13. As shown in Figure 7, in Rt $\triangle ABC$, $AB=3$, $BC=4$, $\angle ABC=90^{\circ}$, a line $BA_1 \perp AC$ is drawn through $B$, and a line $A_1B_1 \perp BC$ is drawn through $A_1$, resulting in the shaded Rt $\triangle A_1B_1B$; then a line $B_1A_2 \perp AC$ is drawn through $B_1$, and a line $A_2B_2 \perp BC$ is drawn through $A_2$, resulting in the shaded Rt $\triangle A_2B_2B_1$; and so on. Please guess the sum of the areas of all the shaded triangles obtained this way.
$13.2 \frac{14}{41}$.
The ratio of the sum of all shaded areas to the sum of all blank areas is equal to the ratio of the area of $\triangle A_{1} B_{1} B$ to the area of $\triangle A_{1} A B$, which is
$$
(3 \times 4 \div 5 \div 3)^{2}=\frac{16}{25} \text {. }
$$

Therefore, the sum of the shaded areas is
$$
3 \times 4 \div 2 \times \frac{16}{16+25}=2 \frac{14}{41} .
$$