1. **Problem Statement:** (i) Starting from a right triangle with sides 1, 1, and hypotenuse $\sqrt{2}$, construct $\sqrt{3}$ using straightedge and compass. (ii) Use this method to construct $\sqrt{n}$ for $n=2,3,4,5,6,\ldots$. 2. **Key Idea and Formula:** The Pythagorean theorem states for a right triangle with legs $a$ and $b$ and hypotenuse $c$: $$c=\sqrt{a^2+b^2}$$ This allows us to construct new lengths by adding squares of known lengths. 3. **Step-by-step Construction for $\sqrt{3}$:** - Start with a right triangle with legs 1 and 1, hypotenuse $\sqrt{2}$. - To get $\sqrt{3}$, construct a right triangle with one leg of length $\sqrt{2}$ and the other leg of length 1. - By Pythagoras: $$c=\sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}$$ 4. **General Construction for $\sqrt{n}$:** - Begin with a segment of length 1. - Construct a right triangle where one leg is $\sqrt{n-1}$ (constructed from previous step) and the other leg is 1. - The hypotenuse will be: $$\sqrt{(\sqrt{n-1})^2 + 1^2} = \sqrt{n-1 + 1} = \sqrt{n}$$ - Repeat this process iteratively to construct $\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6}, \ldots$. 5. **Explanation:** Each step uses the previous hypotenuse as one leg and 1 as the other leg to form a new right triangle. This method uses only straightedge and compass constructions of right triangles and segments. **Final answer:** You can construct $\sqrt{3}$ by forming a right triangle with legs 1 and $\sqrt{2}$. Similarly, $\sqrt{n}$ is constructed by iteratively building right triangles with legs 1 and $\sqrt{n-1}$.