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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.

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.

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.

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. 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.

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.

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.

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.

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 $x$ and $z$, the lengths of the extensions, where $x+z=1$. 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). But I couldn't use this to bound any non-rectangular area! (Not even a rotated square with the given piece.)

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.

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.

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. 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.

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.