Pre Calculus Syllabus
The Pythagorean Theorem
Made plain for the benefit of my students.
Mr. Rives, Sept 2nd, 2005.
Christ Preparatory Academy, Pre Calculus.
Today in class we had a run-in with the distance formula. You will recall I asked you if you understood the formula, or if it was just a mechanical device you had memorized and used with reckless abandon. My conclusion was that it was rather like a weapon in your hands, for which self-inflicted wounds were perpetrated with each new use. So, to teach you how to point the weapon in the right direction (with confidence), I am going to try and demonstrate the facts that lie beneath it and make it what it is -- namely, The Pythagorean Theorem.
I start with the idea of a a right triangle with sides a, b, c, where a is shorter than b. My goal is to show that a2 + b2 = c2. Or, to put that in words, all the sides of a triangle can be turned into squares, and if you add two of those squares, you'll get the area of the third. Our triangle looks like this (using the same letters I just mentioned):
I will try to show that the squares of each side work out such that the sum a2 + b2 makes the area c2 . Here is a picture of the idea:
I hope to make it evident that the green square plus the blue square equal the same space taken up by the yellow square (for any right triangle -- like the gray one here). Think about this for a minute. I am saying that a2 is a square -- the green one to be exact. And b2 results from the notion of taking the b side of the triangle and making a literal square out of it -- the blue one in this case. The proof, then, is one of showing how these three squares relate to one another. When I get done, I will have geometrically demonstrated that any triangle like the grey one here, conforms to the pattern of a2 + b2 = c2.
To get started, I will first take our a, b, c triangle, duplicate it, and then rotate that duplicate into a position next to the original. This is shown in the next picture. You'll see how I just let the letters rotate with the triangle -- I'll do that throughout this web page, so don't let it throw you that my letters are sometimes sideways and sometimes even upside down -- I could have written them upside right, but leaving on their sides or heads shows you how I rotated (which means that the b will be flipped to look like a q at times):
There is the rotation, now I keep moving the duplicate so that the two triangles touch:
There! Two triangles added together. But I want to do it again. In fact, I want to do it so that I have four triangles. I duplicate the original triangle again, rotate it and move into place with the other two:
You get the idea. After I do that one more time with a final duplicate, you see that I have generated a square, like this:
The result is one big square with four equal sides -- the length of every side of this square is a + b. That means that the area of the square is (a + b)2 -- but now I am just telling you the obvious! Notice that inside this bigger square, there is another square formed by the duplicate hypotenuses of each triangle. The area of that inner square is c2 -- colored in yellow below:
Each side of the yellow square is length c. Hmm, that's convenient, because I wanted to turn each side of the triangle into a square, and so far I have succeeded with the hypotenuse. To create a2 and b2, I have some work to do. And I'll start by making a copy of that big square, the one with sides of length (a + b), and erase all of its insides:
This blank square takes up the same space as the one we just made, and now I am going to fill it up with our original triangle and see what happens. I do that by adding my the triangle as follows (notice I moved the letter labels of a and b to the inside the triangle, just to make it easier to read):
See how nicely it fits into the corner!
It is breath-taking, I know, to see a triangle and square working so nicely together.
I hope I have not confused you to this point, so let me recap how we got here. First I created one square with sides of length a + b -- I did that by methodically fitting together four of the same triangle. At this point, I am working on a second square of the same size. To give you a peek at where we are headed with all this, when we get done I will compare the two squares and show you some interesting stuff about the relationship of the sides of the original triangle. Until then, look at the two squares as I have constructed them so far (from this point on, I'll always show them side by side):
Getting back to work on the square I was working on, I am going to add another triangle. It gets inserted like this:
Which results in the following picture:
Hold on, because we are almost done! I am going to add two more triangles so that both of the bigger squares have a total of four triangles in them.
Adding our third triangle:
Check out the results:
The astute observer will see that there is now a square in the top corner, and it has side lengths of size a. Let me color it blue so that what I am talking about will stand out:
Keep that blue square in mind as I finish the construction, because it is no coincidence that we have found an a2 to talk about. We are going to talk about it in relationship to the c2 we already found, and you'll see how c2 is taking up the same amount of area in its the square as a2 and b2 take up in theirs. But I am ahead of myself, I still need to add one more triangle. Remember, I want four of the original triangle in each of the (a + b) squares, and so here is the final addition:
Everything I am doing is easy -- no tricks up my sleeves. You get to see each and every move I make. It's a simple construction game -- one you'll need to be able to play on the test. All you have to do is arrange four triangles twice. When all the arranging is done, we end up with two big squares of the same size, like this:
And, if you look real close (or even not so close), you'll find a b2 is making its appearance. See if you can find it. I will color it for you with green, and so complete my construction:
Behold!
Isn't that nice to view? Wouldn't it be pleasing to make a large version and display it in your home -- a little contemporary art? But the beauty is more than its artistic quality. Notice that the green portion is a square with area b2. If you take out the green square and the blue square (the blue square has an area of a2), you find that what is left (area wise) is four of our original triangles. Likewise, if you take the yellow square out of its container, you see that four of the same size triangles are left behind. The yellow square, which has an area of c2, takes up the same area in its space as a2 + b2 take up in their space (which happens to be spaces of identical size).
To put it in algebraic terms:
Given a right triangle, Green Square + Blue Square = Yellow Square
That is,
a2 + b2 = c2.
Thus I have shown that if you take the square of each side of a right triangle, a relationship between the three squares is in operation -- and that relationship is the Pythagorean Theorem.
My demonstration is over, but I want to make some conclusions for you to consider. Suppose you know the length of two sides of a right triangle. Given what I have said up to this point (defending the Pythagorean Theorem and all), you should be able to find the length of the third side You now have a general principle from which to work, and should be able to find the value of c in the following triangle:
Let me set up the problem for you (the square area of each of the sides relate as follows):
102 + 152 = c2
From here, solving the equation is only as hard as taking a square root. Let me know how that works out for you.
One last note: if you look at all those distance formula exercises in your text book, you'll see that what they are doing is having you use this theorem to find distances between points. The points in the exercises form a kind of triangle on the Cartesian coordinate system, and you are hunting for lengths of hypotenuses! It's the same problem, applied metaphorically to a different domain -- but still the same problem.
I hope you have a good three day weekend, and dream of complex numbers, as I am
Yours in Jesus,
Mr. Rives