smc

Without the loss of generality, let $\mathcal{T}$ have vertices $A$, $B$, $C$, and $D$, with $AB=r$ and $CD=s$. Also denote by $P$ the point in the interior of $\mathcal{T}$. Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Observe that $PX=\frac{8}{r}$ and $PY=\frac{4}{s}$. Now using the formula for the area of a trapezoid yields $$14=\frac{1}{2}\cdot XY\cdot (AB+CD)=\frac{1}{2}\left(\frac{8}{r}+\frac{4}{s}\right)(r+s)=6+2\cdot \frac{r}{s}+4\cdot \frac{s}{r}.$$ Thus, the ratio $\rho :=\frac{r}{s}$ satisfies $\rho +2{\rho }^{-1}=4$; solving yields $\rho =\boxed{2+\sqrt{2}}$. (Observe that the given areas of $3$ and $5$ are irrelevant to the ratio $\frac{AB}{CD}$.)