7:55 -- Second period (again, at high schools "first period" is more like zero period) is the first of the three math classes. It's Basic Math, and the students have a Pizzazz worksheet on factoring -- they must list all the factors of a number.
This worksheet lists all but two factors of a number, and they must find the missing factors. In all but one problem, the two factors don't multiply to the given number. For example, they might be given the number 12 with factors 1, 2, 3, 12, and two missing factors. Since 12/2 = 6, one of the missing factors is 6, and since 12/3, the other missing factor is 4. (The only exception is for a prime number like 13 where neither factor is given, but here it's obvious what the factors should be.)
8:50 -- Second period leaves and third period arrives. This is another math class, except it's Algebra 1B, the second semester of Algebra I spread out to a year for special ed students.
These students are learning about exponents -- this time, they have another worksheet.
9:50 -- Third period leaves for snack.
10:00 -- At this school, tutorial is after snack. I meet the Cross Country runner, who tells me that he broke 16 minutes at yesterday's race. Unfortunately, his team finished in third place, but this is still good enough to advance to the CIF meet in nine days.
10:30 -- Fourth period is the one science class. This time, the students are able to take the test -- mainly because yesterday's referrals resulted in a one-day suspension for the two troublemakers.
11:30 -- Fourth period leaves and fifth period arrives. This is a "Study Skills" class, but today they are finishing Nightmare Before Christmas.
12:25 -- Fifth period leaves for lunch.
1:10 -- Sixth period arrives. This is the final math class, another Algebra 1B. Again, I use the earlier third period class to help me choose which problems to go over.
2:05 -- Sixth period leaves. Once again, this teacher has five straight classes, and so there's no need for me to stay for seventh period.
Today, only sixth period has an aide. But of course, the suspensions make it much easier for me to manage the classes even without an aide. Not only is fourth period much quieter for the test, but so is second period, in which the younger suspended student is also enrolled. Still, I wonder whether I could have managed the class better yesterday without having to suspend anyone. For starters, could I have figured out that the senior wasn't really an aide?
I do wish to compare what the students are learning to my "science" class at the old charter. The three phases of matter (solids, liquids, gases) are also included in the middle school curriculum as part of the unit "Matter and its Interactions." It appears to be part of the seventh grade curriculum under the preferred Integrated Science model.
Today on her Mathematics Calendar 2018, Theoni Pappas writes:
What is the area of this large tangram square, if its small inner square's area is 1/8 square units.
Tangram pieces tend to be symmetrical, hence some symmetry is assumed even if it can't be derived from the given diagram. In particular, the square of area 1/8 has two of its sides along the diagonals of the large square (even though one diagonal isn't drawn all the way, so technically we can't be completely sure that it's a diagonal). One vertex of the small square is at the point where the diagonals of the large square meet, and the opposite vertex of the small square lies on a side of the large square. This info now complete determines the ratio of the areas of the two squares.
We draw in the diagonal of the small square (starting at the center of the large square). Each of these two resulting triangles has the same base and height (hence the same area) as two more small triangles (bounded by side of small square, side of large square, diagonal of large square). This gives a total area of 1/8 + 1/8 = 1/4 for four small triangles. These four triangles comprise one large triangle bounded by one side of the large square and both of its diagonals.
But this triangle is clearly one-fourth of the total area of the square. Therefore the original square is in fact a unit square. The desired area is one square unit -- and of course, today's date is the first.
Notice that some of the symmetry we use in today's problem also appears in today's lesson, so let's hurry and get to it.
Lesson 5-5 of the U of Chicago text is called "Properties of Trapezoids." In the modern Third Edition of the text, trapezoids appear in Lesson 6-6.
This is what I wrote last year in describing this worksheet:
The first theorem that we have is the Isosceles Trapezoid Symmetry Theorem:
The perpendicular bisector of one base of an isosceles trapezoid is the perpendicular bisector of the other base and is a symmetry line for the trapezoid.
Notice that this theorem is similar to the Kite Symmetry Theorem. In many ways, there is a sort of dualism between the kite and the isosceles trapezoid. A kite is defined by having consecutive equal sides, while an isosceles trapezoid has consecutive equal angles. The symmetry line for the kite bisects two of the angles, and the symmetry line for the isosceles trapezoid bisects two of the sides.
Here is the proof of the Isosceles Trapezoid Symmetry Theorem as given by the U of Chicago. This time, since Section 5-5 gives a paragraph proof, let me post a two-column proof here on the blog.
