What is the minimun faces that you need to make a polyhedron with all the faces equilateral triangles, but one(base) square? (not a Pyramid) I'm not a Mathematical person, but I know some things. I can't imagine a polyhedron with those characteristics. That polyhedron do exists? Is it possible to make something like that? (for that is why I ask for the minimum faces, I want to build it for my own)
I'm thinking something like a Icosahedron but with a square base.
I want to build something like that, but I can't find the form, or how I do that.
All right, I hope that this question wasn't in vain.
Thank you for reading, and thank you in advance if you can response
 A: The shape mentioned by achille hui in the comments is the smallest, needing 10 triangles.

Such a shape is necessarily a "Johnson Solid," a (convex) non-uniform polyhedron with regular polygons as faces. Non-uniform just means that all the corners are not "the same", unlike a football (or soccer ball if you prefer), for example.
There happen to be only finitely many Johnson solids (92, in fact), and Wikipedia has the complete list of Johnson solids. 
There are, ignoring the square pyramid, three such shapes made from equilateral triangles and a single square:
Biaugmented triangular prism (10 triangles)
Gyroelongated square pyramid (12 triangles)
Augmented sphenocorona (16 triangles)
A: Expanding some comments into an answer.
@achillehui and @pjs36 have submitted the biaugmented triangular prism, with $10$ triangular faces, as the minimal figure. @pjs36 shows that, under the not-unreasonable assumption of (strict) convexity, it is minimal. It's worth investigating non-convex candidates with fewer triangles.

First, some basic polyhedral arguments.
Suppose there are $n < 10$ triangles, and the $1$ square. (Note that, since each side of the square is connected to a distinct triangle, we have $n \geq 4$.) If we count the edges on each face ($3$ for each triangle, $4$ for the square), then we will have counted each edge of the polyhedron twice. Therefore, writing $e$ for the total number of edges,
$$2e=3n+4 \tag{1}$$
Consequently, $n$ must be even. Let's say that $n=2m$ (with $2\leq m \leq 4$), so that $(1)$ becomes
$$e = 3m+2 \tag{1a}$$
Now, assuming that the polyhedron is "topologally equivalent to a sphere" (a looser condition than convexity that nevertheless excludes self-intersections and other pathologies), we can invoke Euler's "characteristic" formula
$$ v - e + f = 2 \tag{2}$$
where $v$ is the total number of vertices, and $f$ is the total number of faces. By $(1a)$ and $f=n+1=2m+1$, this becomes
$$v = m + 3 \tag{2a}$$
The square "base" accounts for four vertices, so there are $x:=m-1$ (that is, $1$, $2$, or $3$) "extra" vertices. By the Pigeonhole Principle, at least two vertices must share a common extra vertex. Geometrically, equilateral triangles attached to a square base can only share an extra vertex located at the apex, $A$, of a square pyramid with that base. Let's consider cases:


*

*$x=1$. All four base-adjacent triangles share $A$. This is the square pyramid, which OP has dismissed.

*$x=2$. Exactly three base-adjacent triangles share $A$. This gives us a square pyramid with a triangular hole attached to a triangular "flap". Just two more triangles attach to the flap to cover the hole. This is a square pyramid augmented with a regular tetrahedron.

*$x=3$. Exactly two base-adjacent triangles share $A$; the other two triangles cannot share a vertex (well ... they could share the reflected apex, but I think we can reasonably exclude this scenario). If the $A$-adjacent vertices are attached to opposite sides of the square, then we have a square pyramid with two holes; if the $A$-adjacent vertices are attached to adjacent sides of the square, we have a square pyramid with a one large hole. In either case, we cover the holes with the four left-over triangles, and the two non-$A$-adjacent "flaps", to get a square pyramid augmented with two opposite-or-adjacent regular tetrahedra. (The "adjacent" case is non-convex.)


The augmented pyramids are subject to what I described as a "notable (and notorious) geometric coincidence": some tetrahedron faces end up being coplanar with pyramid faces, forming rhombus-shaped faces. The reader can decide whether a rhombus counts as a single (disqualifying) quadrilateral face, or as an acceptable arrangement of two triangles. (This ambiguity arose in a 1980 PSAT Exam question about counting faces. The official answer counted triangles; a geometrically-savvy test-taker counted rhombi.)

Note: The (non-convex) square pyramid augmented with three regular tetrahedra ties the biaugmented triangular prism in being constructed from $10$ triangles, three of which are coplanar. 

Is there a non-convex polyhedron with $1$ square and $10$ non-coplanar triangles?

