1. **Problem statement:** Find the value of $x$ in a right triangle where the legs are 6 and 10, and the hypotenuse is $x$. Then determine if the side lengths form a Pythagorean triple. 2. **Formula used:** In a right triangle, the Pythagorean theorem states: $$x^2 = a^2 + b^2$$ where $x$ is the hypotenuse and $a$, $b$ are the legs. 3. **Apply the formula:** $$x^2 = 6^2 + 10^2$$ $$x^2 = 36 + 100$$ $$x^2 = 136$$ 4. **Solve for $x$:** $$x = \sqrt{136}$$ 5. **Simplify the square root:** $$x = \sqrt{4 \times 34} = \sqrt{4} \times \sqrt{34} = 2\sqrt{34}$$ 6. **Calculate decimal approximation:** $$x \approx 2 \times 5.8309518948 = 11.7$$ (to the nearest tenth) 7. **Check if the side lengths form a Pythagorean triple:** A Pythagorean triple consists of three positive integers $a$, $b$, and $c$ such that $a^2 + b^2 = c^2$. Here, the sides are 6, 10, and approximately 11.7 (not an integer), so they do not form a Pythagorean triple. **Final answer:** - $x \approx 11.7$ - The side lengths do not form a Pythagorean triple.