Since $|AD| = 100 = 12 + 50 + 38 = |AB| + |BC| + |CD|$, the points $A$, $B$, $C$, $D$ lie on a line in this order. Since
$$|BE|^2 + |EC|^2 = 30^2 + 40^2 = 900 + 1600 = 2500 = 50^2 = |BC|^2,$$
the triangle $EBC$ has a right angle at $E$.

Let $Q$ be the projection of $E$ on the hypotenuse $BC$ of this triangle. Then twice the area of this triangle is equal to
$$|EQ| \times |BC| = |EC| \times |EB|,$$
hence
$$|EQ| = \frac{|EC| \times |EB|}{|BC|} = \frac{40 \times 30}{50} = 24.$$
By Pythagoras,
$$|QC| = \sqrt{|EC|^2 - |EQ|^2} = \sqrt{40^2 - 24^2} = \sqrt{1600 - 576} = \sqrt{1024} = 32,$$
and so $|QD| = |QC| + |CD| = 32 + 38 = 70$, hence
$$|ED| = \sqrt{|QE|^2 + |QD|^2} = \sqrt{24^2 + 70^2} = \sqrt{576 + 4900} = \sqrt{5476} = 74.$$