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A polyhedron has 32 edges and the number of its faces is 5 less than twice the number of its vertices. What is the number of its faces?
Given:
A polyhedron has 32 edges and the number of its faces is 5 less than twice the number of its vertices.
To do:
We have to find the number of its faces.
Solution:
Let the number of faces be $F$ and the number of vertices be $V$.
Number of faces is 5 less than twice the number of vertices.
This implies,
$F = 2V - 5$
$F + 5 = 2V$
$V = \frac{F + 5}{2}$.............(i)
Number of edges $E = 32$
We know that,
According to Euler's formula $F - E + V = 2$
$F = 32 - \frac{F + 5}{2} + 2$
$F = 34 - \frac{F + 5}{2}$
$2F=68-F-5$
$2F+F = 63$
$F = \frac{63}{3}$
$F=21$
The number of faces of the polyhedron are 21.
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