Given: ZOID is an isosceles trapezoid with angles I and D equal in measure.
m is the perpendicular bisector ofID
Prove: m is the perpendicular bisector ofZO
m is a symmetry line for ZOID
Proof:
Statements Reasons
1. m is the perp. bis. ofID 1. Given
2. D' = I, I' = D 2. Definition of reflection
3. angle I = angle D 3. Given
4. ray DZ reflected is IO 4. Reflections preserve angle measure.
5. Z' lies on ray IO 5. Figure Reflection Theorem
6. ZOID is a trapezoid 6. Given
7.ZO | | DI 7. Definition of trapezoid
8.ZO perpendicular to m 8. Fifth Postulate
9. Z' lies on line ZO 9. Definition of reflection
10. Z' = O 10. Line Intersection Theorem
11. O' = Z 11. Flip-Flop Theorem
12. ZOID reflected is OZDI 12. Figure Reflection Theorem
13. m symm. line of ZOID 13. Definition of symmetry line
14. m is the perp. bis. of OZ 14. Definition of reflection
This proof is quite long and can be intimidating for students. A teacher can break it down by actually folding the isosceles trapezoid along line m. The teacher can ask, "Where does D fold to?" The students will probably answer I, only to have the teacher ask "Why?"
Then the tougher part is to show that Z folds to O. We do it one step at a time -- first we show that since reflections preserve angle measure, Z' must be somewhere on ray OI -- although not necessarily O (but of course students will want to jump to that conclusion).
Now we have another theorem, the Isosceles Trapezoid Theorem:
In an isosceles trapezoid, the non-base sides are equal in measure.
In other words, our definition of isosceles trapezoid implies what we usually think of when we hear the word isosceles. The text states that this is merely a corollary of the previous theorem -- since the reflection of one of the non-base sides is the other.
Returning as a text, another corollary given is the Rectangle Symmetry Theorem:
Every rectangle has two symmetry lines, the perpendicular bisectors of its sides.
This corollary follows from the Isosceles Trapezoid Symmetry Theorem in the same way that the Rhombus Symmetry Theorem (which shows that a rhombus also has two symmetry lines) follows from the Kite Symmetry Theorem. It shows again the beauty of using inclusive definitions.
So far in this section, we've mentioned the symmetries of the kite, rhombus, isosceles trapezoid, as well as the rectangle. A general trapezoid, as we've mentioned before, has no symmetry. But there's one special quadrilateral that's missing -- the parallelogram. So does a parallelogram have symmetry?
Also, so far in this section we've mentioned reflections and reflection symmetry, but there's another transformation that we've learned about -- rotations and rotational symmetry. So do any of our special quadrilaterals have rotational symmetry.
We can answer both of these questions at once: A parallelogram has rotational symmetry!
Notice that the U of Chicago text doesn't discuss rotational symmetry of any figure, much less that of the parallelogram. Yet we see in the Common Core Standards:
CCSS.MATH.CONTENT.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe therotations and reflections that carry it onto itself.
(emphasis mine)
[2017 update: The rotational symmetry of the parallelogram appears in the Third Edition. Therefore, the many of the changes I made three years ago can also be viewed as "fixing" my Second Edition to include material from the Third Edition. The tricky part is that in adhering to the Second Edition order, the students don't actually see rotations until Chapter 6. So my old worksheet mentions rotations, but that means there's Chapter 6 material on a Chapter 5 worksheet. I'll keep this discussion of rotations here on the blog, but teachers may want to save rotations for Chapter 6.]
And so we should discuss rotational symmetry in general and that of the parallelogram in particular.
But let's begin with the rhombus and rectangle before going on to the general parallelogram. We recall that our definition of rotation is the composite of reflections in intersecting lines -- and if those lines are symmetry lines of a figure, each one maps the figure to itself. Therefore, the entire rotation must map the figure to itself. It follows that any figure that contains two lines of symmetry must automatically have rotational symmetry as well! And the Two Reflection Theorem for Rotations tells us exactly what the center and magnitude of the rotation are. The center is where the two symmetry lines intersect, and the magnitude is double the angle between the intersecting lines.
Corollary to Two Reflection Theorem for Rotations:
If l and m are intersecting symmetry lines of a figure, then it also has rotational symmetry, where the magnitude is twice the non-obtuse angle between l and m and the center is the point of intersection of l and m.
