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Placing a Fire Hydrant

Placing a Fire Hydrant

We’ve used the Illustrative Mathematics task Placing a Fire Hydrant for several years now. Each year, the task plays out a bit differently because of the questions that the students ask and the mathematics that students notice. Which is, honestly, why I continue to teach.

I set up our work for the day as practicing I can make sense of problems and persevere in solving them and also I can attend to precision. If you don’t know how to start at Level 3, use Levels 1 and 2 to help you get there.

SMP1 #LL2LU

SMP5 #LL2LU

In an effort not to articulate all of the requirements ahead of time, I simply asked: where would you place a fire hydrant to serve buildings A, B, and C. Students dropped a point at the location they thought best.

It was then obvious from the students’ choices that they thought equidistant was important.

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This year I didn’t put out tools that students might choose to use. Instead, I set the timer for them to work alone on paper for a few minutes and told them to ask for what they needed. Before I could get from the front of the room to the back, almost every hand was raised to request either a ruler or a protractor. (No one asked for a compass this year. Last year, when I had them out on the tables, lots of students used them.)

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I gave students a few more minutes to work individually with the option, this time, of working with the TI-Nspire software to show their thinking. And at the end of that, I added a few more minutes, asking students to focus on how they could justify that their solution always works. Then I gave them a few minutes to discuss their thinking with a partner.

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I watched (or monitored, according to Smith & Stein’s 5 Practices) while they worked using the Class Capture feature of TI-Nspire Navigator. During that time I also selected and sequenced for our whole class discussion. I wanted some of the vocabulary associated with special segments in triangles to come out of our discussion, so I didn’t immediately start with the correct solution.

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We started with Autumn, who had constructed the midpoints of the sides and then created both a midsegment of the triangle and some medians of the triangle. She could tell that the intersection of the midsegment and medians was “too high”.

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C chimed in that she had constructed lots of midsegments. In fact, she had created several midsegment triangles, one inside the other.

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Next we went to Addison, who not only had created all three medians of the triangle but had also measured to show that the medians weren’t the answer.

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That led to S, who had been trying to figure out when the intersection of the medians would be a good location for the fire hydrant.

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Arienne told us about her approach next. She had placed a point inside of the buildings, measured from the point to each building, and she was moving the point around to a location that would be equidistant from the buildings.

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Reagan talked with us about her solution next. She had constructed the perpendicular bisectors and measured from their intersection to each vertex to show that it always worked.

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I wonder what that point has to do with the vertices. What do you see in the diagram? (I was expecting students to “see” a circle. But they didn’t. They saw a triangular prism.) I wasn’t ready to show them the circle, though. How could I help make the circle visible without telling them? A new question came to me: What if we had a 4th building? Where could we place the building so that the fire hydrant served it, too?

I quickly collected Reagan’s file and sent it out to all of the students so that they could create a 4th building that was the same distance from the fire hydrant as A, B, and C.

While they were working, Janie said, “I have a 4th building the same distance, but how do I place it so that it always works?” (On the inside, I was thrilled that Janie asked this question. It is exciting for students to realize this early on in the course that we are about generalizing and proving so that something always works and not just for one case.)

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How do you place the 4th building so that it always works? What is significant about the location of the 3 buildings and the fire hydrant?

Sofia volunteered that her 4th building always works. (I have to admit that I was skeptical, but I made her the Live Presenter and asked how she made it.) Sofia had rotated building C about the fire hydrant to get d. (How many degrees? Does the number of degrees matter? Would rotating always work? Why would it work?) She rotated C again to get a 5th building between A and B. What is significant about the location of the 5 buildings and the fire hydrant?

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And then they saw it. It wasn’t yet pictured, but it had become visible. All of the buildings would form a circle around the fire hydrant! The fire hydrant is the circumcenter of ∆ABC. The circle is circumscribed about the triangle.

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And so the journey continues … every once in a while finding a more beautiful question.

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Area between Curves

Our learning goals for the Applications of Definite Integrals unit in calculus are the following:

I can calculate and use the area between two curves.

I can use the disc and washer methods to calculate and use the volume of a solid.

I can use the shell method to calculate and use the volume of a solid.

