Semicircles inscribed in a Trapezoid!

Using the diagram above $\triangle{O'MO} \sim \triangle{NOD} \sim \triangle{APO'}$

Using the Pythagorean theorem on $\triangle{OMO'} \implies \overline{RT} = \overline{O'M} = \sqrt{\dfrac{25}{4}d^2 - \dfrac{9}{4}d^2} = 2d$

$\triangle{O'MO} \sim \triangle{NOD} \implies 1 = \dfrac{2\overline{ND}}{3d} \implies \overline{ND} = \dfrac{3}{2}d \implies \boxed{\overline{CD} = \overline{CN} + \overline{ND} = \dfrac{7}{2}d}$

$\triangle{APO'} \sim \triangle{O'MO} \implies 5 = \dfrac{3d}{2\overline{O'P}} \implies \overline{O'P} = \dfrac{3}{10}d \implies$

$\boxed{\overline{AB} = \overline{RO'} - \overline{O'P} = \dfrac{1}{5}d}$

and $5 = \dfrac{2d}{\overline{AP}} \implies \overline{AP} = \dfrac{2}{5}d = \overline{BR}$

$\implies \boxed{\overline{BC} = \overline{BR} + \overline{RT} + \overline{TC} = \dfrac{22}{5}d}$

$\implies A_{ABCD} = \dfrac{1}{2}(\overline{AB} + \overline{CD})\overline{BC} = \dfrac{407}{50}d^2 \implies \dfrac{A_{ABCD}}{d^2} = \dfrac{407}{50} = \dfrac{a}{b}$

$\implies a + b = \boxed{457}$ .