<p>Let \(n\) be the number of sides of a regular polygon. If the number of diagonals of the polygon is 54, then the number of sides \(n\) is:</p>
Step-by-Step Solution
Key Concept: The number of diagonals in an n-sided polygon equals the total segments connecting any two vertices minus the sides: $\frac{n(n-1)}{2} - n = \frac{n(n-3)}{2}$. Set this equal to 54 and solve the resulting quadratic equation.
<p><strong>Step 1:</strong> Recall that from n vertices of a polygon, we can choose any 2 vertices to form a line segment: $\binom{n}{2} = \frac{n(n-1)}{2}$ total segments.</p><p><strong>Step 2:</strong> Among these segments, n are sides of the polygon. The remaining segments are diagonals: Number of diagonals $= \frac{n(n-1)}{2} - n = \frac{n(n-1) - 2n}{2} = \frac{n(n-3)}{2}$</p><p><strong>Step 3:</strong> Set the number of diagonals equal to 54: $$\frac{n(n-3)}{2} = 54$$</p><p><strong>Step 4:</strong> Multiply both sides by 2: $$n(n-3) = 108$$</p><p><strong>Step 5:</strong> Expand and rearrange: $$n^2 - 3n - 108 = 0$$</p><p><strong>Step 6:</strong> Factor the quadratic: $$(n-12)(n+9) = 0$$</p><p><strong>Step 7:</strong> Solve for n: $n = 12$ or $n = -9$. Since n must be positive, $n = 12$.</p><p><strong>Verification:</strong> A 12-sided polygon has $\frac{12(12-3)}{2} = \frac{12 \times 9}{2} = 54$ diagonals ✓</p><p>∴ Answer: <strong>n = 12</strong></p>
Correct Answer: B