\begin{tabular}{lll}\( \mathbf{1} \quad \begin{array}{l}\text { An } n \text {-sided figure has } \frac{1}{2} n(n-3) \text { diagonals. } \\ \text { (i) How many diagonals does an octagon have? } \\ \text { (ii) How many sides does a figure with } 54 \text { diagonals have? }\end{array} \) & \( \mathbf{2} \) marks \\ & & \( \mathbf{4} \) marks \\ \hline\end{tabular}

To solve the given problems, we'll use the formula for the number of diagonals \( D \) in an \( n \)-sided polygon: \[ D = \frac{1}{2} n (n - 3) \] ### (i) Number of Diagonals in an Octagon An **octagon** has \( n = 8 \) sides. Plugging this into the formula: \[ D = \frac{1}{2} \times 8 \times (8 - 3) = \frac{1}{2} \times 8 \times 5 = \frac{1}{2} \times 40 = 20 \] **Answer:** An octagon has **20 diagonals**. --- ### (ii) Number of Sides in a Polygon with 54 Diagonals We are given \( D = 54 \) and need to find \( n \). Using the formula: \[ 54 = \frac{1}{2} n (n - 3) \] First, multiply both sides by 2 to eliminate the fraction: \[ 108 = n (n - 3) \] Expand the equation: \[ n^2 - 3n - 108 = 0 \] This is a quadratic equation in the form \( ax^2 + bx + c = 0 \). To solve for \( n \), we can factor or use the quadratic formula. **Factoring Method:** We look for two numbers that multiply to \(-108\) and add up to \(-3\). The numbers \( -12 \) and \( 9 \) satisfy this: \[ n^2 - 12n + 9n - 108 = 0 \\ (n - 12)(n + 9) = 0 \] Setting each factor equal to zero: \[ n - 12 = 0 \quad \Rightarrow \quad n = 12 \\ n + 9 = 0 \quad \Rightarrow \quad n = -9 \] Since the number of sides cannot be negative, we discard \( n = -9 \). **Answer:** The polygon has **12 sides**.