The problem asks for the sum of the areas of an infinite sequence of squares. Each new square is formed by joining the midpoints of the sides of the previous square. We are given that the side length of the first square is 4 cm.

Understanding the Problem: Sum of Areas of Squares

We start with a square of a known side length. We then create a new square inside it by connecting the midpoints of its sides. This process is repeated indefinitely, creating smaller and smaller squares. We need to find the total area covered by all these squares combined.

Calculating the Area of the First Square

The side length of the first square is given as $s_1 = 4$ cm.

The area of the first square, $A_1$, is calculated using the formula for the area of a square, which is side multiplied by side:

$\qquad A_1 = s_1^2 = (4 \text{ cm})^2 = 16 \text{ cm}^2$.

Finding the Side and Area of the Subsequent Squares

To find the side length of the second square, consider the right-angled triangles formed at each corner of the first square when its midpoints are joined. The legs of these triangles are half the side length of the first square.

Leg length = $\frac{s_1}{2} = \frac{4}{2} = 2$ cm.

The side of the second square ($s_2$) is the hypotenuse of such a triangle. Using the Pythagorean theorem ($a^2 + b^2 = c^2$):

$\qquad s_2^2 = (2 \text{ cm})^2 + (2 \text{ cm})^2 = 4 \text{ cm}^2 + 4 \text{ cm}^2 = 8 \text{ cm}^2$.

The area of the second square, $A_2$, is:

$\qquad A_2 = s_2^2 = 8 \text{ cm}^2$.

Notice that the area of the second square is exactly half the area of the first square.

Let's check for the third square. The side of the third square ($s_3$) is found by joining the midpoints of the second square. The legs of the corner triangles of the second square will have length $\frac{s_2}{2} = \frac{\sqrt{8}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$ cm.

Using the Pythagorean theorem for $s_3$:

$\qquad s_3^2 = (\sqrt{2} \text{ cm})^2 + (\sqrt{2} \text{ cm})^2 = 2 \text{ cm}^2 + 2 \text{ cm}^2 = 4 \text{ cm}^2$.

The area of the third square, $A_3$, is:

$\qquad A_3 = s_3^2 = 4 \text{ cm}^2$.

The area of the third square is half the area of the second square ($4/8 = 1/2$).

Identifying the Geometric Progression of Areas

The areas of the squares form a sequence: $16, 8, 4, \dots$

This is a geometric progression (GP) because each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

The first term is $a = A_1 = 16$.

The common ratio is $r = \frac{A_2}{A_1} = \frac{8}{16} = \frac{1}{2}$.

We can verify this with the next term: $\frac{A_3}{A_2} = \frac{4}{8} = \frac{1}{2}$.

Sum of an Infinite Geometric Series

Since the process continues indefinitely, we need to find the sum of an infinite geometric series. The formula for the sum of an infinite geometric series is $S_{\infty} = \frac{a}{1-r}$, provided that the absolute value of the common ratio $|r|$ is less than 1. In this case, $r = 1/2$, and $|1/2| < 1$, so the sum exists.

Using the formula with $a=16$ and $r=1/2$:

$\qquad S_{\infty} = \frac{16}{1 - \frac{1}{2}} = \frac{16}{\frac{1}{2}}$.

To divide by a fraction, we multiply by its reciprocal:

$\qquad S_{\infty} = 16 \times 2 = 32$.

Conclusion: Total Sum of Areas

The sum of the areas of all the squares, continuing indefinitely, is 32 cm$^2$. This matches the first option.