 Hello friends, welcome again to this session on elements of Indian mathematics. In the previous session, we saw a proof of Pythagoras' theorem as mentioned in Indian Sulva Sutras. In today's session, we are going to discuss another very interesting feature of Indian mathematics where it has been mentioned or the process has been given to find out the area of a square, which is n times the area of any other given square. So what was the use of this particular process or construction method? Basically as we discussed in the previous few sessions, there were methods described for designing and constructing fire altars for performing rituals in post Vedic period. Now in this particular session, we are going to discuss that method which describes that if a given fire altar which has a square interface or square cross section of let us say one particular area, now if we want to design another fire altar whose cross section area is n times that of the given square or the previous or the other fire altar. So this is what we are going to discuss and in the process we will also learn this particular method gives us a process of finding out root of any number geometrically and that is very very interesting. So let us begin. So what I am going to do is I am going to do the construction here in this GeoGebra software and what is the first step? So let us first define the size of the square. So I am going to define it using a slider and since the side cannot be negative, so let it start from 1. So I am going to start or take the size as one you can take lesser value also you can start from 0 also no problem. So let us start from 0 and this is where the side of the square is defined. Now let us now also define a number which is of or let us say the factor by which we want to multiply the area of the given square. So let us say that number is n and I am again defining another variable n and this is where or this is the variable which is the factor by which I want to multiply the given area and find the new square. So this is n and then later on you will also see that we intend to find out the square root of this n itself. So first thing is we have to draw a square of side a. So there are multiple roots of or multiple ways of doing it. You can use construction through circles and on or I am adopting a method of coordinate geometry. So I will be finding out the vertices of the four square and then I will be just joining the vertices. So let us say the first is 0 comma 0. So this is the first vertex a. The second vertex b can be given as a comma 0 is it not a comma 0. So this is b, c can be a comma a. So let us say this is a comma a comma a and the fourth one can be 0 comma a is it a d. So d is let us say 0 comma a, fantastic. So this is or these are the vertices which are going to be joined to make a square. So let us say a, b, c and d, a, b, c, d. So this is my square of side a. So you can see if I increase a the size of the square goes up and down. So let us fix it at 1. Now I am going to give you a word of caution that now the construction is going to be little cluttered. So do not worry I will try to make it as decluttered as possible but just bear with me on this. So let me just shift this point a here and I can, the second step now is to draw a isoscelium strangle. So first I will show you through a rough drawing and then we will construct using geogibra. So let us say I am going to draw a isoscelium strangle like this. So what are the sides? The sides are and later on we are going to prove this as well so do not lose hope. So let us say if this was a then we are going to draw a isoscelium strangle with base n minus 1 a. This is the base and with the other two sides as half times n plus 1 a. So there are two sides with the half n plus 1 a. So the construction is very simple so you draw a line segment with length n minus 1 a. Then taking let us say this point is p and this point is q taking p as the center and the length of radius being half into n plus 1 a we have to draw an arc. And similarly from q as well with the same this you know radius we have to draw another arc intersecting the previous one at let us say point r and this strangle p q r will be our desired isoscelium strangle. And then how do we get the required new area we will see later ok. So let me just put it here and now keeping that rough diagram in mind let us now draw n minus 1 a base. So what I am going to do is I am going to take a circle with center a and radius as n minus 1 a simple n minus 1 times a fantastic. So but you know I do not see here ok so let me do it once again so a is the center and radius is n ok. So here n is 1 so obviously the radius is 0. So let me take n little bit higher ok so you can see yes. So let us say I am taking 1.8 n is 1.8 so hence now you can see the circle to be appearing over there but I do not need the circle I just need this point is it. So let us say this point is e and I can now switch off this circle I do not need this ok. Now I have a point the base is ready a e is the base and I need to now draw 2 other sides here as you can see on the right side I have to just do the 2 sides and what is the length length is half n plus 1 a let us draw the circle once again. So from a as center radius being 0.5 that is half and then n plus 1 and into a is it. So this is my yeah so I got the first circle with center a I have to repeat the same process using center as e and then 0.5 again n plus 1 times a perfect. So here is my point of intersection so right this is the now if I just do away with these circles let us say I do not need these so let me just hide them ok so I do not need them anymore so f point is what I was looking for and let me join or let me draw a polygon. So a e and f is the required isosceles triangle ok. So now that the isosceles triangle is done what is next so let me just switch it off let me delete everything here and now this is the isosceles triangle you will be astonished to know that if I join let us say drop a perpendicular from f on to this line. So this is the perpendicular bisector of side a e and let us name this point g. Now f g represents actually so now if I f g represents the side of the square which was our target square so now if I complete the square using side f g then the new square is going to be n times in area than the original square so that is the beauty ok. Now let me just so how do I draw the square so what I am going to do is I am going to draw a few circles so do not worry so this is one circle that means this point here is another vertex of the square is it not so let me just you know this is another vertex from here again I can draw another circle or rather I can draw you know circle from here with the same radius ok so let me draw a circle from f passing through g ok and h passing through g right so why did I do that all these circles intersect you know add the vertex of the given square or the required square now I can again switch off or hide all these circles I do not need them anymore ok so my construction is over I hope you understood so basically from f g as side I drew some circles of equal side equal radius f g on the three sides and I am now getting the desired polygon so this is my desired polygon f g which is a square so f g hi ok f g hi so this is my square now interestingly the area of this f g hi will be n times that of a b c d understood so area of f g hi square is n times that of a b c d how is that we will you know do it in the proving section right now this is what so you can validate by just measuring the area let us measure the area so clearly the area of this one is a b c d is one and this one is one point eight so you can see n was one point eight so the area of f g hi is one point eight times that of one ok so now you can you can imagine if a b c d is one then f g f g is so one point eight root over one point eight is it if the area of f g hi is one point eight here then one side will be root over one point eight so basically g f represents root over one point eight let us measure g f to understand that so if I measure the distance of f g you can check this comes out to be one point three four and one point three four actually will give you or the square of one point three four will be approximately one point eight right so this is what is the theorem so basically what did we do so in the ancient time what was the need for all this so they had a fire altar whose top view would appear like a b c d and then there were there were rituals wherein you have to draw the or you have to construct fire altars whose area will be n times of other given area a b c d so if a b c d is one square unit and n is one point eight let us say you want to go one point eight times or you know you can change the n we will see we will be seeing by changing and what impacts does it have then you can always find out with this method the area of a square which is n times that of the original square and and if you consider a to be one then you will be able to find out the square root as well is it so let us say if I do not touch a if I change n so let us take n equals to two so you can see now the area is two times the previous one or the original one and you will clearly see f g length over here is one point four one which is approximately root over two is it so this is a very interesting method of finding the square root of any number using geometry now let us also increase so if I increase n beyond two it's not giving me any result why because you will see in the proof the base is and anyways I showed it to you that base was n minus one a so in this case if I increase n the n minus one a goes you know beyond this maybe so hence your construction is not visible here so if I increase a let's say so let's say this is to a is to and let me just zoom out and then let's say if I have n so you can now see as I reduce you know so you can now see that if this area a b c d is four then this is one point eight times of four is seven point two and the new area f g h i is seven point two so this was this is mentioned in Katya in Katya in Sulva Sutra and again it's an ancient mathematics and we can see that this is another very interesting way of finding root over any number using geometry so now let us go to the proof of this particular theorem