I can calculate and use volumes of solids created by known cross sections.

During the lesson focusing on the first goal, we used a scenario from a TImath activity The Area Between to start our conversation.

I rarely send the TNS documents as is to my students or give them a copy of the printed student handout (even though I learn from both in my own planning of how the lesson will play out). This activity gave the following information on the first two pages:

Suppose you are building a concrete pathway. It is to be 1/3 foot deep.

To determine the amount of concrete needed, you will need to:

– calculate area (the integral of the top function minus the bottom function

– calculate volume (area multiplied by depth)

The borders for the pathway can be modeled on the interval -2π ≤ x ≤ 2π by

f(x)=sin(0.5x)+3

g(x)=sin(0.5x)

On the next page, graph the functions. Use the Integral tool to calculate the area under f1 and f2. Then, use the Text and Calculate tools to find the volume of the pathway.

Which takes away any opportunity for students to engage in productive struggle.

I shared this instead:

Suppose you are building a concrete pathway that is to be 1/3 foot deep. The borders for the pathway can be modeled on the interval -2π ≤ x ≤ 2π by f(x)=sin(0.5x)+3 and g(x)=sin(0.5x).

(I’m fully aware that giving them even this much information takes away from the modeling process … but there is always give and take, and for this lesson, the learning goal wasn’t whether they could determine functions for modeling the sidewalk.)

They decided to graph the functions.

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And talked about how they could calculate the area between the curves.

They had never used the Integral tool for graphs, much less the Bounded Area tool, so they oohed and aahed gasped in amazement.

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Sydney asked: Is that the only way to get the area between the curves?

(I knew that she was looking for and making use of structure, composing and decomposing the sidewalk into regions with equal area).

I answered: Is it?

We made Sydney the Live Presenter, and she used the Integral tool to calculate the area between f(x)=sin(0.5x)+3 and the x-axis from -2π to 2π.

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So how can we calculate the amount of concrete needed? The integral and bounded area tools are helpful for visualizing what you’re calculating, but you can’t use those tools on the AP Exam.

And so the students decided to calculate the area between the curves and then multiply by 1/3 to get the volume of the pathway.

Because they were able to tell me what to do, I almost didn’t send a Quick Poll to collect a definite integral that would calculate the volume. I wanted to hurry up and get to a card-matching activity similar to Michael Fenton’s that I knew would be helpful, but instead I eased the hurry syndrome and sent the poll.

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What I saw and heard was well worth the time that it took.

Can you spot the students’ misconception?

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Several students were multiplying the definite integral by and by 1/3, to represent height times base times depth, instead of recognize that the definite integral represented height times base (area), and not just height. (They knew this … we had summed the areas of an infinite number of rectangles for a certain base to calculate area under the curve. But they obviously didn’t know this like they needed to.)

When we calculated their integral, we didn’t get (1/3)*37.699, as expected.

 

Next I purposefully choose a region for which the upper and lower boundaries changed.

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We had a nice look for and make use of structure discussion about different ways to write a definite integral for calculating the area of the region.

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Many of you might notice that there is more opportunity to look for and make use of structure for the concrete pathway. I never asked whether you really need calculus to calculate the volume of the pathway. Nevertheless, I feel like I found two good problems/items/tasks to push and probe student thinking. And there’s always next year, as the journey continues …

 
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Posted by on February 25, 2015 in Calculus

 

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Dilations

Are you aware of all that NCSM has to offer? Many resources are free on their website, and there are additional resources with a paid membership when you login to My NCSM.

I learned about their Illustrating the Standards for Mathematical Practice modules at the NCSM conference a few years ago. While I haven’t reviewed all of them, I’ve been using ideas from their Congruence & Similarity module ever since both with my students and in professional development with teachers.

One of my favorite tasks from the module is Hannah’s Rectangle Problem.

Students determine which rectangles are similar to rectangle a and explain how they know.

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Tracing paper, scissors, and straightedges were available for students to use. (Although my straightedges really are rulers, I did ask my students to refrain from measuring.)

I set the class mode to individual and sent a Quick Poll to collect their results.

I think CS shared with the whole class first. (I’m writing the post two months after the lesson & can’t remember for sure.) CS had traced rectangle a & transformed the rectangle so that it shared a center with each rectangle.

