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.
Best Answer
To follow the argument, a labelled diagram needs to be drawn. The argument is exactly the same as yours, except that everything is given a name.
Let our family of parallelograms have base $b$ and area $A$. So all of our parallelograms have height $h$ where $bh=A$. In particular, the height $h$ is determined.
Note that two of the sides of the parallelogram are given. So to minimize the perimeter, we must minimize the other two sides.
Let $PQRS$ be such a parallelogram, where the vertices $P$, $Q$, $R$, and $S$ are listed in counterclockwise order, and the base is $PQ$. Then the perimeter of the parallelogram is $2(PQ)+2(PS)$. This is $2b+2(NS)$. So to minimize the perimeter, we must minimize $NS$.
Draw a perpendicular from $P$ to the line through $R$ and $S$. Suppose that this perpendicular meets the line through $R$ and $S$ at $N$. Draw a perpendicular from $Q$ to the line through $R$ and $S$. Suppose that this perpendicular meets the line through $R$ and $S$ at $M$.
Note that $PQMN$ is a rectangle with base $PQ$ and height $h$, so it is in our family of parallelograms. Suppose that $N\ne S$. We will show that the rectangle $PQMN$ has perimeter less than the perimeter of $PQRS$. To do this, we need to show that $NS \gt h$.
Note that by the Pythagorean Theorem, $(PS)^2=(PN)^2+(NS)^2=h^2+(NS)^2$. Since $NS\ne 0$, we have $(PN)^2\gt h^2$, that is, $PS \gt h$.
This shows that if $N\ne S$, then $PQRS$ cannot have minimum perimeter in our family. So for minimum perimeter, we must have $N=S$, meaning that we have a rectangle.
Remark: The argument can be greatly shortened. There is only one idea here: If $PQRS$ is not a rectangle, then the side $PS$ is greater than $h$, and therefore our parallelogram has perimeter greater than $2b+2h$, which is the perimeter of the rectangle.