### CAT Coaching – Understanding Geometry – Part 2

All those properties, one is you can simply memorize them like this. This is a right angle triangle, this is cyclic so it will add up to 90. This is a right angle triangle where this will be equal, these two are equal. You drop it there is another right angle triangle. This is a tangent public theorem, drop it, these two are equal. Then this is 2 by 1, these two are equal. And these two are equal like this you can memorize. This is one approach. You can memorize as much as you can. And practice, if possible 500 geometry questions. You will do very well. You will do better if you are not doing it and if you take it at this level. You think this way, think of this way. If you’re looking at maximum quadrilaterals you can also have hexagons and all in a circle. We’ll see them later. And it’s not that it’s difficult because there are 6 points and not 4. It won’t become difficult because it’s still only 6 points. So, when you think of all the diagrams, what the properties can be related like this. Suppose you start like this. One is, 4 points are defined. I’ll just start with 4 because using which I’ll explain both quadrilaterals as well as triangles. And this is just to make you think because all those properties are more or less interrelated. Here, first you can prove them like this. If this is AB, this is AB, you can prove all these theorems by using these 4 words. That is just in the diagram itself, without bringing in the variables or anyhow. If I select this and if I just make it upside down, it will still be AB right? Now if you want to call it a new property write CD and tell that AB equal to CD. I am just proving the properties in the diagram itself. Now if you assume that here itself if you actually consider right angle triangle, right angle triangle now when you are making it upside down, this will be right angle triangle right? this will be right angle triangle and if you drop this line AB parallel and take A1B1 same length and same length here, rotate it 90 degrees, it will become like this and now it will become a square and a circle. Because we were considering only 90 degree angles in the original diagram. So we know that over here this will be equal to this. you can relate this in the way you want. There is not rule here. You can connect 1 to 9 and 9 to 1, if I had discussed 9 now.

So these two are equal. Or if I take this and a. Okay better will be like I will show it at much lower level. You take this and take 4 points. Name it ABCD, name it ABCD. Now I told you it’s solving is nothing but join the points. What is the meaning of solving? If your aim is to solve questions just nothing but join the points. And think of all the basic properties. So one which you can think of is when there are all the 4 points are on the circle, let me call it ABCD. One you can join the lines like this where, without crossing each other you are joining the 4 points. We will call it cyclic. So this is one property which you can think of. See it should be the other way around. When you see a circle and 4 points you should relate to this property or if you are crossing the lines like this then you know that these two are equal because you’re taking about the same arc here. AB is fixed. So these two are equal or you can also take it as, right, these two triangles are nothing but, these two are similar triangles. We will talk about similar triangles also but these two are similar triangles. That is, AXB and DXC, they are similar triangles. And we can also write that this, that is, BX into DX will be equal to AX into CX. It’s all based on similar triangles. Now these are all properties which you can just scan through when you see ABCD on the circle. That is, when 4 on the circle. Because you’ve already considered all the possible cases. One taking like this. And one taking like this. Now, special cases anyway we can define later. Now think of a scenario where, think of a scenario where you consider 3 of them on the circle. See first we considered 4 points on the circle. I am just asking you to think. Next think of 3 on the circle, 3 on 1 in. And that inner point, the one which can be easily defined is the center. It can be based on any point also, any point, but the properties are all which, properties are the ones which will work anywhere. Then only you can call it as a property. So now think of 3 on the circle next is 3 on and 1 inside the circle. One where one is nothing but the center. And connect all the properties here. You can even connect it with the previous ones where 4 are on the circle. That’s the best part. So if you consider this. I’m just defining 3 points here. One is here and 3 like this. Now, now join all the points. When you join all the points, this is one obvious one which you’ll take. This is 2 by 1. If somebody is asking why is it 2 by 1 then you can prove it by taking. Keep this AB same. Don’t change AB. This is C. Now, bring C somewhere here. Take it as C1. And join the points because I am not changing A and B. So what will happen is if this is 1, this is 1, this will be 2. This is why it is 2 by 1. Right? 