Unlocking geometry's toughest unsolved riddle.

This is not an easy one. Yes, but we can do it together. Welcome back to my channel. First, this is not drawn to scale. It's just a freehand sketch. In this particular diagram, the length from here to here is 30, from here to here is 28, from here to here is 26. And this is a circle, okay? A circle. And we are asked to find the area of this circle. Note that this line is tangent to this circle at this point, this line is tangent to this circle at this point, and this line is tangent to this circle at this point. Alright, now first let's name this triangle ABC, right? And we are going to start by finding the area of this triangle. We are going to make use of this Heron's method. The area of a triangle is the square root of S bracket S minus A bracket S minus B bracket S minus C, where your S is the semi-perimeter, okay? S is the semi-perimeter. So S, being the semi-perimeter, is equal to 26 plus 30 plus 28 divided by 2. Alright, now 26 plus 30 will give us 56. 56 plus 28, this is 14, carry 1, will give us 84, okay? So we have 84 divided by 2 as S, which will give us 42. So it means that our S is 42. Okay, therefore the area of triangle ABC will be equal to the square root of 42 bracket 42 minus, now A is the side facing this place, right? Facing the angle A. So it means that A is 26 bracket 42 minus B is the side facing angle B, which is 30 bracket 42 minus C. And our C is the side facing angle C, which is 28. Alright, so area of triangle ABC will be equal to the square root of 42 bracket, now 42 minus 26 will give us 16, right? Yeah, it should be 16 bracket 42 minus 30 will give us 12 multiplied by 42 minus 28. 12 minus 8 is 4 and 3 minus 2 is 1. So we have 14. Alright, now triangle ABC, sorry, area of triangle ABC is equal to the square root of 42 bracket, now 16 times 12 times 14, I think I need to get a calculator so that this will be faster, okay? So 16 times 12 times 14 will give us 16 times 12 times 14 will give us 2688. So let's multiply it by 42 at once. 2688 multiplied by 42 will give us 112. So we have 112 comma 896. Now let's find the square root, okay? So square root of 112, 896 will give us 336. So it means that area of triangle ABC is equal to 336 square units, okay? Now, we are going to write this somewhere, even though you can't see it, I know, but I need to write it somewhere, okay? Area of triangle ABC is 336. Now, if this is the first time you have seen this amazing channel, hit the Yes, so that you don't miss our amazing videos. We upload our videos as often as possible. Tune in at your convenience. You must see something new to watch and give us a thumbs up if you love what we are doing. And I did something, I turned off my auto-dubbing. Let me know your experience in the comments. Do you like it or should I continue with auto-dubbing? All right. Okay. Now look at this. For those that don't know what auto-dubbing is, it means translating my voice into different languages so that everybody will be able to understand what I'm saying. So if, let me know your experience with that on the comments, okay? I've switched it off here because people were complaining that it's actually, some complained that it gives you wrong translation. So I had to switch it off. So let me know what you think or your experience with that. Okay. Now we have found area of triangle ABC. It leads me to clean everything and let us find a way around this thing. Remember that to find the area of this circle, we need to get the radius and that is all I'm trying to do. Okay. So all this, my running around is to get the radius of this circle. All right. Now let us call the center of this circle O. And if we draw a line from the center to the point of contact with this tangent, okay? Remember that this will give us 90 degrees. I will explain why. So, and this is the radius. If we draw another straight line from the center to this point, remember this is another point of contact. Okay. Okay. I'm sorry about that. Oh, what I'm trying to do here is for you to note that this line is tangent to this circle. Okay. The line is tangent to this circle. So this place is also 90 degrees and this is also the radius. Now also this place to this place, we have another 90 degrees. Okay. And this is also the radius of this circle. Now, remember that in a circle of center zero, if you have a tangent to the circle, any line, any straight line from the center to the point of contact between the circle and the tangent, you are going to form 90 degrees. And any line from the center to the circumference is radius. All right. So that is why this is 90, this is 90, and this is 90. Now using external tangent theorem, excuse me, using external tangent theorem, it says that if in a circle, you have two tangents that meet at point P, let's call this tangent AP and call this tangent BP. This external tangent theorem says that from their point of contact, okay, these two lines are equal. Yes. So we are going to apply that here. Now let us call this point M. Okay. And this is C. Remember that this line from C to B is 26. So I'm going to shift my hand to write it, but know that this is for C to B. All right. All right. Now let's call this point P. Let's call this point P. Remember that the line is 26 units. If we call this MC A. All right. Now it simply means that PC is also A. Yes. Remember that this is a freehand sketch. So you may not really understand or see it clearly. But if MC is A, then CP is A. Using this theorem, whereby C is the external point, MC is the first tangent, and CP is the second tangent. All right. Similarly, remember that BP will now be 26 minus A, right? Yeah. So right here, this place is 26 minus A. And that means that from here to here, let's call this place K. So it means that KB is also 26 minus A. All right. Using this particular theorem. Yes. Because B becomes their point of contact. BP and KB are tangents. Okay. Now, having said that, you can also see that MA is a tangent and KA is also a tangent. So using this theorem, MA is equal to KA. All right. Now, MA is A plus 30. A plus 30 is equal to KA is 26 minus A plus 28. All right. So if we solve this, we will have A plus 30 is equal to 26 plus 28. We give us 8 plus 6 is 2. 