 Hi, I'm Zor. Welcome to Unizor Education. I would like to go a little bit deeper into area. And as usually I will use some problems to basically demonstrate the educational material which I wanted to present here. Okay, so I have six problems here which I have symbolically described as this, actually mathematicians like symbols. For instance, this is basically my symbolical representation of the problem number one, which I'm going to do right now, that the area of a circumscribed polygon is equal to half of the perimeter times radius of inscribed circle. So, let's just do this type of drawing. This is my circumscribed perimeter polygon, or if you wish, a circle which is inscribed into a polygon. Every side of the polygon is a tangent. So, the area of this particular polygon is half of the perimeter times radius of the inscribed circle. So, if these are the size of the polygon, half of the perimeter is a plus b plus c plus d plus e over 2 times r. That's what basically it means. Now, how can that be proved? Oh, by the way, I forgot to tell you, if you did not really try to solve these problems yourself, please stop listening to me and try to do it right now. And only then, after you familiarize yourself with the problems, go back to this particular presentation. So, how can that be proven? Actually, it's a very easy problem. Connect every vertex with center of the circle. Now, obviously, the area of the polygon is equal to sum of the areas of these triangles. Now, every triangle has the area which is half of the base times altitude, right? So, let's draw an altitude. Now, this is the perpendicular from the center of the circle to a tangent, which means it's always a radius. Because perpendicular to a tangent is always the point, it goes to a point of tangents. Same thing here. So, these are all radiuses. r, r, r, r, r. So, the area of this triangle, for instance, is a times r divided by 2. Area of this triangle is b as a base times altitude plus b times r over 2 plus, et cetera. The last one is e times r over 2. And if you will summarize this factor out, r, you will get exactly this. So, that's an easy problem. Wiping it out. Next problem. Using my symbolism is, given a triangle and we have to construct a square, the area of which is equal to the area of triangle. Alright, how can we do this? So, we have a triangle. Well, which means we have the base and the altitude using which we can always calculate what the area is. And the area is equal to a times h divided by 2, right? Now, what we would like to do is, we have to construct a square with the side x. And the area of which is equal to x squared, right? x times x. So, we have to find x, which is equal to a times h over 2. Or, considering we know h, we know h divided by 2, let's call it b. So, I'm sorry, that's square. So, we have to find x, that x squared is equal to a b. Now, this is the problem which we have solved many times before. And the way how to do it, if you remember, if you have the right triangle and you have an altitude dropped to hypotenuse, then we have basically proven many times that considering the similarity of these triangles, a times b is equal to h squared. Or, if you wish, a divided by h is equal to h divided by b from the similarity of these triangles. The proportionality of the sides. Which is actually the same thing which we have to do. So, we have to find x, we have to construct the right triangle where the altitude divides the hypotenuse into known segments. Now, how to do this? Again, we did it before many times. First, you have to use a plus b as a hypotenuse and draw a circle on this as a diameter because this is the locus of all points from which this particular segment is viewed at 90 degrees. So, these are all different right triangles which can be built with a plus b as a hypotenuse. And then, this point, you know this point since you know a and b. You just draw the perpendicular and these two points on the circle give you two different triangles which both have exactly the same altitude, which is equal to the x. That's simple. Now, not all of these problems are simple by the way. There is one complex one but we didn't really reach it yet. It will be problem number five. Okay, so now we are number three. So, if you have a polygon, now we have to construct a triangle which has the same area as a polygon. So, that's the test. Alright, so here is our polygon. And I think we must actually say that we have a convex polygon. Now, how can I construct a triangle which will have the same area as this particular polygon? What I will do is I will construct another polygon with the number of vertices one less than the one which we have right now. Fortunately, we have some construction. So, sorry about the noise. So, anyway, so how can I reduce the number of vertices without changing the area? Okay, very simple. A, B, C, D. Let's jump over one particular vertex and connect A to C. And the vertex which we jumped over will draw a parallel line to A, C. Now, any triangle which has A, C as a base and the third vertex on this parallel line like this one, for instance, will have exactly the same area as A, B, C. Why? Because A base is the same and altitude is the distance between these two parallel lines, it's also the same. So, the area is the same. So, if the area is the same what I will do, I will choose one particular point, not this one, but this one. So, I'm making this line, Ae, continue until it crosses my parallel line. Now, this triangle, let's call it B' This triangle, A'A, B'C, is exactly the same area as A, B, C. As I said, the same base and the altitude is also the same because it's the distance between two parallel lines. But, since B'Ae is one straight line, we have basically reduced the number of vertices from 1, 2, 3, 4, 5 to 1, 2, 3, 4. A is no longer a vertex, right? Because B'E is the straight line. So, now we have this which has exactly the same area as before, but now it has one less vertex. Well, we can do exactly the same thing again. Let's say I connect C and E now, jump over B, so now I draw a parallel line. Now, every triangle which has exactly the same base, Ce, and exactly the same altitude as the distance between these two parallel lines as