There are also rectangles e.g.:
Oh hmm.
So this one has 1S + 5T + 5P.
It also is made up of 6 little squares, visually.
The smaller red square I outlined above was 1S + 3T + 3P and made up of 4 little squares, visually.
Looking at the picture, it looks like 1T + 1P = 1 square, and this is consistent with that (we have 2T + 2P more in the big rectangle vs the red-outlined square, and we have two more little squares in the picture).
New rectangle: 1S + 5T + 5P = 400 + (T + P)
Iâm going through the thread so far and organizing the information in the form of a draft blog post to see if that helps.
notes or tree would be better
Figured out something else:
(note that the blue triangle is supposed to be within the purple triangle)
The big purple triangle has an area of 400. Itâs composed of 1S + 4T + 4P. So 1S + 4T + 4P = 400. We knew that already.
The little blue triangle has area 100. Itâs composed of 2T + 1P. So 2T + 1P = 100.
If we cut the blue triangle out of the purple triangle, we have a purple âbandâ where 1S + 2T + 3P = 300.
Thatâs a good thing to notice, but unfortunately, I think that one is not going to turn out mathematically convenient to get an answer from.
I think your main problem is some kind of suppression of brainstorming.
But if the question in the book was to find every square in my pic, of any size, and count them, I think you could do it. And then write down a formula for each different type of square.
Ah doh.
That sounds plausible.
I looked it up, and my assumptions about their intended solution were wrong, btw.
What they do is closer to what Elliot did than what I did.
I think thereâs something a bit weird about what you do & do not block. Iâm not really sure whatâs going on.
I think there are things you know that you didnât list in brainstorming.
And you are doing things like not using pythagorean theorem, because you think thatâs not part of the intended solution, but you are using equations in a way that looks like it could be leading to systems of equations. But I donât think those would be part of the intended solution either (I think itâs a pre-algebra book).
One thing you could do is try to solve it any way you know how, including with pythagorean theorem, and then go back and try to find other ways to solve it too.
With your brainstorming, you could try to write down everything you can figure out about it, without actually trying to put it all together, and then go back through & look at that. I donât think you ever did a step like that.
I actually there may actually be an important point here.
I think something like this may be happening: you are not allowing yourself to use pythagorean theorem because you think the book didnât intend pythagorean theorem to be used. So you want to solve it the way the book intended. You think that the students reading the book could solve it without using pythagorean, so you should be able to too.
Are you doing something like that?
(I have more to say, I just wanted to check if this is happening before I write more.)
The square in the previous post has the formula 4T + 4P. If we relate it to one of the other squares with the black square, like the one in my earlier example, we could say 4T + 4P = 1S + 3T + 3P. That simplifies to 1T + 1P = 1S. We can connect this to the information from the original square (that 400 = 1S + 4T + 4P and get 5S = 400 and S = 80. So the shaded area is 80.
Youâre still avoiding anything resembling brainstorming. How many squares are in my pic, and how many different types are there? (Or donât answer but introspect about why you wonât do this or something.)
I started writing a reply to your earlier question about identifying the squares but then noticed the center square and wanted to follow my thinking about it - like i got excited about that thought or something. Iâm going to answer ingrackeâs question then come back to thinking about the squares.
That sounds about right, but I would also add the following: I think that maybe the book makers maybe thought using the pythagorean theorem would make the problem too easy, and itâs supposed to be a challenge problem. I have a personal issue with doing things âthe easy wayâ where I donât learn as much, so Iâm trying to respect the challenge of the problem .
Ok, that sounds like another problem, besides the one I identified, and could explain part of your blocking brainstorming.
The actual solution was much easier than you were making it. It is possible you were systematically blocking anything you thought was âtoo easyâ, in order to try to make the problem âhardâ enough. But the actual solution was simple. So you failed to brainstorm the things required for the actual solution.
And, btw, I solved the problem multiple different ways, and the Pythagorean theorem way was the hardest one, imo. And doing it first didnât block me from discovering the other simpler & more elegant solutions.
By didnât block me from discovering, I mean it didnât even give me âhintsâ towards the other solutions. Like, knowing the answer from pythagorean didnât make it so that I couldnât discover the other solutions, it didnât just give me the answer in a way that spoiled the problem.
Counting only larger squares (so not counting each separate trapezoid + small triangle combination), and counting only actual squares (and not e.g. rectangles) there are 10 squares divided into 4 types:
1 square consisting of the entire picture, whose formula is 4S + 16P + 16T = 1600
4 instances of the original square, whose formula is 1S + 4P + 4T = 400
4 squares whose formula is 1S + 3T + 3P
1 center square whose formula is 4T + 4P.
There are other shapes that almost get to squares but they get âclipped offâ at the edges of the picture.
Why? This is exactly the type of suppression Iâm talking about.
BTW, besides excluding squares, this is correct. The answer you got above was also correct.
