What is the maximum possible area of a parallelogram with one side of

Hi anu1706

You are right in your understanding that the maximum possible area for a given perimeter is a square.
So for a give perimeter of 24, the sides should have been 24/4 = 6 units each.

But here we have an additional constraint that the length of one of the sides is 2.
Now, if the other sides are also 2 each, the perimeter will only be 8 and not 24.

So the only way to have a perimeter of 24 and length of one of the sides as 2 is to take the parallelogram as a rectangle, which will result in the answer explained in the posts above.

Hi anu1706

You are right in your understanding that the maximum possible area for a given perimeter is a square.
So for a give perimeter of 24, the sides should have been 24/4 = 6 units each.

But here we have an additional constraint that the length of one of the sides is 2.
Now, if the other sides are also 2 each, the perimeter will only be 8 and not 24.

So the only way to have a perimeter of 24 and length of one of the sides as 2 is to take the parallelogram as a rectangle, which will result in the answer explained in the posts above.[/quote]

Thanks!! I got it now..

Thank you so much. Can you give as the proof just to increase our understanding?

I think you're asking why does a rectangle have the maximum area for a given set of sides, am I right?
One way is to see that the area of a parallelogram is ABsinQ (Q being the angle between them), so sin is max when Q=90, hence rectangle.

Another way not involving trigonometry would be, imagine a parallelogram with fixed side lengths and we will be varying angles (base and top = 6, and other sides=2)
For any parallelogram, it is true that area = base * perpendicular height from the base to the top

Now the base length is 6 and fixed, the height is the distance between the top and bottom lines (That is how far the two 6 sides are, perpendicular distance), right?
In which case do you get the max height (distance between the sides)?
When the height is equal to 2, ie when the side of length 2 is actually the height! Again giving you that Q must be 90, hence rectangle.

A way to check this is to take the polar opposite, Q=0/180 degrees, the figure collapses into a line and has area 0.

I hope I was able to get my point straight!

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