[color=#999999][color=#999999][color=#999999]This activity belongs to the [i]GeoGebra book[/i] [url=https://www.geogebra.org/m/h3gbmymu]Linkages[/url].[/color][/color][/color][br][br]Bar structures, in space, may appear more rigid than they really are. Let's see an example.[br][br]In this construction (which contains no scripts) we have a triangular prism, half a cube. The orange bars all have length 1 unit. The purple bars measure the same as the diagonals of the cube faces, the square root of 2. Therefore, the gray faces are right triangles (and as such triangles of given measures, rigid).[br][br]If now, in addition to O and U, we fix the point E, it might appear that the prism has 0 degrees of freedom. But it's not like that. We can verify that we can move both A and B while preserving the structure (although with certain limitations, in both cases, if the existence of point D is to be guaranteed).[br][br]Let's follow the construction process. Point A rests on the sphere with center O and radius 1. It therefore has 2 degrees of freedom. Positioned A, point B can then move along the dotted circle, intersection of the unit spheres with centers U and A. Thus, the number of internal degrees of freedom is 3, since the position of the isomer of D (box D') does not add any degrees of freedom.

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]