Which of the following are the faces, edges and vertices of a polyhedron?

# Which of the following are the faces, edges and vertices of a polyhedron?

1. A
Faces =10, edges =20 and vertices =15
2. B
Faces =7, edges =5 and vertices =4
3. C
Faces =4, edges =6 and vertices =4
4. D
Faces =5 edges =4 and vertices =3

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### Solution:

Let us check for each of the options.
We know Euler’s formula for Polyhedron is,
F + V = E + 2.
Where F is the number of faces, V denotes the number of vertices, and E denotes the number of edges.
For the option (1) given,
Number Faces = F = 10, Number of edges = E = 20, Number of vertices = V = 15.
According to Euler's Formula, F + V - E = 2.
By substituting the given values in LHS, we get
LHS = F + V + E
= 10 + 20 -15
= 30 - 15
= 15 ≠ 2
So LHS ≠ RHS
Hence it is not a polyhedron.
For option (2) given,
Number Faces = F = 7, Number of edges = E = 5, Number of vertices = V = 4.
According to Euler's Formula, F + V - E = 2.
By substituting the given values in LHS, we get
LHS = F + V + E
= 7 + 4 - 5
= 11 - 5
= 6 ≠ 2
So LHS ≠ RHS
Hence it is not a polyhedron.
For option (3),
Number Faces = F = 4, Number of edges = E = 6, Number of vertices = V = 4.
According to Euler's Formula, F + V - E = 2.
By substituting the given values in LHS, we get
LHS = F + V + E
= 4 + 4 - 6
= 8 - 6
= 2
LHS = RHS
Therefore, it is a polyhedron.
For option (4),
Number Faces = F = 5, Number of edges = E = 4, Number of vertices = V = 3.
According to Euler's Formula, F + V - E = 2.
By substituting the given values in LHS, we get
LHS = F + V + E
= 5 + 3 - 4
= 8 - 4
= 4 ≠ 2
So LHS ≠ RHS
Hence it is not a polyhedron.
Therefore, the correct option is (3).

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