 Hello and welcome to another session on gems of geometry. So guys today we are going to take up another very interesting and very important formula given by one of the very famous Indian mathematicians Brahma Gupta. So Brahma Gupta existed in the 7th century around 7th century in Christian era and apart from other mathematicians like Bhaskaracharya and Aravata he happens to be well known for his algebra and geometrical results. So one of the formula which is related to finding out the area of a cyclic quadrangle is what we are going to discuss today. So all of you know that cyclic quadrangle is nothing but a quadrangle whose all the four vertices are lying on a circle. So you can see here I have drawn a cyclic quadrangle ABCD and it lies on this circle. So Brahma Gupta has expressed the area of the cyclic quadrangle in terms of its four sides and the formula is something like this. So k which represents the area is equal to under root s minus a, s minus b, s minus c, s minus d. Now many of you would have heard of Heron's formula. So Heron's formula is a formula given by a mathematician, philosopher, scholar, hero, hero of Alexandria and it was somewhere around 2nd century AD. So 5 centuries later Brahma Gupta came up with his own formula where he expressed this formula for finding out the area of a cyclic quadrangle. Now you can all see that Heron's formula becomes one of the special case of Brahma Gupta's formula. That means if you reduce one of the four sides ABC and either of them one of them to 0 then it reduces to Heron's formula. So let us discuss this formula first we will see and we will validate this and then we will try to prove this and we will also demonstrate how we can, what was the proof given by many scholars and in this particular video we will be proving this using trigonometry. So let us begin the demonstration first and then we will be trying to prove this. So what I have done is I have drawn a circle with radius 5 center O and ABCD is the quadrangle ABC and V are four points on the circle and I have already calculated the area you can see on the right side panel. SS happens to be the semi-perimeter okay so A plus B plus C plus V divided by 2 so S is semi-perimeter of this quadrangle. The values of ABC and V in this particular case have been given right it's been shown here if you change the locations of A and B you can see all the values related to you know the change is also changing for example in this case the value of B and C are changing so small a small b small c small d are the sides of this quadrangle. So as you move any of these points and you will see the entire value keeps changing okay so very clearly S has been calculated here 14.05 S minus A S minus B S minus C and S minus B has been calculated as well and then we have found out the area using the formula or using the calculator in built in the software but let us also try to find out the area of this quadrangle from some other sources which are other mechanisms which we have studied. So what we can do is we can always split this quadrangle into two halves so let us say I join the diagonal BD and if I find out individual you know triangle ABDs area and triangle BCD area and add them it should match up with this value 48.65 okay so let us try to do that so how do we do it so for that I know that area of a triangle either I can use the Heron formula itself to find out the area or rather since we have the software here we will be using the you know the formula of half into base into height so for that I have to first find out the altitudes from A and C so let us do that let me drop a perpendicular from A to this line okay so this line yeah so this is a perpendicular line on BD from A let me find out the point of intersection so this is the point of intersection E similarly I can drop another perpendicular from C to this line and again let me call this point as F right so CF is perpendicular to BD you can see and AE is perpendicular to BD okay now since I don't require these lines so I can just switch them off so I don't need them okay and now I can simply join the points using a simple line segment so AE and FC so basically now we just need to find out the base this BD length length AE length FC and find out the area of the triangle so for that we need the formula panel here so let me do that okay so now the area so keep keep in mind the area which is there already calculated by this formula Brahma Gupta's formula how much is it 46.3 and let's see whether we can match this using our conventional method of you know finding the area of triangles okay so let me let me first define few lengths so I'm defining D1 you can you can see on the right left hand side panel here D1 right so what is D1 I'm defining D1 as so it's already yeah so let me start with D1 is equal to right what is D1 distance okay distance between which two points A and E right A comma E so this is D1 so D1 is 5.27 yeah so D1 is AE distance AE okay or rather I can change the name and put it like this so that it becomes much clearer wrap DAE okay so DAE is this I can put the label maybe yeah so but this label is different let it be oh sorry so yeah so I is the name of this thing but I don't want the label let it be so DAE is distance between A and E similarly D let's say F and C okay so F and so I'm writing FC is equal to what will that be is equal to distance distance you can see on the left hand side corner where I'm writing the formula here dislocation distance what is the distance distance is F and C F and C okay so AE I already calculated now I'm calculating yep DFC and DAE is done now I also need the distance between B and D for the base so I'm writing D underscore BD BD and this will be equal to this will be equal to distance between distance between B and D B comma D okay so 9.97 so I have calculated distance between AE distance between F and C and distance between B and D so let us say the first area A1 the first area or I can write area of triangle ABD so let me write it as ABD okay area of triangle ABD is equal to how much half so 1 by 2 into base what is the base guys so I have to write 1 by 2 1 by 2 and into half into base what is base BD right so I have to write D BD D BD into into which one D AE isn't it so D underscore AE right here is the value 26.24 so area of triangle ABD is 26.24 you can always check it from this tool as well because we have area tool so you can always find out a right 7 C1 oh this is our circles area so we don't need this we don't need this we need the area of area of the triangle okay so let me delete it so I have to find out area area so select polygon circle or conic so I have to basically define a polygon for that so let it be I am not doing that I am again calculating through our usual mechanism now I am interested in finding out area of BDC or BCD so let's say BCD you can again focus here and you will be able to see this BCD is equal to half again so 1 by 2 and then into base base is again BD so D BD okay and then multiplied by the altitude FC D correct and here is the area so 26.24 and 20.06 so total area of the quadrangle would be area total area let's say a underscore abcd okay so this is the area of quadrangle abcd will be nothing but aabd underscore area of abd plus area of area of what area of BCD is it it BCD correct and if you add see whatever what is the value I am getting 46.3 and let's check what was the value here we had got 46.3 see it matches so here you can see 46.3 was through the calculation which calculation this calculation under root s minus a s minus b s minus c s minus b I have done that calculation and here area is coming out if I sum the area of the two parts of the quadrangle I am also getting 46.3 so that means this area does work okay so what I can do is I can always change this value see if I change the location of let's say point a right so you can see yeah 44.72 and let me put it on the same frame so that becomes easier to view so 44.72 so whatever I do so I change the location of v so 45.58 45.58 right so here anywhere you can take d here so 37.4 37.4 so that means what we can very easily see that you know this formula given by Brahma Bhukta in seventh century yeah and and in that century he came up with this formula that area of our cyclic quadrangle can be expressed as root over semi perimeter minus a multiplied by semi perimeter minus b multiplied by semi perimeter minus c multiplied by semi perimeter per meter minus d right so if you do that you will be getting the area which we validated by finding out the area as the sum of the areas of the two triangles divided by one of the dinals of the quadrangle right now next what we are going to do is we are going to prove it so how do you know or how do we know that this formula works and why is it working so let's check that proof out