Can we say that there are more than a type $3\text {D-Figure}$ with perpendicular faces? A cube or a cuboid is a kind of a parallelepiped with adjacent faces being perpendicular. It can clearly be visualised/proved.
Now, is there any other $3\text{D}$ structure with the adjacent faces being perpendicular? 
In my opinion, there isn't any. Because, if we consider a plane $P$ and draw another plane perpendicular to it, and then another pependicular to it, and carry on, then we always get any of those two structures. So I don't think so. 
So, does there exist any other structure (definitely a convex object) with adjacent faces being perpendicular to each other?
 A: I'm going to assume that this question is talking about convex polyhedrons only. Then it can be restated:

Every convex polyhedron $P$ whose adjacent faces form right angles is a cuboid.

Note: My proof needs more rigour + pictures!
Proof: Start at any face $F$ of $P$. By convexity, we know this is a convex polygon. We know by our lemma (E) that $P$ must be an extrusion of $F$, meaning there is another copy of $F$ directly opposite and parallel to $F$ and all other sides are rectangles. Choose any one of these other sides $F^*$. We already know it is a rectangle. Now apply the lemma again to this side. We conclude that $P$ is an extrusion of a rectangle, with an identical rectangle opposite and parallel to it. This is exactly the description of a cuboid.

Lemma (E): Given any face $F$ in $P$, we know that $P$ is an orthogonal extrusion of $F$, meaning that there is a copy of $F$ directly opposite $F$ and parallel to it.
Proof: We know by lemma (S) that any side meeting an edge of $F$ must be perpendicular to $F$ and shares an entire edge with $F$. Since this is true of all edges of $F$, and since we can apply convexity at any cross section of our final shape, we know that the extruded shape is the only one possible.

Lemma (S): Given an edge $E$ of some face $F$ in our polyhedron, there is exactly one face that meets $E$ at a right angle.
Proof: By our rule that all adjacent faces form right angles, we know there are only two possible choices for two faces meeting $E$. Either both point in the same direction, or the directions are opposite. If the directions are opposite, then it is possible to draw a line from the interior of one face to the interior of the other which cuts the border of both faces. This is not allowed by convexity. In conclusion then, there cannot be two distinct faces other than $F$ joining the edge $E$.
Corollary: We know that the face joining $F$ at $E$ shares the entirety of $E$ with $F$. Otherwise, we could again break convexity by drawing a line from the interior of this adjoining face to any unshared portion of $E$.
