Right Triangle Side Lengths: Solve for x+9, x+2, and x+10

To find the lengths of the sides of the right triangle, we will apply the Pythagorean theorem, which states $a^2 + b^2 = c^2$ for a right triangle, where $c$ is the hypotenuse.

Given the side lengths are $x + 9$, $x + 2$, and $x + 10$, we assume $x + 10$ is the hypotenuse because it is the largest value and confirm it by checking with the theorem.

Substitute into the Pythagorean theorem:
$(x + 2)^2 + (x + 9)^2 = (x + 10)^2$

Let's expand each side and solve for $x$:
$(x + 2)^2 = x^2 + 4x + 4$
$(x + 9)^2 = x^2 + 18x + 81$
$(x + 10)^2 = x^2 + 20x + 100$

Combine these into a single equation:
$x^2 + 4x + 4 + x^2 + 18x + 81 = x^2 + 20x + 100$

Simplify and combine like terms:
$2x^2 + 22x + 85 = x^2 + 20x + 100$

Rearrange to form a quadratic equation:
$2x^2 + 22x + 85 - x^2 - 20x - 100 = 0$
$x^2 + 2x - 15 = 0$

Factor the quadratic equation:
$(x + 5)(x - 3) = 0$

Solve for $x$:
$x + 5 = 0 \Rightarrow x = -5$ (Not valid as $x > 1$)
$x - 3 = 0 \Rightarrow x = 3$

Therefore, substituting $x = 3$ will provide the side lengths:
$<begin>$ - Short side: $x + 2 = 3 + 2 = 5$
$<same time>$ - Other side: $x + 9 = 3 + 9 = 12$
$<sell>$ - Hypotenuse: $x + 10 = 3 + 10 = 13$

These side lengths $5$, $12$, and $13$ form a well-known Pythagorean triple. Therefore, the solution to the problem is $5, 12, 13$.