# Rigor and Conceptual Understanding: The Common Core Shifts in Mathematics

Hi,

When I was teaching Math to adult students, I often ran into a brick wall when teaching the Pythagorean Theorem.  I came across this video that I would like to share with those teachers instructing in Math classes.  It is from the Teaching Channel and called Challenging Students to Discover Pythagoras. https://www.teachingchannel.org/videos/pythagorean-theorem-lesson You will watch Mr. Ivy inspire his students to reach towards understanding and mastery of this challenging concept.

What do you think?  Does this lesson give you any ideas?

Meryl Becker-Prezocki, SME

### Teaching the Pythagorean Theorem

Hi to all,

Does someone have a really good approach to teaching this concept (Pythagorean Theorem) that you would like to share with other community members?  How do you reinforce it so that the students "get it" and can recall the knowledge later?

Please share and get this conversation going!

Meryl, SME

http://tapintoteenminds.com/2014/06/28/pythagorean-theorem-part1/   has a nice video that visually shows what "a squared" and "b squared" are. (Scroll down for the video; it's just a screen shot on top.)   He shows the "a" and then a square with that as a side... and that many square units... then "B"... then walks 'em around...

Surprise, surprise -- many of our students don't really understand squares or square roots... this presents them visually :)   (Ideally there would be a few other examples, and some practice...)

This is also a place where I bring up the "big idea" of "parts and wholes."   If you're looking for the hypotenuse - that's the big part... so you're going to be adding to get it.   If you're looking for one of the legs... they're the smaller parts, so you're going to be subtracting the part from the whole (the hypotenuse squared) to get it.   For my visual thinkers, that helps with the overwhelming nature of working through the formula.

Hi Susan and others,

Thanks for contributing the video and offering your wise explanation.  I liked the video.  I think that it is one that I would find useful to share with a class.  I really liked the fact that the formula was not introduced.  This provides the opportunity for the students and class to come up with rules to describe the math procedure.

Meryl, SME

What I especially like is that it complements the lesson you shared nicely.   So often our students see what math teachers recognize as "the same thing" but... it's not, to them... I like the idea of figuring out the relationship with crackers, too... Cheezits are even a square inch....

Another site I really like for math videos is www.mathantics.com --  he utterly ignores "common core" (and therefore, things don't have grade levels stamped on them, which can be disheartening to my adults) ;)

Words like Pythagorean and Theorem bring out in students either the recitation of a formula with no connections, or a strong sweat.  With adults, I would not introduce the idea with the name of the idea like in the video.  I just say that we’re working with something special about right triangles.  We also know that through the process of discovery, memorable connections are made.

One approach is to give each student or student pair a cut-out, cardstock right triangle measuring 3” by 4” by 5” and a bag of baked cheese square crackers.  Ask them to use the crackers as a measuring tool and record the measurement of the legs, leaving the crackers lined up alongside the sides.  Next, ask them to “square” the sides using additional crackers and share how many crackers it took.  Ask them to estimate and then lay out the number of crackers it would take to measure the length and then the square of the hypotenuse.  Lastly, ask students to explain a relationship between the legs and the hypotenuse by writing a rule.  Follow up by creating a model with graph paper and colored pencils.  Here are some student answers:

L2 + L2 = H2

(A x A) + (B x B) = h 2

"multiply the sides together to get the square add the legs together it equals the hypotenuse"

They came up with a different formula and they own it.  They understood the relationship between the sides!

Connie,

I think that the students will really gain an understanding when they are actually handling and manipulating the materials.  I really like the idea of using the cheese square crackers in this situation.  Thanks for providing the student answers too.  You are right.  They own it now.  I am sure that many others reading your comments are going to put your idea in their tool bag.  Thanks for contributing your valuable technique.

Meryl, SME

It seems to me that any teacher hoping to help students understand the Pythagorean Theorem (PT) and its current uses should at least point out the following:

1. Pythagoras was a Greek from the island of Samos who lived 570 -- 495 BC.  He was a philosopher and is claimed by some to have made some contributions to mathematics.  The basic geometric fact that is now referred to as the PT is a result known earlier by both Babylonians and Indians among others.

2. PT is a synthetic geometric result -- an observation about the areas of 3 squares on the sides of any right triangle.  As seen in Euclid's volumes on geometry, areas of plane regions could be combined (added), dissected and compared -- all without any mention of numbers used to "measure lengths and areas".  PT: The area of the square on the hypotenuse is the sum of the areas on the legs.

3. Currently we mainly are interested in the analytic geometry version of the PT that says if a right triangle has legs with lengths A , B, and the hypotenuse has length C, then A^2 + B^2 = C^2.  (Notice, no explicit mention of areas, just a relation among lengths.)   We often have two points P and Q in a Cartesian plane and we can easily measure the horizontal displacement |Q1 - P1| = A and vertical displacement |Q2 - P2| = B and we then expect  to compute the actual distance from P to Q as  sqrt(A^2 + B^2) = C. Just draw the picture to see why: the right angle can be at (P1, Q2) or (Q1, P2).

