Initial square side: 16 cm. Diagram 1: Midpoints form a square and four triangles, legs \( x_1 = \frac{16}{2} = 8 \) cm.

Area: \( A_1 = \frac{1}{2} \times 8 \times 8 = 32 \text{ cm}^2 \).

For \( n=2 \): Inner square’s diagonal = 16 cm. Pythagorean theorem:

\[ 8^2 + 8^2 = 128 \]

\[ \text{Side} = \sqrt{128} = 8\sqrt{2} \text{ cm} \]

Triangle legs: \( x_2 = \frac{8\sqrt{2}}{2} = 4\sqrt{2} \text{ cm} \).

Area: \( A_2 = \frac{1}{2} \times 4\sqrt{2} \times 4\sqrt{2} = 16 \text{ cm}^2 \).

For \( n=3 \): Square side \( 8\sqrt{2} \). Diagonal of triangle with legs \( 4\sqrt{2} \):

\[ (4\sqrt{2})^2 + (4\sqrt{2})^2 = 64 \]

\[ \text{Side} = 8 \text{ cm} \]

\[ x_3 = \frac{8}{2} = 4 \text{ cm} \]

Area: \( A_3 = \frac{1}{2} \times 4 \times 4 = 8 \text{ cm}^2 \).

Table:

Method 1: Areas: \( A_1 = 32 \), \( A_2 = 16 \), \( A_3 = 8 \). Geometric sequence, ratio \( r = \frac{16}{32} = \frac{1}{2} \).

\[ A_n = 32 \cdot \left(\frac{1}{2}\right)^{n-1} \]

\[ A_6 = 32 \cdot \left(\frac{1}{2}\right)^5 = 1 \text{ cm}^2 \]

Method 2: Side lengths: \( x_1 = 8 \), \( x_2 = 4\sqrt{2} \), \( x_3 = 4 \). Pattern: \( x_n = 8 \cdot \left(\frac{1}{\sqrt{2}}\right)^{n-1} \).

\[ x_6 = 8 \cdot \left(\frac{1}{\sqrt{2}}\right)^5 = \sqrt{2} \text{ cm} \]

Area: \( A_6 = \frac{1}{2} \times \sqrt{2} \times \sqrt{2} = 1 \text{ cm}^2 \).

Method 1: Shaded area: four triangles per stage. First triangle: \( x_1 = \frac{k}{2} \), \( A_1 = \frac{1}{2} \cdot \frac{k}{2} \cdot \frac{k}{2} = \frac{k^2}{8} \).

Four triangles, ratio \( \frac{1}{2} \):

\[ 4 A_n = \frac{k^2}{2} \cdot \left(\frac{1}{2}\right)^{n-1} \]

Infinite sum: \( S = \frac{\frac{k^2}{2}}{1 – \frac{1}{2}} = k^2 \).

Given: \( k^2 = k \).

\[ k (k – 1) = 0 \]

\[ k = 1 \text{ cm} \] (since \( k \neq 0 \)).

Method 2: Shaded area = \(\frac{k^2}{4}\).

Given: \( \frac{k^2}{4} = k \).

\[ k^2 – 4k = 0 \]

\[ k (k – 4) = 0 \]

\[ k = 4 \text{ cm} \] (since \( k \neq 0 \)).

(a)

(b)

(c)