I think I would choose for a cube and a triangle where the triangle is made
of the difference between two cubes. Some trigonometry is needed. The code
could use some functions I suppose but this is just a quick example where
one side is closed.

width=160;
height=100;
thickness=6;
bottom=50;

angle=-15;

cube([width,thickness,height]);
cube([width,bottom+thickness,thickness]);
translate([0,bottom,-sin(angle)*thickness]) rotate([angle,0,0])
cube([width,thickness,height]);

cube([thickness,bottom+thickness,height]);
translate([thickness,bottom+thickness,0])
rotate([0,-90,0])
difference() {
cube([height,abs(tan(angle))*height,thickness]);
rotate([0,0,atan(abs(tan(angle))*height/height)])
cube([sqrt(pow(abs(tan(angle))*height,2)+pow(height,2)),abs(tan(angle))*height,thickness]);
}

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Eric, there are alot of trigonometrical functions in there that i havent seen
since I left school 20 years ago. And to cascade them even confuses me more.
LOL. I need to do some studying now :-(

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Sebastian, sorry I didn't want to confuse you (and it was kind of late when I
posted yesterday). I simplified the code by removing some redundancy and it
looks less horrifying now. Start at the bottom of the code and add an
exclamation mark before the last line and work your way up removing the
previous exclamation mark. This gives a good idea of what’s going on. I
already inserted the first exclamation mark below.

width=160;
height=100;
thickness=6;
bottom=50;

angle=-20;

cube([width,thickness,height]);
cube([width,bottom+thickness,thickness]);
translate([0,bottom,-sin(angle)*thickness]) rotate([angle,0,0])
cube([width,thickness,height]);

cube([thickness,bottom+thickness,height]);
translate([thickness,bottom+thickness,0])
rotate([0,-90,0])
difference() {
cube([height,abs(tan(angle))*height,thickness]);
!rotate([0,0,abs(angle)])
cube([sqrt(pow(abs(tan(angle))*height,2)+pow(height,2)),abs(tan(angle))*height,thickness]);
}

The difference() calculates the triangle. One cube is rotated over a second
one resulting in a triangle. The part
‘sqrt(pow(abs(tan(angle))*height,2)+pow(height,2))’ is Pythagoras to
calculate the length of the hypotenuse of the triangle that is rotated over
the first.

The abs(tan(angle))*height calculates the length of the side opposite of the
variable angle. The larger the angle the larger the opposite side.

I hope this help.

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