E, B'C has, would have the same area. So, I will continue this line and draw the double prime. Now, B''B2, if you wish, Ce triangle and B'Ce have exactly the same base and exactly the same altitude, which is the distance between these two parallel lines. So, the areas are the same, which means my triangle B'Ce has exactly the same area as quadrangle B'Ce. So, I am reducing the same, I am reducing again the number of vertices by 1 having the area the same. So, now I have a triangle. So, that's how I have reduced from the polygon with 5 sides to a polygon with 3 sides. Now, no matter what kind of polygon I have, 6 sides, 27 sides, whatever it is, using this process I will construct eventually a triangle, because in every step I am reducing the number of vertices by 1 and eventually I am getting the triangle, which has exactly the same area as the original polygon. Now, next, we have two squares and we have to build another square which has the area equal to their sum, or they are different actually, I have two problems here. Okay, that's simple. Remember the theory of Pythagorean theorem. If you have the right triangle, then this square plus this square is equal to this square. So, if we have two different squares, initial squares, all we have to do is use their sides as two-catchity of the right triangle, and then the k-partinals would be exactly the side of the square which has the area equal to sum of these two areas. So, basically we are talking about constructing the right triangle by two-catchity, which we know how to do. Now, if you want to do the difference, how to construct a square which has the area equal to the difference between the area of two-given squares. Well, then one given square is supposed to be built on the k-partinals and another is built on the k-catchity. So, again, we have reduced our problem to a problem of constructing the right triangle by k-partinals and k-catchity, and we also know how to do it. So, that's very simple. Alright, and now we are approaching a difficult problem. Now, difficult in, well, not in some ideological sense, difficult because probably we will all know how to do it, but there are a certain number of calculations involved and I will attempt to do it without making any mistakes. Now, I might, but anyway, just bear with me. Alright, so we are talking about number five. Now, this is the famous formula of Hiro of Alexandria, that the area of the triangle is equal to, area of a triangle is equal to square root of P, P minus A, P minus B and P minus C, where P is half of the perimeter. So, that's what we have to prove. That's the formula which expresses the area of any triangle, not in terms we used to, like base and altitude, multiply and divide by two, but in terms of three sides. Alright, so, I mean, he was able basically to come up with this problem and obviously there are, well, there are certain interests actually because triangle is well defined by its three sides, so how the area is defined by these three sides. Alright, so that's exactly what we want to do right now and well, let's do it relatively straightforward, but there are certain number of calculations involved, so that's why I'm just trying to be careful. Alright, so let's say this is A, this is B, this is the whole thing is C, and I will introduce two new variables, this is X and altitude H. Now, we know how to express the area in terms of C and H, it's C times H divided by two, right, but we don't know H. Well, H can be defined from this right triangle, but we don't know X neither. Well, but there is another triangle and we know that this piece is C minus X, right? If this is X, this is C minus X because the whole that length is C. So, let's just put together whatever we know about this and then it will be easier. We have two variables, X and H, unknown variables, but we have two right triangles, which means we can always have two equations. So, one equation is the Pythagorean theorem of this. So, it's X squared plus H squared is equal to H squared, right? That's the Pythagorean theorem of this triangle. Now, this one is C minus X squared plus H squared is equal to B squared, right? So, we have two variables, X and H, two equations. So, we can actually find X and H. If we find X and H, well, particularly H, we can find the area by multiplying it by C and divide by two, right? Now, how can this be solved? That's actually easy because let me change, let me open the parenthesis here and I don't really need this anymore. I still have one more problem, but I remember it by heart. Alright, so, the first we will leave as is. The second one we will C squared minus two C X plus X squared plus H squared equals two B squared. So, this is my new system. Well, from this, we can very easily find X. How? Let's subtract from this, subtract this, X squared plus H squared, X squared plus H squared, which is H squared, which means C squared minus two C X plus instead of this, we can substitute this. Now, I have one linear, by the way, equation with one variable X, from which we can find X, obviously. Two C X is equal to A squared minus B squared plus C squared, am I right? So, two C X, if I will add two C X, it will be there. I will subtract B squared, it will be here with a minus sign. A squared and B squared are with a plus sign. From which X is equal to A squared minus B squared plus C squared divided by two C? Okay? Well, X is found. Now, we have to find H. So, let me wipe out this. And obviously, this is simpler. So, I will put H squared is equal to A squared minus X squared. And X squared is this, which is A squared minus B squared plus C squared, squared, or C squared. Right? The square of this is this. So, that's what we need. We need H squared. We need H actually. Now, the area is equal to C times H divided by two, right? C was our base. H was an altitude from it up. So, let's do A squared, which is equal to C squared, H squared divided by four. Because, you see, A squared would be easier if I will, you know, find this particular expression. I don't really have to deal with square roots. So, I would like to convert this into this. Whatever is under the square root. And again, this is just simple algebra. So, let's try to basically do it