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## 8th grade (Illustrative Mathematics)

### Course: 8th grade (Illustrative Mathematics) > Unit 8

Lesson 11: Extra practice: Pythagorean theorem# Another Pythagorean theorem proof

CCSS.Math:

Visually proving the Pythagorean Theorem. Created by Sal Khan.

## Want to join the conversation?

- 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.(58 votes)

- Why is line b the height? It seems much smaller than line c. So how is it the height?6:40-6:50(20 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_.(8 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.(7 votes)
- There is a version of side b that was rotated 90 degrees. It is used as the altitude of the parallelogram.(6 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)

- 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)

- We shall call this "Sal's proof of the Pythagorean Theorem"(6 votes)
- In the Pythagorean Theorem proofs exercises, there are some statements, "Triangle ABC and Triangle EFG are congruent by SAS". And, "The triangles are congruent by SSS". I've not come across the terms SAS and SSS before. Am I correct in guessing it's Side Angle Side, and Side Side Side?(3 votes)
- you are correct. here is the first video on SSS and the other postulates. =)

https://www.khanacademy.org/math/geometry/congruent-triangles/cong_triangle/v/congruent-triangles-and-sss

you can keep watching through the series to see the rest =)(5 votes)

- my proof would to use a ruler to measure the length of the hypotenuse(3 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)

- What grade is this for(3 votes)
- For most people it is 8th grade. Algebra is in 8th grade but if you don't take summer school or skip a math level then people take it in high school.(4 votes)

## Video transcript

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.