base  height  . Since quadrilateral EDCF was created by simply pruning ADE from the parallelogram and grafting on BCF and ADE  BCF , the area of the parallelogram equals that of the rectangle. Therefore, the area of a parallelogram can also be calculated via the formula Area   base  height  . C
C
D
D
B
A
E
F
B
A
F
E
Let’s move on to deriving the area formulae for other polygons. Page 7 Deriving the Area of a Triangle via Parallelograms (and Brute Force) To begin, let’s start with a new sketch. In this sketch, use the point tool, which looks like three points. Label these as A, B, and C with the label tool. , to create C
A
B
Next, select all three points using the selection arrow tool and, from the Construct menu above, select the Segments option. You will create the triangle shown below. C
A
B
Mark the interior angles of ABC as shown below using the marker tool. C
A
B
Use shortcut commands to select parts of the sketch (Control‐A) and then copy (Control‐C) and paste (Control‐V) the entire sketch. You should have a picture like this. Page 8 Select the pasted copy of
ABC and, from the Transform menu above, choose the Rotate option with the angle of rotation set to 180 . Your diagram will now look like this. Hide the copied version of ABC so it will not distract you. 
Select only the newest version of ABC , i.e. the one created via the rotation, and drag the triangle so that its vertex B lies on the original vertex C and its vertex C lies on the original vertex B. This is possible because the copied and rotated BC is congruent to the original BC . This will create a quadrilateral as shown below. Page 9 Notice that, in this new quadrilateral, opposite angles are congruent, adjacent angles are supplementary, and opposite sides are congruent. Hence, the new quadrilateral is a parallelogram and, thusly, its area can be calculated via the formula Area   base  height  . Since this parallelogram was created using two copies of the original triangle, the area of the triangle must be half that of the 1
parallelogram. Hence, the area of a triangle can be calculated via the formula Area   base  height  . 2
Deriving the Area of a Triangle via Parallelograms (with Elegance) Alternatively, this process can also be performed by selecting our original triangle and rotating it 180 about point C via the Rotate option in the Transform menu. Now use the marking tool to show where the corresponding angles have been relocated. Your diagram should look like this. C
A
B
Now, in order select points C and then B. From the Transform menu, click on the Mark Vector option. A dashed line will travel from point C and to point B and then vanish. This confirms you have marked a vector. Deselect all objects and then select only the rotated portions of the diagram. From the Transform menu, click on the Translate option. A popup menu will appear. Note the marked option is already selected. Click the translate button. Your new diagram will look like this. C
A
B
Again, use the marking tool to show where the corresponding angles have been relocated then hide the rotated copy of the triangle. Page 10 C
A
B
We again notice that, in this new quadrilateral, opposite angles are congruent, adjacent angles are supplementary, and opposite sides are congruent. Hence, the new quadrilateral is a parallelogram and, thusly, its area can be calculated via the formula Area   base  height  . Therefore, once again we see the area of a triangle can be calculated via the formula Area 
1
2
 base  height  . Sketchpad Satisfies the Skeptics If for any reason you are skeptical about the congruencies of the opposite sides or angles, you can use the Measure function of The Geometer’s Sketchpad to verify these. To measure the length of segments, simply select desired segments and then choose the Length option from the Measure menu. The results will be displayed in text boxes with labels automatically added to the measured segments. C
j = 5.28 cm
j'' = 5.28 cm
j
A
j''
B
Similarly, angles can be measured by selecting in order the vertices that define them and then choosing the Angle option from the Measure menu. mCAB = 59.89°
mCA''B = 59.89°
mA''CA = 120.11°
mA''BA = 120.11°
C
A''
A
B
Furthermore, you can verify that the adjacent angles at vertices B and C are supplementary by having The Geometer’s Sketchpad perform the sums of the angle measurements. To do this, click on the Number menu and select the Calculate option. A calculator‐ish display will appear. Select an angle measurement you wish to work with from the sketch, press the “+” button on the display, and then Page 11 select the other angle measurement you wish to work with from the sketch. Finally, click the “ok” button. Your calculations will appear in text boxes. mCAB = 59.89° mCAB + mA''BA = 180.00°
