Like what you saw?
Create FREE Account and:
- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics
Pythagorean Theorem Proofs - ConceptFREE
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
The Pythagorean theorem states that in a right triangle the sum of its squared legs equals the square of its hypotenuse. The Pythagorean theorem is one of the most well-known theorems in mathematics and is frequently used in Geometry proofs. There are many examples of Pythagorean theorem proofs in your Geometry book and on the Internet.
The Pythagorean theorem only applies to right triangles. And what it says is if you have two legs and a hypotenuse where the hypotenuse is the side that's opposite your right angle then a special relationship exists. And that is the square of one of your legs plus the square of the other leg has to equal the square of the hypotenuse. Now there are lots of proofs. There are probably hundreds of proofs out there. And there are even books that are nothing but proofs.
I'm going to show you just one here. And it's going to start with a triangle where we have two legs A and B and we have a hypotenuse C. What I'm going to do is I'm going to draw in four more triangles.
So what I'm going to do is I'm going to draw a Side A, and then I'm going to draw in my hypotenuse C it actually looks like I need to make this a little bit longer and then I'm going to have Side B. And that is going to be a right angle.
So these triangles are congruent, even though they might not look congruent. And then once we have that, then we need to draw in our other triangle.
So this is going to be a right angle. This is going to be B, this is going to be A, and that's going to be C. And then, here is our last triangle. So this is going to be C, that's A, and that is B.
So the key to this proof - because right now it's a little confusing - is what you know about the area of squares and rectangles.
Well, let's start with this large square. So I'm talking about this square right here. How can I calculate the area of that square?
Well first, I need to know what is one of my side lengths. And it looks like one is A+B, A+B, A+B, and A+B. So to calculate the area of this whole square, we can say we're going to take one of our side lengths - which is A+B - and square it. And that has to equal the sum of its parts.
So let's start off with this smaller square here in the middle. I see that I have Side C for all of these sides on the square, so that's going to be C^2. And I'm going to have one, two, three, four congruent triangles. So I'm going to add in four times my triangles.
But how do I calculate the area of a triangle? Well that's going to be base times height. So one is B and one is A, and then we have to divide that by two. This is A times B, divided by two.
So if I took a step up here what I'm saying is the area of the big square has to equal the area of the small square, which is the one with Side C, plus the area of your four triangles.
So that's the general theory behind what we're doing here. We're saying that the big square is equal to the sum of its parts. So if we clean this up, we should end up with our Pythagorean theorem, which says A^2+B^2=C^2.
Let's go back to algebra. If I square this binomial, I'm going to have A^2, my first term. Plus, I'm going to multiply these two together, times two - so that's going to be 2(AB) and then, I'm going to have my second term squared.
So all we did was expand this binomial being squared. We've got A^2+2(AB)+B^2. I'm just going to bring down the C^2. Nothing is going to happen with that.
Four divided by two is two, so we're going to have plus 2(AB).
So if I look at this equation right now, we are not quite at our Pythagorean theorem. So I need to manipulate this equation somehow.
So I'm going to grab a different colored marker. And I see over here that I have a 2(AB) and a 2(AB) on both sides.
So I'm going to subtract 2(AB), subtract 2(AB), and then, now, I can say that we have A^2.
2(AB) minus itself is zero, plus B^2.
Over here we have 2(AB) minus itself, so that's zero. And we end up with our Pythagorean theorem.
So this is one of the more common proofs of the Pythagorean theorem. It's the one that I do in my class.
And the key to this one was writing this first equation, which is the area of the big square has to equal the area of the small square, plus the area of the four triangles inside that big square.
Please enter your name.
Are you sure you want to delete this comment?