Geometry: Is it possible for two geometric shapes to have the same perimeter and area and not be identical?

Question

Yes, this is part of my homework. I've never been good at it, but I am trying. This question really floors me. When I think of all possible shapes, I can't imagine this being even remotely possible at all. However, I don't want to overlook anything.

Please provide your input. I don't believe it is possible, not without the shapes being identical. If you have a different perspective, please do share.

Answer 1

Consider an equilateral triangle with side 1, and a parallelogram with sides 1 and 1/2 with angle 60 degrees in between them.

The triangle has base 1 and height sqrt(3) / 2.
The parallelogram has base 1 and height sqrt(3) / 4.

Each has perimeter 3 and area sqrt(3) / 4.


The good thing about parallelograms is that you can maintain a given perimeter and just tweak the angle until you get the area you desire (with an upper limit of course).

Answer 2

Dr D has the right idea. Imagine that you have a partially filled soft plastic bottle with no air gap, so that it's somewhat crumpled. You will find that it's not hard to deform that plastic bottle into other shapes, even though it contains the same amount of liquid and has the same surface area. Only when the bottle is filled to maximum capacity that it becomes hard to deform. A circle, for example, as a shape of fixed perimeter and area, cannot be deformed (because the circle has the maximum possible area for a given perimeter), but an oval of fixed perimeter and area can be (because an oval does NOT have the maximum possible area for a given perimeter--i.e., other oval-like shapes have same perimeter and area). I've given this question a star because as simple as this question is, it does have significance in structural mechanics, such as in design of spacecraft.

Addendum: Expanding on Zanti3's answer, think of any geometric shape, with a "hole" inside of it, also of any gemoetric shape. Obviously the total perimeter and total area are both constant, but the hole can be moved anywhere inside.

Answer 3

and i thought you could just easily prove that to anyone using a simple tangram.
Dr. D's example is just what i was talking about. the equilateral triangle cut from 1 vertex to the center of opposite side (this is the height). then you flip 1 of the 2 and fix it to the previous height (so you actually put the vertices and centers together), to get the paralellogram(?) with the same area and perimeter.
there's other example, but i'm not good in that kind of puzzle, so you can goggle it for some more.

Answer 4

Just adding a specific example to the
good answers you already have.

An equilateral triangle with side = x,
has the same area and perimeter
as a rectangle with sides
x[3 - √(9 - 4√3)] / 4 and x[3 + √(9 - 4√3)] / 4

Just so you can visualise those numbers,
to two decimal places, they are 0.39x and 1.11x.

I'm sure it's feasible to correlate many other shapes as well.

Answer 5

Just to add to the fun:

Imagine you have two unequal squares and you align the squares so that one side of the smaller side is completely within a side of the larger square. Clearly, you can have the "bump" of the smaller square anywhere along the side of the larger square. Sliding the smaller square up and down the side of the larger square will not change the area or the perimeter, but it will change the shape of the figure that makes up the union of the two squares.

Answer 6

Not as impressive as the other answers but I'll post it anyway. Imagine quadrilateral ABCD (with fixed side lengths) inscribed in a circle. The perimeter and area of such a quadrilateral will be constant regardless of any rearrangements of its sides. By rearranging the order of its sides we can generate 6 distinct quadrilaterals (or 3 if a quadrilateral and it's reflection are not considered distinct).

Answer 7

Yes it is possible.

Let us consider a regular polygon of 'n' sides such that the diagonal distance is 2R. When 'n' increases infinitely, it has the same area and the perimeter as that of a cirle having radius = R.