Right Triangle Side Lengths: Solve for x, x+2, and x+4

To solve this problem, we begin by using the Pythagorean theorem, as the triangle is right-angled. Let's identify the hypotenuse:

• The side lengths are given as $x$, $x + 2$, and $x + 4$.
• Since $x + 4$ is the largest, it will serve as the hypotenuse, so we'll denote the sides as follows: $a = x$, $b = x + 2$, and $c = x + 4$.

Using the Pythagorean theorem, we write:

$x^2 + (x + 2)^2 = (x + 4)^2$

Let's expand and simplify the equation:

$x^2 + (x^2 + 4x + 4) = x^2 + 8x + 16$

Simplifying further:

$2x^2 + 4x + 4 = x^2 + 8x + 16$

Rearrange all terms to one side:

$2x^2 + 4x + 4 - x^2 - 8x - 16 = 0$

Simplifying gives:

$x^2 - 4x - 12 = 0$

This is a standard quadratic equation, which we can solve using factoring. By factoring, we find:

$(x - 6)(x + 2) = 0$

Setting each factor equal to zero gives solutions $x = 6$ and $x = -2$. Since $x > 1$, we discard $x = -2$.

The valid solution is $x = 6$.

Now, substitute $x = 6$ back into the expressions for the side lengths:

• The first side: $x = 6$
• The second side: $x + 2 = 6 + 2 = 8$
• The hypotenuse: $x + 4 = 6 + 4 = 10$

Therefore, the lengths of the sides of the triangle are $6$, $8$, and $10$, which matches choice 4.

Therefore, the correct answer is choice 4: $6, 8, 10$.