Squares and Right Triangles.

$\theta + \lambda = 90^{\circ} \implies \triangle{APD} \sim \triangle{PBQ} \implies \dfrac{4}{3} = \dfrac{a}{b} \implies b = \dfrac{3a}{4}$

$\implies a - b = \dfrac{a}{4}$

Using the Pythagorean theorem on $\triangle{APD} \implies 17a^2 = 256 \implies a = \dfrac{16}{\sqrt{17}} \implies$

$A = a^2 - 6 = \dfrac{256}{17} - 6 = \dfrac{154}{17} = \dfrac{\alpha}{\beta} \implies \alpha + \beta = \boxed{171}$ .