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.