Question

can three be polyhedron with 11 faces , 9 vertices and 30 edges give reason for your answer​

Answer 1

Answer:

No. Such a polyhedron is not possible.

Step-by-step explanation:

According to Euler's formula of Polyhedron;

A polyhedron is possible only if

→ F + V - E = 2

where:

• F stands for Faces
• V stands for Vertices
• E stands for Edges

Applying the above formula for given Faces, Vertices and Edges;

   F + V - E = 2

⇒ 11 + 9 - 30 ≠ 2

⇒ - 10 ≠ 2

∴ The given polyhedron is not possible.

KNOW MORE:

Let us verify this relation for a cube.

Cube has 6 faces, 8 vertices and 12 edges.

   F + V - E = 2

⇒ 6 + 8 - 12 = 2

⇒ 14 - 12 = 2

⇒ 2 = 2

⇒ LHS = RHS

So, this polyhedron (cube) is possibe.

Answer 2

✰ Concept :-

In this question, we will use the concept of "Euler's formula of polyhedron", i.e.

$\bigstar \: \underline{\boxed { \sf {\: F+ V- E = 2}}}$

Where,

• F = faces
• v = vertices
• E = edges

✰ Given :-

• F = 11
• V = 9
• E = 30

Now, we will simply put the provided values in the equation and check whether they satisfy the equation or not.

$:\implies \: 11 + 9 - 30 = 2 \\ : \implies \: 20 - 30 = 2 \: \: \: \: \: \: \: \\: \implies \: - 10 \ne2 \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, since, R.H.S is not equal to L.H.S.

Therefore, no there can't be any polyhedron with the provided dimensions.