Fractals and Visualising Solids | FIO

Question 3

Find the area of the region remaining at the $n$ th step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

Solution

Understand the Question

In each fractal construction, we start with an initial shape of area $1\text{ sq. unit}$ at Step $0$ .
At each subsequent step, the remaining shapes are subdivided, and a fixed fraction of the area is removed, keeping a constant fraction of the previous step's area.
To find the area at the $n$ th step, we identify the fraction of area retained in one step and multiply it repeatedly $n$ times.

(a) Find the area of the region remaining at the $n$ th step for Sierpinski's Carpet.

Step 1 · Find the Remaining Area for Initial Steps

At each step, a square is divided into $9$ equal smaller squares and the central $1$ is removed, leaving $\dfrac{8}{9}$ of the area. Diagram 1

$\text{Area at Step 0} = 1 \text{ sq. unit}$

\begin{aligned} \text{Area at Step 1} &= \dfrac{8}{9} \times \text{Area at Step 0} \\[0.6em] &= \dfrac{8}{9} \times 1 \\[0.6em] &= \dfrac{8}{9} \text{ sq. units} \end{aligned}

\begin{aligned} \text{Area at Step 2} &= \dfrac{8}{9} \times \text{Area at Step 1} \\[0.6em] &= \dfrac{8}{9} \times \dfrac{8}{9} \\[0.6em] &= \left(\dfrac{8}{9}\right)^2 = \dfrac{64}{81} \text{ sq. units} \end{aligned}

Step 2 · Generalize to the $n$ th Step

Since the remaining area is multiplied by $\dfrac{8}{9}$ at every step, the area at Step $n$ is:

\begin{aligned} \text{Area at Step } n &= \underbrace{\left(\dfrac{8}{9}\right) \times \left(\dfrac{8}{9}\right) \times \cdots \times \left(\dfrac{8}{9}\right)}_{n \text{ times}} \\[0.6em] &= \left(\dfrac{8}{9}\right)^n \text{ sq. units} \end{aligned}

Answer

(a) $\left(\dfrac{8}{9}\right)^n \text{ sq. units}$

(b) Find the area of the region remaining at the $n$ th step for Sierpinski's Gasket.

Step 1 · Find the Remaining Area for Initial Steps

At each step, connecting the midpoints of the sides divides each triangle into $4$ equal smaller triangles. Removing the central triangle leaves $\dfrac{3}{4}$ of the area. Diagram 2

$\text{Area at Step 0} = 1 \text{ sq. unit}$

\begin{aligned} \text{Area at Step 1} &= \dfrac{3}{4} \times \text{Area at Step 0} \\[0.6em] &= \dfrac{3}{4} \times 1 \\[0.6em] &= \dfrac{3}{4} \text{ sq. units} \end{aligned}

\begin{aligned} \text{Area at Step 2} &= \dfrac{3}{4} \times \text{Area at Step 1} \\[0.6em] &= \dfrac{3}{4} \times \dfrac{3}{4} \\[0.6em] &= \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16} \text{ sq. units} \end{aligned}

Step 2 · Generalize to the $n$ th Step

Since the remaining area is multiplied by $\dfrac{3}{4}$ at every step, the area at Step $n$ is:

\begin{aligned} \text{Area at Step } n &= \underbrace{\left(\dfrac{3}{4}\right) \times \left(\dfrac{3}{4}\right) \times \cdots \times \left(\dfrac{3}{4}\right)}_{n \text{ times}} \\[0.6em] &= \left(\dfrac{3}{4}\right)^n \text{ sq. units} \end{aligned}

Answer

(b) $\left(\dfrac{3}{4}\right)^n \text{ sq. units}$

Common Mistakes

Removed vs. Remaining Area: Confusing the removed fraction ( $\dfrac{1}{9}$ for carpet, $\dfrac{1}{4}$ for gasket) with the remaining fraction ( $\dfrac{8}{9}$ and $\dfrac{3}{4}$ ). Make sure to use the portion that is kept.
Off-by-One Exponent Error: Writing the power as $n-1$ or $n+1$ . Step $0$ has area $1 = \left(\frac{8}{9}\right)^0$ , so Step $n$ directly corresponds to exponent $n$ .

Question 3

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