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Question Number 3249 by Rasheed Soomro last updated on 08/Dec/15

$$\mathcal{H}owcould\sqrt{5}bedrawnonnumberedlineusing$$$$scaleandcompassonly?(Exactly\sqrt{5}notitsdecimalapproximation.)$$

Answered by prakash jain last updated on 08/Dec/15

$$\mathrm{For}\sqrt{5}$$$$\mathrm{draw}\mathrm{a}\mathrm{horizonal}\mathrm{line}\mathrm{AB}\mathrm{of}\mathrm{length}5+1=6$$$$\mathrm{mark}\mathrm{a}\mathrm{point}\mathrm{at}5\left(\mathrm{C}\right).$$$$\mathrm{bisect}\mathrm{AB}\mathrm{and}\mathrm{get}\mathrm{center}\mathrm{point}\mathrm{D}.$$$$\mathrm{draw}\mathrm{a}\mathrm{semicircle}\mathrm{of}\mathrm{radius}\mathrm{AD}.$$$$\mathrm{draw}{\mathrm{\perp }}^r\mathrm{to}\mathrm{AB}\mathrm{at}\mathrm{C}.\mathrm{Let}\mathrm{it}\mathrm{intersect}\mathrm{semicircle}$$$$\mathrm{at}\mathrm{E}$$$$\mathrm{length}\mathrm{of}\mathrm{CE}=\sqrt{5}$$$$\mathrm{DCE}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angled}\mathrm{triangle}$$$$\mathrm{DE}=3$$$$\mathrm{CD}=2$$$${\mathrm{DE}}^2={\mathrm{CD}}^2+{\mathrm{CE}}^2$$$${\mathrm{CE}}^2=5\mathrm{or}\mathrm{CE}=\sqrt{5}$$

Commented by RasheedAhmad last updated on 08/Dec/15

$$\mathcal{N}ice!$$

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