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Current time:0:00Total duration:9:03

CC Math: 8.G.B.6

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.