mCA''B = 59.89° mCA''B + mA''CA = 180.00°
mA''CA = 120.11°
C
mA''BA = 120.11°
A''
A
B
Most amazingly, the calculations and measurements are live and dynamic. Select any of the original points, i.e. A, B, or C, and move that point. Notice the parallelogram properties and congruencies are preserved even though the relative measurements change. mCAB = 85.31° mCAB + mA''BA = 180.00°
mCA''B = 85.31° mCA''B + mA''CA = 180.00°
mA''CA = 94.69°
C
mA''BA = 94.69°
A
B
A''
Recognizing that Life isn’t that Different for Trapezoids Start with a new sketch. Use the line segment construction tool to create two segments that share a single point. This is the same process we used to begin creating the parallelogram. Go ahead and use the label tool to label the points as A, B, and C. A
B
C
Use the Construct menu and Parallel Line option to construct a line parallel to BC passing through point A. Your diagram may look like this. Page 12 A
B
C
Use the line segment construction tool to create a segment starting at point C and terminating anywhere on the newly created parallel line. Construct the new segment so that it is obvious that it is not parallel to AB . A
B
C
Label the newest point D and use the Segment option from the Construct menu to construct a segment from point A to point D. Finally, hide the parallel line. Voila, a trapezoid. A
D
B
C
Mark the four interior angles with ascending numbers of strokes. A
D
B
C
Page 13 From parallel line axioms, we can see that A and B are supplementary as well as C and D are supplementary. If the reader is skeptical, use the Measure function of The Geometer’s Sketchpad to verify this. Now, use the methods detailed in the elegant version of the triangle area derivation to rotate the trapezoid 180 and mark the locations of corresponding angles. D
A
B
C

Now, translate the rotated image along the vector CD ; mark the locations of corresponding angles, and, finally, hide the rotated trapezoid. A
D
B
C
Note that the adjacent angles at point C are the original C and a rotated and translated copy of D . Since C and D were supplementary, the adjacent angles at point C are also supplementary and thereby form a straight angle. This is also true for the adjacent angles at point D. Thus, our new polygon is a quadrilateral. Furthermore, for a third time we notice that, in this new quadrilateral, opposite angles are congruent, adjacent angles are supplementary, and opposite sides are congruent. Hence, the new quadrilateral is a parallelogram and, thusly, its area can be calculated via the formula Area   base  height  . Here, however, we must notice that the “base” of the parallelogram was constructed via a sum of the parallel sides of the original trapezoid. To make our formula better reflect the original trapezoid, let us modify it to Area   sum of lengths of parallel sides  height  . Unfortunately, this is far too cumbersome. So, we simplify it to Area   sum of bases  height  , where Page 14 we understand that bases are the parallel sides of the trapezoid. Finally, since we needed two copies of the trapezoid to construct our parallelogram, we see the area of a trapezoid can be calculated via the 1
formula Area   sum of bases  height  . 2
Extending this Method to Regular n‐gons Now that we understand the idea of using rotations and transformations to form parallelograms and thereby derive area formulae, let’s see how this would apply to regular n‐gons. For the sake of brevity, step by step directions will be omitted in favor of a general description of the method. Begin with any regular n‐gon. Realize there are two cases to explore, those being case 1, n is even, and case 2, n is odd. Let’s examine case 1, n is even. So assume we have a regular n‐gon with an even number of sides. For our example, we will consider the regular octagon shown below. Since this octagon is regular, note all sides and all interior angles are congruent. Mark them as such. Page 15 Now construct bisectors of each interior angle. Mark the bisected angles as congruent. Note these bisectors intersect at a single point, the circumcenter of the octagon, i.e. the center of the circle that would circumscribe the octagon. This circle will be important for later discussion, but we don’t need it for now. Understanding this, we will remove the circle from the diagram. Now, construct segments from the circumcenter to each vertex of the octagon and hide the angle bisecting rays. Page 16 By the angle‐side‐angle (ASA) postulate, we observe the newly created triangles are congruent. Thus, since corresponding parts of congruent triangles are congruent, all of the angles of these triangles that have the circumcenter as a vertex are congruent. Similarly, the segments that join the octagon vertices to the circumcenter are also congruent. Mark all of these as such. Using rotations and translations, align the congruent triangles as shown below. D
