High school geometry
Bhaskara's proof of the Pythagorean theorem
An elegant visual proof of the Pythagorean Theorem developed by the 12th century Indian mathematician Bhaskara. Created by Sal Khan.
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- How exactly did Sal cut the square into the 4 triangles? At time1:13.(14 votes)
- He just picked an angle, then drew a line from each vertex across into the square at that angle. His angle choice was arbitrary. because as he shows later, he ends up with 4 identical right triangles. By just picking a random angle he shows that it works for any right triangle.(8 votes)
- How does the video above prove the Pythagorean Theorem?(5 votes)
- Think about the term "squared". That simply means a square with a defined length of the base. So when you see a^2 that just means a square where the sides are length "a". The same would be true for b^2. The Pythagorean theorem states that the area of a square with "a" length sides plus the area of a square with "b" sides will be equal to the area of a square with "c" length sides or a^2+b^2=c^2. Bhaskara simply takes his square with sides length "c" defines lengths for "a" and "b" and rearranges c^2 to prove that it is equal to a^2+b^2. Hope that helps. Watch the video again. :)(11 votes)
- why is it still a theorem if its proven? is there a difference between a theory and theorem?(2 votes)
- The word "theory" is not used in pure mathematics. "Theory" in science is the highest level of scientific understanding which is a thoroughly established, well-confirmed, explanation of evidence, laws and facts.
In pure mathematics, such as geometry, a theorem is a statement that is not self-evidently true but which has been proven to be true by application of definitions, axioms and/or other previously proven theorems.(10 votes)
- Can you please mention the original Sanskrit verses of Bhaskara along with their proper reference ? Being a Sanskrit scholar I'm interested in the original source.(7 votes)
- So who actually came up with the Pythagorean theorem? Pythagoras, Bhaskara, or James Garfield?(3 votes)
- Again, you have to distinguish proofs of the theorem apart from the theorem itself, and as noted in the other question, it is probably none of the above.(3 votes)
- At1:50->2:00, Sal says we haven't proven to ourselves that we haven't proven the quadrilateral was a square yet, but couldn't you just flip the right angles over the lines belonging to their respective triangles, and we can see the big quadrilateral (yellow) is a square, which is given, so how can the small "square" not be a square?(3 votes)
- so many steps just to proof A2+B2=C2 it's too hard for me to try to remember all the steps(2 votes)
- You won't have to prove the Pythagorean theorem, the reason Sal runs through it here is to prove that we know that we can use it safely, and it's cool, and it strengthens your thinking process.(2 votes)
- I know a simpler version, after drawing the diagram, it is easy to show that the area of the inner square is b-a. So adding the areas of the four triangles and the inner square you get 4*1/2*a*b+(b-a)(b-a) = 2ab +b^2 -2ab +a^2=a^2+b^2 which is c^2. It is much shorter that way. I figured it out in the 10th grade after seeing the diagram and knowing it had something to do with proving the Pythagorean Theorem. Did Bhaskara really do it this complicated way?(2 votes)
- i agree with you. I learned that way to after googling. Why do it the more complicated way? That way is so much easier. i 100 percent agree with you!(1 vote)
- Why can’t we ask questions under the videos while using the Apple Khan academy app? I am on my iPad and I have to open a separate Google Chrome window, login, find the video, and ask you a question that I need. Then go back to my Khan Academy app and continue watching the video.