I know that c, b, e, and g work because the space between the dilated rectangles has the same “width” all the way around. CS showed us what he meant, talking about the “frame” that forms.

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AC added that she did that, too, but she eyeballed the length between corresponding vertices to check for similarity.

Another student noted that h works because the two rectangles are congruent, which of course made us think about why similarity does not imply congruence but why congruence does imply similarity.

What about one that doesn’t work?

CS shared what he did with rectangle d, and showed how the horizontal width of the “frame” isn’t the same as its vertical width.

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Can we check whether one rectangle is a dilation of rectangle a using a different center?

At this point, we watched the video that accompanies the NCSM learning module, so that students could see Randy’s method. Had one of my students come up with Randy’s method, of course we wouldn’t have watched the video, but they didn’t, and so I was glad for a different source of learning other than me.

(Randy draws the diagonals of the rectangles from a shared vertex to show whether or not the rectangles are similar.)

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I love that the rectangles not only gave my students the opportunity to remember how we define similarity from grade 8:

CCSS-M 8.G.A.4. Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.

But also gave them the opportunity to practice one of our high school geometry standards as well:

CCSS-M G-SRT.A.2. Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar.

Students then dilated a figure using our dynamic geometry software and played. I watched using Class Capture.

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What happens when you move the center of dilation?

What happens when the scale factor is 3?

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What changes when the scale factor is –3?

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What is significant about a scale factor of –1?

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And students transferred what they had learned to paper:

Dilate ΔXYZ about point C by a scale factor of 2.

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Dilate points A and B about point O by a scale factor of 3.

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And so the journey continues, with many thanks to NCSM for all of the work they do to provide worthwhile resources to the mathematics education community.

 
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Posted by on January 10, 2015 in Dilations, Geometry

 

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Midsegments

Midsegments

CCSS-M-G-CO.C.10

Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.

What do you do when the standard for the day gives away what you want students do explore and figure out on their own?

I’ve made a deal with my administrator to post the process standard for the day (Math Practice) instead of the content standard.

In many of our geometry classes, our learning goals include look for and make use of structure and look for and express regularity in repeated reasoning.

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We defined midsegment:

A midsegment of a triangle is a segment whose endpoints are the midpoints of two sides of the triangle.

The midsegment of a trapezoid is a segment whose endpoints are the midpoints of the non-parallel sides.

Then we constructed the midsegment of a trapezoid. Students observed the trapezoid as I changed the trapezoid.

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I sent a Quick Poll: What do you think is true about a midsegment of the trapezoid?

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It creates both a triangle and a trapezoid. 1

all the midpoints form a similar triangle    1

mn is parallel to yx   2

((1)/(2))the size of the origin        1

parallel to base of triangle   1

all the midsegments make a similar triangle upside down          1

parallel to the base   1

cre8

Δ

and a trap      1

creates triangle and trapezoid        1

It will form the side of a triangle that is similar to the original.   1

The midsegment is parallel to the side not involved in making the midsegment.        1

it would be a median            1

MN is parallel to YX  1

parallel to base          1

mn parallel to yx       1

MN is parallel to the bottom line     1

XMN=XNM     1

cuts the tri into a trap and tri          1

all the segments will make a similar triangle tothe original         1

all of the midpoints connected make a similar triangle to the original one       1

creates ∆ on top + trap. on bottom 1

Triangle XMN is similar to triangle XYZ.

Line MN is parallel to line YX.          1

2/3 the largest side  1

2/3 o  1

side parallel to the midsegment is a           1

it makes a triangle and a tra            1

mn is ll to yx, mnx is congruent to triangle mon   1

In order to change things up a bit, I quickly printed the students conjectures, cut them up, and distributed a few to each team. Now you decide whether the conjectures you’ve been given are true. And if so, why?

I used Class Capture to monitor while the students talked and worked (and played/explored beyond the given conjectures).

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Then I asked what they figured out through a Quick Poll:

seg mn parallel to both seg ab and seg dc 1

parallel to both of the bases            1

it cre8 2 trap 1

mn is parallel to dc and ab  2

Trapezoid ABCD is similar to ABNM.

