Can a polyhedron have 10 faces, 20 edges and 15 vertices?

If F = 10, V = 15 and E = 20.
Then, we know Using Euler’s formula, F + V – E = 2
L.H.S. = F + V – E
= 10 + 15 – 20
= 25 – 20
= 5
R.H.S. = 2
L.H.S. ≠ R.H.S.
Therefore, it does not follow Euler’s formula.

Number of faces = F = 10

Number of edges = E = 20

Number of vertices = V = 15

Any polyhedron satisfies Euler’s Formula, according to which, F + V − E = 2 For the given polygon,

F + V − E = 10 + 15 − 20 = 25 − 20 = 5 ≠ 2

Since Euler’s formula is not satisfied, such a polyhedron is not possible.

No.

Let us use Euler’s formula

V + F = E + 2

15 + 10 = 20 + 2

25 ≠ 22

Since the given polyhedron is not following Euler’s formula, therefore it is not possible to have 10 faces, 20 edges and 15 vertices.

Since, F + V = E + 2

As 10 + 15 ≠ 20 + 2

∴ A polyhedron cannot have 10 faces, 20 edges and 15 vertices.

No,

Using Euler’s formula

V + F = E + 2

15 + 10 = 20 + 2

25 ≠ 22

Since the given polyhedron is not following Euler’s formula, therefore its not possible.

Euclid formula
faces+vertices-edges should equal to 2
F+V-E=10+2-15=5
this polyhedral not satisfy euclid formula.