So far we have seen the geometric
construction that allows us to find the 
square roots of natural numbers
but it was a recursive process
that is to know the square root of a given natural number it was necessary to
know the square root of the previous one.
Now we're going to look at
another construction which is more powerful and
direct as well
So it will allow us to find
in one step
the square root of
an arbitrary number
say 3.14
this is based on the geometric property
of similarity
Similarity in is nothing but
scaling
either scaling down or scaling up of a geometric figure. For example
here we have a triangle
and we have scaled it up
by the factor say 1.5
and then once again
we are going to scale up
by the same factors to get this third triangle
Now if we take the corresponding lengths
of these three triangles - say these green
lengths: 16, 24 and 36
then they bear the same ratio
and therefore they are in
proportion. So to say 16:24
is 24:36
16:24 = 24:36
such triplet of numbers is said to be in geometric progression
and the middle number
can be had
as the square root (as shown here)
of the product of the first and last
number. So 16 (the first number)
multiplied by 36 (the last number)
will give us
the squared of 24 (the middle number)
and that's the route
we can take to find square roots!
Now such triplets in geometric progression can be had
using right angle triangles
so we will start with
this one over here
and we are going to rotate it
so that this
base of the triangle
comes and aligns with the 
height of it
Then we are going to scale
this copy
that we just made
about this point so that height of the copy
becomes the same as the original
So now we have
two right angle triangles which are necessarily similar because they are just
rotated and scaled 
and they have the same height
Now the base of 
this triangle, then this height
and the base of the original triangle
will be
in geometric progression
Because this base upon height
will bear the same ratio as this height upon this base
and therefore the product of these two
bases
will give us the square of this height
Now how does that help us in finding the square root?
imagine a situation where
one of of these bases - say this one - is equal to unity
and in that case, the height
will be the square root of this base! 
Let's see how -
Suppose we want to find the square root of say 3.14 So i will draw a 
segment of 
length 3.14 units
then extended it with a co-linear segment of 
one unit
these are
the two bases
one is unity
and one if a number whose square root we want to find
Now we want to construct this 
larger
right angled triangle whose hypotenuse is this compound segment.
A quick way of constructing right angle triangles with given hypotenuse
is to use that hypotenuse as
the diameter of a semi circle
because the included angle in a semi circle is ninety degrees
then from their junction
we'll be drawing this line which is going
to contain this height here
and then trim it
up to the circle
and we are actually done!
but just to make it look like our previous figure I'm going to construct
these
two lines
and under the circumstances
we can get the square root of this red length
as this yellow length. Let's actually measure it
this is
3.14 because we plotted it so
while this
is 1.772
this should be the square root. Let's check if it is indeed so
Punch in 1.7720
square of that
is giving us 3.1399
almost 3.14
So this is the way we can obtain square roots.
Now these three triangles that we constructed,
have a strange connection with an
interesting
ancient tool called the Arbelos
An Arbelos is
a cobbler's knife
and if you look at its blade
you'd realize it is made up of three semi circles really. 
So there is this large one
and from which
these two smaller semi-circles of half the size are subtracted
When Archimedes saw this he was
absolutely fascinated
and he quickly did
two things that mathematicians like to
do
1. He generalized this shape
by making these two smaller semicircles of unequal size
and
2. He started thinking about propertys of the shape
like its area
Here the area is shown in blue and this
is how the Arbelos is related to the triangles
that we have constructed
Archimedes came up with
a very impressive statement. He said that if
you drawl a circle
whose diameter is given by this height
then the area of this circle is equal to the area of this Arbelos.
Let's see how her proved that
We've already seen that area of a geometric figure
is proportional to any of its linear dimensions
so we can express the area of this larger semi-circle
in terms of its diameter (a + c) squared
of course multiplied by some constant factor which will filter out because
it's going to be there for all our expressions
So this is the area of the larger semi-circle
from it let us subtract this
semi-circle of diameter (a)
soI'll subtract (a) squared
Then the second semi-circle
i'm going to subtract so a minus (c) squared
Now algebraically speaking this is the
expansion of
binomial (a + c) squared
So if you subtract a squared and c squared from that
you'd be left with only
2 x a.c
this a x c
is the product of these two
bases
which we are already shown to be the square of 
this height
so this expression 2.a.c
will be equal to
two b squared. And this b squared
will be the area of 
this green semi-circle because (b) is its diameter
and there are two of them
so that makes
the whole of
this green circle
and thus Archimedes showed
that the area of this Arbelos
is equal to the area 
of this circle
This is called
Archimedes's Lemma
