 Hi and how are you all today? I am Priyanka and let us discuss this question. It says parallelogram A,B,C,D and rectangle A,B,E,F are on the same base A,B and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle. Now in order to prove the given question we need to draw a diagram to model this situation. Let us have a diagram. Now this is a diagram that will be helping us in the solution. Here there is A,B,C,D is a parallelogram and A,B,E,F is a rectangle on the same base A,B. Let us start with our solution. We are given in the question that A,B,C,D is a parallelogram, A,B,E,F is a rectangle and their common base and the common base is A,B. Now we need to prove that the perimeter of A,B,C,D that is the parallelogram is greater than the perimeter of the rectangle that is A,B,E,F. Let us start with our proof. Now we know that since A,B,C,D is a parallelogram then we can say that A,B is equal to C,D. Let this be the first equation. Again A,B,E,F is a rectangle therefore we can say that A,B is equal to E,F because opposite sides of a rectangle are equal to each other. Let this be the second equation and therefore we can say that C,D is equal to E,F as they both are equal to the same side. From 1 and 2 we can say that C,D is equal to E,F. Now from 1,2 and let this be the third one. From 1,2 we have A,B plus C,D is equal to A,B plus E,F and let this be the fourth equation. Now we know that from all the line segments drawn from a point to a given line the perpendicular line distance is the shortest. Therefore we can say that BE is less than BC because the perpendicular distance is the shortest and also AF is less than AD. They both are because of the same reason that the perpendicular distance between two lines is the shortest. Let this be the fifth equation. This be the fifth and this be the sixth equation. Further from equation 4,5 and 6 we have that is on adding them we have A,B plus C,D plus BC plus AD is greater than A,B plus E,F plus BE plus AF. Now if you carefully notice A,B plus C,D plus BC plus AD is the perimeter of the parallelogram and that is greater than the perimeter of the rectangle that is A,B,E,F. So in the end we can write down that hence the perimeter of the parallelogram is greater than the perimeter of the rectangle. So this completes our session. I hope you enjoyed the session. Take care of your observations. Bye for now.