1. **Problem Statement:** We have a sequence of figures made by joining polygons with sides of unit length. The first three figures have perimeters 5, 8, and 11 respectively. We need to find the number of outer lines and perimeter for figure 6, find the figure number for perimeter 65, and express the general formula for figure $n$. Also, show no figure can have perimeter 100. 2. **Understanding the pattern:** The perimeter increases by 3 units each time a polygon is added: from 5 to 8 (+3), 8 to 11 (+3). 3. **Formula for perimeter:** Since the perimeter increases by 3 for each added polygon, starting at 5 for figure 1, the perimeter for figure $n$ is: $$P_n = 5 + 3(n-1) = 3n + 2$$ 4. **Number of outer lines:** The table shows the number of outer lines as $n + \frac{2}{2n}$ for figure $n$. This matches the pattern: - Figure 1: $1 + \frac{2}{2} = 1 + 1 = 2$ - Figure 2: $2 + \frac{2}{4} = 2 + 0.5 = 2.5$ - Figure 3: $3 + \frac{2}{6} = 3 + \frac{1}{3} = 3.333...$ So for figure 6: $$\text{Outer lines} = 6 + \frac{2}{12} = 6 + \frac{1}{6} = 6.166...$$ 5. **(i) For figure 6:** - Outer lines: $6 + \frac{2}{12} = 6 + \frac{1}{6}$ - Perimeter: Using $P_n = 3n + 2$, for $n=6$: $$P_6 = 3 \times 6 + 2 = 18 + 2 = 20$$ 6. **(ii) Find $n$ when perimeter is 65:** $$65 = 3n + 2$$ $$65 - 2 = 3n$$ $$63 = 3n$$ $$n = \frac{63}{3} = 21$$ 7. **(iii) General formula for figure $n$:** - Outer lines: $$\text{Outer lines} = n + \frac{2}{2n} = n + \frac{1}{n}$$ - Perimeter: $$P_n = 3n + 2$$ 8. **Show no figure can have perimeter 100:** Assume perimeter 100: $$100 = 3n + 2$$ $$100 - 2 = 3n$$ $$98 = 3n$$ $$n = \frac{98}{3} = 32.666...$$ Since $n$ must be an integer (number of polygons), no figure has perimeter 100. **Final answers:** - (i) Figure 6: Outer lines = $6 + \frac{1}{6}$, Perimeter = 20 - (ii) Perimeter 65 corresponds to figure $n=21$ - (iii) General formulas: Outer lines = $n + \frac{1}{n}$, Perimeter = $3n + 2$ - No figure has perimeter 100 because $n$ is not an integer.