Next I joined two pieces together and showed how they, together with the connecter, equalled the length of a larger piece. We wrote this as Z = 2X + L. (L = linker or 1.25″. This is the part of the linking piece that does not overlap with any of the sticks).  We made a right angle triangle. The two sides could be X, but the third one was a different piece, not Z. So we named it Y and found that  So we figured out what it was Y = X+L. Therefore Z = X + Y.

We wrote out some of the formulas for substituting one piece for a combination of pieces. Then we rebuilt our first right angle triangle and  we made a second triangle without the connectors. We used Y instead of the edges that were once X+L and we thought that Z could be used instead of Y + L. It looked good, but then logically, if Z = Y+L and Y = X + L, then we should be able to make a piece corresponding to the orange piece out of one white piece and two linkers (Z = X + 2L). If Z = 2X + L AND it equalled 2L + X, then X and L would have to be the same thing. But they aren’t. So we went back and realized that when making the triangle the orange Z is actually larger than the hypotenuse of the triangle we were trying to copy. Z does NOT equal Y + L. The experience was a good lesson in double checking one’s work. While we were trying to figure out where our mistake was we considered the possiblity that we would have to use K to stand for the relevant distance of the angle connecter pieces. We had been relying on the notion that we could consider the triangle extending just over the quadrilatoral part of the angle connector and halfway through the connector’s circle. Thus two angle pieces would equal L.
M compared the experience to an incident in The Number Devil where the number devil is trying and trying to figure out a math problem he couldn’t solve. Mention of The Number Devil reminded me of another aspect to that book, and I quickly put together a little K’nex demonstration of the length of the square root of two relative to one. Picture the distance of the white piece plus the relevant part of the angle connectors (X + L, aka Y). This is going to be “one unit.” The square made up of the white pieces (and connectors) is therefore one unit squared. We can see that the area of the larger (yellow outlined) square is two units. It is twice the size of one unit squared. So each side of it (Y + L) would equal the square root of two.

• ### MaryAnne

I love this visual introduction to Algebra!

• ### Marie

This is a terrific lesson I hope to use when I’m teaching geometry to my 6th & 7th grade students. Manipulative a are never too “babish” for anyone. Thanks for the idea.

• ### Marie

Amazing lesson! I hope to use this in the Fall with my math students! I love manipulatives whenever I can. Thanks

This site uses Akismet to reduce spam. Learn how your comment data is processed.