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The theorem of Pythagoras is a
well known theorem.

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

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

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ancient Babylonians and to the
ancient Egyptians.

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

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squared equals C squared, but
the actual statement of the

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theorem is more to do with

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

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The theorem talks about
a right angle triangle.

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

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

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

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Now that means we have to
identify this side the

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

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

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

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hypotenuse. So the theorem then
says the square on the

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

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there we've got our square.

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

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of the areas.

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Of the squares on the
two shorter sides.

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So these are all squares.

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All the same, all got right
angles in them.

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This is a square as well.

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

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

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C A+B equals C.

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

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see how that actually works.

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Got here a
single red triangle.

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It's a very special triangle.

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

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

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bit of Blu tack.

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

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round it. So that
we've drawn around that

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triangle, now I want
to fix some squares

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on the sides of

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

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

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

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

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Now I'm going to put the other

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half on. Here.

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Lastly, debate.

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I will make it up with
this bit.

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

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

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

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hypotenuse. Let's see how right

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

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

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here, so I can put one on there.

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That one
on there.

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Now put

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the square.

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In there.
That

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one
in

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there.
And then finally put

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this one in there.

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

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

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

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

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how do we get the traditional?

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Schoolboy schoolgirl theorem. A
squared plus B squared equals C

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squared. So draw our
right angle triangle.

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Will put on the

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squares. On the sides.

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And
we

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label
these

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squares

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

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lay. It'll be a little C.

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

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

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

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

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

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its area is B squared.

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

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traditional result. Always when
answering questions on

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Pythagoras Theorem, make sure
that you read the question.

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

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

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

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or two examples.

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

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sides are of length three
and of length 4.

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

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

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

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

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Is C squared?

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A is 3 three squared
plus B is 4, four

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squared equals C squared.

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

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

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

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

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

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

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

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want to know is what's the other

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

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and this one, why I'm call this

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

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

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

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

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

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

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said. Is 13, so that's 13

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squared. So 25
plus Y squared

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is equal to

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169. And so
Y squared must be

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equal to 144.

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

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

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

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12 Will do.

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

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

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

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angle triangle, the hypotenuse
is still the side that is

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

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triangle. Let's label our sides.

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And again the labelings
different but the relationship

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

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So let's substitute in our

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numbers. P squared that's

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

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

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17 squared. Now we need some

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numbers. Q

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squared equals.
15 squared, that's

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

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

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matter which. Now let's find Q

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

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

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

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been whole number values.

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And these whole number triples
or pythagorean triples as we

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call them. Occur quite

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

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questions you in getting an
exact answer, you almost

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

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

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

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

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

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

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to work them out.

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Using a Calculator.

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

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

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it's 3 by 4.

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

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Three wide.

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For long. And.

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12 high

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Let's complete our box.

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

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it look a little bit
3 dimensional by putting in.

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

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

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

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

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

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

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

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

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

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

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

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

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

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the diagonal of this base
rectangle here? Well, pythagoras

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

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so that's 9 + 16.

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

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

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

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

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

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

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the square root of 169, which

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

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

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in 3 dimensional objects.

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What about making use of it in

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other ways? What about

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cartesian geometry? Cartesian
geometry is geometry that set

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

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.22, and so on.

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One of the other things about
Pythagoras theorem is this.

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It tells us if we
have a particular triangle, it

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tells us the relationship
between the sides.

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And it works in reverse if.

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It's.

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The area.
Of the

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square. On
the longest

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side. Is

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equal.

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00:17:06,050 --> 00:17:11,724
Tool.
Area of

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the squares.

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00:17:15,870 --> 00:17:19,542
On the two

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shorter sides.
Of

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a

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00:17:26,876 --> 00:17:29,319
triangle.

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Then
The triangle

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is right

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

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

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00:17:51,100 --> 00:17:56,692
it's saying is if this is so,
then the triangle is right

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

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coordinate geometry to find out
whether or not a triangle is

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right angled or not so.

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Let's take the three

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points. 34

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26
One note and let's see

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if they do give us a
right angle triangle.

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

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mark off on the axes.

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Roughly equal spaces.

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The
same

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markings.

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On the Y axis, again
roughly equal spaces.

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

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Will call that the point a.

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

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

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and four that will be roughly

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00:19:19,014 --> 00:19:21,878
there. Let's join up.

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These points.

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To give us a triangle.

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

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These were only rough markings.

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

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

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00:19:45,358 --> 00:19:47,854
will help us to be exact.

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00:19:47,870 --> 00:19:53,148
So I need to work out the
lengths of the sides of the

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00:19:53,148 --> 00:19:55,178
triangles, so let me surround.

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00:19:57,900 --> 00:19:59,829
Is triangle with.

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00:20:00,330 --> 00:20:01,890
A rectangle.

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00:20:06,240 --> 00:20:10,490
And because I've surrounded by a
rectangle, these are all.

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00:20:11,050 --> 00:20:12,460
Right angles here.

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00:20:13,140 --> 00:20:16,716
And I can see now what the
lengths of these little bits

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00:20:16,716 --> 00:20:22,431
are. So let me workout AB
squared to begin with. a B

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00:20:22,431 --> 00:20:27,723
squared. Now I can do that
because this side here is of

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00:20:27,723 --> 00:20:32,574
length 6, so that's 6 squared
plus this length squared here,

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00:20:32,574 --> 00:20:34,779
which is a blank one.

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00:20:35,700 --> 00:20:40,500
That's 36 plus one is 37, so
I've used Pythagoras in this

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00:20:40,500 --> 00:20:46,500
right angle triangle to find out
what a B is, or at least to find

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00:20:46,500 --> 00:20:51,700
out what AB squared is. Let me
do that again here, this time

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00:20:51,700 --> 00:20:56,900
for the side BCBC squared is and
I'm going to use this right

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00:20:56,900 --> 00:21:02,100
angle triangle here, and again I
can see that this length here is

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00:21:02,100 --> 00:21:06,500
one, so that's one squared plus
the square of this length.

