Can a rectangle have the same numerical value for its perimetar and its area?

Answer 1

If the area of a rectangle is equal to

LengthxSide

and perimeter of the same rectangle is equal to

2(Length)+2(side)

and we set these to equal each other, the result is

LxS=2L+2S

If we solve this equation for L we get

L=2S/(S-2)

We could likewise solve for S to get


S=2L/(L-2)

And we see it is possible under this circumstance for a rectangle to have the same numerical value for its perimeter and area. For example L=3, S=6 is a simple solution.
The problem is arguably not very meaningful. because perimeter and area are different physical quantities; one is a length (say in metres) and the other an area (in square metres). If the numbers are the same it's just a numerical coincidence with no geometrical significance.
However, one can make the problem more interesting by considering L getting very large. Then S becomes close to 2, and we have a long thin rectangle with area approximately 2L and perimeter also approximately 2L (because the two long sides are much larger than the short ones).
Exercise for the reader : what happens to the rectangle when L becomes just very slightly above 2?

As a further addendum, we can ask if it's possible to get a square with the area equal to the perimeter. A square has L=S, so the required condition becomes L=2L/(L-2)
which can be manipulated to give L(L-4)=0. Ignoring the zero solution, this gives L=4 and the square has area 16 and perimeter 16.