Not sure what the question means, unless it is meant to refer to 3-dimensional shapes.
If so, some answers are:
a cylinder,
a cone,
a section of a sphere,
an ellipsoid with two equal axes intersected by a plane defined by those axes,
a symmetric paraboloid intersected by a plane perpendicular to its axis of symmetry,
a torus (doughnut) intersected by a plane perpendicular to its "main" radius.
No, two-dimensional shapes do not have faces in the way three-dimensional objects do. Faces are flat surfaces that make up the boundaries of 3D shapes. In contrast, 2D shapes, such as squares or circles, have only length and width, existing on a single plane without depth.
A polyhedron.
They are closed 3-dimensional shapes with polygonal faces.
A rhombus is a two dimensional figure while the concept of {faces, vertices and edges} is relevant to 3-dimensional shapes.
They are both 3 dimensional shapes with faces that are polygons.
No, two dimensional shapes do not have faces
No, two-dimensional shapes do not have faces in the way three-dimensional objects do. Faces are flat surfaces that make up the boundaries of 3D shapes. In contrast, 2D shapes, such as squares or circles, have only length and width, existing on a single plane without depth.
A polyhedron.
Bases are faces but faces are not necessarily bases.
Cylinder
No. A sphere is a three dimensional shape which has no polygonal faces. Similarly an ellipsoid, a torus, a paraboloid, hyperboloid etc are 3-D shapes with no polygonal faces.
They are closed 3-dimensional shapes with polygonal faces.
a cone
triangular pyramaid
A three dimensional shape bounded by plane (flat) faces is a polyhedron.
the answer really is 6there are 6 faces in a three-dimensional model
A rhombus is a two dimensional figure while the concept of {faces, vertices and edges} is relevant to 3-dimensional shapes.