 So, here is another theorem which says that the area of a triangle is equal to the product of the semi-perimeter and the in-radius. So, I believe you already know what a semi-perimeter and in-radius is, but nevertheless let me explain once again. So, semi-perimeter is nothing but the half of the perimeter of the triangle. So, in the given triangle if you see, this side is A and this side is let's say B and this side is C, then semi-perimeter is given by a symbol S, symbol S and it is believed that this was popularized by the famous mathematician Euler who said S is equal to A plus B plus C by 2. So, A plus B plus C by 2 is S and in-radius is nothing but the radius of the inscribed circle. So, you can see there is a circle inscribed triangle ABC and the radius of that circle is R. So, R is in-radius of triangle ABC. So, hence we have to prove that area of triangle ABC is equal to S into R and let's see how we got it. So, first of all let's join BI, I happens to be the in-center, then IC and then AI, okay? We are doing this. So, area of triangle ABC will be equal to area of triangle AIB plus area of triangle BIC plus area of triangle CIA, isn't it? These are the three, if you see triangles which added together will give you the area. This is the first one, this is the second one, so these are the two one and the third one is the one which is left in white, okay? So, this is the sum of area of triangle ABC is given by the sum of three areas like that, okay? Now, you know that radius is always perpendicular to the tangent to the tangent at the point of contact or at the point of tangency, point of tangency, tangency at the point of tangency. So, hence what do we learn? We learn that area of AIB, area of triangle AIB will be nothing but if you see AB is the tangent and IZ is the radius. So here, this is 90 degree, here also this is 90 degree and here it is 90 degree, so hence I can say that area of triangle ABC will now be equal to half into AIB into AB into IZ or simply R, correct? Similarly, this will be half into BC into R plus half into CA into R, isn't it? Let's take half R common, half into AB plus BC plus CA into R, correct? That means this is nothing but S into R, why? Because if you see S was nothing but AB plus BC plus CA upon 2, right? So hence we got the desired result, right? Which is nothing but equal to semi-perimeter, semi-perimeter, semi-perimeter into in radius. This is what we achieved, okay? So hence always remember area of a triangle is always equal to the semi-perimeter multiplied by its in radius.