Squares and Triangles

$\Delta$ ADE ~ $\Delta$ ABC

Therefore, given x is square length, we have $\frac {x}{3-x} = \frac {4}{3}$

and, finally, x=12/7

If we were to have catheti a and b, respectively, x would be given by formula

$x=\frac {ab}{a+b}$

Let the length of the square be $x$ . $AC=\sqrt{AB^2+BC^2}=\sqrt{3^2+4^2}=5$ . Employing similar triangles,

$AE=\frac{5x}{4}$ while $CE=\frac{5x}{3}$ , therefore:

$\frac{5x}{4}+\frac{5x}{3}=5$

$\frac{x}{4}+\frac{x}{3}=1$

$3x+4x=3\times4$

$(3+4)x=12$

$7x=12$

$x=\frac{12}{7}$

Therefore, $a+b=12+7=\boxed{19}$