 Okay, six people have joined. So, let's start the class. What we need to do, I think we discussed that we need to do surface area and volume and all those things. Can you just remind me once? Can you guys hear me? Write your name, why are you not writing your name in the chat box? Okay, let's start with the class. Nobody is replying. Okay, fine, good. Sraddha, we have done enough question on logarithms, I think. And it's anyway there in the second, this thing. We have already covered your paper once, syllabus. So, revision I do just before your examination. This particular lesson, let's donate to doing something which is required for your next paper. Anyway, when you are asking logarithm questions, let me just find out a few logarithm questions. Okay, write down this question. Log 10, 100 plus log three, 27 plus log five base 25 is equal to x, find the value of x. Okay, fine, I will do area of parallelogram. After solving two, three questions, I will be solving all of you. Okay, answer is 5.5, so this is log 10, and this is 10 square, this is log three, this is three cube, this is log five square five. So, this is nothing but this two comes out here, this three comes out here, and one and a half comes out here. This is nothing but five and a half. So, that's the answer. Okay, let's solve next question. Okay, let me first teach you because area of parallelogram will take a lot of time. So, let me first teach you area of parallelogram. So, it's chapter number 11 in your book perhaps, chapter number 10 or 11, just look at it, RS Agarwalits, areas, and there are few axioms given over there, just go through it. I'll start theorems once you go through axioms. Okay, so now let me go to the first, I hope you have gone through the axioms, area axioms. The first theorem is, theorem one which I'm trying to prove is a diagonal of a parallelogram divides it into two equal, into two equal divides it into two triangles equal areas. So, let me draw a parallelogram for you. Do you want some time to prove it? Do you want to apply your brain? It's just similarity question. Take one minute to solve this, try to prove similarity. Okay, take two minutes, no issues. Yes, that is a question of congruency, huge congruency. In congruent triangle, the fundamental concept is what do you mean by congruent triangle? Congruent triangle means when we say that one triangle or two congruent figures means when one figure can superimpose second figure, they become congruent figure. What do you mean by superimposition? It means that exactly same figure, two exactly same figure. So if we have two exactly same figures, it means that area of those two figures would also be same. So as soon as the word congruency is getting associated with any two figures, one thing which lies by default is that the area of those two figures would be same. So wherever in two triangles I'm proving congruency, I'm also proving that area of those two triangles are same. So in parallelogram, what happens is, so A, B, C, D is a parallelogram and I have to prove that area of triangle A, B, D is equal to area of triangle B, C, D. So how will I prove that? So let me take these two triangles. In those two triangles, A, B is equal to D, C because opposite sides are parallel and equal. B, C is equal to A, D. In both the case, opposite sides equal. B, D is equal to B, D. This is common. So I have side, side, side congruency. If two triangles are congruent, then the area will also be equal. So that's how I prove it. Okay, let me go to another theorem. So now by taking this theorem up, now what is altitude? So altitude anywhere, anywhere if you want altitude, suppose it's like this, parallelogram is like this. Y, D is altitude, anywhere altitude is a line drawn from the vertex to the opposite side and it should be perpendicular to the side. So perpendicular drawn to the opposite side from a vertex is known as altitude. And why I'm telling you the meaning of altitude because in parallelogram or in triangle, while finding out the area of any particular triangle or parallelogram, we sometimes utilize the concept of altitude. So I'll let you know what's the formula of parallelogram, area of parallelogram or triangle. So once I have done this, the second theorem is that parallelograms on the same base between two same parallel, between the same parallel lines equal in area. So guys try to understand how do I prove it? So I take this as base and this is, and this is, so suppose this is AB and this is CF. And what I do is that AB is the base of the two parallelograms which I'm trying to, which I'm going to construct and AB and CF are parallel lines between which these parallelograms will exist. So one parallelogram is like this. I'm making here. And one parallelogram is like this. And suppose this is D, this is E. Because in your book, this is E, so I'm making a D. So this is D. So what are the two parallelograms? Try to understand. The two parallelograms are enough to prove that area of parallelogram, AB, E, F is equal to area of parallelogram, AB, CF. Now try to understand area of parallelogram, AB, E, F is this area, area of quadrilateral, I'm writing quad, AB, E, D plus area of triangle, AD, F. And area of quadrilateral, AB, CD would be, again, area of quadrilateral, AB, E, D plus area of triangle, B, C, E. Now you see here on both the sides, area of quadrilateral, AB, ED is common. So if I have to prove that both the areas are same, then area of ADF and area of BC has to be proved correct. Sorry, has to be proved same. How do I prove that