Applying this to the rhombus, we see that both its diagonals are symmetry lines. Because a rhombus is a kite, its diagonals are perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.
And now let's look at the rectangle. Its symmetry lines are the perpendicular bisectors of its sides, and we can use the Two Perpendiculars Theorem and the Fifth Postulate to show that these symmetry lines are also perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.
But as it turns out, all parallelograms -- not just the rhombuses and rectangles -- can be shown to have 180-degree rotational symmetry. The U of Chicago doesn't prove this, but Hung-Hsi Wu does. In his Theorem 4, he proves that the opposite sides of a parallelogram are equal, by showing that the parallelogram has rotational symmetry.
Here is Wu's proof of his Theorem 4, in paragraph form (as Wu himself gives it):
Given: ABCD is a parallelogram
Prove: AD = BC
Proof:
Let M be the midpoint of the diagonalAC and we will use Theorem 1 -- Wu's First Theorem (that a 180-degree rotation maps a line to a parallel line, already proved on this blog) to explore the implications of the 180-degree rotation R around M.
Because MA = MC and rotations preserve distance, we have C" = A, so that R maps line BC to a line passing through A and parallel to line BC. Since the line AD has exactly the same two properties by assumption, Playfair implies that R maps line BC to line AD. Similarly, R maps line AB to CD. Thus since B lies on both AD and CD, its image B" is a point that lies on both AD and CD. By the Line Intersection Theorem, this point is exactly D.
Recall we also have C" = A. Therefore R maps the segmentBC to the segment joining D (which is B") to A (which is C") by the property that a rotation (like a reflection) maps segments to segments. The latter segment has to be the segment DA, by the Unique Line Assumption. Thus by the Figure Reflection Theorem, R maps ABCD to CDAB, so ABCD has rotational symmetry. Since rotations preserve distance, BC = AD, as desired. QED
This proof is very similar to the Isosceles Trapezoid Symmetry Theorem. Both theorems depend on what Wu calls "Lemma 6" -- identifying the reflection or rotation image of a point by finding two lines that intersect at the preimage point and noting that the images of these two lines must intersect at the image of the point.
Like the Isosceles Trapezoid Symmetry Theorem and its corollary the Isosceles Trapezoid Theorem, we should call part of the proof the Parallelogram (Rotational) Symmetry Theorem and then derive that opposite sites are equal as a corollary. Wu also derives as a corollary that opposite angles are equal -- but we can also derive this by applying the Trapezoid Angle Theorem twice -- since a parallelogram is a trapezoid. Even the third major property of parallelograms -- that their diagonals bisect each other -- can be derived as a corollary (since rotations preserve distance, we must also have MB = MD in the above proof).
So we can prove all the major properties of parallelograms using only rotations and not using triangle congruence at all. The U of Chicago derives the properties of other figures using reflection symmetry but reverts to triangle congruence for the parallelogram properties (in Chapter 7).
Finally, notice that since rhombuses and rectangles are parallelograms, we can show that they have 180-degree rotational symmetry without looking at their lines of symmetry. But I still like using the Corollary to the Two Reflection Theorem for Rotations because it can be used to determine the rotational symmetry for figures other than quadrilaterals. We earlier proved that an equilateral triangle has three lines of symmetry -- so it also has rotational symmetry.
Today, only sixth period has an aide. But of course, the suspensions make it much easier for me to manage the classes even without an aide. Not only is fourth period much quieter for the test, but so is second period, in which the younger suspended student is also enrolled. Still, I wonder whether I could have managed the class better yesterday without having to suspend anyone. For starters, could I have figured out that the senior wasn't really an aide?
I do wish to compare what the students are learning to my "science" class at the old charter. The three phases of matter (solids, liquids, gases) are also included in the middle school curriculum as part of the unit "Matter and its Interactions." It appears to be part of the seventh grade curriculum under the preferred Integrated Science model.
Today on her Mathematics Calendar 2018, Theoni Pappas writes:
What is the area of this large tangram square, if its small inner square's area is 1/8 square units.
Tangram pieces tend to be symmetrical, hence some symmetry is assumed even if it can't be derived from the given diagram. In particular, the square of area 1/8 has two of its sides along the diagonals of the large square (even though one diagonal isn't drawn all the way, so technically we can't be completely sure that it's a diagonal). One vertex of the small square is at the point where the diagonals of the large square meet, and the opposite vertex of the small square lies on a side of the large square. This info now complete determines the ratio of the areas of the two squares.
