None. Using Euler's formula v - e + f = 2, where v is vertices, e is edges, and f is faces, we see that for your question f = 3. No solid figure can have less than 4 faces (a tetrahedron).
The "Surface Area" of the solid figure. Note, the word "total" in the answer above is not correct/needed - there can not be anything less than a surface area of a solid figure.
There are infinitely many sets. For example, a cube, cuboid, parallelepiped, rhombohedron and their less regular counterparts all have 6 quadrilateral faces, 12 edges and 8 vertices. There are similar sets for polyhedra with a different number of faces.
A rectangular prism has 6 faces. A pentagon prism has 7 faces. Therefore a rectangular prism has one less face than a pentagonal prism.
Each face of a polyhedron is a polygon which has at least three straight sides; all the polygons of the polyhedron join along one of their sides to another polygon to form the edges of the polyhedron.Now consider when you have two parallel faces of a polyhedron:They do not meet (as they are parallel) and so have no side in common to form an edge of the polyhedron;There must be at least three sides to each polygon which means there must be at least three more polygons to join the two faces together (one for each side of the parallel faces);Thus there are at least 2 + 3 = 5 faces to the polyhedron, ie the polyhedron has 5 or more faces.A tetrahedron has 4 faces which is less than 5 faces.Thus two of the faces of a tetrahedron cannot be parallel.The smallest polyhedron which can have a pair of parallel faces is a pentahedron with two triangular parallel faces and three quadrilateral faces joining them (for example a tetrahedron with one corner cut off by a plane parallel to the opposite face).
None. Using Euler's formula v - e + f = 2, where v is vertices, e is edges, and f is faces, we see that for your question f = 3. No solid figure can have less than 4 faces (a tetrahedron).
The "Surface Area" of the solid figure. Note, the word "total" in the answer above is not correct/needed - there can not be anything less than a surface area of a solid figure.
Since a cube has 6 faces, you would be looking for a solid shape with only 2 faces and no such solid exists.
According to the Euler characteristic which applies to all simply connected polyhedra,# edges + 2 = # vertices + # faces. So the answer is 2 fewer.
tetrahedron
A pyramid with a triangular base. It has 6 edges, or a pyramid with a square base, which has 7 edges
There are infinitely many sets. For example, a cube, cuboid, parallelepiped, rhombohedron and their less regular counterparts all have 6 quadrilateral faces, 12 edges and 8 vertices. There are similar sets for polyhedra with a different number of faces.
a polygon has 6 sids and 6 faces because it is a closed figure and any closed figure has 8 sides or less
6 faced more or less.. !! Counting faces and one of the bottom = 7 .!! See yaa kisees . ;D hope my ansewer help u a lot.!
Yes, an example of this is a sphere which does not have any edges. If you had intended to ask if there are any polyhedra with less than three edges, the answer to that would be no, as the only figure constructable from three distinct lines is a triangle.
cube
A rhombus is a two-dimensional shape, so it does not have any vehicles. However, it does have edges and faces. A rhombus is a quadrilateral with four sides of equal length. Each pair of opposite sides is parallel and the opposite angles are equal. A rhombus has the following properties: It has four edges or sides of equal length. It has four vertices (corners) where the edges meet. It has four angles, each of which measures less than 180 degrees. It has two diagonals that bisect each other at right angles. Since a rhombus is a flat shape, it has only one face, which is a quadrilateral with four sides. Therefore, a rhombus has 4 edges, 1 face, and 0 vehicles. Hope this helps you!