The theorem of Pythagoras is a
well known theorem.

It's also a very old one. Not
only does it bear the name of

Pythagoras in ancient Greek,
but it was also known to the

ancient Babylonians and to the
ancient Egyptians.

Most school boys school girls
learn of it as a square plus B

squared equals C squared, but
the actual statement of the

theorem is more to do with

areas. So let's have a look at
the statement of the theorem.

The theorem talks about
a right angle triangle.

So then we have a right angle
triangle and we show that with

this little mark here, so we
have a square in that corner.

What the theorem says is that
the square on the hypotenuse.

Now that means we have to
identify this side the

hypotenuse, and it's always the
longest side in the right angle

triangle and it's always the one
that is opposite to the right

angle. So it's this side here
which bears the name the

hypotenuse. So the theorem then
says the square on the

hypotenuse, so if they draw a
square on this hypotenuse, so

there we've got our square.

And it says that the area of
that Square is equal to the sum

of the areas.

Of the squares on the
two shorter sides.

So these are all squares.

All the same, all got right
angles in them.

This is a square as well.

All sides are equal. Got a
right angle. What this says

is that this area A plus this
area B is equal to that area.

C A+B equals C.

OK. Let's have a look at
a special case and see if we can

see how that actually works.

Got here a
single red triangle.

It's a very special triangle.

This side along the bottom is
one unit long and that side

along there is 2 units long, so
I'm just going to fix it. Little

bit of Blu tack.

On there I'm just going to fix
it so that in fact I can draw

round it. So that
we've drawn around that

triangle, now I want
to fix some squares

on the sides of

this triangle. So there's
the square on that side, but the

square on this side is going to
look in fact just a tiny bit

different. It's going to be an
assembly of triangles, each one

identical to this original
triangle, so there's half of it.

Now I'm going to put the other

half on. Here.

Lastly, debate.

I will make it up with
this bit.

OK, now remember what the
theorem says. It says that if we

take the area of this square and
add it to the area of that

square, we should get the area
of this square on the

hypotenuse. Let's see how right

that is. Each one of these red
triangles that I've got here and

that I'm putting on is exactly
the same as these triangles

here, so I can put one on there.

That one
on there.

Now put

the square.

In there.
That

one
in

there.
And then finally put

this one in there.

What we can see there is that
quite clearly the area of this

square is made up of the area of
that square plus the area of

that square. OK, if this theorem
is true, and certainly it seems

to be true for this, and it has
been proved that it's true, then

how do we get the traditional?

Schoolboy schoolgirl theorem. A
squared plus B squared equals C

squared. So draw our
right angle triangle.

Will put on the

squares. On the sides.

And
we

label
these

squares

A&B&C. We label
the lengths of the sides that

lay. It'll be a little C.

Then the theorem tells us that
if we add together the areas of

the squares on the two shorter
sides, the result is the area of

the square on the longest side
of the triangle the hypotenuse.

So A plus B is equal to see, but
this is a square.

And so it's area is A squared
and this is also a square, so

its area is B squared.

This is a square and its area is
C squared and so we have the

traditional result. Always when
answering questions on

Pythagoras Theorem, make sure
that you read the question.

Don't just write down this
result Willy nilly. Make sure

you've actually read the
question and that you're

thinking about which sides are
which. Let's have a look at one

or two examples.

First of all, let's take
a triangle whose two shorter

sides are of length three
and of length 4.

And the question is, how long is
the longest side the hypotenuse?

So we can label our sides, let's
call this one a.

Call this one B and then the one
we want to find is see.

So using the theorem, we know
that A squared plus B squared.

Is C squared?

A is 3 three squared
plus B is 4, four

squared equals C squared.

3 squared is 9 and 4 squared. Is
16 equals C squared, and so 25

is equal to C squared, and so we
now need to take the square root

of both sides, and so five is
equal to see.

That's the length of the
hypotenuse. The longest side of

the triangle. Let's take another
example. This time one where we

know the hypotenuse. Let's say
it's of length 13 and we know

one of the shorter sides. Let's
say that's of length 5, but we

want to know is what's the other

shorter side. Let's label these.
If I call this One X, let's say,

and this one, why I'm call this

one zed. Because I've label them
differently, doesn't make the

relationship any different. It
still X squared, plus Y squared

equals zed squared. The sum of
the squares on the two shorter

sides is equal to the square on
the hypotenuse. Let's

substituting. Now X is 5, so
that's 5 squared plus Y squared.