We draw in the diagonal of the small square (starting at the center of the large square). Each of these two resulting triangles has the same base and height (hence the same area) as two more small triangles (bounded by side of small square, side of large square, diagonal of large square). This gives a total area of 1/8 + 1/8 = 1/4 for four small triangles. These four triangles comprise one large triangle bounded by one side of the large square and both of its diagonals.
But this triangle is clearly one-fourth of the total area of the square. Therefore the original square is in fact a unit square. The desired area is one square unit -- and of course, today's date is the first.
Notice that some of the symmetry we use in today's problem also appears in today's lesson, so let's hurry and get to it.
Lesson 5-5 of the U of Chicago text is called "Properties of Trapezoids." In the modern Third Edition of the text, trapezoids appear in Lesson 6-6.
This is what I wrote last year in describing this worksheet:
The first theorem that we have is the Isosceles Trapezoid Symmetry Theorem:
The perpendicular bisector of one base of an isosceles trapezoid is the perpendicular bisector of the other base and is a symmetry line for the trapezoid.
Notice that this theorem is similar to the Kite Symmetry Theorem. In many ways, there is a sort of dualism between the kite and the isosceles trapezoid. A kite is defined by having consecutive equal sides, while an isosceles trapezoid has consecutive equal angles. The symmetry line for the kite bisects two of the angles, and the symmetry line for the isosceles trapezoid bisects two of the sides.
Here is the proof of the Isosceles Trapezoid Symmetry Theorem as given by the U of Chicago. This time, since Section 5-5 gives a paragraph proof, let me post a two-column proof here on the blog.
Given: ZOID is an isosceles trapezoid with angles I and D equal in measure.
m is the perpendicular bisector of
Prove: m is the perpendicular bisector of
m is a symmetry line for ZOID
Proof:
Statements Reasons
1. m is the perp. bis. of
2. D' = I, I' = D 2. Definition of reflection
3. angle I = angle D 3. Given
4. ray DZ reflected is IO 4. Reflections preserve angle measure.
5. Z' lies on ray IO 5. Figure Reflection Theorem
6. ZOID is a trapezoid 6. Given
7.
8.
9. Z' lies on line ZO 9. Definition of reflection
10. Z' = O 10. Line Intersection Theorem
11. O' = Z 11. Flip-Flop Theorem
12. ZOID reflected is OZDI 12. Figure Reflection Theorem
13. m symm. line of ZOID 13. Definition of symmetry line
14. m is the perp. bis. of OZ 14. Definition of reflection
This proof is quite long and can be intimidating for students. A teacher can break it down by actually folding the isosceles trapezoid along line m. The teacher can ask, "Where does D fold to?" The students will probably answer I, only to have the teacher ask "Why?"
Then the tougher part is to show that Z folds to O. We do it one step at a time -- first we show that since reflections preserve angle measure, Z' must be somewhere on ray OI -- although not necessarily O (but of course students will want to jump to that conclusion).
Now we have another theorem, the Isosceles Trapezoid Theorem:
In an isosceles trapezoid, the non-base sides are equal in measure.
In other words, our definition of isosceles trapezoid implies what we usually think of when we hear the word isosceles. The text states that this is merely a corollary of the previous theorem -- since the reflection of one of the non-base sides is the other.
Returning as a text, another corollary given is the Rectangle Symmetry Theorem:
Every rectangle has two symmetry lines, the perpendicular bisectors of its sides.
This corollary follows from the Isosceles Trapezoid Symmetry Theorem in the same way that the Rhombus Symmetry Theorem (which shows that a rhombus also has two symmetry lines) follows from the Kite Symmetry Theorem. It shows again the beauty of using inclusive definitions.
So far in this section, we've mentioned the symmetries of the kite, rhombus, isosceles trapezoid, as well as the rectangle. A general trapezoid, as we've mentioned before, has no symmetry. But there's one special quadrilateral that's missing -- the parallelogram. So does a parallelogram have symmetry?
Also, so far in this section we've mentioned reflections and reflection symmetry, but there's another transformation that we've learned about -- rotations and rotational symmetry. So do any of our special quadrilaterals have rotational symmetry.
We can answer both of these questions at once: A parallelogram has rotational symmetry!
