Answer: 5.

Solution. Let $C D=B C=a, A B=b$. Drop a perpendicular from point $D$ to $B C$, and apply the Pythagorean theorem: $(a-2)^{2}+b^{2}=a^{2}$. From this, $b^{2}=4(a-1), a=\frac{b^{2}}{4}+1$. Suppose $A D=2$ is the smaller base of the trapezoid. The perimeter $P=2 a+b+2a$, so $b>2$. Considering that $a=\frac{b^{2}}{4}+1$, $b$ must be even, yielding the permissible values of $b: 4,6,8,10,12$.

Suppose $A D=2$ is not the smaller base of the trapezoid. Then $a=1$ or $a=2$. In the first case, there is a contradiction with the relation $a=\frac{b^{2}}{4}+1$. In the second case, the quadrilateral $A B C D$ becomes a square, which is not considered a trapezoid.

Comment. A correct and justified solution - 7 points. Proven that $b \leq 12$ - 4 points. Odd values of $b$ not eliminated - 3 points deducted. The value $b=2$ not eliminated - 1 point deducted. The correct answer obtained by selecting numbers that satisfy the condition, but not shown that other answers are impossible - 3 points, if there is an incomplete justification - 4-5 points. Some answers found by selection - 1-2 points. The problem is not solved or solved incorrectly - 0 points.
10.2. How many rectangular trapezoids $A B C D$ exist, where angles $A$ and $B$ are right angles, $A D=2$, $C D=B C$, the sides have integer lengths, and the perimeter is less than 100?