Maths?

Obviously your -- very valid -- question is all about the sum of convergent infinite series and the triangles are but an example. But first things first:
You're given the base b and height h of a triangle and have to work out its area; as the area is the only property of the triangle we're interested in and "a" is a word in the English language and "t" is not, I will refer to the area of the triangle by t:
t = 1/2 b h.
Furthermore we are given a set of rules. Applying these rules to the triangle, we get the area of another triangle. And then we do it again: The rules never change but as we apply them to each and every subsequent triangle, we get an infinite series of different triangle-areas. One rule modifies b, which I will call B and the other one which modifies h I will call H, respectively. In order to get a feeling where we are heading, we sum up the first five values. And finally we go infinity for dessert!

To keep track of the steps I introduce indices. So, the ground state -- the red triangle -- will become t.0, the green one t.1, purple will be t.2 and so on.
t,0 = 1/2 b h ;; Now we apply the rules:
t.1 = 1/2 b B h H ;; And we do it again:
t.2 = 1/2 b B B h H H ;; And again:
t.3 = 1/2 b B B B h H H H
...
A pattern pops up and two things become quite clear:
1. There are constant parts and repetitive parts.
2. The repetition can be described in an exponential representation. Please note: if n is NOT 0, then n^0 = 1.
Thus:
t.0 = 1/2 b h (BH)^0
t.1 = 1/2 b h (BH)^1
...
and somewhere down the road the k-th triangle will have the area t.k = 1/2 b h (BH)^k.
Now it makes sense that I chose 0 as the index for the ground state, the first (red) triangle. The index works as a counter for our steps, serves as exponent in each instance of the iteration and if we were to write a computer programme we wouldn't have to mess around with the internal indexing of a sequence related data structure of the programming language. Furthermore the first triangle isn't a special case anymore, as (BH)^0 = 1 and 1 is the identity in relation to multiplication.

In order to add up the areas of the first five triangles t.0 + t.1 + ... + t.4 we extract the constant part and write:

1/2 b h * sum (BH)^k ,for k=0 to 4

Let's plug in the values:
b = 96cm
h = 64cm
t = 1/2 * 96cm * 64cm = 3072cm²

B = 1/2 , as a factor
H = 3/4 , as a factor
BH = (1/2)(3/4) = 3/8 = 0.375

Thus the sum of the areas of the first five triangles in our series equals

3072cm² * (0.375^0 + 0.375^1 + 0.375^2 + 0.375^3 + 0.375^4) <=>
3072cm² * (1 + 0.375 + 0.140625 + 0.052734375 + 0.01977539 ) <=>
3072cm² * 1.588134766 = 4878.75cm²

This was the initial trick, but the second follows quick.
The series looks pretty much convergent: The terms decrease by an order of magnitude in just two steps. But pretty much is not good enough. Well, the criterion for convergence of a series of the form
sum s^k ,for k=0 to infty. is:

If |s| < 1, then the series converges to 1/(1-s).

Our s equals 3/8. As 3/8 = 0.375 < 1 the condition is met and our series converges to

1/(1 - 3/8) = 1/(8/8 - 3/8) = 1/(5/8) = 8/5 = 1.6

We must not forget the constant that we extracted earlier, so the whole lot converges to

3072cm² * 1.6 = 4915.2cm².

Now, that was fun ;-)