 Hello and welcome to the session. Let us discuss the following question that says in figure 9.17, PQRS and ABRS are parallelograms and x is any point on side Br show that first area of PQRS is equal to area of ABRS and second is area of AXS is equal to half area of PQRS. So before solving this problem let us first learn a simple theorem which is theorem 9.1 of your book which says parallelograms same base between the same parallel lines are equal in area. We will learn a simple property which is if a triangle in the parallelogram same base between the same parallel lines then area of triangle is equal to the area of parallelogram prove this. So first let us draw a parallelogram suppose this is a parallelogram ABCD and a triangle with base DC we have such that it intersects AB at a point E and the triangle is between the same parallel lines AB and CD and also the parallelogram have the base DC then area of triangle DEC is equal to half base into height so base is DC and let the perpendicular distance between the two parallel lines be H so we have into H. Now DC into H is the area of parallelogram since the area of parallelogram is the product of base and the corresponding height so we have half into area of parallelogram ABCT so that is the area of triangle is equal to half area of parallelogram so with the help of these two we are going to solve the above problem so this is our key idea. Let us now start to prove the first one where we have to prove that area of PQRS is equal to area of ABRS. Now parallelograms QRS and ABRS are on the same base SR are between the same parallel lines PBR so this in place of RS is equal to area of ABRS and this is by theorem 9.1 of your NCRT textbook so this completes the first part and now we have to show that area of AXS is equal to half area of PQRS. Now let us see the figure the triangle ASX lies on the base AS and is between the parallel lines AS and BR and the parallelogram ABRS also if we consider AS as base then it lies between the two parallel lines AS and BR thus we can apply the above property which is our second key idea that area of triangle AXS is equal to area of triangle ABRS upon AXS. So let us write down first now in triangle AXS parallelogram ABRS we have area of triangle AXS is equal to half times area of triangle ABRS since they have the same base between the same parallel lines thus we have area of AXS is equal to area of ABRS. Now in the first part we have proved that area of ABRS is equal to area of PQRS so as we can write area of AXS is equal to half times area of PQRS and this is by part one just above we have solved and as we can say that area of RS is equal to area of ABRS and AXS is equal to half times area of PQRS this completes the session take care and have a good day.