area of ADF and area of BC would be same? There can be only one possibility. Whenever you have to prove that area of two triangles are same and nothing has been, I mean, no numerical parameters have been given to you. Then only one thing is left out in your hand that is proving that the two triangles are congruent. So I have to prove that the two triangles are congruent. So see here, in triangle ADF, AF is equal to BD of triangle BDC and in triangle ADF, AD is equal to BC. And look at here, this side is parallel to this side and this side is parallel to this side. So angle between two same sets of parallel lines would always be same. So angle BCE would be equal to angle, sorry, not BC, I'm sorry, angle CDE would be equal to angle DAF. Why I'm taking this angle equal? I'm taking this angle equal because there are angles between two same sets of parallel lines. So angle between two same sets of parallel lines. Is it okay? So now try to understand when you have proved that what kind of congruency is this? This is SAS congruency. So I have proved that area of ADF and area of BCE are equal and this is common to both of them. So hence I'm proving that on the same base and between two same parallel lines whichever parallelogram you will make, the area of those parallelograms would be equal. Now let me go to few extensions of the theorems we have already proved. We have two theorems and those theorems are, first theorem was that diagonal divides parallelogram into two equal parts and second theorem is on same base and between two same parallel lines. If you draw any number of parallelogram, the area of all those parallelograms would be same. So now let me go to the third concept. So this concept is to find out area of parallelogram and how do I find area of parallelogram? So try to understand. Here I have to prove that this is A, B, C, D and I have to prove suppose I am drawing a perpendicular or altitude from vertex A to line D, C and this is L. So a parallel to L I am drawing a perpendicular from B and this is M. So try to understand. I have to prove that area of parallelogram A, B, C, D is equal to A, B into A, L and how do I prove it? So try to understand I have a rectangle and what is this rectangle? So try to look at the figure very properly. The rectangle is, sorry, not parallelogram, rectangle is B, M, L. So what is this rectangle? This rectangle A, B, M, L, the area would be A, B into A, L. Is it okay? And if this is A, B into A, L then try to understand on the same base and between two same parallel lines. If you draw two parallelograms so you have to understand that rectangle is a type of parallelogram. So area of parallelogram A, B, C, D would be equal to, this is area A, B, M, L would be equal to area parallelogram A, B, C, D. Why? Because they are between same base and same parallel line. So that's how you can prove it. Now how can you prove that? Now try to understand. Just go through the theorem. I don't know which theorem it is in your book. So there is one theorem in your book which is triangles on the same base between the same parallel lines are equal in area. So go to this theorem and try to prove it. Triangles on the same base and between two same parallel lines will have same area. Try to prove this theorem. Have you proved it? Okay fine. Have you proved it? Can you guys reply quickly? Are you guys listening to me or not? So suppose this is one parallel line and this is another parallel line. So I make one triangle like this. So my one triangle is A, B, C. My another triangle is B, D, C. So triangle A, B, C and triangle B, D, C. So how do you solve it? So try to understand. Let me try to make a parallelogram. So one parallelogram, one line I'll draw parallel to this. So I draw parallel to this is construction and I am drawing this BF construction is such that BF is parallel to DC. And if BF is parallel to DC, then what happens is that now you should understand that which this BF and DC parallel will make a parallelogram B, C, D, F. So this is one parallelogram and I'm doing one construction here. So this is my E. And this construction is such that CE is parallel to AB and my parallelogram is A, B, C, E. So what I do is that, what I do is that I know that both the parallelograms B, C, D, F area of parallelogram B, C, D, F is equal to area of parallelogram A, B, C, E. Why? Because same base and between same parallel lines. So what happens? I have already proved in first theorem that diagonal will divide the area into two equal parts. So let's find out half of this area, half of area B, C, D, F would be equal to half of area A, B, C, E. And you see here in this particular parallelogram BD is the diagonal. So if BD is diagonal, half of the area would be area of triangle BDC. So this becomes area triangle BDC and here AC is the, in this particular case AC is the diagonal. So half of the area would be area of triangle ABC. So this is what you needed to prove. Now how do you prove that area of a triangle is 1 by 2 into height into, these are all same things which I have been discussing for last half an hour. So I have to prove that, suppose this is my triangle ABC and I draw perpendicular here which is AM. So how do I prove? So I make a parallelogram and if I make a parallelogram suppose this is D. So area of parallelogram ABCD would be equal to AM into BC. Altitude into base which I just proved. So area of triangle would be area of triangle ABC would be half of area of parallelogram ABCD. So it is equal to 1 by 2 AM into BC. So that is how you prove it. Now