Interpreting images representing geometric series

You are correct, in the first image the largest gray square is $\frac 12 \times \frac 12$ of the original square area. Now what is the area of the next gray square? It might help to continue the sides of the first gray square to the other side of the original square. The idea is that the sides of the gray squares form a geometric progression, as does the area of the gray squares. What is the sum of the progression of the sides? What is the ratio of that progression? What is the ratio of the progression of the areas?

The same ideas will solve the others.

In your first example, with the squares, you color in $(\frac{1}{4})^n$ with each new square. These squares add to $\sum_{n=1}^\infty (\frac{1}{4})^n = \frac{\frac14}{1-\frac14}=\frac{1}{3}.$

In your second example, with the rectangles, the first rectangle is $\frac{1}{2}$ of the square, but your second rectangle is $\frac{1}{4}\cdot\frac{1}{2}$, so this sum is $\frac{1}{2} \sum_{n=0}^\infty (\frac{1}{4})^n = \frac12\cdot\frac{1}{1-\frac14}=\frac12\cdot\frac43=\frac{2}{3}.$

Your triangles also sum in the same way to $\frac{2}{3}.$

You can solve these without geometric series.

Say the proportion of coverage of the first image is $s$. Split the first image up into four squares. The topleft and bottomright are not covered at all. The bottomleft square is fully covered. And the topright square? Well it's covered in exactly the same way as $s$ itself. So we find:

\[s = \frac{1}{4}(0 + s + 1 + 0)\] \[4s = 1 + s\] \[s = \frac{1}{3}\]

Similarly for the other images we find equations $s = \frac{1}{4}(1 + s + 1 + 0)$ and $s = \frac{1}{4}(\frac12 + s + 1 + \frac12)$.