A
E
B
F
C
Page 17 Note that at points A, B, C, D, E, and F the three angles (two with double strokes and one with triple strokes) are congruent to the original three angles of each triangle and thus have a sum of 180 . Hence, the figure is a quadrilateral. Furthermore, since again we see that opposite angles are congruent, adjacent angles are supplementary, and opposite sides are congruent, the quadrilateral is a parallelogram. So once more its area can be calculated via the formula Area   base  height  . D
E
F
height
A
B
C
Here, however, we note that the base consists of half of the sides from the original octagon. Therefore, the length of the base is half of the perimeter of the original octagon. Taking this into account, we 1
modify our area formula to be Area   perimeter  height  . Furthermore, note that the height of the 2
parallelogram is also the perpendicular distance from any edge of the original octagon to its center, which is called the apothem. Therefore, the area of the octagon, or more generically any n‐gon with n 1
being even, is Area   perimeter  apothem  2
apothem
Now we examine case 2, n is odd. We will work this case more quickly since the general method has already been demonstrated. Assume we have a regular n‐gon with an odd number of sides. Let’s use a regular pentagon for the example. As before, let’s mark all of the congruent sides and internal angles. Page 18 Again, we will construct the angle bisectors of the pentagon’s internal angles and thereby find the circumcenter. Furthermore, re‐mark the angles to reflect the new congruencies. Like before, let’s hide the circumscribed circle and angle bisectors and instead construct segments from the vertices of the pentagon to the circumcenter. Again, the angle‐side‐angle (ASA) postulate establishes that the newly created triangles are congruent. As such, we mark the corresponding parts as congruent. Page 19 Using rotations and translations, align the congruent triangles as shown below. C
A
B
Note that at points A, B, and C the three angles (two with double strokes and one with triple strokes) are congruent to the original three angles of each triangle and thus have a sum of 180 . Hence, the figure is a quadrilateral. This time, however, with an odd number of triangles, the resulting quadrilateral is not a parallelogram. We can resolve this problem in two different ways. Here is the first method. Clearly, the addition or removal of one triangle to either end of the quadrilateral would give us an even number of triangles and, as before, the quadrilateral would be a parallelogram. As an example of this we add one triangle to the right end of the quadrilateral below. Since by definition opposite sides of parallelograms are parallel, the top and bottom of our quadrilateral are parallel. This fact remains true if we remove the “extra” dashed triangle and return to the quadrilateral formed only from the triangles that constituted the original pentagon. Clearly, the left and right sides of the quadrilateral formed from the triangles of the pentagon, for simplicity and clarity let’s call it the “original quadrilateral”, are not parallel. Thus, the “original quadrilateral” has exactly one pair Page 20 of parallel sides and is therefore a trapezoid. Fortunately, we have already found a formula for the area 1
of a trapezoid, i.e. Area   sum of bases  height  . 2
height
Observe that the two bases are merely a realignment of the sides from the original pentagon. Therefore, the sum of the bases is the sum of the sides of the original pentagon, i.e. its perimeter. 1
Accordingly, we can update our area formula to be Area   perimeter  height  . Furthermore, note 2
that the height of the trapezoid is also the apothem of the original pentagon. Therefore, the area of the 1
pentagon, or more generically any n‐gon with n being odd, is Area   perimeter  apothem  . 2
apothem
Alternatively, i.e. as the second method, we could create a duplicate copy of the entire quadrilateral. Then we use rotations and translations to align the two quadrilaterals as shown below. Page 21 Now, we see that opposite angles are congruent, adjacent angles are supplementary, and opposite sides are congruent. Hence, the new quadrilateral is a parallelogram and as such its area can be calculated via the formula Area   base  height  . It is important to observe that the base of our parallelogram is five of the sides from the original pentagon. Therefore, the length of the base is the perimeter of the original pentagon. Accordingly, we can update our area formula to be Area   perimeter  height  . height