That's what we're trying to find
is equal to zed squared and

said. Is 13, so that's 13

squared. So 25
plus Y squared

is equal to

169. And so
Y squared must be

equal to 144.

And now if we just turn over the
page and take that last line

again, Y squared is equal to
144. We need to take the square

root of both sides. So why must
be the square root of 144?

12 Will do.

One more. Let's take our
right angle triangle here, like

this. And let's say this is
17 and this is 15.

This one is the one we want to
know now. It's still a right

angle triangle, the hypotenuse
is still the side that is

opposite to the right angle.
Still, the longest side in the

triangle. Let's label our sides.

And again the labelings
different but the relationship

remains the same P squared plus
Q Squared is equal to R-squared.

So let's substitute in our

numbers. P squared that's

15 squared. Q is what
we're trying to find sales plus

Q squared is equal to R-squared
and R is 17. So we have

17 squared. Now we need some

numbers. Q

squared equals.
15 squared, that's

225. And 17 squared
is 289. You can work these two

out on a Calculator or you can
use long multiplication. Doesn't

matter which. Now let's find Q

squared. That's equal to 299
takeaway 225, which is 64. And

so taking the square roots again
Q is equal to 8.

In each of these three, the
answers of being exact, they've

been whole number values.

And these whole number triples
or pythagorean triples as we

call them. Occur quite

frequently. But you probably
won't be lucky when you do

questions you in getting an
exact answer, you almost

certainly have to use a
Calculator for many of them, and

you have to decide to shorten
the answer to a given number of

decimal places or a given number
of significant figures. I've

used whole numbers 'cause it's
easier for me to work with them

on this page, but do remember
they won't always be exact. In

fact, more than likely they
won't be exact and you will have

to work them out.

Using a Calculator.

Let's have a look at another
application of Pythagoras.

Supposing we know the dimensions
of a box, let's say we know that

it's 3 by 4.

By 12 so that means
we know that it's.

Three wide.

For long. And.

12 high

Let's complete our box.

So it's 3 by 4
by 12 high. Let's make

it look a little bit
3 dimensional by putting in.

In the dotted line, the bits
that we can't see.

Question. How big is the
diagonal of the box? The

diagonal not that runs across a
face but runs from one corner

across to another corner. How
can we work this out?

Which is a cuboid. Each of these
sides is a rectangle. All the

joints are at right angles.
Let's put this diagonal in.

So there it goes spanning the
box well across here, down to

here and across here we've got
another right angle triangle.

So we know that if we knew that,
we could find that. But this

gives us a right angle triangle
down here. And we do know these

two sides, so let's see if we
can work that out. Let's.

Make clear we've got a right
angle triangle here and one

here. So first of all, how
long is that? How long is X

the diagonal of this base
rectangle here? Well, pythagoras

tells us that 3 squared +4
squared is equal to X squared,

so that's 9 + 16.

Is 25 and that must mean that
X is equal to 5.

So we've now got X is 5 and
we can use Pythagoras again in

this triangle. 5 squared plus
12 squared is equal to the

length of the diagonal squared.
Let's call that Y equals Y

squared, so we have 25 +
144 is equal to Y squared,

so that's 169 is equal to
Y squared. So why must be

the square root of 169, which

is 13? So we can use the
theorem of Pythagoras in three

dimensions. We can use it to
solve problems that are set up

in 3 dimensional objects.

What about making use of it in

other ways? What about

cartesian geometry? Cartesian
geometry is geometry that set

out on a plane that's got
Cartesian coordinates 11 the

.22, and so on.

One of the other things about
Pythagoras theorem is this.

It tells us if we
have a particular triangle, it

tells us the relationship
between the sides.

And it works in reverse if.

It's.

The area.
Of the

square. On
the longest

side. Is

equal.

Tool.
Area of

the squares.

On the two

shorter sides.
Of

a

triangle.

Then
The triangle

is right

angled. So this is turned
the theorem round. Instead of

saying if the triangles right
angled, then this is So what

it's saying is if this is so,
then the triangle is right

angled and we can make use of
this in areas like Cartesian

coordinate geometry to find out
whether or not a triangle is

right angled or not so.