there is another theorem. So that theorem is, no that's not a theorem. So this I have already proved. Now there is one thing more that you need to do and this is a trapezius. I have to prove that. So this is my trapezius. In trapezius two sides are parallel to each other. That's the only condition given and what happens is I have to prove that area of trapezius would be equal to 1 by 2 into sum of the parallel side, sum of the parallel sides multiplied by altitude. So I am drawing an altitude from this place here and suppose this is CM. So I have to prove that it is equal to 1 by 2 AB plus CD into CM. And how do I prove it? So again I will draw perpendicular to this. What I'll do is I'll draw perpendicular to this. I'll draw, I'll make a construction. This is 90 degrees. Suppose this is L where AL is parallel to CM. So how does it help me out? So you should understand that from here what I'll do is I'll draw a diagonal. Another construction is diagonal AC. So area of trapezius ABCD is equal to, look at here guys, that would be equal to area of triangle ABC plus area of triangle ACD. What will be area of triangle ABC? So it would be 1 by 2 into AB into AM plus an area of triangle ACD would be base DC and height AL. So 1 by 2 into CD into height AL. Now AL is equal to CM because and two same parallel lines. So distance between parallel lines, perpendicular distance between two parallel lines will always be same. So AL is equal to CM. So suppose AL and CM is equal to H. So I'm writing here 1 by 2 AB plus CD. CM I have assumed to be H, so H. So take H common. So 1 by 2, these both are H. So I'm taking 1 by 2 H common in bracket you get AB plus CD. So that's how you prove it. Now your homework is, now you will prove, I will not prove it. You will prove that area of the rhombus is equal to 1 by 2 into product of its two diagonals. You will prove that area of the rhombus is equal to 1 by 2 into product of its two diagonals. Have you done it? Okay. So let me just prove it for you guys. And then I'll start with solving examples. So suppose I make a diagonal and diagonal is AB, CD. And this is 90 degree. This is 90 degree. This is 0.0. So area of diagonal, sorry, area of rhombus. I'm sorry. Area of rhombus AB, CD is equal to area of triangle ABC plus area of triangle BCD. Now area of triangle would be equal to area of triangle ABC would be equal to 1 by 2 into OA into BC. And area of triangle BCD would be equal to 1 by 2 into OD into BC. I take 1 by 2 BC common. I have OA plus OD. OA plus OD is equal to AD. So that is equal to 1 by 2 BC into AD. So that's how that's what your answer should be. So once you have done this question. Let me go to the first question of the example. First question of the example is suppose a parallelogram is here. ABCD. So I'm saying that ABCD is equal to 1 parallelogram. I have midpoint EF. So an F are midpoints AB and CD. You have to prove that area AEFC is equal to area EBCF. Just try proving it. Have you done it? Guys you guys are not writing anything on chat. I'm just speaking from this side. Okay fine. So what in this particular question we need to do is that we need to show for this. I show that AE is equal to DF and AD is equal to EF. So it is a parallelogram. Similarly this EF is equal to BC and FC is equal to EB. So this is also a parallelogram. If I prove that both of them are parallelograms. So there is a theorem which I discussed that same base and same parallel line. Two parallelograms will have equal area. So that's what you have to do. Next is let me go to the second. So go to the second example of your book. I'm not writing it anymore. So go to second example of your book and start solving it. And let me know. Once you solve the question let me know. So that I can move ahead to the third question, fourth question likewise. All the examples of this chapter are very important. Are you done with second question? I'm done with all the theorems. Now only thing left out in the class is doing as many questions as possible. And I'm focusing on examples because good questions are there in the examples which we cannot neglect. Nobody is saying me anything. Have you done second question? Guys, you guys are not responding at all. Okay. Okay, fine. I'll do this some. So A, B, C, D is a parallelogram. The question tells that in the adjoining figure A, B, C, D is a parallelogram whose diagonals A, C and B, D intersect at point O. Line segment through point O meet A, B at P and D, C at Q. I have to show that area of A, P, Q, D is half of area of parallelogram A, B, C, D. It's the same question as the second first one. So this is my parallelogram. And these are my diagonals. So A, B, C, D. And this is O. And through it, a line passes through. And this gives me Q, P. And the shaded part I have to show that it is half of area of parallelogram A, B, C, D. So how do I prove it? So try to understand. Actually, if you look at this question, they have tried to prove that area of this part, this P, Q. And they have tried to prove that area of parallelogram A, P, Q, D, or area of quadrilateral, I'll say, is equal to area of triangle A. What they are trying to do over here is, okay, let me do one thing. Let me not do it with this method. What I have is try to understand. If I prove that, if I prove a correlation, so these three lines divides this quadrilateral into six parts. So this is part one, part two, part three. And this is part four, part five, part six. If I prove that one is equal to four, two is equal to five, and three is equal to six. So I will