8 plus 6 is 14. Sorry. Right 4, carry 1. Then 2 plus 2 is 4 plus 1, 5. So we have 54. Therefore, A will be 54 minus 30. And that will give us 24. Okay. So it means that, sorry, we have A plus 30 is equal to, so it's going to be 54 minus A. So we have that, excuse me. So this A will cross to become A plus A, right? So we have 2A is equal to 54 minus 30. We give us 24. Okay. So we now have that A is equal to 12. Now, if A is equal to 12, then permit me to replace this accordingly. So it means that MC is 12. So we have 12 here. Then CP is 12 as well, right? Now PB, BP will be 26 minus A. So BP will be 26 minus 12, which will give us 14. Okay. So I'll replace here with 14. Also that means this place is 14 as well. All right. Now we are going to take a different dimension. I tell you that this is not an easy one, but just pay attention. You will understand what I'm doing because we are getting closer to our answer. Once you're able to get the radius of this particular circle, then we have gotten the area, right? So let us go ahead and do some addition and subtractions and get the value of our radius. Now, if you look at what we have on the board, I want you to look at OMCP. If we decide to divide it into two equal parts, okay, you can see that we have triangle OMC. So in triangle OMC, the area will be half multiplied by the base. The base is 12 and the height is our radius. You can see that. So you can see that we have 6R. That is area of triangle OMC is equal to 6R. Now, since area of triangle OMC is equal to 6R, and you know that OCP and OMC, that they are congruence. Therefore, area of triangle OCP will also be equal to 6R. Similarly, area of triangle, let us come to this side. Let's divide this into two again. So area of triangle OPB will be equal to half multiplied by the base is 14 and the height is our R as well. So you divide this, you have 7R. So this is 7R. That is area of triangle OPB. And since it is congruent to triangle OKB, area of triangle OKB will also be equal to 7R. So if we add everything here, the area of this side, that is from here, everything. So we are going to have that area of, I think we do it this way. Now, area of this particular quadrant A, K, M, K, O, M, A. Okay. Area of the quadrant K, O, M, A. Let us write like that. We will now be equal to, we have gotten the area of this particular triangle. That is area of triangle ABC plus all these areas that we got from here, right? So it's going to be plus 6R. 6R plus 6R. That is two brackets 6R plus two brackets 7R. This 7R and this 7R. Okay. I just hope you are with me. Then we have area of this quadrant K, O, M, A will be equal to area of triangle ABC is 336. And this will give us 12R plus 14R. So we have area of the quadrant of the quadrilateral K, O, M, A is equal to 336 plus 26R. So we are going to keep this somewhere. Okay. So I'm going to clean everything, clean everything and write it somewhere. All right. Area of the quadrant K, O, M, A is equal to 336 plus 26R. Very important. All right. I know what you are thinking. This particular video seems to be long, but it's worth it. I told you the question is not that easy. Okay. So let's try to learn together. I actually saw this in a page and I decided to share it with you. All right. Okay. Now this is what we have, but I want us to clean all these things now. Let us clean them. Let's clean them. All right. All right. Now, having done that, let us take from here to here. Let's draw a line that separates this side from this side, or let's divide this particular quadrant, quadrant around into two. So we're going to draw a diagonal. All right. Now, having drawn this line, I want you to notice something. We have shared this quadrant around into two, but even though we shared it into two, we have two right angle triangle that are equal. Okay. So we have triangle O, M, A. Now let's find its area. So area of triangle O, M, A will be equal to half multiplied by the base. Now the base is 12 times 30, 12 plus 30, which is 42, multiplied by the height. The height is R. Okay. Okay. Now, so the area of O, M, A, triangle O, M, A will be, this will divide this to give us 21. So we have 21 R. Remember that triangle O, M, A is congruent to triangle K, O, A. Therefore, the area of the quadrant, everything now will be 21 R multiplied by two. So that is for area of the quadrant K, O, M, A will be equal to two times 21 R. Yes, because triangle O, M, A is equal to triangle K, O, A. Therefore, their areas are equal as well. All right. Now, but you already have that the quadrant around K, O, M, A, the area is also this. So if we call this equation one and this equation two, and both of them are equal to the same thing, it simply means that 336 plus 26 R is equal to two times 21 is 41, 42 R. Okay. If this is equal to this, then this is equal to this. Now let's take this to the other side. So we'll have 336 is equal to 42 R minus 26 R. We're almost there. So if we solve for that, we have 336 is equal to 12 minus six is six and three minus two is one. So we have 16 R. We are there. So to get R, we divide 336 by 16. So when 16 divide this, we have two, two remainder one, divide 16, we have one. So it means that our radius is 21. Therefore, area of the circle. So area of the circle is equal to pi R squared, which is equal to pi multiplied by 21 squared, which is equal to 441 square, 441 pi square units, if I'm not mistaken, because 21 squared is 441, right? Yeah, I think so. All right. Thank you so much for watching and see you in my next video. Bye.