1, 1 and 2. Now the best part is this now so the only question is this C1, sorry, AC1B and these two are anyway equal right? That we know. That is a previous property where we considered the 4 on the circle. So 4 on the circle can be directly related to 3, 1. 3 on the circle and 1 inside by connecting this. So this is 2 by 1. And by learning the property anyway we normally take it as diagram is like this 2 by 1. Now, 3 on the circle. Now the same 2 by 1, this is the best part, from this 2 by 1, let’s prove the next one. This is all I am just giving you proof in the diagram itself and which are very easy to interesting to understand. Now, move A and B away from each other in such a way that so push A and B away from each other and stop at a point where it’s taking its maximum length. We call it a diameter so it’s nothing but a special case of the previous property where this 2 will turn out to the 180 and this 1 will become 90. So these are not new properties. You try relating the way you want. That’s when you’ll really learn this. You can, you don’t need to really follow this order. Next class I’ll give you different order, but you don’t need to wait for that. You can learn 1 to 9, 9 to 1, 5 to 9 back to 1. Because they are all interrelated. And all these properties what I discussed till now are based on basically 4 points. And what is that you can think of? 4 on the circle, 3 on the circle 1 inside, 3 on the circle 1 outside. You’ll prove everything. Now, so right now we are discussing 3 on the circle and 1 inside right? 3 on the circle and 1 inside. So this, this is 90, this is 180, so this will be 90. There is nothing new in that. It’s the same 2 by 1. But special case where we know that it’s 180 because it’s a straight line. Now from the same diagram, from the same diagram if you think further. I am not teaching anything, I am just. This is how you should learn ideally. And then, now we were talking about, so A here, B here, C here right? This is 3 on the circle, 3 on the circle as of now. As of now it’s 3 on the circle. Now take 1 outside. Now that how it’s, how you can take that is this 3. Now this, this is the word. Move B closer to C. And we know that there won’t be any change because we are not changing this arc here. This will be same, B1 will be same, AB2C will be same. Now as of now this BC, B1C, B2C are, they are all secants. It’s cutting the circle at 2 points. Stop at a point where it’s, where, this is what you’re trying to do, make B equal to C. Then what will happen is the same line, so this is not changing right? This will be, this is still same. This is what we discuss as tangent-secant theorem. So this is tangent-secant theorem where we just in a question you will identify by looking at 1 outside and 3 on the circle. That’s how we should relate. If you just try learning through lines there’s an issue because lot of questions just to make it look difficult all the points will be crisscrossed while making the question. Just to make it look difficult. It can never be difficult. If the question is about circles and points. If the question is about circles and points we’re talking about all these basic properties, you call it 9, 10, 20, 30 that’s basically 1 or 2. Not even 1 or 2. You’re looking at 4 points and the circle. And different ways of placing them. There are not many right? 4 outside, 4 on the circle, 3 on the circle, 2 on the circle. 2 on the circle is. This is the other one right? Now if you make just 2 on the circle, 1 inside and 1 outside. All these things you’ve already seen. Right? And now if you have, if you’re looking at 4 outside the circle, I’ve already showed you all 4 outside the circle. Might look at like this. if it’s only 3 outside the circle, might look at like this. This might look like this. No, so, whether it’s length based, angle based, everything can be, angle based means you’ll have to bring this into the picture and further divide into, divide them into these points. Right? All the points wherever it’s touching these 3 are 3 kites. Angle based questions you’ll further divide and take angle bisectors. Then start solving them. Most importantly you can start making questions. And start predicting questions. Diagram based also its easy right? If you bring all of them together also, so, properties are all interrelated. Finally when you solve questions you should start seeing the diagrams at the point level. Start looking at the diagrams at the point level. You will start solving them really fast. Any number of properties you don’t need to worry. Because they are all interrelated whether you know it or not you can connect and start solving them. That’s it. Most of the angle based questions you can solve in multiple ways. So, now when you are solving questions also this is another, now all the 4 diagrams if this one ratio you can note down for while solving questions I will discuss that.

