# I have 6 faces, 8 vertices, and 12 edges. Which figure am l?

There is no unique formula for getting the figure. However, according to Euler's Polyhedral Formula, in a convex polyhedra, if $V$ is the number of vertices, $F$ is number of faces and $E$ is number of edges than $V - E + F = 2$.
It is apparent that with $6$ faces, $8$ vertices, and $12$ edges, then $8 - 12 + 6 = 2$, hence it is a valid polyhedra.
However, it is evident that the figure is a cuboid or quadrilaterally-faced hexahedron, as it too has $6$ faces, $8$ vertices, and $12$ edges.