Right Square Pyramids

For the square base:

I generalized the problem using $M$ and $N$ as shown above. At the end of the problem I will replace the given values $(M = 5, N = 12)$ .

Using the above diagram $\triangle{APD} \sim \triangle{PBQ} \implies \dfrac{N}{M} = \dfrac{a}{x}$ $\implies a = \dfrac{N}{M}x \implies$

$y = \dfrac{N}{M}x - x = (\dfrac{N - M}{M})x \implies$ $(\dfrac{(N - M)^2}{M^2} + \dfrac{N^2}{M^2})x^2 = N^2 \implies$

$x = \dfrac{NM}{\sqrt{(N - M)^2 + N^2}} \implies \boxed{a = \dfrac{N^2}{\sqrt{(N - M)^2 + N^2}}}$

$2\pi r = 4a \implies r = \dfrac{2a}{\pi} \implies s = \sqrt{\dfrac{4a^2}{\pi^2} + \dfrac{a^2}{4}} =$ $\dfrac{a}{2\pi}\sqrt{16 + \pi^2} \implies$

The lateral surface area $A = 4(\dfrac{1}{2})(a)(\dfrac{a}{2\pi})\sqrt{16 + \pi^2} =$ $\dfrac{a^2}{\pi}\sqrt{16 + \pi^2}$

$= \dfrac{N^4 \sqrt{16 + \pi^2}}{((N - M)^2 + N^2)\pi}$

Using $M = 5$ and $N = 6 \implies A = \dfrac{20736\sqrt{16 + \pi^2}}{193\pi} =$ $\dfrac{a\sqrt{b + \pi^2}}{c * \pi}$

$\implies a + b + c = \boxed{20945}$ .