See question number, let’s solve some questions. What are the possible questions? Question types are: you can have length based questions, or, you know, area based questions. I am talking about plane geometry here. That is only up to 2-d right? One dimensional, two dimensional as of now. One dimensional questions, two dimensional questions. Can say specific questions as angle based questions or some more specific questions particular you can say heights and distances. Heights and distances are nothing but right angle triangles. Another word for this is right angle triangles. So, they can never be difficult if it’s about right angle triangles. This is its own chapters, the way it’s given in most of the books. This is a chapter right? Heights and distances. You need 5 minutes to complete the chapter. So, length based questions. Let’s look at what are the ways you can expect length based questions right? Maximum number of questions will be given by using this logic. Length into length is equal to some other ways of finding the same area. And while making the question, and while making the questions, so by solving this is what you should expect. Just one be left as an unknown. This is very, very important. Length based questions. This kind of a classification. Now instead of giving L3 into L4 on the right hand side, area can be given directly also. For a particular diagram. So using this you can make a length based question. It can also be same thing can also be like this. In words I can write it as a diagram can be cut into parts, so sum of parts equal to total area. So this looks very simple right? But if I put this logic in a question the question will look very difficult. It will look difficult actually, as far as CAT level is concerned. So, again, all these difficult terms are relatively difficult. It won’t be for you. It’s as per CAT level. So sum of parts equal to overall area is another logic. So these are very basic things. This is a typical question which you’ll never get because it’s so easy. If I write 25, 15 here, finding this distance you’ll never get. That’s 12. How do we find this out? By equating 2 different methods of finding the area. Half into 25 into h is half into 20 into 15. So, you can find h. There is an example for L1, L2, L3, L4 out of which one is not given. Now, it’s too easy a question for you to get in the paper. So, what they’ll do is when they are making the questions they will put the same diagram and put two extra circles which are actually not required just to make it look difficult. This is an actual CAT question. This is how the questions are made, how is it made difficult? Because it’s all based on 10th standard math, right? Max 11th standard. It’s made difficult by giving you things which are not necessary. Not just in math. It’s everywhere it’s like that, identifying what is relevant. What is important is all about solving, it’s all about cracking the exam. So, now the same question you can write it as distance between two centers is 25, where the circles are with radii 20 and 15. Find this common secant. What is it called? Common secant. So, because it’s cutting both the circles at two points. Finding this distance, this common secant is, it’s an actual CAT question. But what is different here? There is nothing different right? Just two circles are unnecessarily given. That’s the only difference here. So, answer is 12 here, 12 here. I am not making my questions. Unfortunately this is most of the times this is CAT level, most of the times. So, unfortunately or fortunately. Now this 12 plus 12 is 24 right? Now the second logic of sum of parts equal to overall area right? It’s actually the same logic, sum of parts equal to overall area. This is what you should expect. Question will be made like this. I am just making, or this is how you should make a question. Give x, y, z for a triangle. You need to give x, y, z. And put a circle, or a semi-circle over there. Now, find the radius. Not an easy question. It is easy but just made to look difficult. So, if x, y, z is given, finding this, this is r, is as easy as, what is the method? Cut the diagram. The method is cut the diagram. While solving, making questions this is what you’ll think of. Sum of parts equal to overall area. But solving questions is easy. Just cut the diagram and equate the same thing. Then how will you solve? See, we don’t need to solve questions here. You can expect them. If x, y, z is given, indirectly what is given? Area is given. What area is given? Overall area of this triangle, overall area of this triangle ABC is given, or you can find out. I just need to give these 3 sides. x, y, z given means you can you know how to find the area of a triangle right? Root of s into s minus a s minus b s minus c. Now, so after getting the area so you have the area on the right hand side. What is not given? x is given, y is given. What is not given? r is not given. That’s what you need to find out. And how do you find out? Just cut the diagram. This is, you cut the diagram, it’s over. This is what we call solving. You might say, just to make it look math you can say join A to this point. But, now how is it over? This is just logic right? It’s not cont it’s not math, it’s just logic. It’s over because you have on the left hand side you have half into x into r plus half into y into r. x is given, y is given. Only r is left out. This is how you make a slightly difficult question. This is not easy. Easy questions are not worth mentioning anywhere. This is what you call difficult. You write this in words it’s a difficult geometry question in CAT. Don’t tell me it’s an easy one. If you see a paper with 4 geometry questions, this will be the most difficult in that paper. Finding this out. Finding the radius of a, the maximum semi-circle which can be fit into a triangle with given sides x, y, z. Or you can think about circle and a square, sorry, triangle and a square. They are all possible questions. But finally see when you see a diagram what is the most important thing is you being able to read this as the triangle and another triangle inside. So they are similar and it’s over. Or if I show you a question like this. These are all similar triangles, all of them right? You’ve something like this, something like this, something like this. They are all similar triangles. Now to make it look difficult some circles and all will be put here and there and all. But seeing it at this level is what matters. Triangle in a triangle in a triangle and all. They are all similar. So similar triangles also I’ll discuss like that only in just through diagrams not through SAS, SSS and all. This is how you get similar triangles right? Either like this or like this. They are all similar triangles. So that is part of your one of the coming sessions. And it’s just a 10-15 minute discussion. Your similarity what we have wasted a lot of time is over no time. Only in diagrams. So, but here let me just. I am discussing how to expect length based questions. So L1L2 equal to L3L4. So see question number 1. That is an example for what I have discussed just now. All the initial sessions are mainly from a conceptual point of view just to make you think in multiple directions. Completing the sheet will happen in the coming sessions. So, more than solving the questions made by me, it’s better to make your own questions or to reach that level.

