First Solution: Let $|EB| = x$ and $|AE| = y$, then $|DC| = x + y$,
$$|BC| = \frac{32}{x}, \quad |AF| = \frac{24}{y}, \quad \text{and so } |FD| = \frac{32}{x} - \frac{24}{y}.$$

$$30 = \frac{1}{2} \left( \frac{32}{x} - \frac{24}{y} \right) (x + y) = 4 + 16 \frac{y}{x} - 12 \frac{x}{y}.$$
Multiply across by $\frac{y}{x}$ and rearrange to get the equation
$$16 \left(\frac{y}{x}\right)^2 - 26 \left(\frac{y}{x}\right) - 12 = 0$$
which has only one positive solution $\frac{y}{x} = 2$. If $T$ denotes the area of triangle $EFC$, the area of $ABCD$ is
$$16 + 12 + 30 + T = \frac{32}{x}(x+y) = 32\left(1+\frac{y}{x}\right) = 96$$
and so $T = 96 - 58 = 38$.

Second Solution: (after Adam Kielthy) Let
$$a = |AE|, \quad b = |EB|, \quad c = |AF|, \quad d = |FD|$$
and denote the triangle areas as follows
$$\begin{aligned} Y &= |EBC| = \frac{b(c+d)}{2} & Z &= |CDF| = \frac{d(a+b)}{2} \\ W &= |AEF| = \frac{ac}{2} & X &= |ECF| \end{aligned}$$
from which we obtain $ac = 2W$, $ad = 2Z - bd$, $bc = 2Y - bd$.

The area of the rectangle is equal to
$$X + Y + Z + W = (a+b)(c+d) = ac + ad + bc + bd = 2(Y + Z + W) - bd,$$
which implies $bd = Y + Z + W - X$. On the other hand, from $YZ = bd(a+b)(c+d)/4$, we obtain $bd = 4YZ/(Y + Z + W + X)$. Therefore, $Y + Z + W - X = 4YZ/(Y + Z + W + X)$, i.e. $(Y + Z + W)^2 - X^2 = 4YZ$. This implies $X^2 = (Y + Z + W)^2 - 4YZ = (16+30+12)^2 - 4 \cdot 16 \cdot 30 = 1444$, hence $X = 38$.