2009 Grade 8 CEMC Gauss Problems/Problem 8

Problem

The number of faces ( $F$ ), vertices ( $V$ ), and edges ( $E$ ) of a polyhedron are related by the equation $F + V - E = 2$ . If a polyhedron has $6$ faces and $8$ vertices, how many edges does it have?

$\text{ (A) }\ 12 \qquad\text{ (B) }\ 14 \qquad\text{ (C) }\ 16 \qquad\text{ (D) }\ 18 \qquad\text{ (E) }\ 10$

Solution 1

We can use the equation provided in the problem, and plug in $6$ for $F$ , and $8$ for $V$ :

$6 + 8 - E = 2$

We can combine $6$ and $8$ to get:

$14 - E = 2$

Adding $E$ to both sides, we get:

$E + 2 = 14$

Subtracting $2$ from both sides of the equation, we get:

$E = \boxed {\textbf {(A) } 12}$

~anabel.disher

Solution 2

We can remember that a rectangular prism has $6$ faces, $8$ vertices, and $\boxed {\textbf {(A) } 12}$ edges, without doing any calculation.

~anabel.disher

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2009 Grade 8 CEMC Gauss Problems/Problem 8

Problem

Solution 1

Solution 2