Revise vertices, edges and faces. 1. Identify and count the: i. vertices ii. edges iii. faces

Here are the solutions to the problems:

1. Identify and count the:

• Object 1 (Dark brown hexagonal prism):

  • i. faces: $\boxed{\textbf{8}}$
  • ii. edges: $\boxed{\textbf{18}}$
  • iii. vertices: $\boxed{\textbf{12}}$

• Object 2 (Layered square pyramid):

  • i. vertices: $\boxed{\textbf{5}}$
  • ii. edges: $\boxed{\textbf{8}}$
  • iii. faces: $\boxed{\textbf{5}}$

• Object 3 (Brown rectangular prism/cuboid):

  • i. faces: $\boxed{\textbf{6}}$
  • ii. edges: $\boxed{\textbf{12}}$
  • iii. vertices: $\boxed{\textbf{8}}$

• Object 4 (Stack of 3 cubes, forming a rectangular prism):

  • i. faces: $\boxed{\textbf{6}}$
  • ii. edges: $\boxed{\textbf{12}}$
  • iii. vertices: $\boxed{\textbf{8}}$

• Object 5 (Green hexagonal prism):

  • i. faces: $\boxed{\textbf{8}}$
  • ii. edges: $\boxed{\textbf{18}}$
  • iii. vertices: $\boxed{\textbf{12}}$

• Object 6 (House-like structure: rectangular prism base + triangular prism roof):

  • i. vertices: $\boxed{\textbf{10}}$
  • ii. edges: $\boxed{\textbf{17}}$
  • iii. faces: $\boxed{\textbf{9}}$

2. Find or draw pictures of objects with:

• a. 8 vertices, 12 edges and 6 faces: This describes a rectangular prism (also known as a cuboid or a cube).

  • Drawing description: Draw a rectangle. Draw another identical rectangle slightly above and to the right of the first. Connect the corresponding vertices of the two rectangles with straight lines.

• b. 4 vertices, 6 edges and 4 faces: This describes a triangular pyramid (also known as a tetrahedron).

  • Drawing description: Draw a triangle for the base. From a point above the center of the triangle, draw lines connecting to each of the three vertices of the base.

3. Can a 3-D object have equal numbers of vertices, edges and faces?

No, a 3-D object (specifically a polyhedron) cannot have an equal number of vertices, edges, and faces. This is governed by Euler's formula for polyhedra, which states that $V - E + F = 2$, where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces. If $V = E = F$, then substituting this into the formula would give $V - V + V = 2$, which simplifies to $V = 2$. A polyhedron cannot have only 2 vertices, 2 edges, and 2 faces in a non-degenerate form.

The answer is $\boxed{\textbf{No}}$.