Certainly! To solve this problem, we can use the concept of an infinite geometric series.
Let's start by calculating the area of the first square:
[tex]\[ \text{Area of the first square} = (\text{side length})^2 = 16^2 = 256 \text{ cm}^2 \][/tex]
Each subsequent square is constructed by joining the midpoints of the sides of the previous square. When you join these midpoints, the new side length is equal to the length of the diagonal of the smaller square created inside. Because the smaller square is turned at a 45-degree angle inside the larger one, the diagonal of the smaller square is equal to the side length of the larger one.
For a square, the diagonal (d) can be calculated using the side length (s) as follows:
[tex]\[ d = s \cdot \sqrt{2} \][/tex]
This is derived from the Pythagorean theorem because the diagonal forms a right-angled triangle with the sides of the square.
Now, when we create the new square, its side length is half the diagonal of the previous square:
[tex]\[ \text{New side length} = \frac{\text{Old diagonal}}{2} = \frac{\text{Old side length} \cdot \sqrt{2}}{2} \][/tex]
The new area will be:
[tex]\[ (\text{New side length})^2 = \left(\frac{\text{Old side length} \cdot \sqrt{2}}{2}\right)^2 = \frac{\text{Old side length}^2 \cdot 2}{4} = \frac{\text{Old side length}^2}{2} \][/tex]
This means that each subsequent square has an area that is half the area of the previous square. Hence, the ratio of the areas of successive squares is [tex]\( \frac{1}{2} \)[/tex].
Given that the area of the first square is 256 cm², the total sum of the areas of all the squares can be found by summing an infinite geometric series:
[tex]\[ S = a + ar + ar^2 + ar^3 + \dots \][/tex]
Here:
- [tex]\( S \)[/tex] is the sum of the series,
- [tex]\( a \)[/tex] is the first term, which is the area initial (256 cm²), and
- [tex]\( r \)[/tex] is the common ratio between each term (in this case [tex]\( \frac{1}{2} \)[/tex]).
The sum of an infinite geometric series where [tex]\( |r| < 1 \)[/tex] is given by:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Now we can apply this formula:
[tex]\[ S = \frac{256}{1 - \frac{1}{2}} = \frac{256}{\frac{1}{2}} = 256 \times 2 = 512 \text{ cm}^2 \][/tex]
Therefore, the sum of the areas of all the squares is [tex]\( 512 \text{ cm}^2 \)[/tex].
