Ireland

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

$$30=\frac{1}{2}\left(\frac{32}{x}-\frac{24}{y}\right)(x+y)=4+16\frac{y}{x}-12\frac{x}{y}.$$ 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 $$a=|AE|,b=|EB|,c=|AF|,d=|FD|$$ and denote the triangle areas as follows $$\begin{array}{rlrl}Y& =|EBC|=\frac{b(c+d)}{2}& Z& =|CDF|=\frac{d(a+b)}{2}\\ W& =|AEF|=\frac{ac}{2}& X& =|ECF|\end{array}$$ 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 $X=38$.