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Let $a$ denote the length of a diagonal opposite adjacent sides of length $14$ and $3$, $b$ for sides $14$ and $10$, and $c$ for sides $3$ and $10$. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain: $$\begin{array}{rl}{c}^2& =3a+100\\ c^2& =10b+9\\ ab& =30+14c\\ ac& =3c+140\\ bc& =10c+42\end{array}$$ Using equations $(1)$ and $(2)$, we obtain: $$a=\frac{c^2-100}{3}$$ and $$b=\frac{c^2-9}{10}$$ Plugging into equation $(4)$, we find that: $$\begin{array}{rl}\frac{c^2-100}{3}c& =3c+140\\ \frac{c^3-100c}{3}& =3c+140\\ c^3-100c& =9c+420\\ c^3-109c-420& =0\\ (c-12)(c+7)(c+5)& =0\end{array}$$ Or similarly into equation $(5)$ to check: $$\begin{array}{rl}\frac{c^2-9}{10}c& =10c+42\\ \frac{c^3-9c}{10}& =10c+42\\ c^3-9c& =100c+420\\ c^3-109c-420& =0\\ (c-12)(c+7)(c+5)& =0\end{array}$$ $c$, being a length, must be positive, implying that $c=12$. In fact, this is reasonable, since $10+3\approx 12$ in the pentagon with apparently obtuse angles. Plugging this back into equations $(1)$ and $(2)$ we find that $a=\frac{44}{3}$ and $b=\frac{135}{10}=\frac{27}{2}$. We desire $3c+a+b=3\cdot 12+\frac{44}{3}+\frac{27}{2}=\frac{216+88+81}{6}=\frac{385}{6}$, so it follows that the answer is $385+6=\boxed{391}$