 Hello friends, so in the last session we saw Heron's formula and we learned that Heron's formula is a formula to find out the area of a triangle whose three sides ABC and R given and we learned that that formula is area of triangle ABC is equal to under root SS minus A, S minus B and S minus C. Now it is one thing to know the formula but other thing to actually ask how this formula is arrived at and we will address this inquisitiveness today and see how we can prove this Heron's formula. Now there are multiple proofs already given for it. There are trigonometry based proofs, geometry based proofs and in this case we are going to prove it algebraically and we will be using one theorem which is pretty popular and you know it quite well and that formula is or that concept is Pythagoras theorem. So let's see how do we prove it. So what I have done is I have shown you a triangle here which is ABC. I have dropped up a perpendicular CD which is whose height is H and I have called AD as small d so you can see it here this is small d and the remaining part DB is C minus D. So with this let's start the proof. So what do we know about you know there are two triangles. What do we see? We see two triangles, two right-angled triangles ADC and CBD both right-angled at D. So I am writing in triangle let's say first let's take ADC. So what can I say? We can say B square is equal to H square plus D square isn't it? Let us say this is 1. Then in triangle CDB you can say A square is equal to H square plus C minus D whole square. This is 2 and both of this are by Pythagoras theorem Pythagoras theorem isn't it? So now what? We will do 2 minus or 1 minus 2. Subtract the second equation from the first one so you will get B square minus A square on the left hand side you will get H square plus D square and then minus H square minus C minus D whole square isn't it? Look carefully you will be getting this and then let's expand the right hand side you will get H square plus D square minus H square minus within bracket C square plus D square minus 2CD. Now this H square this H square will go so you will get D square minus C square minus D square plus 2CD. So this D square this D square will go so what do we see? We see that B square minus A square is equal to C square or rather 2CD minus C square that means D will be equal to B square minus A square plus C square divided by 2C isn't it? Or after rearranging I can call it or I can say minus A square plus B square plus C square upon twice of C. This is a good looking result. So the small D is minus A square plus B square plus C square divided by 2C. Now in the same triangle ADC what do we see? Let us now try to use the first equation this one. So from first equation what is H square my friend? So H square is clearly B square minus D square. Let me just show you the figure once again so that you are not confused. So I am saying here H square is equal to B square minus D square by Pythagoras theorem again I can say that and I have just found out the value of D. This is D so let us do the calculation looks like a tedious one but never mind we will do it. So minus D square will be minus A square plus B square plus C square divided by 2C whole square isn't it? This is what is H square. Now if you see this is A square minus B square from there are two squares difference of two squares we can see clearly there is A square and there is minus B square. So what is the expansion like? So it is nothing but A plus B A minus B so hence we will write B plus minus A square plus B square plus C square divided by 2C. This is the first term and the second term is B minus minus A square plus B square plus C square divided by this divided by 2C. A plus B A minus B form. Now let us simplify it further so what will you get if you see this is 2BC 2BC minus A square minus B square minus C square divided by 2C and this term is 2BC plus A square minus B square minus C square by 2C isn't it? Now what? So if you see 2BC term is there so hence can I again write this as this will be so this is 2BC term here and wait a minute so this will be minus A square oh so this should be plus and plus I'm sorry this is wrong so plus A plus B square plus C square and this looks fine fair enough. Now what? So if you see there is a there is some pattern coming out so 2BC and B square plus C square if you club together you know what it is. Similarly here 2BC and minus B square minus C square I can club together and see what will happen. So let's do this so this is nothing but B square plus C square plus 2BC which is nothing but B plus C whole square minus A square upon 2C divided by and oh sorry next term is A square again and you take minus common and you write B square plus C square minus 2BC upon 2C I hope you understood the calculations I just rearranged the previous expression. Now what is it? It is nothing but B plus C whole squared minus A squared divided by 2C and this is A squared minus B minus C whole squared upon 2C is it it? So again we are getting a difference of square thing before that what we'll do is let's club this 2C and this 2C together and we can write this as 1 upon 4C square and if you see my friend we can write the first term in the above statement as B plus C minus or plus A first and B plus C let me just take away some brackets or let me rewrite it again so that you know you're not confused once again I'm writing please look at it carefully so I'm writing B plus C plus A and B plus C minus A why am I writing this as because if you notice again this is square of or difference of two squares again A square minus B square form so A plus B A minus B will be applied so I can write this