Let the sides of the right triangle be $6$, $a$, and $c$, where $a$ is one of the legs, and $c$ is the hypotenuse. By the Pythagorean Theorem, we have $36+a^2=c^2$. Thus, $c$ is the hypotenuse of a right traingle with legs $6$ and $a$. Since $6$ and $a$ are positive integers, $c$ must be an integer. We use casework to find out the different possible values of $c$:
Case 1: $c < 10$. The smallest possible value of $c$ is $7$. Since $a > 0$, the values of $c$ are $7,$ $8$, and $9$. Clearly, $7$ and $9$ ($c=9$ since the problem asked for all non-congruent triangles) will not result in a right triangle since $6^2+7^2,6^2+9^2 < 9^2$. However, if we use $c=8$, $6^2+8^2=10^2$, so this triangle works.
Case 2: $11 > c > 10$. The values of $c$ are $11$ and $12$. For $c=11$, $6^2+11^2=13^2$, so this triangle works. $6^2+12^2 > 13^2$, so $c=12$ will not work.
Case 3: $14 > c > 12$. Using the pythagorean numbers, we see that there are no triangles that satisfy this condition.
Thus, there are a total of $\boxed{3}$ triangles that fit this condition.
Find the number of non-congruent right triangles, where all the sides are positive integers, and one of the legs is $6$.
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