Question

Could a polyhedron exist with the given number of faces, vertices, and edges? Drag yes or no to each combination

Answer 1

Answer:

no

NO

NO

Step-by-step explanation:

good luck have a nice day

Answer 2

Answer:

(a) No

(b) No

(c) No

Step-by-step explanation:

Given

See attachment

Required

Select Yes or No for each

To do this, we make use of Euler's formula

$F + V - E = 2$

Where

$F \to Faces; V \to Vertices; E \to Edges$

$(a):\ Faces = 8; Vertices = 12; Edges = 6$

Using: $F + V - E = 2$

$8 + 12 - 6 = 2$

$14 = 2$

The above equality is false. Hence, (a) does not exist

$(b):\ Faces = 6; Vertices = 6; Edges = 4$

Using: $F + V - E = 2$

$6 + 6 - 4 = 2$

$8 = 2$

The above equality is false. Hence, (b) does not exist

$(c):\ Faces = 20; Vertices = 30; Edges = 12$

Using: $F + V - E = 2$

$20 + 30 - 12 = 2$

$38 = 2$

The above equality is false. Hence, (c) does not exist