term as this and for this we can write this as A plus B minus C and next term next term is A minus B plus C isn't it? So A plus B A minus B form once again fair enough I hope if you have if you have any trouble in understanding I would suggest pause it for a minute and then you know look at the calculations carefully you'll get it so primarily what I'm using is A square minus B square is A minus B A plus B okay now again this will be 1 by 4C square okay so what is now if you see the terms are B plus C plus A or let me just rewrite it in a arranged manner so I can write it as A plus B plus C then this can be written as minus A plus B plus C isn't it then this can be written as A minus B plus C and A plus B minus C so hence what I'm doing is this is this term this one comes here just to bring an order so and then rearranging it here and this is here I hope it's clear to all of you right so this you get another you know the value of let's say the H what was H guys we are all calculating here we are calculating H square so this is H squared isn't it so what was H square the square of this altitude and I'll tell you why we are doing this because we have to find altitude and then I'm going to apply half into base into height to find the area correct now how do I convert into convert it into the given form so if you see it what is it A plus B plus C so hence I am saying S is equal to A plus B plus C by 2 right this implies A plus B plus C is equal to 2S right similarly A plus B plus C is equal to S 2S sorry so A plus B or rather minus A plus B plus C okay plus 2A is equal to 2S so have I written minus A and added plus 2A so that again becomes A plus B plus C look carefully if you don't understand pause it and understand it once again so can I not say it as minus A plus B plus C is 2S minus 2A which is 2S minus A right similarly again A plus B plus C is 2S so hence can I not say A minus B plus C plus 2B is equal to 2S so adding minus B and or sorry subtracting B and adding 2B so you'll get again the same thing so check these two are same and now from here I can get A minus B plus C is equal to 2S minus 2B which is equal to 2 times S minus B and similarly guys A plus B minus C will be equal to 2S minus C so just remember these three terms and then entire expression what can I say about it let me write it here then so that we get a clear vision clear picture now so what do we say we are saying H square was this now A plus B plus C can be written as 2S so let me write it as 2S minus A plus B plus C can be written as 2S minus A so let me write it as 2S and S minus A sorry this is 2S minus A so 2S minus A then the third term A minus B plus C is here which is equal to 2S minus B right so again I write 2S minus B and finally it will be S minus C 2 2 times S minus C 2 times S minus C and in the denominator we had 4C square if you see yep yeah so what is this finally then so if you see I can just strike off these two and this four so this term becomes H square is equal to S four times because this two and this two will get multiplied to get 4S four and this S and then S minus A S minus B S minus C and divided by C squared so what is H my friend H is 2 by C under root S S minus A S minus B S minus C now the same expression can be written as half into H into C so I take this two here denominator and take this C in here numerator so it becomes half into H into C is equal to under root S S minus A S minus B and S minus C now let's pay attention to what is this half into H into C and you will be happy to see that this is nothing but area of the triangle ABC so area of triangle triangle ABC area of triangle ABC is equal to half into C into H half into base into height so half H C and here we got an expression for half H C what is that this one half H C is equal to under root S S minus A S minus B S minus C so this is how we can prove here on formula so we arrived at the triangles area is either half into H into C that is half into base into height or if you know the three sides it will be under root semi perimeter times semi perimeter minus first side into semi perimeter minus second side into semi perimeter minus third side this is what is the case let's take a quick example and understand let's say there is a triangle whose sides are 3 4 and 5 you would be knowing that this is the right angle triangle anyways because it this is the R it is Pythagorean triplet so let's say A B and C so what is S guys S is A plus B plus C by 2 which is equal to 3 plus 4 plus 5 upon 2 which is equal to 6 right now let's find out what is S minus A S minus A is 6 minus 3 which is 3 what is S minus B is 6 minus 4 which is 2 and S minus C is 6 minus 5 which is 1 so this area triangle so we are I'm denoting it by symbol delta is equal to under root S which is 6 S minus A which is 3 S minus V which is 2 S minus C which is 1 and hence the answer is 6 square units okay let's check whether we have got it right so if you know this was a right angle triangle isn't it so 3 4 5 happens to be a Pythagorean triplet so this angle is right let's say this is 3 this is 5 and this is 4 so clearly the area of this triangle was nothing but half into 4 into 3 base into height which is equal to 6 and hence it matches with our formula here as well so please note and I'm writing Heron's formula Heron's formula is applicable to to all types of triangles okay there is no restriction as to this is only for right angle triangle or an equilateral triangle or something like that it is true for it holds for all types of triangles okay so I hope you got the proof of this formula and in the next sessions we will be taking up application of this Heron's formula