OpenSAT

We can use the Pythagorean theorem to solve this. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, $10^2 = 6^2 + x^2$, where $x$ is the length of the other leg. Simplifying, we get $100 = 36 + x^2$. Subtracting 36 from both sides, we get $64 = x^2$. Taking the square root of both sides, we get $x = \sqrt{64} = 8$. However, the problem asks for the length of the *other* leg, which is the hypotenuse minus 6, so the answer is $\sqrt{136}$.