Lines AB and CD are parallel to the midsegment. 1

MN is parrellel to AB and DC           1

It creates a line that is parallel to the bases and forms two trapezoids. 1

nidsegment is parallel to top and bottom sides    1

((AB+DC)/(2))          2

it would be parallel to the sides above and below it        1

MN is parallel to DC and AB 1

(AB+DC)/(2)=MN     1

its parallel to DC        1

It makes two trapezoids       1

it is // 2 ab and dc    1

it forms two trapozoids        1

It is parallel to the sides above and below it.         1

makes 2 trap. 1

It makes a similar trapezoid.            1

they make 2 trapezoid, ab+dc/2     1

parellel to base of trapezoid            1

ab ll to mn ll to dc. when a parallelogram, creates 2 congruemt trapezoids      1

trapezoid ABNM is similar to trapezoid MNCD      1

all midsegments make a diala         1

And then we talked about how they knew these statements were true.

Jameria had a lot of measurements on her trapezoid. I made her the Live Presenter. What conjectures can we consider using this information?

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I made Jared the Live Presenter.

What does Jared’s auxiliary line buy us mathematically?

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I made Landon the Live Presenter.

What conjectures can we consider using this information?

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I sent a Quick Poll to formatively assess whether students could use the conjecture we made about the length of the midsegment compared to the length of the bases of the trapezoid.

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What about the midsegments of a triangle?

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And why?

We had not yet started our unit on dilations, and so there was more to the why in a later lesson.

And so the journey continues, even making deals with my administrators as needed, to create a classroom where students get to make and test and prove their own conjectures instead of being given theorems from our textbook to prove.

 
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Posted by on January 4, 2015 in Angles & Triangles, Geometry, Polygons

 

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Proving Segments Congruent First

CCSS-M.G-CO.B.8 Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions.

Proving triangle congruence from rigid motions has been one of our most challenging new standards. Which is exciting for me as a teacher, because I’m always up for learning about something that all of the textbooks from which I’ve taught geometry have let slide into our deductive system as postulates with no need of proof.

SAS

SSS

So after two years of teaching these standards, it occurred to me that maybe we shouldn’t start with proving triangle congruence using rigid motions. Instead, why don’t we see what happens when we start with two segments.

What set of rigid motions will show that segment AB is congruent to segment CD?

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Students started creating a plan (sequence of rigid motions) on paper. But before we moved to the technology to test the plans, we talked about attend to precision. Instead of saying that you’ll use a translation and a rotation, let’s be specific about what translation and what rotation. A translation of what segment by what vector? A rotation of what segment about what point using what angle measure?

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I began to see the specifics on paper, but while students were pretty confident about the translations they had named, they were not totally confident about the rotations they had named. We needed our technology to help us see.

One team translated segment AB using vector BC.

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Then they rotated segment A’B’ about B’ using angle A’B’D.

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We can see that it works: the blue pre-image is now black. And we can move the original segment to see that it sticks.

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Did this team prove that segment AB is congruent to segment CD? Or did they prove that segment AB is congruent to segment DC? Does it matter?

How many times have we told students that saying segment AB is congruent to segment CD is the same thing as saying that segment AB is congruent to segment DC? It occurred to me in the midst of this lesson that we have actually shown why those endpoints are interchangeable in our congruence statement for segments.

Another team used a translation (segment AB using vector BD) and was trying to use a reflection. When I discussed their work with them, they said, “we know where to draw the line, but we don’t know how to describe it”.

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That team presented their work to the class using Live Presenter.

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They drew in a line and reflected segment A’B’ about the line.

 

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It didn’t work, but they moved the line into the right place.

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So what’s significant about the line of reflection that works?

Someone in the class suggested that it’s the angle bisector of angle DCB”.

Is it?

We don’t have to wonder. We can verify using our Angle Bisector tool.

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What else is significant about the line?

This year in geometry we are often going to have to see what isn’t pictured (look for and make use of structure).

What if we draw in the auxiliary segment B’D? What else is significant about the line of reflection?

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It’s the perpendicular bisector of segment B’D. And again, we don’t have to wonder. We can verify using our Perpendicular Bisector tool.

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Can you show that segment AB is congruent to segment CD using only reflections?

