10. (6 points) As shown in the figure, two people, A and B, start from point $A$ at the same time and walk along the sides of the square $A B C D$ in the direction of the arrow. The side length of the square $A B C D$ is 100 meters. A's speed is 1.5 times B's speed. The two people meet for the first time at point $E$. Then, the area of triangle $A D E$ is larger than the area of triangle $B C E$ by square meters.
1000
【Solution】Since the speed of A is 1.5 times the speed of B, the speed ratio of the two is: 1.5:1 = 3:2. Therefore, when the two meet at point E, A has traveled: $(100 \times 4) \times \frac{3}{3+2} = 240$ (meters); B has traveled: $400 - 240 = 160$ (meters);

Thus, $EC = 240 - 100 \times 2 = 40$ (meters), $DE = 160 - 100 = 60$ (meters);
The area of triangle $ADE$ is larger than the area of triangle $BCE$ by:
$$
\begin{array}{l}
60 \times 100 \div 2 - 40 \times 100 \div 2 \\
= 3000 - 2000, \\
= 1000 \text{ (square meters). }
\end{array}
$$

The answer is: 1000.