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.