1. **Problem Statement:** (i) Starting from a right triangle with sides 1, 1, and $\sqrt{2}$, 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:** Use the Pythagorean theorem: in a right triangle with legs $a$ and $b$, the hypotenuse is $\sqrt{a^2 + b^2}$. 3. **Step (i) Construction of $\sqrt{3}$:** - Start with the given right triangle with legs 1 and 1, hypotenuse $\sqrt{2}$. - To get $\sqrt{3}$, construct a right triangle with legs $\sqrt{2}$ and 1. - By Pythagoras, hypotenuse = $\sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}$. - So, draw a segment of length $\sqrt{2}$ (the hypotenuse of the original triangle). - At one endpoint, construct a perpendicular segment of length 1. - Connect the free ends to form the hypotenuse of length $\sqrt{3}$. 4. **Step (ii) Construction of $\sqrt{n}$ for $n=2,3,4,5,6,\ldots$:** - Start with segment length 1. - Construct $\sqrt{2}$ as the hypotenuse of a right triangle with legs 1 and 1. - Then construct $\sqrt{3}$ as above. - To get $\sqrt{4}$, construct a right triangle with legs $\sqrt{3}$ and 1, hypotenuse $\sqrt{4} = 2$. - Continue this process iteratively: for $\sqrt{n}$, construct a right triangle with legs $\sqrt{n-1}$ and 1. - The hypotenuse will be $\sqrt{n}$. 5. **Summary formula:** $$\text{Hypotenuse} = \sqrt{(\sqrt{n-1})^2 + 1^2} = \sqrt{n-1 + 1} = \sqrt{n}$$ This method uses the Pythagorean theorem repeatedly to construct $\sqrt{n}$ for any natural number $n$. **Final answer:** The construction of $\sqrt{3}$ and $\sqrt{n}$ is done by iteratively constructing right triangles with legs 1 and the previous hypotenuse length.