Can we surround a non-rectangular area with Lego fences? My children have some Duplo fences, these you have to put down on two points, and at both ends they extend a little where you can connect several to surround some area.
So a fence is described by a triple, $x, y, z$, where $y$ is an integer, the distance between the two points, while for $x$ and $z$, the lengths of the extensions, $x+z$ has to be a positive integer.
Moreover, it is natural to assume that $x$ and $z$ are also integers to allow an axis-parallel rectangle to be constructed.
Apparently, Lego knows how to use the Pythagoras theorem, because the two points where you have to put them down are at distance $y=5$ from each other, so it is also possible to put down a fence diagonally, like at (0,0) and (3,4).
Also, they have $x=z=2$.
From here an easy calculation and checking a few cases should show that it's not possible to bound any interesting area, but only axis-parallel rectangles.
So my question is that supposing you have some identical fences as mentioned above, is it possible to surround a non-rectangular area?
Instead of distance 5, the base distance can be something else, and you can also pick how much the fence extends, so I'm interested in any related results, primarily in small, realizable examples, that Lego could make.
For example, if $x=y=z=5$, then it's easy to make a rotated square, or a rhombi, of side length 15, as pointed out by Gerhard, but this I consider a trivial solution.
Note on update: In the first version of the problem I've missed the simple construction by jwim, which works if $x=z=0$, as I've posed the problem badly first.
 A: If I understand the geometry of the fences correctly, it should be possible to surrounded a triangle of dimension 15 by 20 by 25, each of which is a multiple of 5. In general, if 5 is replaced with $n$, $n$ must be hypotenuse of a Pythagorean triplet $(a,b,n)$, to get interesting possibilities. Then, a triangle $(na,nb,n^2)$ can be formed. The set of possible shapes that can be surrounded then includes anything built up out of $(n,n)$ squares and this $(na,nb,n^2)$ triangle. These are, of course, the base triangle and square polygons. Its possible that there are base pentagons, hexagons, etc. that can be made as well.
Edit:
There are 6 vectors that can be formed with the single fence, ignoring multiples:


*

*(0,5)

*(5,0)

*(3,4)

*(4,3)

*(3,-4)

*(4,-3)


A fenced in area is a linear combination of these, with coefficients in Z, that add to the zero vector. That probably is the best way to systematically study the problem.
A: I guess I don't understand the geometry either, as the problem seems simple to me.  Take a length which is the distance between two grid points.  If this distance $y$ needs to be integral, choose it thusly.  Now one can arrange $x$ and $z$ so that $x+z=y$, and also so that the endpoints lie on a horizontal line and on a vertical line, but not necessarily both.  Now (assuming nontriviality of the choice), one can form rhombi, polyiamonds, and other nonrectangular areas using just translates and reflections of the original (non-horizontal and non-vertical) fence segment.  In order to have nontrivial rotations allowed, one must impose restrictions on $x$ and $z$ which may not be compatible with $x+z$ being integral.
If the question is what manner of shapes can be so constructed, then one needs $x$ and $z$ to be distances from some point to two grid points, and arranged so that there can be more than one possible angle measure between the implied rays.  I suspect two such angles will be rationally independent, but do not have even a heuristic at this time to convince others of my suspicion.
Gerhard "Is This Post Rationally Dependent?" Paseman, 2016.08.29.
