Scale it!!

Using Phythagoras Theorem, $AD = \sqrt{AE^2 + ED^2} = \sqrt{7^2 + 24^2} = 25 = BC$ $FC = \sqrt{BC^2 - FB^2} = \sqrt{25^2 - 15^2} = 20$

Now draw a line parallel to $AB$ and passing through $F$ that meets $BC$ at $G$ and $AD$ at $H.$

We know that, Area of $BFC = \dfrac 12 \times FG \times BC = \dfrac 12 \times FG \times FC \Rightarrow FG = 12.$

$HD = GC = \sqrt{20^2 - 12^2} = 16$

Using similarity of triangles, $\dfrac {HF}{HD} = \dfrac {7}{24} \Rightarrow HF = \dfrac {14}{3}$

$AB = HG = FG + HF = 12 + \dfrac {14}{3} = \dfrac {50}{3} = \dfrac ab$

Therefore, $a + b = 50 + 3 = \boxed{53}$