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Question Number 3838 by Rasheed Soomro last updated on 22/Dec/15

Show that the construction of        the rectangle of minimum         perimeter when its area  is ab         where a=AB and b=CD are  given  is possible with ruler and compass.

$${Show}\:{that}\:{the}\:{construction}\:{of} \\ $$$$\:\:\:\:\:\:\mathrm{the}\:\mathrm{rectangle}\:\mathrm{of}\:\mathrm{minimum}\: \\ $$$$\:\:\:\:\:\:\mathrm{perimeter}\:\mathrm{when}\:\mathrm{its}\:\mathrm{area}\:\:\mathrm{is}\:\boldsymbol{\mathrm{ab}}\: \\ $$$$\:\:\:\:\:\:\mathrm{where}\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{AB}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{CD}}\:\mathrm{are}\:\:\mathrm{given} \\ $$$${is}\:{possible}\:{with}\:{ruler}\:{and}\:{compass}.\:\:\:\:\: \\ $$$$ \\ $$

Commented by prakash jain last updated on 22/Dec/15

For a given area square has minimum  permeter among rectangles.  side of square=(√(ab))

$$\mathrm{For}\:\mathrm{a}\:\mathrm{given}\:\mathrm{area}\:\mathrm{square}\:\mathrm{has}\:\mathrm{minimum} \\ $$$$\mathrm{permeter}\:\mathrm{among}\:\mathrm{rectangles}. \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{square}=\sqrt{{ab}} \\ $$

Commented by Rasheed Soomro last updated on 22/Dec/15

To  Sir Prakash  Question is then ′to show that a rectangle  can be squared with ruler and compass.′  But you don′t need to do so, because in  answering some other questions you have  already proved that ab can be constructed  when a and b are given and also squareroot  of any number...   If anybody else wants to try,he can.

$$\mathcal{T}{o}\:\:\mathcal{S}{ir}\:\mathcal{P}{rakash} \\ $$$${Question}\:{is}\:{then}\:'{to}\:{show}\:{that}\:{a}\:{rectangle} \\ $$$${can}\:{be}\:{squared}\:{with}\:{ruler}\:{and}\:{compass}.' \\ $$$$\boldsymbol{\mathrm{But}}\:{you}\:{don}'{t}\:{need}\:{to}\:{do}\:{so},\:{because}\:{in} \\ $$$${answering}\:{some}\:{other}\:{questions}\:{you}\:{have} \\ $$$${already}\:{proved}\:{that}\:\boldsymbol{\mathrm{ab}}\:{can}\:{be}\:{constructed} \\ $$$${when}\:\boldsymbol{\mathrm{a}}\:{and}\:\boldsymbol{\mathrm{b}}\:{are}\:{given}\:{and}\:{also}\:{squareroot} \\ $$$${of}\:{any}\:{number}...\: \\ $$$$\mathcal{I}{f}\:{anybody}\:{else}\:{wants}\:{to}\:{try},{he}\:{can}. \\ $$

Commented by prakash jain last updated on 22/Dec/15

Answers to previous question already show  that ab and (√(ab)) can be drawn.

$$\mathrm{Answers}\:\mathrm{to}\:\mathrm{previous}\:\mathrm{question}\:\mathrm{already}\:\mathrm{show} \\ $$$$\mathrm{that}\:{ab}\:\mathrm{and}\:\sqrt{{ab}}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}. \\ $$

Commented by Rasheed Soomro last updated on 22/Dec/15

THanK^S !

$$\mathcal{TH}{an}\mathcal{K}^{\mathcal{S}} ! \\ $$

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