1. **State the problem:** We have a right triangle where one leg is 4 m less than the hypotenuse, and the other leg is 2 m less than the hypotenuse. We need to find the length of the hypotenuse. 2. **Set variables:** Let the hypotenuse be $x$ meters. 3. **Express legs in terms of $x$:** - First leg = $x - 4$ - Second leg = $x - 2$ 4. **Use the Pythagorean theorem:** For a right triangle with legs $a$, $b$ and hypotenuse $c$, the relation is: $$a^2 + b^2 = c^2$$ 5. **Apply the theorem:** $$(x - 4)^2 + (x - 2)^2 = x^2$$ 6. **Expand the squares:** $$(x^2 - 8x + 16) + (x^2 - 4x + 4) = x^2$$ 7. **Combine like terms:** $$x^2 - 8x + 16 + x^2 - 4x + 4 = x^2$$ $$2x^2 - 12x + 20 = x^2$$ 8. **Bring all terms to one side:** $$2x^2 - 12x + 20 - x^2 = 0$$ $$x^2 - 12x + 20 = 0$$ 9. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-12$, $c=20$. 10. **Calculate the discriminant:** $$\sqrt{(-12)^2 - 4 \times 1 \times 20} = \sqrt{144 - 80} = \sqrt{64} = 8$$ 11. **Find the roots:** $$x = \frac{12 \pm 8}{2}$$ 12. **Calculate each root:** - $$x = \frac{12 + 8}{2} = \frac{20}{2} = 10$$ - $$x = \frac{12 - 8}{2} = \frac{4}{2} = 2$$ 13. **Check for valid solution:** Hypotenuse must be longer than legs, so $x=2$ is invalid because legs would be negative or zero. 14. **Final answer:** The length of the hypotenuse is $\boxed{10}$ meters.