prove that this part is half of the other part. So I'm just trying to prove that. I'm trying to prove that one is equal to four. So I'm taking triangle OAP and I'm taking triangle OQC. If I take these two triangles, so try to understand OA will be equal to OC. Why? Because they are diagonals bisect each other. So parallelogram diagonal will bisect each other. Then this angle is equal to this angle. So angle QOC is equal to angle AOP. Why? Because they are opposite angles. And you see here, this is a transversal line cutting two parallel lines AB and DC. So this angle, angle OQC is, one second guys, my handwriting is, angle OQC would be equal to this angle, which is angle, angle APO, so they are alternate angles. So I have congruency. So ASA congruency I have here. So what I do is that I prove that these two areas are congruent. Is it okay? So these two areas are congruent. It means that one is equal to four. Similarly, I can prove that OD is equal to, in three and six, I can prove that OD is equal to OB. And this angle would be equal to this angle and this angle would be equal to this angle. So similarly, I can prove that three is equal to six. And here I don't need to prove anything in two and five. This is equal to this. This is equal to this. And these two sides are equal. So side, side, side congruency will come. So two will be equal to five. So all three congruencies can be proved. And as soon as you prove congruency one, two, three is one is equal to four, two is equal to five, three is equal to six. So it means that one plus two plus three plus four plus five plus six is equal to area of parallelogram. And this is equal. So I can write two times one plus two plus three is equal to area of parallelogram. So one plus two plus three would be half of area of parallelogram. So that's how you can prove it. Now solve third question. If you did not understand, let me know. So third question, let me know if there is any difficulty. Have you done question number three? Okay, great. Go to question number four then. Question number four, start solving. Have you done question number four? Question number four, any doubt? Let me know. I'll solve question number four. Question number four. Okay. Fine, I'll do it. No issues. So question number four is, where is question number four? Okay, in a quadrilateral, I am solving example four. Okay. In a quadrilateral ABCD, it is being given M is the midpoint of AC. So some kind of quadrilateral has been made over here. There is one line going on this side. One line going another side. And this is something like this. And this is kept over here. And this is M here. So one would be like this. And one would be like this. So this is ABCD and this is M. And I have to prove that area of ABMD, area ABMB is equal to area BDMBC. Okay, great. So try to understand all of you. There are this, this particular figure is divided into four parts. I'm marking it one, two, three, four. So I have to prove that one plus two is equal to three plus four. And how do I prove it if I'm able to prove that area of one is equal to any one of them? So like if I'm proving that area of one is equal to area of four, suppose. So yes, how do I prove it? Okay, great. So that can be proven here. So I'm trying to prove that area one is, this is area ABM. And area four is, I'm trying to prove that area four is BMC. So I'm trying to prove this here. So this is the midpoint. So AM would be equal to MC. MB is common between them. And what about the other thing? Anyone can tell me. Nice. Tell me. Okay, before this, you prove that in this question, I'm giving you extension of the question. Okay, which question do you want me to solve? Okay, fine. I understood I'm solving the wrong question. Which question do I need to solve? Can anyone send me the picture of that question that I need to solve? Okay, this is example six I understood. But which question do I need to solve? Can you send me the picture? In my book, this is question four. I have a different RS Agarwal edition than yours. I understand I have a different RS Agarwal edition. I'm solving the author itself. The editions are different. Can you send me the picture of that question? Example, Shradha, can you send me? I understood guys. Example four, I don't have that question in my book. Send it to me on WhatsApp. I'll solve. Okay. Example four tells me that in an adjoining figure, AD is one of the medians of ABC and P is the point on AD. So I have to prove that area of BPD is equal to area of CDP. So guys, that's what I was telling you. Question example six is similar to example four. What you need to do is I'll tell you one thing which is very, very important. Try to prove that or prove this theorem that medians divide triangles into two equal parts. Prove the theorem. Try to prove it that medians divide triangles into two equal parts. Try to prove it. If you are not proving it, then let me know. Okay. So if you know that, if you have proved that triangles divide or in a median triangles divide, sorry, in a triangle, medians divide it into two equal parts, then what's the problem in solving the other thing? How do you prove that? Can anyone send me? How did you prove? Can those who are sending me that they have done it? How do you prove it? I'm solving it for you guys. Those who have not done. Suppose this is triangle ABC and I draw a median D here. Median is what divides the opposite side into two equal parts. So if these two sides are opposite, I'm trying to prove that area of triangle ABD is equal to area of triangle