1. **State the problem:** We have a right triangle where one leg is 1 millimeter longer than the shorter leg, and the hypotenuse is 2 millimeters longer than the shorter leg. We need to find the lengths of all sides. 2. **Define variables:** Let the length of the shorter leg be $x$ millimeters. 3. **Express other sides in terms of $x$:** - The other leg is $x + 1$ millimeters. - The hypotenuse is $x + 2$ millimeters. 4. **Use the Pythagorean theorem:** For a right triangle, the sum of the squares of the legs equals the square of the hypotenuse: $$x^2 + (x + 1)^2 = (x + 2)^2$$ 5. **Expand and simplify:** $$x^2 + (x^2 + 2x + 1) = x^2 + 4x + 4$$ $$x^2 + x^2 + 2x + 1 = x^2 + 4x + 4$$ 6. **Combine like terms:** $$2x^2 + 2x + 1 = x^2 + 4x + 4$$ 7. **Bring all terms to one side:** $$2x^2 + 2x + 1 - x^2 - 4x - 4 = 0$$ $$x^2 - 2x - 3 = 0$$ 8. **Factor the quadratic:** $$ (x - 3)(x + 1) = 0 $$ 9. **Solve for $x$:** $$x - 3 = 0 \Rightarrow x = 3$$ $$x + 1 = 0 \Rightarrow x = -1$$ Since length cannot be negative, $x = 3$ millimeters. 10. **Find the other sides:** - Other leg: $x + 1 = 3 + 1 = 4$ millimeters. - Hypotenuse: $x + 2 = 3 + 2 = 5$ millimeters. **Final answer:** The lengths of the sides are 3 millimeters, 4 millimeters, and 5 millimeters.