Let's take the three

points. 34

26
One note and let's see

if they do give us a
right angle triangle.

So first of all, let's plot
these points. First of all, will

mark off on the axes.

Roughly equal spaces.

The
same

markings.

On the Y axis, again
roughly equal spaces.

And now we're plot the points.
The point one note will be here.

Will call that the point a.

The Point B will be the .26 that
will be roughly there.

And the Point C will be the .3
four and so that will be three

and four that will be roughly

there. Let's join up.

These points.

To give us a triangle.

Now it doesn't look right angle,
but we don't really know 'cause.

These were only rough markings.

Is it a right angle triangle or
not? You mustn't believe what

our eyes tell us, especially
when we can do calculation which

will help us to be exact.

So I need to work out the
lengths of the sides of the

triangles, so let me surround.

Is triangle with.

A rectangle.

And because I've surrounded by a
rectangle, these are all.

Right angles here.

And I can see now what the
lengths of these little bits

are. So let me workout AB
squared to begin with. a B

squared. Now I can do that
because this side here is of

length 6, so that's 6 squared
plus this length squared here,

which is a blank one.

That's 36 plus one is 37, so
I've used Pythagoras in this

right angle triangle to find out
what a B is, or at least to find

out what AB squared is. Let me
do that again here, this time

for the side BCBC squared is and
I'm going to use this right

angle triangle here, and again I
can see that this length here is

one, so that's one squared plus
the square of this length.

So that's 2 squared.

That's 1 + 4 gives
main 5. Finally, let's do

a C squared.

And again, I can see that this
length here is of length 4, so

that's 4 squared plus the square
of this length here. That's 2

squared is equal to 16 + 4, and
that gives me 20.

Now, if this is a right angle
triangle, if I take the squares

of the lengths of the two
shorter sides, that's a CNBC.

Add the squares of those
lens together. I should get

the square of the length of
the third side. If it's

right angle and if I don't
then it's not right angle,

so let's do that AC squared
plus BC squared is 5 + 20.

5 for BC squared, 20 for AC
square, and that gives me 25.

But a B squared is
37. That does not equal

37 and therefore the triangle
ABC is not right angle.

The fact that Pythagoras theorem
is about squares is fairly well

known. What's not as well known,
although it's fairly obvious

once you've seen it.

Is that if you take any regular
figure or similar figures and

place them on the sides of a
right angle triangle?

Then the area of the figure
on the hypotenuse is equal to

the sum of the areas of the
same figures on the other two

shorter sides.

Let's make it perhaps as
dramatic as we can. Supposing

report semi circles on the sides
of this right angle triangle.

Is it the case that the area
of the semi circles on the two

shorter sides add up to the area
of the semicircle on the

hypotenuse? Is that so? Well,
let's label the sides.

Play, it'll be, it'll

see. And workout the area area
of a is now the area of a

semicircle is half the area of
the circle and the area of a

circle is pie are squared π
times the square of the radius.

Now the diameter of the circle
is a, so the radius is a over

2, so that's π.

Times a over 2 squared. If we
work that out, that's Pi over 8A

squared, over 2 all squared is a
squared over 4. That's where the

a square comes from and the two
times by the four gives us the

8. Be what's the area
of this semicircle here? Well,

again. The diameter of the
semicircle is B, so the radius

is be over 2, so the area
is 1/2 Pi R-squared and are. We

just said was be over 2, so
we square that and we have π

over 8B squared, this one.

Same again, C equals 1/2.
Thai R-squared and R is

1/2 the diameter 1/2 of
CC over 2 all squared,

and so that's Pi over

8C squared. So let's add
these together A&B and see what

we get a is π over 8A
squared. B is π over 8B squared,

and there's a common factor here
of π over 8. We can take

out, leaving us with A squared
plus B squared, but.

The theorem of Pythagoras tells
us that in a right angle

triangle, a squared plus B
squared is C squared, and so

that's π over 8 times by C

squared. Which is just say.

And so the question that we
asked at the top is true. This

does work for semi circles. It
would work for equilateral

triangles. It would work for
hexagons. It would work for

similar isosceles triangles and

so on. Pythagoras theorem is
quite a wonderful theorem, and

there's quite a lot to be gained
from exploring pythagorean

triple, such as three 455-1213
and the other one that we saw in

this lecture, eight 1570.