We left this exercise for Problem Solving Points, as of course by now we were running out of class time.

One student shared her work with me the next day.

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And the next.

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We still have work to do. But that’s good … this was the first week of school.

We did this task at my CMC-S session recently, and I asked about using at least one reflection since, like my students, most everyone translated and then rotated.

I was expecting to hear: Reflect segment AB about the perpendicular bisector of segment AC.

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Then reflect segment A’B’ about the perpendicular bisector of segment A’D.

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Some day I’m going to learn to not be surprised by solutions that are different from mine. One of the participants suggested extending lines AB and CD until they meet at point I. Then reflecting segment AB about the angle bisector of angle BIC. Then they translated segment A’B’ using vector B’C. Several weren’t satisfied with proving segment AB congruent to segment DC, so we noted that we could reflect segment B’’A’’ about its perpendicular bisector to show that segment AB is congruent to segment CD.

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Another participant asked whether students had been confused moving into proving triangles congruent by changing the order of the vertices, since we can’t do that in a congruence statement. We did this at the beginning of the rigid motions unit, and I didn’t notice any issues moving into congruence statements for triangles where the order of the vertices matters.

And so the journey continues … every once in a while figuring out a task that will help us along the way towards meeting our learning goals.

 
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Posted by on November 9, 2014 in Geometry, Rigid Motions

 

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A-S-N-T-F

In geometry, we often use Always, Sometimes, or Never:

A trapezoid is ___ a parallelogram.

A parallelogram is ___ a trapezoid.

(Be careful how you answer those if you are using the inclusive definition of trapezoid.)

And in geometry, we often use True (A) or False (S or N):

A trapezoid is a parallelogram.

A parallelogram is a trapezoid.

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(Apparently we were talking about squares and trapezoids, not parallelograms and trapezoids, when we were figuring out which had to be true.)

And in geometry, we often use implied True (A):

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So when a few students asked about this question, we asked whether you could draw any parallelogram that doesn’t have four right angles. Since you can, we don’t say that the statement is (A) true.

[Note: the green marks indicate the number of students who answered both and only rectangle and square.]

In geometry, we are still learning the implications of the inclusive definition of trapezoid. Another of our questions was the following.

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And thank goodness, in geometry, I have students who question me, even those with a voice so quiet we have to lean in to hear. “But I can draw a trapezoid that doesn’t have exactly one pair of parallel sides. I don’t think the trapezoids should be marked correct.”

And of course, he is right. In our deductive system, we don’t name any quadrilaterals with exactly one pair of parallel sides.

So how could I recover our lesson and be sure that my students understood both what we mean by (A) true and our inclusive definition of trapezoid?

I asked students to look ahead to a graphic organizer (borrowed from Mr. Chase, who borrowed from mathisfun.com), review it, and answer the new question. I took the question from the opener and changed “exactly one” to “at least one”. I asked students to work alone.

Graphic Organizer

Here’s what I got back.

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Without showing them the results, I asked students to talk with their teams and answer the poll one last time. All 31 students answered correctly.

So what next?

I’ve been determined over the past three years to stay away from the quadrilateral checklist. You remember the one, right? This is mine from the first 18 years of teaching geometry. I didn’t complete the list for them – each team had a different figure, and they measured (with rulers and protractors before we had the technology with measurement tools) and figured out which properties were always true. But still – how effective is it to complete a checklist, even when you and your classmates are figuring it out?

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We wanted to get at how the quadrilaterals are the same and how they are different in a way that was more engaging than just showing a few figures and asking students to calculate a missing measurement.

So another Quick Poll to get the conversation going. Students immediately began talking with their teams. For which figure(s) will the one angle measure be enough for us to determine the remaining interior angle measures?

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And decent results.

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We went to figure B. Why isn’t one angle measure enough?

And then figure C. Why isn’t one angle measure enough?

And then figure D. Why is one angle measure enough?

Student justifications included words like “rotation”, “reflection”, “decompose into triangles”, “isosceles triangle”. We talked about how we knew the triangles in the kite pictured were not congruent, and in fact not similar either, when decomposed by the horizontal diagonal. Informal justifications … but justifications, nonetheless … and hopefully ammunition for students to realize they can make sense out of these exercises transformationally without having a list of properties for each figure memorized.