So ABC is a, PQR is a triangle. Equilateral triangle. You can take any point inside and then this is what you need to find out. Right? Diagram is given. Side is given as 5. That means the right hand side is already given right? That’s easy. √3 by 4 52. Now, looks a very different question than what I have just explained right now but logically you can think on the same lines. Cut the sum of parts equal to overall area. It’s over. So the only thing which you are doing is cut the diagram. You cut the diagram it’s over. Now you can think of √3 by 4 52 as sum of, sum of these three triangles right? You have three triangles here. And you are supposed to find out this OL, OM and OR. And when you are equating this sum of parts what is POR? POR is half into 5 into OL, on the left hand side. QOR is half into 5 into OR and POR is half into 5 into OM. What is the only unknown on the left hand side? Half into 5 into OL, OM and ON. And this is the required answer. Your aim is to find the sum of these 3 distances right? OM, ON and OL. So, the logic is the same. You cut the diagram, solve the question. By? What is the logic? Sum of parts. Cut the diagram means it’s nothing but sum of parts equal to overall area. And while making the question you either need to give the area or you either give the area or you can give enough data so they can find the area using some other method. And, don’t give everything else then it won’t be a question. Just give one of the L values as an unknown. One of the unknown, L values as an unknown value. So this length into length equating with some other area this is another supposedly difficult question. Area of a triangle is 80. So one side is given as 20, the other side is given as 10. Find the 3rd side. This is also a, this is about equating different methods of finding the area. But in a diagram if you do that, half into 20 into h is 80. So h will be 8. h is equal to 8. This is an actual question. A better one I will say. So, now this is a right angle triangle, so 10, 8, this will be 6. So this will be replaced by 14. This is another right angle triangle so x is nothing but root of 196 plus 64. This is a difficult question. What is difficult? Most of the people as soon as they read they won’t visualize. What they will try doing is they will try writing root of s into s minus a into s minus b into s minus c. And will take s as 30 plus x by 2. And then you will get the answer but you will end up wasting a lot of time. Which is as easy as including the people who write your books, they do that. This is an actual question and that’s how the solution is given in all the books. Even though the question is as easy as 20 here, 10 here, x here. This dropping a perpendicular is what people don’t even think. Because it’s very difficult. And when we know that, see, there is nothing much we need to expect in a triangle right, there are only few lines which you need to drop. It’s called altitudes, medians, perpendicular bisectors and angle bisectors, and nothing else. At the level in which you get questions you don’t need to expect anything else. Think of all those points, all those lines this will become a direct question. This is not an easy one but it’s very easy now. Area is 80. The data given was the area was given, these two lengths were given. So once you drop a perpendicular you’ll get this as 8. You’ll immediately write this as 6 because 10, 8, 6 is a, 10, 8, 6 will match with x2 + y2 = z2. And remaining distance is 14 right? So this is not basic very basic Pythagoras theorem question. 8th standard level? So 260. There is one more good question based on the same logic right? It’s a length based question L1L2, L3L4. L3L4 can also be directly given as area. Now one more, I am just giving you questions how can think of. Now there is a triangle. Same triangle which I have given before. 25, 20 and 15. Now this is a supposedly difficult question. So we have two triangles here. Two more right angle triangles, ABC is one triangle, ADC is another triangle. Find the sum of inradius in these two triangles. I will say this is one of the most difficult questions, in a, from a previous paper. Time when I, difficult is wrong, time consuming ones. But the fact is now this r1, r2 is again by, what is r1? That’s again length right? Some length. So equate r into s with, we know how to find the area right, of these triangles. Because we know this line is 12. If that is 12, this will be 9 and this will be 16. So all the, everything is given now. Both are right angle triangles so it’s nothing but equating, equating two different methods of finding the area. And it’s all the same right? I am showing you so many questions across, they are all based on the logic is the same, equating two different methods of finding the area. So now how do we find this inradius here? Inradius how will you find out? We know how to find the area. Using half into product of perpendicular sides. Half into 9 into 12. So that’s 54. That is 54 right? 54 is also equal to r1 into s, or r1 into s1. What is r1? That is inradius. So, what is s1? 