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00:21:06,540 --> 00:21:08,708
So that's 2 squared.

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00:21:09,300 --> 00:21:16,110
That's 1 + 4 gives
main 5. Finally, let's do

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a C squared.

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00:21:19,230 --> 00:21:25,796
And again, I can see that this
length here is of length 4, so

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

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00:21:31,424 --> 00:21:37,052
squared is equal to 16 + 4, and
that gives me 20.

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00:21:37,580 --> 00:21:43,612
Now, if this is a right angle
triangle, if I take the squares

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

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00:21:50,030 --> 00:21:54,320
Add the squares of those
lens together. I should get

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00:21:54,320 --> 00:21:59,039
the square of the length of
the third side. If it's

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00:21:59,039 --> 00:22:03,758
right angle and if I don't
then it's not right angle,

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00:22:03,758 --> 00:22:09,335
so let's do that AC squared
plus BC squared is 5 + 20.

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00:22:11,150 --> 00:22:17,520
5 for BC squared, 20 for AC
square, and that gives me 25.

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00:22:19,130 --> 00:22:26,730
But a B squared is
37. That does not equal

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00:22:26,730 --> 00:22:34,330
37 and therefore the triangle
ABC is not right angle.

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00:22:34,340 --> 00:22:41,534
The fact that Pythagoras theorem
is about squares is fairly well

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00:22:41,534 --> 00:22:48,074
known. What's not as well known,
although it's fairly obvious

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00:22:48,074 --> 00:22:50,690
once you've seen it.

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

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00:22:57,864 --> 00:23:03,284
place them on the sides of a
right angle triangle?

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00:23:03,970 --> 00:23:10,390
Then the area of the figure
on the hypotenuse is equal to

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00:23:10,390 --> 00:23:17,345
the sum of the areas of the
same figures on the other two

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00:23:17,345 --> 00:23:18,415
shorter sides.

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00:23:19,620 --> 00:23:24,930
Let's make it perhaps as
dramatic as we can. Supposing

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

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00:23:31,610 --> 00:23:38,666
Is it the case that the area
of the semi circles on the two

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00:23:38,666 --> 00:23:44,714
shorter sides add up to the area
of the semicircle on the

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00:23:44,714 --> 00:23:49,250
hypotenuse? Is that so? Well,
let's label the sides.

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00:23:49,890 --> 00:23:53,306
Play, it'll be, it'll

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00:23:53,306 --> 00:24:00,766
see. And workout the area area
of a is now the area of a

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00:24:00,766 --> 00:24:06,642
semicircle is half the area of
the circle and the area of a

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00:24:06,642 --> 00:24:12,066
circle is pie are squared π
times the square of the radius.

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00:24:12,066 --> 00:24:18,394
Now the diameter of the circle
is a, so the radius is a over

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00:24:18,394 --> 00:24:20,202
2, so that's π.

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00:24:21,040 --> 00:24:27,424
Times a over 2 squared. If we
work that out, that's Pi over 8A

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00:24:27,424 --> 00:24:33,352
squared, over 2 all squared is a
squared over 4. That's where the

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00:24:33,352 --> 00:24:39,736
a square comes from and the two
times by the four gives us the

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00:24:39,736 --> 00:24:46,416
8. Be what's the area
of this semicircle here? Well,

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00:24:46,416 --> 00:24:52,700
again. The diameter of the
semicircle is B, so the radius

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00:24:52,700 --> 00:24:59,896
is be over 2, so the area
is 1/2 Pi R-squared and are. We

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00:24:59,896 --> 00:25:07,092
just said was be over 2, so
we square that and we have π

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00:25:07,092 --> 00:25:09,662
over 8B squared, this one.

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00:25:10,590 --> 00:25:18,160
Same again, C equals 1/2.
Thai R-squared and R is

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00:25:18,160 --> 00:25:25,730
1/2 the diameter 1/2 of
CC over 2 all squared,

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00:25:25,730 --> 00:25:29,515
and so that's Pi over

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00:25:29,515 --> 00:25:36,845
8C squared. So let's add
these together A&B and see what

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00:25:36,845 --> 00:25:44,615
we get a is π over 8A
squared. B is π over 8B squared,

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00:25:44,615 --> 00:25:51,830
and there's a common factor here
of π over 8. We can take

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00:25:51,830 --> 00:25:57,380
out, leaving us with A squared
plus B squared, but.

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00:25:58,000 --> 00:26:02,059
The theorem of Pythagoras tells
us that in a right angle

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00:26:02,059 --> 00:26:06,118
triangle, a squared plus B
squared is C squared, and so

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00:26:06,118 --> 00:26:08,701
that's π over 8 times by C

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00:26:08,701 --> 00:26:11,900
squared. Which is just say.

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00:26:12,580 --> 00:26:18,417
And so the question that we
asked at the top is true. This

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00:26:18,417 --> 00:26:22,907
does work for semi circles. It
would work for equilateral

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00:26:22,907 --> 00:26:27,397
triangles. It would work for
hexagons. It would work for

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00:26:27,397 --> 00:26:29,193
similar isosceles triangles and

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00:26:29,193 --> 00:26:34,714
so on. Pythagoras theorem is
quite a wonderful theorem, and

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00:26:34,714 --> 00:26:39,644
there's quite a lot to be gained
from exploring pythagorean

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00:26:39,644 --> 00:26:46,053
triple, such as three 455-1213
and the other one that we saw in

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00:26:46,053 --> 00:26:48,025
this lecture, eight 1570.