4. There are some very easy ways to see why the area version of the PT is correct.   Here is one of them.

Take a square sheet of paper whose sides have length A+B. Put a single mark along each edge so that it divides that side into lengths A and B.

(a)  And do it so that if you go around the square clockwise the sequence of lengths is A, B, A, B, A, B, A, B.  Now connect the marked spots with a line segments going around clockwise.  The result is that the paper is now marked with 4 right triangles with leg lengths A, B and hypotenuse C, and these triangles are arranged around the central square whose side has length C.  You could cut along the lines if you like (dissect the [A+B]-square).

(b)  Take the same size square sheet of paper and mark each edge to divide that side into lengths A, B as before, except now make the sequence of lengths A, B, B, A, B, A, A, B going around clockwise.  Draw 2 lines across the sheet perpendicular to each other so that in one corner of the sheet we have a B-square and in the opposite corner we have an A-square, and in the other corners we have two AXB rectangles.

Now compare these two dissections of our [A+B]-square sheet.  Two of the triangles in (a) together have the same area as one of the rectangles in (b), so the 4 triangles in (a) have the same total area as the two rectangles in (b).   Thus the remaining area in (a), namely that of the C-square, must be the same as the remaining area in (b), namely the total of the area of the A-square plus the area of the B-square.  That completes the argument.!!

Note that if we use the fact that we can compute area using number lengths of sides, then (a) corresponds to the numerical property (A+B)^2 = C^2 + 4*[A * B]/2   while (b) corresponds to the property (A+B)^2 = A^2 + B^2 + 2 * [A * B].  From which we get  C^2 = A^2 + B^2.

If we look at (a) again, we can fold each of the 4 triangles inward toward the center of the sheet and then we see that we have the C-square covered by (dissected into) these 4 triangles arranged around a central square of side length |B-A|.  This corresponds to the numerical property C^2 = 4 * [A*B]/2 + (B-A)^2.   After expanding we get C^2 = 2*A*B + A^2 +B^2 -- 2*A*B = A^2 + B^2.

I am a bit skeptical of the claims about several of the videos about the PT that were referred to as being useful for helping students understand it.

First note that the geometric argument I gave for the PT only used a square piece of paper and simple dissections to easily convince everyone -- and this was for any lengths of legs A and B.  The videos only look at the case where A and B are integer lengths so that the areas involved are also integers.  And one of the videos only looked at the very special case where all three sides of the triangle were integers, 3,4,5.  If A and B are integers but C isn't, I guess that students will not find it easy (even using geoboard or grid paper) to count up exactly the area of the C-square.  Unless they are familiar with Pick's Theorem, or they cut off the corners of the C-square as I mentioned before and then added the areas of the 4 triangles and the area of the inner square. (They will likely need suggestions for that.)  So just what are they "discovering"?

It seemed to me that mostly what the videos were concerned with was convincing students that the area of an integer A-square is in fact just the count of the number of 1X1 unit squares that can minimally cover the A-square, namely, A^2 of them.  I think that should have been worked on before being applied to the PT.  Also, they need to be convinced and comfortable with the generalization: how to compute area for AXB rectangles with A and B no longer integers -- it's still the minimal number of 1X1 stamps that are needed to cover the rectangle (we are allowed to cut up stamps as needed to cover  small spaces with no gaps and no overlap).

I thoroughly enjoyed your number-free dissection description and think that it would "turn the light bulb on" for some students.   I'm a little curious about your number 3.   If I draw the picture, I don't "see why" I could compute the PT.

In my experience, many students benefit greatly from seeing simple, integer examples of more abstract concepts.   They struggle with generalizing and making connections between the observations of dissecting things and the symbols we use to describe those relationships.

Most of my students have a strong tendency to feign that they have been convinced of a concept, because they do not believe they can understand it, and then memorize what they have to in order to answer a certain kind of problem.  While this got lots of 'em through K-12, it tends to fail miserably for them at our college.   One of the things they need to discover, again and again, is that math really does work on real things and isn't just a bunch of formulae.

My earlier note had an item 3 which S.Jones asked about. The point is that given two points P and Q in a Cartesian coordinate plane, we often want to find the distance from P to Q.  Consider the rectangle  with horizontal and vertical sides which has the points P and Q as opposite corners of the rectangle.  The other pair of opposite corners are R = (P1,Q2) and S = (P2, Q1), so we have the congruent right triangles PRQ and PSQ, either of which we could use along with the Pythagorean Theorem to compute the length of the segment from P to Q, the hypotenuse of each of the triangles.

And don't forget that we use exactly the same idea when we generalize to computing the distance between points in 3-space, 4-space, ...etc.

Hi Everyone,

You all brought up great points relevant to this challenging concept.  My main idea was to explore some ways to guide students to discover the Pythagorean Theorem vs. memorizing something that they really do not understand.  Your comments all reflect the move for higher student achievement.  Thanks for your rich contributions.

Meryl, SME