Successive Successful Squares

Let the side length of the largest square be $a$ . Consider the 3-4-5 triangle up top of the largest square. We note that

$\frac {3-a}a = \dfrac 34 \implies 12-4a = 3a \implies a = \frac {12}7$

Let $a_0 = a$ and the subsequent side lengths of the squares on top of the largest square be $a_n$ . Then we note that, for $n \ge 0$ , $\dfrac {a_{n+1}}{a_n} = \dfrac {\frac {12}7}4 = \dfrac 37$ . The area of squares of the largest square and those on top is:

$\sum_{n=0}^\infty a_n^2 = \sum_{n=0}^\infty \left(\frac 37\right)^2a_0^2 = \frac 1{1-\frac 9{49}} \cdot \frac {144}{49} = \frac {144}{40}$

Similarly, let $b_0 = a$ and the subsequent side lengths of the squares on the left of the largest square be $b_n$ . Then we note that, for $n \ge 0$ , $\dfrac {b_{n+1}}{b_n} = \dfrac {\frac {12}7}3 = \dfrac 47$ . The area of squares of the largest square and those on the right is:

$\sum_{n=0}^\infty b_n^2 = \sum_{n=0}^\infty \left(\frac 47\right)^2b_0^2 = \frac 1{1-\frac {16}{49}} \cdot \frac {144}{49} = \frac {144}{33}$

Then the area of all squares within the triangle:

$A = \sum_{n=0}^\infty a_n^2 +\sum_{n=0}^\infty b_n^2 - a^2 = \frac {144}{40} + \frac {144}{33} - \frac {144}{49} = \frac {13542}{2695}$

Therefore the required answer $13542+2695=\boxed{16237}$ .

First we find the largest square's side, x.
(3 - x) / x = x / (4 - x)
(3 - x)(4 - x) = x² = x² - 7x + 12
7x = 12 ==> x = 12/7

After the first square, we'd have created two new triangles. The upper vertical side would be 3 - x = 9/7 with a scale of (9/7)/3 = 3/7 from the original 3-4-5 triangle and the lower horizontal side would be 4 - x = 16/7 with a scale of (16/7)/4 = 4/7 from the original 3-4-5 triangle. Using the infinite sum formula, S = a / (1 - r), we get :
Total squares' area
= (12/7)² x [ 1/(1 - (3/7)²) + 1/(1 - (4/7)²) - 1** ]
= (144/49) x [ 49/40 + 49/33 - 1 ]
= 13542 / 2695

Answer
= 13542 + 2695
= 16237

**Minus 1 because we used the same a for both series but in effect the first square is one and the same, and has been double counted by the previous sums.