You can see how this can be inconvenient for students. Would you please add the feature on the Apple app so that we can ask questions under the videos? Please don’t disregard my request and pass it on to a decision maker. We appreciate you!(2 votes)
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I will now do a proof for which we credit the 12th century Indian mathematician, Bhaskara. So what we're going to do is we're going to start with a square. So let me see if I can draw a square. I'm going to draw it tilted at a bit of an angle just because I think it'll make it a little bit easier on me. So let me do my best attempt at drawing something that reasonably looks like a square. You have to bear with me if it's not exactly a tilted square. So that looks pretty good. And I'm assuming it's a square. So this is a right angle. This is a right angle. That's a right angle. That's a right angle. I'm assuming the lengths of all of these sides are the same. So let's just assume that they're all of length, c. I'll write that in yellow. So all of the sides of the square are of length, c. And now I'm going to construct four triangles inside of this square. And the way I'm going to do it is I'm going to be dropping. So here I'm going to go straight down, and I'm going to drop a line straight down and draw a triangle that looks like this. So I'm going to go straight down here. Here, I'm going to go straight across. And so since this is straight down and this is straight across, we know this is a right angle. Then from this vertex on our square, I'm going to go straight up. And since this is straight up and this is straight across, we know that this is a right angle. And then from this vertex right over here, I'm going to go straight horizontally. I'm assuming that's what I'm doing. And so we know that this is going to be a right angle, and then we know this is going to be a right angle. So we see that we've constructed, from our square, we've constructed four right triangles. And in between, we have something that, at minimum, looks like a rectangle or possibly a square. We haven't quite proven to ourselves yet that this is a square. Now the next thing I want to think about is whether these triangles are congruent. So they definitely all have the same length of their hypotenuse. All of the hypot-- I don't know what the plural of hypotenuse is, hypoteni, hypotenuses. They have all length, c. The side opposite the right angle is always length, c. So if we can show that all the corresponding angles are the same, then we know it's congruent. If you have something where all the angles are the same and you have a side that is also-- the corresponding side is also congruent, then the whole triangles are congruent. And we can show that if we assume that this angle is theta. Then this angle right over here has to be 90 minus theta because together they are complimentary. We know that because they go combine to form this angle of the square, this right angle. And this is 90 minus theta. We know this angle and this angle have to add up to 90 because we only have 90 left when we subtract the right angle from 180. So we know this has to be theta. And if that's theta, then that's 90 minus theta. I think you see where this is going. If that's 90 minus theta, this has to be theta. And if that's theta, then this is 90 minus theta. If this is 90 minus theta, then this is theta, and then this would have to be 90 minus theta. So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle, so at minimum, they are similar, and their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length, b. So the longer side of these triangles I'm just going to assume. So this length right over here, I'll call that lowercase b. And let's assume that the shorter side, so this distance right over here, this distance right over here, this distance right over here, that these are all-- this distance right over here, that these are of length, a. So if I were to say this height right over here, this height is of length-- that is of length, a. Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared. Now, what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's, and hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste. So this is our original diagram. And what I will now do-- and actually, let me clear that out. Edit clear. I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much-- well, the way I drew it, it's not that-- well, that might do the trick. I want to retain a little bit of the-- so let me copy, or let me actually cut it, and then let me paste it. So that triangle I'm going to stick right over there. And let me draw in the lines that I just erased. So just to be clear, we had a line over there, and we also had this right over here. And this was straight up and down, and these were straight side to side. Now, so I moved this part over down here. So I moved that over down there. And now I'm going to move this top right triangle down to the bottom left. So I'm just rearranging the exact same area. So actually let me just capture the whole thing as best as I can. So let me cut and then let me paste. And I'm going to move it right over here. While I went through that process, I kind of lost its floor, so let me redraw the floor. So I just moved it right over here. So this thing, this triangle-- let me color it in-- is now right over there. And this triangle is now right over here. That center square, it is a square, is now right over here. So hopefully you can appreciate how we rearranged it. Now my question for you is, how can we express the area of this new figure, which has the exact same area as the old figure? I just shifted parts of it around. How can we express this in terms of the a's and b's? Well, the key insight here is to recognize the length of this bottom side. What's the length of this bottom side right over here? The length of this bottom side-- well this length right over here is b, this length right over here is a. So the length of this entire bottom is a plus b. Well that by itself is kind of interesting. But what we can realize is that this length right over here, which is the exact same thing as this length over here, was also a. So we can construct an a by a square. So this square right over here is a by a, and so it has area, a squared. Let me do that in a color that you can actually see. So this has area of a squared. And then what's the area of what's left over? Well if this is length, a, then this is length, a, as well. If this entire bottom is a plus b, then we know that what's left over after subtracting the a out has to b. If this whole thing is a plus b, this is a, then this right over here is b. And so the rest of this newly oriented figure, this new figure, everything that I'm shading in over here, this is just a b by b square. So the area here is b squared. So the entire area of this figure is a squared plus b squared, which lucky for us, is equal to the area of this expressed in terms of c because of the exact same figure, just rearranged. So it's going to be equal to c squared. And it all worked out, and Bhaskara gave us a very cool proof of the Pythagorean theorem.