We spent a little more time on rhombi using a Math Nspired document for exploration, after which I sent another Quick Poll:

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How did ten students get 144˚?

The students figured out the error: a misreading of which angle is 36˚ … not a misunderstanding of angle measure relationships in rhombi.

And then more about kites using the same Math Nspired activity, during which time a student asked to be made the Live Presenter so that he could show his concave kite to the class. What properties do concave and convex kites share? (More than I expected. I’m not the what-can-I-do-to-break-the-rule kind of person. But I am surrounded by students and daughters who are.) And I am still amazed that SC asked to be the Live Presenter since that was usually the time that he excuses himself to go to the restroom.

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So what information is enough angle-wise in the kite for you to determine the rest?

We ended with a bit of closure with two final Quick Polls & results that provide evidence of student learning.

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And so the journey continues … always rethinking and revising lessons and questions to get the most out of our time and conversations together.

 
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Posted by on November 5, 2014 in Geometry, Polygons, Rigid Motions

 

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Origami Regular Octagon

We folded a square piece of paper as described in the Illustrative Mathematics task, Origami Regular Octagon. I didn’t want students to know ahead of time that they were creating an octagon, so I changed the wording a bit. We folded (and refolded … luckily, there was not a 1-1 correspondence between paper squares and students). Students worked individually to write down a few observations and then we talked all together.

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It’s an octagon.

There are 8 equal sides.

There are 8 equal angles.

It’s a regular octagon (this is the first year my students have come to me knowing what it means for a polygon to be regular).

How do you know there are 8 equal sides and 8 equal angles?

Because we folded it that way.

How do you know there are 8 equal sides and 8 equal angles?

Because one side is a reflection of its opposite side about the line that we folded.

What is the significance of the lines that you folded?

They are lines of symmetry.

There are 8 of them.

The opposite sides are parallel.

How can you tell?

This took a while. Maybe longer than it needed to.

Another student raised his hand.

I figured out that the sum of the angles in the octagon is 540˚.

(I don’t have the sum of the interior angles of an octagon memorized since I can calculate it, but I did know that 540˚ was too small.)

How did you get that?
I made an octagon and measured the angle. Then I multiplied by 8.

On your handheld?
Yes.

Okay. Let’s see what you have. I made him Live Presenter.

He showed us the angles he measured that were 67.5˚.

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It might help if we can see the sides of your angles. Will you use the segment tool to draw them?

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Other students argued that we needed to double 540 to get the sum of the angles in the octagon, 1080˚.

What else do you notice?

Triangles.

Congruent triangles.

Right triangles.

Students noticed different numbers of triangles.

And they recognized that we knew about congruence because of reflections.

Somehow we asked the question about the value of the angle (x).

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I set up a Quick Poll to collect student responses.

Almost everyone got the correct answer of 22.5˚.

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Can you tell me how you got your answer?

One student used the ¼ square with a 90˚ angle that had been bisected by the folded line to be 45˚ and the bisected again by the folded line to argue that x was 22.5˚.

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Did anyone do something different? Hands went up all around the room.

AC hasn’t talked to the whole class yet today, so I asked what she did.

I saw a circle with 360˚ and divided by 16.

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Then DC’s hand went up. 360/16 is equivalent to 180/8. I saw a line divided into 8 equal parts (or straight angle).

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Then TC showed us the isosceles triangle she used with the 62.5˚ base angles.

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Then someone else showed us the right triangle he used with the complementary acute angles.

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Before we knew it, we had spent almost an hour talking about a regular octagon. And learning math using quite a few Math Practices: construct a viable argument and critique the reasoning of others, look for and make use of structure, use appropriate tools strategically.

I’ve wondered before how much longer we will need to talk about generalizing relationships for interior and exterior angles in polygons. Today I got a glimpse of students being able to figure out those relationships by looking for and making use of structure. The only concern that remains is the length of time it would take to do that on a high stakes standardized test such as the ACT or SAT. And so the journey to do what is best for my students continues …

 
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Posted by on September 21, 2014 in Geometry, Rigid Motions, Tools of Geometry

 

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