# Which geometric figure (polyhedron) has 15 quadrilateral faces?

I am looking for a polyhedron which consists only out of 15 quadrilateral faces? Does such a thing exist?

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If you don't mind some of the faces lying in the same plane, you can take a cube and cut three of the sides into four smaller squares each. – Barry Cipra Aug 8 '13 at 16:52

Let $ABCDE$ be a regular pentagon inscribed inside the unit circle on the x-y plane. Let $P = (0,0,1)$ and $Q = (0,0,-1)$ be two points on the $z$-axis.
The convex hull of $A,B,C,D,E$ and $P,Q$ is a pentagonal bipyramid.
Let $A'$ and $B'$ be the mid-point of $AB$ and $BC$ respectively. If one construct a vertical plane containing $A'$ and $B'$, this plane will intersect with the pentagonal bipyramid above in a small rhombus near vertex $B$. If one "chop off" the vertex $B$ along this rhombus and repeat the same thing for the remaining 4 vertices, one will obtain a convex polyhedron with 17 vertices, 30 edges and 15 quadrilateral faces as shown at end.