Points $B$, $D$, and $J$ are midpoints of the sides of right triangle $ACG$. Points $K$, $E$, $I$ are midpoints of the sides of triangle $JDG$, etc. If the dividing and shading process is done 100 times (the first three are shown) and $AC=CG=6$, then the total area of the shaded triangles is nearest

$\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$
Solution 1
Since $\triangle FGH$ is fairly small relative to the rest of the diagram, we can make an underestimate by using the current diagram.  All triangles are right-isosceles triangles.
$CD = \frac {CG}{2} = 3, DE = \frac{CD}{2} = \frac{3}{2}, EF = \frac{DE}{2} = \frac{3}{4}$
$CB = CD = 3, DK = DE = \frac{3}{2}, EL = EF = \frac{3}{4}$
$[CBD] = \frac{1}{2}3^2 = \frac{9}{2}$
$[DKE] = \frac{1}{2}(\frac{3}{2})^2 = \frac{9}{8}$
$[ELF] = \frac{1}{2}(\frac{3}{4})^2 = \frac{9}{32}$
The sum of the shaded regions is $\frac{9}{2} + \frac{9}{8} + \frac{9}{32} = \frac{189}{32} \approx 5.9$
$5.9$ is an underestimate, as some portion (but not all) of $\triangle FGH$ will be shaded in future iterations.  
If you shade all of $\triangle FGH$, this will add an additional $\frac{9}{32}$ to the area, giving $\frac{198}{32} \approx 6.2$, which is an overestimate.
Thus, $6 \rightarrow \boxed{A}$ is the only answer that is both over the underestimate and under the overestimate.

Solution 2
In iteration $1$, congruent triangles $\triangle ABJ,  \triangle BDJ,$ and $\triangle BDC$ are created, with one of them being shaded.
In iteration $2$, three more congruent triangles are created, with one of them being shaded.
As the process continues indefnitely, in each row, $\frac{1}{3}$ of each triplet of new congruent triangles will be shaded.  The "fourth triangle" at the top ($\triangle FGH$ in the diagram) will gradually shrink, 
leaving about $\frac{1}{3}$ of the area shaded.  This means $\frac{1}{3}\left(\frac{1}{2}6\cdot 6\right) = 6$ square units will be shaded when the process goes on indefinitely, giving $\boxed{A}$.

Solution 3
Using Solution 1 as a template, note that the sum of the areas forms a [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series):
$\frac{9}{2} + \frac{9}{8} + \frac{9}{32} + \frac{9}{128} + ...$
This is the sum of a geometric series with first term $a_1 = \frac{9}{2}$ and common ratio $r = \frac{1}{4}$ This is the easiest way to do this problem.
The sum of an infinite geometric series with $|r|<1$ is shown by the formula. $S_{\infty} = \frac{a_1}{1 - r}$ Insert the values to get $\frac{\frac{9}{2}}{1 - \frac{1}{4}} = \frac{9}{2}\cdot\frac{4}{3} = 6$, giving an answer of $\boxed{A}$.

Solution 4
Find the area of the bottom triangle, which is 4.5. Notice that the area of the triangles is divided by 4 every time. 4.5*5/4≈5.7, and 5.7+(1/4)≈5.9. We can clearly see that the sum is approaching answer choice $\boxed{A}$.