15 plus 12 plus 9 by 2, which is 18. So r1 is equal to 3. Same way you can find out r2 also. Time consuming but it’s not difficult anyway. Because you know that it’s a length based question so you know that it’s about equating areas. Same area but in different ways. So how do you find r2? You have an in circle over there also. Here r2s2 equal to, what is the area? Half into 12 into 16, which is 96. What is s2? What is s2? Semi perimeter right? So 20 plus 16 plus 12. That is 48 by 2 is 24. So what is r2? r2 is 4. So answer is 3 plus 4 equal to 7. This is an actual question. Told you sometimes you get questions at this level also. But I don’t think you’ll consider this difficult. Now, these patterns are very ee, you won’t solve it like this right? Because these are all direct patterns. Finally you’ll learn this fast that 3, 4, 5. r will be 1 and R will be 2.5. This we’ll do it only towards the end. And if it is 6, 8, 10, that means its 2 and it’s 5. And if it is 9, 12, 15, these are all Pythagorean triplets with the same ratio of 3 is to 4 is to 5. Then this will be 3 and this will be 7.5. And if it is 12, 16, 20. Inradius will be 4, circumradius will be 10. That’s with a multiple of 2.5. And in 25, that is 15, 20, 25, this will be 5 and this will be 12.5. So, by the time you take the paper you’ll take only this much time. You see this triangle, see this triangle, 3 plus 4, answer is 7. Faster than the guy who is making up the paper. 100%. 99% because some of them will anyway know these patterns. So, but most of them won’t. Answer is 3 plus 4, 7. Now, I am talking about 3 is to 4 is to 5, right angle triangles. They are not for beginning. Finally in pattern workshop this is exactly what we will do. And the scope for explaining in more and more patterns is very limited because it’s much easier than what you think in terms of predicting. If I tell you that this triangle itself, this is the this is what you will see maximum number of times in all papers put together. The triangle where even the numbers are fixed. This is the most frequently asked right angle triangle values in previous CAT papers. Most frequently asked. I’ve already discussed enough 2-3 questions based on these values itself. There is a reason because once you take a triangle like this, 25, 20, 15, even the inner ones are right angle triangles with the same ratio. So while making questions lot of scope. But this is not what you need to learn. I am not teaching any pattern. I am teaching you how to actually predict questions at a much lower level. So that it will cover everything. Now, so length based question is what I am discussing right? Now one more thing, which is going back to question number 1. This is question number 1, right? Have we got the answer after doing all these predictions? Even that’s not required because that’s a question which you don’t need to predict. Because of this word. Effectively, what is given you read that question once more then you will see this word explicitly given. Telling you do whatever you want. It means do whatever you want. So that is why, and now you see with the same set of options which was given, marking the answer you might take more time than getting the answer. Because you will simply push this point here, question is over. This is, this is also common sense. It’s not, nothing more than common sense. You are not, there is not much scope for too much thinking at all. This is given explicitly. And then if you are trying to follow all those predictions and all, what we have discussed you are wasting time. You can take any point right. That’s the question, any point. So the best point to be taken is, take P equal to O. So that, finally you see, always ask the question in words. What is the question? You need to find the cumulative distance from this point to all the 3 sides. Now this is a point on this line so there is no distance, only that’s zero. It’s on this line also, so that’s zero plus zero. And to this distance we know this is a standard distance. What is that distance, it is called h, and what is h? √3 by 2 into side. This is 100% faster than the guy who is making your paper, if you are solving like this. And don’t tell me it’s difficult to think like this. There is nothing to think. So, it’s because it’s a flexible question. Now why is it a flexible question? Because you can take any point. Yeah you might tell that you can’t take this point because it’s theoretically wrong because it’s given that it’s inside the triangle and all. I agree. That’s what some of you are thinking right now. I have an answer for those who have that particular doubt. That’s the only doubt which you can have. There is no scope for any doubt. Don’t ask, please don’t start asking doubts in geometry and all. There are areas where you can but not in geometry. Because you can approximately take P equal to O. It’s, you are sitting so far from me, it’s not P equal to O. It’s infinitely close to P. That is why answer is infinitely close to zero, infinitely close to zero, and infinitely close to √3 by 2 into 5. So we will mark the answer as √3 by 2 into 5. I agree it’s a compromise but that’s what is required. So. Okay so that’s an easy question, but a flexible question. And both its one and the same. Flexible means easy.