ADC. So how it is done? So I draw a perpendicular from here. Suppose this is perpendicular AL. So try to understand for this ABD area would be 1 by 2 into BD into AL. And for this, it would be 1 by 2. AL would be same because on this DC base, AL is the perpendicular. So AL into DC. And I know that BD is equal to DC. Hence both the areas would be same. So it means that median divides the triangle into two equal parts. Now any question, which is let me go to the question that you guys have sent me. So what happens over here is this is ABC and this is the median D. So you should know that you should directly write that area of triangle ABD would be equal to area of triangle ADC. We have already proved this theorem. So by theorem that median will divide the triangle into two equal parts. Now what happens is I am drawing a line. So I'm drawing a line. So from this point also this is median. This whatever point is PD is median of line BC in triangle PBC. So if it is median, then area of triangle PBC, sorry, PBD would be equal to area of triangle PDC. That's what they have written over here time and again. So what I have to prove. Then you have to prove that this area is equal to this area. I have already written that this is equal to this and PBD is equal to PDC. Substract from here. So ABD minus PBD is equal to ADC minus PDC. And what do you get? So this complete area minus this would be you will get APB and here you will get APC. So you get both the sides same. Did you guys understand this? Okay, let's solve other questions. Question number 5, 6 all those questions. Question number 6, have you done guys? Okay, great. Okay, keep on posting because my book is different. So if you have doubt in any problem, send it to me. Okay, question number 7 guys. If you have doubt you can send it to me guys. Any doubt you can post, post guys, post your doubts. Solve, which question you guys are solving just post in the chat box so that I can know how many people are there first of all who are solving questions. Example 7, Sraddha, Varsha, which question Varsha, I know you are solving which question you are solving. Dhruti is on question number 7, question number 9, Varsha is on question number 7. Anchita which question are you solving? Okay, great. Do I need to solve example 7? It's a very simple question. It's a very simple question. So when it's constitutionally it's a very structured program where since the lab side it will be around to come on please. So in a way we have to post perhaps more than one piece. It's part of the current kind of a program. Every month is quite important in terms of what are the expenses for any of the issues. What is the study? So we work and we change. Question number 8, 9, 10, we will solve 2-3 more questions and then we will stop. So that's our paper call. The leader of the group said to us that we don't understand the place where we live. This is actually the game that we use in our time. It's an expression of what we are doing. Okay Anchita, no issues. Where are you going? In between the class you are not supposed to leave. Okay Anchita. No issues. Where are you going? In between the class you are not supposed to leave. Okay Anchita, no issues. Where are you going? In between the class you are not supposed to leave. No issues, go. Anyway I am leaving after 2-3 questions. No issues Anchita, go. 2-3 questions more I will solve and then I will stop the class. Those who have not joined the session, Anirudh has not joined the session. Anirudh hardly joins online sessions. I have not got the attendance. Who else? Not got the attendance from Tamaya, not got the attendance from... Diksha. Diksha. Okay, fine. Who is this guy? Listen to the voice by Stray Kids, I am crying. Can you write your name? I am asking your name, how will I come to know your name? Oh Sanjana, I got your attendance, perhaps you have written your name. I am talking about Anirudh, Tamaya is here, okay great. So I got confused between Diksha is not here and Anirudh, Diksha and Tamaya are not here. I think. So we gave our project to a company called Sanjana, which is full of people. And Sanjana was actually founded after that, last year we were doing this. Most of the time, in fact, sometimes you will see other people here, but I do not know if they are doing it or not. Oh Sanjana, sorry about that. Okay guys, what I am understanding is that I request you to do questions offline and keep on sending me doubts if you have anything in particular. And probably if you have any doubt, probably I will send it to you by WhatsApp or something. And let's wrap up the session and once again we will meet next week. Most probably by online medium only and we will try to move ahead. So probably next class, this is what's the day today. So maybe on 15th or something, 15th, 17th whenever I am in whichever way you guys are okay with. So I will have the next class on Maths and that class would be particularly with respect to religion of what we have already done. So that you can. So next class, do you guys have school on Saturday? Can anyone post in the chat box? Do you know, do you guys have school on Saturday? Okay, no issues. So depending on that test of Maths next Saturday, which has already been planned and getting delayed because what to say. Okay, no issues. We'll have a math test before that. I'll plan it somehow and we'll have a class next Saturday for sure. If not offline in your school then online for sure. So thank you so much for joining the session and we'll meet again. Thank you. Thank you.