Question 6

We will find the perimeter of the Koch Snowflake at each step by observing how the side length and number of sides change.

Step 1 — Perimeter at Step 0

The starting shape is an equilateral triangle. Let the side length of this triangle be $L_0$. The problem states that $L_0$ is 1 unit. An equilateral triangle has 3 equal sides. So, the number of sides at Step 0 is $N_0 = 3$. The perimeter is the total length of all its sides.

$P_0 = N_0 \times L_0$

$P_0 = 3 \times 1$

$\boxed{P_0 = 3 \text{ units}}$

Step 2 — Perimeter at Step 1

To create the shape at Step 1, we change each side of the Step 0 triangle. Each side is divided into three equal smaller parts. The middle part is then removed. Two new segments are added to form an outward-pointing equilateral triangle. This means one original segment becomes four new segments. The length of each new segment is one-third of the original segment's length. Let the new side length be $L_1$.

$L_1 = L_0 / 3$

$L_1 = 1 / 3 \text{ units}$

The number of sides also changes. Each of the 3 sides from Step 0 is replaced by 4 new sides. Let the number of sides at Step 1 be $N_1$.

$N_1 = N_0 \times 4$

$N_1 = 3 \times 4$

$N_1 = 12 \text{ sides}$

Now, we calculate the perimeter for Step 1.

$P_1 = N_1 \times L_1$

$P_1 = 12 \times (1/3)$

$\boxed{P_1 = 4 \text{ units}}$

We can also see that the perimeter is multiplied by $4/3$ from the previous step.

Step 3 — Perimeter at Step 2

We apply the same rule to each segment of the Step 1 shape. Each segment of length $L_1$ is divided into three equal parts. The middle part is removed and replaced by two new segments. Let the new side length be $L_2$.

$L_2 = L_1 / 3$

$L_2 = (1/3) / 3$

$L_2 = 1/9 \text{ units}$

The number of sides changes from $N_1 = 12$ to $N_2$. Each of the 12 sides from Step 1 is replaced by 4 new sides. Let the number of sides at Step 2 be $N_2$.

$N_2 = N_1 \times 4$

$N_2 = 12 \times 4$

$N_2 = 48 \text{ sides}$

Now, we calculate the perimeter for Step 2.

$P_2 = N_2 \times L_2$

$P_2 = 48 \times (1/9)$

$\boxed{P_2 = 16/3 \text{ units}}$

Again, we observe that the perimeter is multiplied by $4/3$ from the previous step: $P_2 = P_1 \times (4/3) = 4 \times (4/3) = 16/3$.

Step 4 — Perimeter at Step n

We can see a clear pattern in how the perimeter changes at each step. At each step, the number of segments multiplies by 4. At each step, the length of each segment divides by 3. So, the total perimeter is multiplied by a factor of $4/3$ at each step. Let $P_n$ be the perimeter at step $n$. We started with $P_0 = 3$ units. For Step 1, we multiplied $P_0$ by $(4/3)^1$. For Step 2, we multiplied $P_0$ by $(4/3)^2$. Following this pattern, for Step $n$, we multiply $P_0$ by $(4/3)^n$.

$P_n = P_0 \times (4/3)^n$

$P_n = 3 \times (4/3)^n$

$\boxed{P_n = 3 \times (4/3)^n \text{ units}}$

Answer

Perimeter of Koch Snowflake:

Step 0: 3 units Step 1: 3 * (4/3)^1 units Step 2: 3 * (4/3)^2 units Step 3: 3 * (4/3)^3 units Step 4: 3 * (4/3)^4 units ... Step n: 3 * (4/3)^n units. Perimeter of step n = 3 * (4/3)^n units.