 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says in figure 8.42 that is this figure, A, B, C, D is a trapezium in which A, B is parallel to C, D and A, B is twice of C, D. Find the ratio of area of triangles A, O, B and triangle C, O, D. So let us start with the solution to this question. It is given to us that A, B is twice of C, D and A, B is parallel to C, D. We have to find area of triangle A, O, B divided by area of triangle C, O, D. That is the ratio of areas of triangles A, O, B and C, O, D. So let us see the proof to this now. First of all we draw a line M, O parallel to C, D as well as to A, B, meeting A, D in M. So let us draw that line that is M, O. This is parallel to A, B as well as to C, D. Now in triangle A, B, D we see that M, O is parallel to A, B. Therefore, D, O divided by O, B is equal to D, M divided by M, A by basic proportionality theorem. So we can say that in triangle A, B, D, M, O is parallel to A, B. This implies D, O over O, B is equal to D, M over M, A. This we get by basic proportionality theorem and we name this one. Now let us consider another triangle that is triangle C, A, D. Here also we see that M, O is parallel to C, D. Therefore, C, O divided by O, A is equal to D, M divided by M, A. So we write that in triangle C, A, D, M, O is parallel to C, D. This implies C, O divided by O, A is equal to D, M by M, A. This again by basic proportionality theorem we name this two. Now from equations one and two we get D, O divided by O, B is equal to C, O divided by O, A because both of them are equal to D, M divided by M, A. Now we consider the two triangles, triangle A, O, B and triangle C, O, D. Here we see that this angle is equal to this angle because they are vertically opposite angle. Also we have C, O by O, A is equal to D, O by O, B. This we have just proved. So this shows that triangle A, O, B is similar to triangle C, O, D. So let us write what we have just seen in triangle A, O, B and triangle C, O, D. First angle A, O, B is equal to angle C, O, D. This we get because they are vertically opposite angles. And another thing that we have just proved is this. D, O by O, B is equal to C, O by O, A. Now from these two things we conclude that triangle A, O, B is similar to triangle C, O, D. Now we know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. So applying this theorem we have area of triangle A, O, B divided by area of triangle C, O, D equal to A, B square divided by C, D the whole square twice of C, D. So we can write twice of C, D the whole square divided by C, D the whole square D square divided by C, D square. It is cancelled from numerator denominator here left with 4 divided by 1. So we see that area of triangle A, O, B divided by area of triangle C, O, D is 4 is to 1. So required ratio is our answer to the question that is 4 is to 1. So I hope that you understood the question and enjoyed the session. Have a good day.