Furthermore, note that the height of the parallelogram is also the apothem of the original pentagon. Modifying our current area formula to reflect this we find Area   perimeter  apothem  . But wait, we needed two copies of the triangles from our original pentagon to create this parallelogram. This means our formula is not calculating the area of our original pentagon, but rather twice the area. Therefore, the area of the pentagon, or more generically any n‐gon with n being odd, is 1
Area   perimeter  apothem  . 2
apothem
In summary, regardless of whether n is even or odd, we can now say the area of any regular n‐gon is 1
given by the formula Area   perimeter  apothem  . 2
Page 22 Concerning Circles, Infinity, and Area The method for using parallelograms to find the area of a circle combines the n‐gon method with the concept of limits. First, however, we must recall a few facts about circles, radii, diameters, and  . The original Greek definition of  is that it is the ratio of the circumference to diameter of any circle, i.e. C C
C
  . Then since diameter is defined to be twice the radius, we see    . Thusly, C  2 r . d 2r
d
Now begin with any circle and divide it into an even number of congruent sectors. For the first example, let’s use four sectors. Mark the corresponding congruent parts as shown below. Now, as we did with the n‐gons, use rotations and translations to rearrange the sectors as shown below. Does this remind you of anything? Not yet, let’s try again with ten sectors. Page 23 Now let’s try with 24 sectors. Finally, let’s try with 60 sectors. For added clarity, the triangle vertices are hidden in this diagram. Notice that as the number of sectors increases, the composite figure created by the rotated and translated sectors more closely resembles a parallelogram. We can explain this in a number of ways. We could simply attribute this to the property of local linearity, i.e. the closer you zoom in on any curve, the more it resembles a line. Alternatively, we could notice that the greater the number of sectors, the shorter the arc on each sector and, thusly, the closer together the endpoints of each arc. Thinking about Page 24 this from a calculus perspective, if we constructed a line passing through the endpoints of the arc of one of the sectors, it would be a line passing through two points on the original circle, i.e. a secant line to the circle. As we increase the number of sectors, the distance between the endpoints of a sector, i.e. the points creating our secant line, decreases. As we decrease the distance between the points creating a secant line, the secant line more closely approximates a tangent line. Further recall that a line tangent to a circle is perpendicular to the radius of that circle. Thus, as the number of sectors increases and our secant lines transform into tangent lines, the radius of the circle, which we could think of as the height of the sector, becomes the height of the triangles and, thereby, the height of the parallelogram. So thinking of the composite figure created by the rotated and translated sectors as a parallelogram, its area is generated by the formula Area   base  height  . Note that the base consists of half of the arcs from the sectors and hence would be half of the circumference of the circle. Furthermore, recall that we observed earlier that the height of the parallelogram is the radius of the circle. Incorporating these details into our area formula yields a satisfying and concurring result. Area   base  height  
1
2
1
 C  r    2 r  r    r 2 2
Final Thoughts The parallelogram: viewable now as the foundation of area formulae for circles, squares, triangles, rectangles, trapezoids, and regular n‐gons. The parallelogram: the epitome of simplicity, potency, and versatility. So what do you say? Isn’t it time we give the parallelogram just a little more respect? Join the revolution and raise the call! Viva la Parallelogram! Viva la Parallelogram! Viva la Parallelogram! Page 25 Arenas Where the Manifesto is Best Delivered Possible TEKS correlations could include: 6th Grade: 6B, 8B 7th Grade: 6A, 6B, 9A Geometry: 2A, 2B, 8A, 9A, 9B Precalculus: 1E Materials Needed to Support the Cause The only materials required to implement this activity into a K‐12 classroom are a computer (or computer lab) and the The Geometer’s Sketchpad software. Please visit the website of Key Curriculum Press (http://www.keypress.com/sketchpad/) for information on pricing of the software. Note also some features utilized in this presentation require version 5 of The Geometer’s Sketchpad and may not be available in earlier versions. Page 26