Square Problem.

$\triangle{ FEC}$ is a right triangle since $(3,4,5)$ is a pythagorean triple.

Using the above diagram we see that $\triangle{FDC} \sim \triangle{FAE} \implies$ $\dfrac{4}{3} = \dfrac{x}{x - y} \implies y = \dfrac{x}{4} \implies$

$x^2 + \dfrac{x^2}{16} = 16 \implies 17x^2 - 256 \implies x = \dfrac{16}{\sqrt{17}} \implies y = \dfrac{4}{\sqrt{17}}$

$\implies x - y = \dfrac{3}{4}(\dfrac{16}{\sqrt{17}}) = \dfrac{12}{\sqrt{17}} \implies$

$d^2 = 9 - \dfrac{144}{17} = \dfrac{9}{17} \implies d = \dfrac{3}{\sqrt{17}} \implies x - d = \dfrac{13}{\sqrt{17}} \implies$

$A_{\triangle{EBC}} = \dfrac{1}{2}(\dfrac{13}{\sqrt{17}})(\dfrac{16}{\sqrt{17}}) = \dfrac{104}{17} = \dfrac{a}{b} \implies a + b = \boxed{121}$ .