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I was helping a home-schooled student with her homework when we came upon several images of figures that we were supposed to find the perimeter of. In several of the figures, some of the lengths were not given; yet, with some thought, the perimeter could still be figured out. For example, the perimeter of the above figure is 28. She could not "see" this.

Her arithmetic skill are admirable. She is neat, organized, and is capable of looking over her work and finding where she made her mistake. Yet, drawing the square and showing what equalled what, did not impress her. I longed for a box of toothpicks. I got the feeling that she was getting area and perimeter confused in some way but I ran out of time before we could resolve her problem.

Does anyone have any ideas what conceptual problems she wasn't resolving?

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  • $\begingroup$ If you just drew a square, with two adjacent sides labeled with the length 7, is she able to get the perimeter? $\endgroup$ – Joel Reyes Noche Jul 2 '17 at 7:47
  • $\begingroup$ Nope. And I want to repeat this. She seems to be a very brigjht student otherwise. There is a concept here that is obvious to me that she is just not seeing. $\endgroup$ – Steven Gregory Jul 2 '17 at 8:26
  • $\begingroup$ Postscript. It's been a while but she finally came back to my station. She has demonstrably figured it out on her own. The odd thing is, she can't explain to me what her problem was. $\endgroup$ – Steven Gregory Jul 13 '17 at 19:00
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This is pure speculation — I don't have enough information to know what the student was thinking — but they may have only known how to compute the perimeter in one way: taking the lengths of each edge and adding them up. (They may have even unconsciously identified the perimeter as exclusively being the result of that procedure.) However, in the given figure, we only know two of the six edge lengths; the other four edges have to be split up into two pairs for which we only know the sum of the edge lengths.

So, it's plausible that the missing observation was something like this: to compute the perimeter of a polygon, it's enough to know the sums of edge lengths for any partition of the edges; it's not necessary to know each edge length individually. Once stated outright, it's not hard to see this is true — it follows immediately from associativity of addition — but many observations that are obvious after the fact aren't always obvious beforehand.

Putting this observation into practice also requires more reasoning than adding up a bunch of known edge lengths, since you have to think about which edges can be grouped together so that their total length can be computed. It's a problem that can't be resolved by purely mechanical application of a fixed procedure; one typically has to play around with it a bit and try out a few things. Students who are less experienced or comfortable with problems that require use of reasoning, trying out ideas that aren't guaranteed to be successful, and don't conform to a fixed procedure will often have difficulty with such problems.

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    $\begingroup$ To add to this: I would expect this student would also have difficulty with a traditional geometric argument that, for example, two angles are congruent, without knowing how many degrees the angles measure. In other words, this student may still be reasoning concretely, and needs multiple examples of situations where you can know something about a relationship without knowing the numbers -- for example, a problem that needs you to know that two angles measure exactly the same, or two angles that sum to 180 deg, or (like this one) two lengths that sum to a fixed amount. $\endgroup$ – Opal E Jul 1 '17 at 22:07
  • $\begingroup$ My best guess is that she knew algebraically what the answer was but not geometrically what the answer was. This was why I wanted a box of toothpicks. I wanted her to see that the perimeters of the above figure and and of the corresponding square were both 28 toothpicks. I thought being able to physically manipulate the toothpicks would help her visualize the more abstract nature of perimeter. $\endgroup$ – Steven Gregory Jul 2 '17 at 4:09
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Perhaps a targeted illustration, a "proof without words," might suffice to break the grip of algebra and computations, which might be confusing her...?


          Perimeter


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    $\begingroup$ This was the first thing I tried. I admit I was a bit at a loss when it didn't work. Graph paper was the next thing I tried; and she managed to get the correct answer. But she couldn't see how getting an answer to a specific problem generalized to the answer to an unspecified problem. $\endgroup$ – Steven Gregory Jul 2 '17 at 4:02
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Give her a collection of 5 such shapes on graph paper, all of which have the same perimeter, and have her compute. Then see if you can talk about why these shapes all have the same perimeter. See if she can draw another shape on her own which has the same perimeter. Make sure that all of these use "the same trick".

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"Adding up the lengths of the edges" really is an adequate definition for the perimeter of a figure. The concept she might be missing is that it doesn't matter how long the individual edges are, the perimeter is always the same.

I would try explaining this to her using a challenge. Give her some grid paper (if she's used to it), tell her the length of all the edges except two, and ask her to find all the possible perimeters for the figure. That isn't a rigorous mathematical proof, but it might help convince her.

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    $\begingroup$ Towards the end of the hour, we were using grid paper. We ran out of time before any "aha"s occurred. $\endgroup$ – Steven Gregory Jul 2 '17 at 9:44
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    $\begingroup$ What were you asking her to do? My idea is that you ask her to draw the shape in all the possible ways she can think of, and then calculate the perimeter of every single one. Not just calculating the perimeters of shapes that you draw for her. $\endgroup$ – user7868 Jul 11 '17 at 11:01
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Your approach sounds a bit stiff to me. Put a picture inside the shape. "Adding up the lengths of the edges" might be "How far is it round the edge?" for a start. It doesn't take long before the picture can be dropped and better mathematical language can be introduced, unless your pupil has limited learning skills.

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    $\begingroup$ I'd prefer to think of it as a blind spot. Most students have prejudices and misconceptions about algebra and geometry; e.g. $\sqrt{x^2+y^2} = x+y$ or "How can $|x|=-x$ if you said the absolute value is always positive?". In those cases I know what their problem is and I can lead them through the analytic and synthetic processes necessary to adjust their paradigms. In this case, I had no idea what she was thinking. $\endgroup$ – Steven Gregory Jul 11 '17 at 13:48
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Is it just a simple matter of pointing out two pairs of congruent segments? It is evident that the perimeter of the rectangle is 4x7=28.

convex

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  • $\begingroup$ It is just to you and me. Not to everyone. $\endgroup$ – Steven Gregory Jul 3 '17 at 18:10
  • $\begingroup$ I like this approach. I gathered that the student in question can calculate the perimeter of rectangles, so all that remains is that she "accept" that this transformation of the concave shape into a rectangle preserves the perimeter, $\endgroup$ – yoniLavi Jul 9 '17 at 2:03
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I often see a "shift" among my fourth graders when we move into geometry units. The students who have been quite successful with computation and even algebra concepts begin to struggle to "see" things in geometry...similar to what you describe. What's helped is putting the area and perimeter piece on hold until we've done a thorough investigation of the attributes of two-dimensional figures. Once the students are truly comfortable identifying and classifying figures based on their attributes they have a much easier time "seeing" with problems like this.
The go-to strategy most of my students would use here would be decomposition of this shape into two rectangles. They'd then be able to apply what they know about congruent opposite sides of rectangles to "see" that the lengths of the two parallel unknown sides must equal 7.

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