Question Number 3249 by Rasheed Soomro last updated on 08/Dec/15
![How could (√5) be drawn on numbered line using scale and compass only? (Exactly (√5) not its decimal approximation.)](https://www.tinkutara.com/question/Q3249.png)
$$\mathcal{H}{ow}\:{could}\:\sqrt{\mathrm{5}}\:\:{be}\:{drawn}\:{on}\:{numbered}\:{line}\:{using} \\ $$$${scale}\:{and}\:{compass}\:{only}?\:\left({Exactly}\:\sqrt{\mathrm{5}}\:{not}\:{its}\:{decimal}\:{approximation}.\right) \\ $$
Answered by prakash jain last updated on 08/Dec/15
![For (√5) draw a horizonal line AB of length 5+1=6 mark a point at 5 (C). bisect AB and get center point D. draw a semicircle of radius AD. draw ⊥^r to AB at C. Let it intersect semicircle at E length of CE=(√5) DCE is a right angled triangle DE=3 CD=2 DE^2 =CD^2 +CE^2 CE^2 =5 or CE=(√5)](https://www.tinkutara.com/question/Q3256.png)
$$\mathrm{For}\:\sqrt{\mathrm{5}} \\ $$$$\mathrm{draw}\:\mathrm{a}\:\mathrm{horizonal}\:\mathrm{line}\:\mathrm{AB}\:\mathrm{of}\:\mathrm{length}\:\mathrm{5}+\mathrm{1}=\mathrm{6} \\ $$$$\mathrm{mark}\:\mathrm{a}\:\mathrm{point}\:\mathrm{at}\:\mathrm{5}\:\left(\mathrm{C}\right). \\ $$$$\mathrm{bisect}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{get}\:\mathrm{center}\:\mathrm{point}\:\mathrm{D}. \\ $$$$\mathrm{draw}\:\mathrm{a}\:\mathrm{semicircle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{AD}. \\ $$$$\mathrm{draw}\:\:\bot^{{r}} \:\mathrm{to}\:\mathrm{AB}\:\mathrm{at}\:\mathrm{C}.\:\mathrm{Let}\:\mathrm{it}\:\mathrm{intersect}\:\mathrm{semicircle} \\ $$$$\mathrm{at}\:\mathrm{E} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{CE}=\sqrt{\mathrm{5}} \\ $$$$\mathrm{DCE}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angled}\:\mathrm{triangle} \\ $$$$\mathrm{DE}=\mathrm{3}\: \\ $$$$\mathrm{CD}=\mathrm{2} \\ $$$$\mathrm{DE}^{\mathrm{2}} =\mathrm{CD}^{\mathrm{2}} +\mathrm{CE}^{\mathrm{2}} \\ $$$$\mathrm{CE}^{\mathrm{2}} =\mathrm{5}\:\mathrm{or}\:\mathrm{CE}=\sqrt{\mathrm{5}} \\ $$
Commented by RasheedAhmad last updated on 08/Dec/15
![Nice!](https://www.tinkutara.com/question/Q3259.png)
$$\mathcal{N}{ice}! \\ $$