8. (10 points) Five points $A, B, C, D, E$ on a plane satisfy: $A B=16$ cm, $B C=8$ cm, $A D=10$ cm, $D E=2$ cm, $A C=24$ cm, $A E=12$ cm. If the area of triangle $E A B$ is 96 square cm, then the distance from point $A$ to $C D$ is $\qquad$ cm.
【Analysis】Determine the positional relationship of the five points. $A B+B C=16+8=24=A C$, so, $A 、 B 、 C$ are on the same straight line. Similarly, $D$ is between $A 、 E$;

Since the area of $\triangle E A B$ is 24 square centimeters, and only when angle $A$ is a 90-degree right angle, its area is, so, angle $A$ is a right angle; then $\triangle C A D$ is also a right triangle, and according to the Pythagorean theorem, $C D=13$ centimeters can be calculated;

Let: The distance from point $A$ to $C D$ (which is the height on side $C D$) be $X$, and set up an equation to solve for $X$.

【Solution】E
Solution:
According to the problem, it can be known that $A B C$ is on the same straight line, otherwise it would not form $A C=12$ centimeters, similarly, $A D E$ is also on the same straight line.

Since: The area of $\triangle E A B$ is 24 square centimeters, and only when angle $A$ is a 90-degree right angle, its area is: $A B \times(A D+D E)$ $\div 2=8 \times 6 \div 2=24$, so, angle $A$ is a right angle.
If $A$ is a right angle, then $\triangle C A D$ is also a right triangle, and according to the Pythagorean theorem $C D \times C D=A D^{2}+A C^{2}$, solving gives $C D=13$ centimeters.

Let: The distance from point $A$ to $C D$ (which is the height on side $C D$) be $X$, set up the equation:
$$
13 \times X / 2=5 \times 12 \div 2
$$

Therefore: $X \approx 4.62$ centimeters