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Question Number 3647 by Rasheed Soomro last updated on 17/Dec/15

$$Drawalinesegmentequalto$$$$\mathit{a}\mathit{b}unitswhenAB=\mathit{a}unitsandCD=\mathit{b}units$$$$aregiven.Onlyrulerandcompassmaybe$$$$used.$$

Commented by prakash jain last updated on 17/Dec/15

$$\mathrm{Draw}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}\mathrm{CD}\mathrm{with}\mathrm{center}\mathrm{C}.$$$$\mathrm{Draw}\mathrm{a}\mathrm{line}\parallel \mathrm{to}\mathrm{AB}\mathrm{passing}\mathrm{thru}\mathrm{C}.\mathrm{Let}$$$$\mathrm{us}\mathrm{say}\mathrm{it}\mathrm{cuts}\mathrm{the}\mathrm{circle}\mathrm{at}\mathrm{E}.$$$$\mathrm{CE}=\mathrm{CD}=\mathrm{radius}=b$$$$\mathrm{CE}\parallel \mathrm{AB}$$$$\mathrm{draw}{\mathrm{\perp }}^r\mathrm{to}\mathrm{AB}\mathrm{at}\mathrm{B}\mathrm{of}\mathrm{unit}\mathrm{length}(\mathrm{BF}=1)$$$$\mathrm{connect}\mathrm{AF}.△\mathrm{AFB}\mathrm{is}\mathrm{right}\mathrm{angle}\mathrm{triangle}$$$$\mathrm{with}\mathrm{\angle }\mathrm{B}=90°.$$$$\tan \mathrm{\angle }\mathrm{AFB}=\frac{\mathrm{AB}}{\mathrm{BF}}=a$$$$\mathrm{Draw}\mathrm{a}\mathrm{line}l\parallel \mathrm{to}\mathrm{AF}\mathrm{passing}\mathrm{thru}\mathrm{C}.$$$$\mathrm{Draw}\mathrm{a}\mathrm{line}{\mathrm{\perp }}^r\mathrm{at}\mathrm{CE}\mathrm{at}\mathrm{E}.$$$$\mathrm{Let}\mathrm{us}\mathrm{sat}\mathrm{it}\mathrm{cut}l\mathrm{at}\mathrm{G}.$$$$△\mathrm{GCE},\mathrm{\angle }\mathrm{GEC}=90°,$$$$\mathrm{AF}\parallel \mathrm{GC},\mathrm{AB}\parallel \mathrm{CE}$$$$\mathrm{\angle }\mathrm{AFB}=\mathrm{\angle }\mathrm{CGE}$$$$\tan \mathrm{\angle }\mathrm{CGE}=\frac{\mathrm{GE}}{\mathrm{CE}}=a\Rightarrow \mathrm{GE}=ab$$$$\mathrm{Same}\mathrm{construction}\mathrm{as}\mathrm{used}\mathrm{in}\mathrm{computing}$$$$\mathrm{square}.\mathrm{unit}\mathrm{length}\mathrm{is}\mathrm{assumed}.$$

Commented by Rasheed Soomro last updated on 19/Dec/15

$$\mathcal{T}h\alpha n\mathbb{k}\mathcal{S}!$$$$First\mathcal{I}willtrymyownthen\mathcal{I}willread$$$$youranswer\left(inordertotrainmyself\right).$$

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