Question

Which of the following cannot be true for a polyhedron?

(a) $V=4,F=4,E=6$

(b) $V=6,F=8,E=12$

(c) $V=20,F=12,E=30$

(d) $V=4,F=6,E=6$

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

The correct option is D

(d)

Explanation for the correct option:

Option (d): Determine the polyhedron that cannot hold true for the given value.

Euler’s formula for any polyhedron is given by $\mathrm{Face}\left(F\right)+\mathrm{Vertices}\left(V\right)-\mathrm{Edge}\left(E\right)=2$.

Consider $V=4,F=6,E=6$.

Substitute the above values into the formula.

$\begin{array}{rcl}\mathrm{Face}\left(F\right)+\mathrm{Vertices}\left(V\right)-\mathrm{Edge}\left(E\right)& =& 2\\ & \Rightarrow & 6+4-6=2\\ & \Rightarrow & 4\ne 2\end{array}$

Hence, option (d) is correct.

Explanation for the incorrect options:

Option (a): Consider $V=4,F=4,E=6$.

Substitute the above values into the formula.

$\begin{array}{rcl}\mathrm{Face}\left(F\right)+\mathrm{Vertices}\left(V\right)-\mathrm{Edge}\left(E\right)& =& 2\\ & \Rightarrow & 4+4-6=2\\ & \Rightarrow & 2=2\end{array}$

Hence, option (a) is incorrect.

Option (b): Consider $V=6,F=8,E=12$.

Substitute the above values into the formula.

$\begin{array}{rcl}\mathrm{Face}\left(F\right)+\mathrm{Vertices}\left(V\right)-\mathrm{Edge}\left(E\right)& =& 2\\ & \Rightarrow & 8+6-12=2\\ & \Rightarrow & 2=2\end{array}$

Hence, option (b) is incorrect.

Option (d): Consider $V=20,F=12,E=30$.

Substitute the above values into the formula.

$\begin{array}{rcl}\mathrm{Face}\left(F\right)+\mathrm{Vertices}\left(V\right)-\mathrm{Edge}\left(E\right)& =& 2\\ & \Rightarrow & 12+20-30=2\\ & \Rightarrow & 2=2\end{array}$

Hence, option (c) is incorrect.

Therefore, options (a), (b), and (c) are incorrect.

Hence, option (d) is correct.