High school geometry
Visually proving the Pythagorean Theorem. Created by Sal Khan.
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- How con we use this proof and theorem in are daily life's? I don't see how we could use it.(3 votes)
- The real value of teaching proof in geometry class is to teach a valuable life skill. You learn to think logically, step-by-step, to learn to distinguish what you think is true from what can be shown to be true. We call these skills "critical thinking". These skills can keep you from being deceived.
You don't have to try very hard to find an advertisement or other claim that may or may not be true. There is great advantage in having the skills to examine a claim one piece of evidence at a time until you know the claim if false or know that it is well-supported and may be true.
So, geometry is a class where the basics of that skill are taught. Will you need to know in most professions whether two triangles are congruent? No, you won't. Will you need to be able to analyze claims to see if they are rumor, myth, wild speculation, impossible, or well-supported and likely true? Yes, you will. And you use the same kind of skills (different details, of course) to do both.(61 votes)
- Why is line b the height? It seems much smaller than line c. So how is it the height?6:40-6:50(21 votes)
- When Sal flipped the parallelogram so that b_ was the base, I found it helpful to flip the line _b that was on the triangle to the right of the parallelogram as well in my head. Since they are at a right angle to each other, it showed me that the hight of the parallelogram was in fact _b_.(9 votes)
- At6:40I don't understand how he found out that the height was b. What are the processes involved in finding the height. It just doesn't seem logical either.(8 votes)
- There is a version of side b that was rotated 90 degrees. It is used as the altitude of the parallelogram.(7 votes)
- whats this mans iq?(10 votes)
- What is the first known proof of the Pythagorean Theorem?(5 votes)
- Euclid first mentions it. I think in the beginning of High School Geometry, Sal mentions something about Euclid then.(2 votes)
- Who proved this theorem?(10 votes)
- The Pythagorean Theorem was known long before Pythagoras, but he may well have been the first to prove it.(5 votes)
- my proof would to use a ruler to measure the length of the hypotenuse(5 votes)
- If you have a 45-45-90 triangle, such as 2-2-2√2, how do you measure 2√2? It works for Pythagorean Triples such as 3-4-5 or 5-12-13.(5 votes)
- Who created this proof? It is a lot like Bhaskara's proof.(6 votes)
- A person named Euclid made this proof. He was a Greek mathematician, who is often referred to as the "father of geometry(2 votes)
In case you haven't noticed, I've gotten somewhat obsessed with doing as many proofs of the Pythagorean theorem as I can do. So let's do one more. And like how all of these proofs start, let's construct ourselves a right triangle. So I'm going to construct it so that its hypotenuse sits on the bottom. So that's the hypotenuse of my right triangle. Try to draw it as big as possible, so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We can have two sides that are equal. But I'll just draw it so that it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle. So maybe it goes right over there. That's side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90-degree angle. So the first thing that I'm going to do is take this triangle and then rotate it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale as much as I can eyeball it. This side of length a will now look something like this. It'll actually be parallel to this over here. So let me see how well I can draw it. So this is the side of length a. And if we cared, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case. That's going to be 90 degrees. That's going to be 90 degrees. Now, let me draw side b. So it's going to look something like that or the side that's length b. And this and the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now, what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially-- and let me label. So this is height c right over here. Let me do that white color. This is height c. Now, what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. It's maintained the same distance. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now, the next question I have for you is, what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is sitting on the ground. So this is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. I'll use this green color. The height of the parallelogram is given right over here. This side is perpendicular to the base. So the height of the parallelogram is a as well. So what's the area? Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now, let's do the same thing. But let's rotate our original right triangle. Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get? So the side of length c if we rotate it like that, it's going to end up right over here. I'll try to draw it as close to scale as possible. So that side has length c. Now, the side of length of b is going to pop out and look something like this. It's going to be parallel to that. This is going to be a right angle. So let me draw it like that. That looks pretty good. And then the side of length a is going to be out here. So that's a. And then this right over here is b. And I wanted to do that b in blue. So let me do the b in blue. And then this right angle once we've rotated is just sitting right over here. Now, let's do the same exercise. Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel. So this is length b down here. This is length b up there. Now, what is the area of this parallelogram right over there? What is the area of that parallelogram going to be? Well, once again to help us visualize it, we can draw it sitting flat. So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. That's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives us right there. We know that this is 90 degrees. We did a 90-degree rotation. So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now, things are starting to get interesting. And what I'm going to do is I'm going to copy and paste this part right over here, because this is, in my mind, the most interesting part of our diagram. Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I'm going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a few parts of it. I want to do that in black so that it cleans it up. So let me clean this thing up, so we really get the part that we want to focus on. So cleaning that up and cleaning this up, cleaning this up right over there. And actually, let me delete this right down here as well, although we know that this length was c. And actually, I'll draw it right over here. This was from our original construction. We know that this length is c. We know this height is c. We know this down here is c. But my question for you is, what is the area of this combined shape? Well, it's just a squared plus b squared. Let me write that down. The area is just a squared plus b squared, the area of those two parallelograms. Now, how can we maybe rearrange pieces of this shape so that we can express it in terms of c? Well, it might have jumped out at you when I drew this line right over here. I want to do that in white. We know that this part right over here is of length c. This comes from our original construction. Whoops. I lost my diagram. This is of length c. That's of length c. And then this right over here is of length c. And so what we could do is take this top right triangle, which is completely congruent to our original right triangle, and shift it down. So remember, the entire area, including this top right triangle, is a squared plus b squared. But we're excluding this part down here, which was our original triangle. But what happens if we take that? So let me actually cut. And then let me paste it. And all I'm doing is I'm moving that triangle down here. So now, it looks like this. So I've just rearranged the area that was a squared b squared. So this entire area of this entire square is still a squared plus b squared. a squared is this entire area right over here. It was before a parallelogram. I just shifted that top part of the parallelogram down. b squared is this entire area right over here. Well, what's this going to be in terms of c? Well, we know that this entire thing is a c by c square. So the area in terms of c is just c squared. So a squared plus b squared is equal to c squared. And we have, once again, proven the Pythagorean theorem.