 Hello and welcome to the session. In this session we discussed the solving question that says, from each of the solving figures, that is these two figures, find the value of x. Before we move on to the solution, let's discuss one theorem that says, if internally, externally, then the product of the length segments, this is the key idea that we use for this question. Let's proceed with the solution now. The first figure that the quads A, B and C, D intersect each other internally is equal to 4 cm, P, B is equal to 6 cm, P, C is equal to 5 cm and P, D is equal to A. We are supposed to find the value of x. If two quarters of a circle intersect internally or externally, then the product of the length of the segments are equal. So, according to this theorem, product of the lengths of the segments of the quads A, B and C, D would be equal. That is, now the segment of the quads A, B are A, P and P, B. So, product of A, P and P, B would be equal to the product of the segments of the quads C, D which are P, C and P, D. So, A, P into P, B is equal to P, C into P, D. Now, substituting the values, we have A, P minus P, A which is 4 cm into P, B which is 6 cm is equal to P, that is 5 cm into P, D which is, so from here we have x is equal to 4 into 6 upon 5 is equal to 24 upon 5 and this is equal to 3.8 cm. That is, we have x is equal to 4.8 cm. So, this is the answer for the first part of the question. Let's consider the second part and C, D each other is also given that A, B is equal to P is equal to 6 cm, C, D is equal to x is equal to 4 cm. Now, from the theorem we have that is 2 quads of a circle intersect internally or externally when the product of the lengths of the segments are equal. That is, in this case the 2 quads A, B and C, D are intersecting each other externally at point P. So, the product of the segments of the 2 quads would also be equal. That is, product of the segments of the quads A, B which is into P, B would be equal to the product of the segments of the quads C, D which is P, D. Substitute the value P, D plus A, D cm plus A, D that is n cm and this is equal to 16 cm. 2 cm into P, D which is 6 cm is equal to P, C. From the theorem we have P, C is equal to P, D plus P, D now P, D which is x and so this is equal to 4 plus x cm cm into P, D that is 4 cm. This gives us 96 cm is equal to 4 into 4 plus x that is 96 is equal to 16 plus this gives us 4 x is equal to 96 minus 16. Now, from here we have x is equal to 80 upon now 4, 20 times is 80, x is equal to 20 cm. So, the answer for the second part is x is equal to 20 cm. See session where we have understood the solution of this session.