Question

Can a polyhedron have 8 faces, 14 edges and 6 vertices? Specify reason.​

Answer 1

Can a polyhedron have 8 faces, 14 edges and 6 vertices?

Given -

• No. of faces = 8 faces
• No. of edges = 14 edges
• No. of vertices = 6 Vertices

Formula -

• F + V - E = 2

LHS -

${\sf{:\implies \: F + V - E }}$

${\sf{:\implies \: 8 + 6 - 14 }}$

${\sf{:\implies \: 14 - 14 }}$

${\sf{:\implies \: 0 \bold{ }}}$

RHS -

${\sf{:\implies \: 2 }}$

and,

→ 2 ≠ 0

→ RHS ≠ LHS

No , polyhedron can't have 8 faces, 14 edges and 6 vertices

Answer 2

Answer: No.

Explanation:

We're given that,

No. of faces = 8

Edges = 14

No. of vertices = 6

Euler's formula: F + V = E + 2

LHS: F + V = 8 + 6 = 14

RHS: E + 2 = 14 + 2 = 16

Therefore, LHS ≠ RHS.