There is a 3×3 dot grid. The distance between one point to another is 1. How many different non-congruent polygons can you make on the grid?
Rules:

1. All vertices of the polygon must be on the grid
2. Only non self intersecting polygons
3. Only polygons with non-empty interior (<=> positive area)
4. The area of each polygon is 2
5. At least one side of the polygon is 1.5 or bigger
   
   please help me!

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Im trying to solve with picks theory, i need to solve with the most creative way not just to draw the pentagons 

Rule 3 is redundant, given rule 4. 

And if rule 4 is correct, then it seems to me that the answer is 1.

Sorry you dont need rule 3 I forgot to remove it

20 minutes ago, Strange said:

And if rule 4 is correct, then it seems to me that the answer is 1.

I hadn't heard of Pick's theorem before. Based on that, it looks like the answer is at least 3. 

I can't think of any other way of solving it other than drawing all the possibilities, guided by Pick's theorem. Maybe someone more imaginative will have a better idea.

But Pick's theorem does mean that the problem can be restated as: find all closed paths that touch exactly 6 points.

7 hours ago, Dani052 said:

There is a 3×3 dot grid. The distance between one point to another is 1. How many different non-congruent polygons can you make on the grid?
Rules:

1. All vertices of the polygon must be on the grid
2. Only non self intersecting polygons
3. Only polygons with non-empty interior (<=> positive area)
4. The area of each polygon is 2
5. At least one side of the polygon is 1.5 or bigger
   
   please help me!

I am not sure whether you mean your grid to be 3 spaces and four dots (points) or two spaces and 3 dots.

If you only mean 3 dots x 3 dots it is only possible to fit one polygon answering your description unless the polygons can overlap.

(This is because the area of a 3 point x 3 point grid is 4 units so fitting the first one means that the only disjoint areas or one congruent to the first are left.

It can't be true that the distance between any one point and another or even its neighbours is always 1 since that would rule out diagonals.

Without diagonals there is only one solution, thought it can be positioned in different ways.

With diagonals I thought I had found four solutions, but then I realised that one was a flip of another so I think there are 3.