accurately and hopefully we will not make any mistakes. So, here A squared is equal to C squared times H squared divided by four. And instead of H squared, we do this. All right. So, it's C squared over four times A squared. That's times this first one. And then the second one, minus. Now, C squared over four, again, four, not two, times this. S squared minus B squared plus C squared divided by four C squared. A squared and C we can reduce. Okay. It would be easier for me if we will multiply and divide this by four. So, I will have 16 here as well. And here 16. So, A squared is equal to one-sixteenth times A squared C squared minus. I'm sorry. I forgot this. Minus A squared minus B squared plus C squared squared. Like this. Right? I realized I missed two because C squared times A squared is basically the fourth power of four. And without this two, I have power of two. So, which means definitely something is wrong. Actually, no, I made another mistake. I made four, but I put 16. So, it's four actually. I have to have four here. Like this, right? Now that's correct. One-sixteenths. So, it's four C squared A squared A squared C squared, right? Okay. This seems to be like fun. All right. Let's just transform it into this particular expression and I'll show you how. Equal to one-sixteenths. Now, this is the difference between two squares. This is two AC squared, and this is A squared minus B squared plus C squared again squared. Now, you know that X squared minus Y squared is equal to X minus Y times X plus Y, right? This is a simple formula. Just multiply it if you don't believe me. That's, you have to basically know by heart. So, the difference between these two squares, I will represent as the first one minus the second one times the first one plus the second one. Okay. Now, I don't like these parentheses, so I will, since this is the minus, I will change the sign of every one of them. So, A would be with minus, B would be with plus, and C would be with minus. Equals. All right. Now, let's think about this. These three members minus A squared minus C squared and plus two AC. Now, it's obviously A plus C, sorry, A minus C. A minus C squared. Correct? What is A minus C squared? A squared minus two AC plus C squared. Now, I have a minus in front, so this would be with minus, this was minus, and this was plus. And B squared remains. Here, two AC, this is plus, this is plus, and this is plus, so this is A plus C squared minus B squared. Right? Equals. Each of them, I will also represent as difference between two squares. So, this one will be this minus this, which is B minus A plus C, right? Because this is minus, this is minus, so B minus A minus C would be this. Times B plus A minus C. This would be difference between two squares. This is A plus C minus B, A plus C plus B. And what I will do is, instead of having one-sixteenths here, I will put divided by two, divided by two, divided by two, and divided by two. That's the same thing, right? Okay. We are almost there. Because A plus C plus B, this is half of the, divided by two, it's half the perimeter. So, this thing is equal to P. Now, A plus C minus B over two, A plus C minus B over two is equal to P minus B, because P is A plus B plus C over two minus B. If you will convert it into one fraction, you will have A plus B plus C minus two B divided by two, which is B minus two B, A plus C minus B over two, right? So, A plus C minus B is P minus B. Now, from symmetrical consideration, this is P minus C, and this is P minus E. And we have proven the formula. A square, the square of the area is equal to multiplication of these four elements. And we've got the formula, which was first obtained by a hero of Alexandria. The formula of a hero, or sometimes is hero, depending on some kind of etymology of this word. Sometimes his name is spelled as hero, sometimes as hero. All right. And the last problem is, okay, we have this formula that the area of any triangle is equal to square root of P, P minus A, P minus B, P minus C. And my last problem is, after we have gotten through all these calculations, it would be nice to check it in some trivial case. Okay, so let's say we have an equilateral triangle. So, A and A and A, they're all the same. Is the formula correct in this particular case? Well, let's just calculate the formula using this, and we will have what? We will have half of the perimeter is 3A divided by two, right? That's half of the perimeter. So, we have square root of 3A divided by two. Now, P minus A is 3A over 2 minus A, which is A over 2, right? So, it's A over 2. And same thing, P minus B because B is equal to A and C is equal to A. And what do we have? We have square root of 3 times A over 2 to the fourth degree, right? Which is A square over 2 times square root of 3. Square root of something in the fourth degree is, I'm sorry, it must be 2 square, which is actually 4. It's A over 2 square, that's why it's S square over 4. So, that's how it's supposed to be. This is the area of a triangle. Now, can we check it? Well, again, that's very easy actually, because in this particular case, my altitude is very easy to calculate. This is A over 2. So, H square is equal to A is equal to hypotenuse, which is A square minus this calculus, A over 2 square, which is A square minus A square over 4, which is 3 quarter of A square, right? Now, A we know, H we also know, so H is A over 2 square root of 3, right? You just take the square root of this. Now, the area is A times H divided by 2, which is A over 2 times A over 2 times square root of 3, which is A square over 4 square root of 3. So, as you see, in this trivial case of equilateral triangle, the formula obviously gives us the same result. So, this is just a checking, just a verification that in some trivial case, we do have exactly the same value, just to be on the same side. Okay, that actually concludes this particular lecture. I would actually recommend you to go through the Heron's formula again, I mean the proof of this formula again, because other problems are really trivial. I do recommend you to go through them, but that's actually trivial. But the formula of Heron is really very, very important. So, that's it for today. Thank you very much.