Question

If a polygon has 54 diagonals, find the number of sides of the polygon.

• (A) 10
• (B) 11
• (C) 12
• (D) 15

Hint:

Use the formula \(D = \frac{n(n-3)}{2}\) directly and quickly verify by substituting the value of \(n\) to avoid solving full quadratic in exam conditions.

Answer 1

The Correct Option is C

Concept: A polygon is a closed geometric figure with three or more straight sides. In any polygon with \(n\) sides, diagonals are line segments joining non-adjacent vertices. The formula for the number of diagonals in an \(n\)-sided polygon is: \[ D = \frac{n(n - 3)}{2} \] Derivation: From \(n\) vertices, total line segments formed are: \[ \binom{n}{2} = \frac{n(n - 1)}{2} \] Out of these, \(n\) are sides of the polygon. Therefore diagonals are: \[ D = \frac{n(n - 1)}{2} - n = \frac{n(n - 1 - 2)}{2} = \frac{n(n - 3)}{2} \]

Step 1: Using the given information \(D = 54\).
\[ 54 = \frac{n(n - 3)}{2} \]

Step 2: Simplify the equation.
Multiply both sides by 2: \[ 108 = n(n - 3) \] \[ 108 = n^2 - 3n \]

Step 3: Form quadratic equation.
\[ n^2 - 3n - 108 = 0 \] Now factorize. We need two numbers whose product is \(-108\) and sum is \(-3\). The pair is \(-12\) and \(9\). \[ n^2 - 12n + 9n - 108 = 0 \] Grouping: \[ n(n - 12) + 9(n - 12) = 0 \] \[ (n - 12)(n + 9) = 0 \]

Step 4: Find valid value of \(n\).
\[ n - 12 = 0 \Rightarrow n = 12 \] \[ n + 9 = 0 \Rightarrow n = -9 \] Since number of sides cannot be negative, we reject \(n = -9\). Therefore, \[ n = 12 \]