Notice that the U of Chicago text doesn't discuss rotational symmetry of any figure, much less that of the parallelogram. Yet we see in the Common Core Standards:
CCSS.MATH.CONTENT.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe therotations and reflections that carry it onto itself.
(emphasis mine)
[2017 update: The rotational symmetry of the parallelogram appears in the Third Edition. Therefore, the many of the changes I made three years ago can also be viewed as "fixing" my Second Edition to include material from the Third Edition. The tricky part is that in adhering to the Second Edition order, the students don't actually see rotations until Chapter 6. So my old worksheet mentions rotations, but that means there's Chapter 6 material on a Chapter 5 worksheet. I'll keep this discussion of rotations here on the blog, but teachers may want to save rotations for Chapter 6.]
And so we should discuss rotational symmetry in general and that of the parallelogram in particular.
But let's begin with the rhombus and rectangle before going on to the general parallelogram. We recall that our definition of rotation is the composite of reflections in intersecting lines -- and if those lines are symmetry lines of a figure, each one maps the figure to itself. Therefore, the entire rotation must map the figure to itself. It follows that any figure that contains two lines of symmetry must automatically have rotational symmetry as well! And the Two Reflection Theorem for Rotations tells us exactly what the center and magnitude of the rotation are. The center is where the two symmetry lines intersect, and the magnitude is double the angle between the intersecting lines.
Corollary to Two Reflection Theorem for Rotations:
If l and m are intersecting symmetry lines of a figure, then it also has rotational symmetry, where the magnitude is twice the non-obtuse angle between l and m and the center is the point of intersection of l and m.
Applying this to the rhombus, we see that both its diagonals are symmetry lines. Because a rhombus is a kite, its diagonals are perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.
And now let's look at the rectangle. Its symmetry lines are the perpendicular bisectors of its sides, and we can use the Two Perpendiculars Theorem and the Fifth Postulate to show that these symmetry lines are also perpendicular. So the magnitude of the rotation is twice 90, or 180 degrees, about the point where the diagonals intersect.
But as it turns out, all parallelograms -- not just the rhombuses and rectangles -- can be shown to have 180-degree rotational symmetry. The U of Chicago doesn't prove this, but Hung-Hsi Wu does. In his Theorem 4, he proves that the opposite sides of a parallelogram are equal, by showing that the parallelogram has rotational symmetry.
Here is Wu's proof of his Theorem 4, in paragraph form (as Wu himself gives it):
Given: ABCD is a parallelogram
Prove: AD = BC
Proof:
Let M be the midpoint of the diagonal
Because MA = MC and rotations preserve distance, we have C" = A, so that R maps line BC to a line passing through A and parallel to line BC. Since the line AD has exactly the same two properties by assumption, Playfair implies that R maps line BC to line AD. Similarly, R maps line AB to CD. Thus since B lies on both AD and CD, its image B" is a point that lies on both AD and CD. By the Line Intersection Theorem, this point is exactly D.
Recall we also have C" = A. Therefore R maps the segment
This proof is very similar to the Isosceles Trapezoid Symmetry Theorem. Both theorems depend on what Wu calls "Lemma 6" -- identifying the reflection or rotation image of a point by finding two lines that intersect at the preimage point and noting that the images of these two lines must intersect at the image of the point.
Like the Isosceles Trapezoid Symmetry Theorem and its corollary the Isosceles Trapezoid Theorem, we should call part of the proof the Parallelogram (Rotational) Symmetry Theorem and then derive that opposite sites are equal as a corollary. Wu also derives as a corollary that opposite angles are equal -- but we can also derive this by applying the Trapezoid Angle Theorem twice -- since a parallelogram is a trapezoid. Even the third major property of parallelograms -- that their diagonals bisect each other -- can be derived as a corollary (since rotations preserve distance, we must also have MB = MD in the above proof).
So we can prove all the major properties of parallelograms using only rotations and not using triangle congruence at all. The U of Chicago derives the properties of other figures using reflection symmetry but reverts to triangle congruence for the parallelogram properties (in Chapter 7).
Finally, notice that since rhombuses and rectangles are parallelograms, we can show that they have 180-degree rotational symmetry without looking at their lines of symmetry. But I still like using the Corollary to the Two Reflection Theorem for Rotations because it can be used to determine the rotational symmetry for figures other than quadrilaterals. We earlier proved that an equilateral triangle has three lines of symmetry -- so it also has rotational symmetry.
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