8. As shown in Figure 2, each segment of the broken line $A-B-C-D$ is parallel to the sides of the rectangle, and it divides the rectangle into two equal areas. Point $E$ is on the side of the rectangle such that segment $A E$ also bisects the area of the rectangle.
Given that segment $A B=30, B C=$
$24, C D=10$. Then $D E=$
$\qquad$
8.12.

Solution 1: Let line segment $A E$ intersect $B C$ at point $M$, and draw $E P \perp B C$, with the foot of the perpendicular being $P$.
Let $E D=x, B M=y$. Then $M P=24-x-y$.
By the problem, $S_{\triangle B B Y}=S_{\text {quadrilateral } U C D E}$, then
$15 y=120-5 y+5 x$.
Thus, $4 y=24+x$.
Also, $\triangle A M B \backsim \triangle E M P$, so $\frac{A B}{E P}=\frac{B M}{M P}$.
Therefore, $\frac{30}{10}=\frac{y}{24-x-y}$.
Hence, $4 y=72-3 x$.
From equations (1) and (2), we get $x=12$.
Solution 2: On both sides of the original rectangle, subtract equal-area smaller rectangles from each part so that $C D$ lies on the edge of the remaining rectangle (as shown in Figure 10). At this point, the broken line $A-B-C$ also divides this smaller rectangle into two equal-area parts. When both are further reduced by the area of the shaded part, they are still equal, i.e.,
$$
30 \times 24-10 \times 24=A G \times 40 \Rightarrow A G=12 .
$$

Since $A E$ also divides this smaller rectangle into two equal-area parts, we have $A G=E D=12$.