I was stuck (not exactly) at this problem for about a month and finally decided to ask my doubt here. This problem is:
Given a family of parallelograms '$R$', all of which are on equal bases and have the same area, Prove that among all the parallelograms it is rectangle which has the least perimeter.
And for this problem my approach was:
It is given that all parallelograms have equal area, and the area of a parallelogram $=$ base $\times$ height. Since all the bases are equal, the height will also be, and thus all parallelograms will be situated between two parallels.
Also, if I have to prove that the perimeter is least, then I will have to find some relation involving perimeter, and thus it will be like:
perimeter $=$ base $+$ opposite side parallel to base $+$ sum of two other sides
The two other sides are parallel to each other but not necessarily perpendicular to base. Now, since we know that the least length possible of last two parallel lines will be when they are perpendicular to the base (according to Pythagoras that in a right triangle the hypotenuse is longest), we get least perimeter figure in the family of parallelograms as the rectangle.
But, I do not know whether this method that I have used is right or not. It all seems like I did not use much mind to prove this "obvious result", and just wrote the statement, as it is. So, if you guys have any suggestions or an alternatives, a correct and elegant solution, or any advice on where I went wrong, please share it here. Thanks.