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Question Number 40764 by mondodotto@gmail.com last updated on 27/Jul/18
a piece of wire 40cm long is cut into two parts and each part is then bent into a square.if the sum of these squares is 68cm^2 find the lengths of the two pieces of wire.
$$\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{piece}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{wire}}\:\mathrm{40}\boldsymbol{\mathrm{cm}}\:\boldsymbol{\mathrm{long}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{parts}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{each}}\:\boldsymbol{\mathrm{part}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{bent}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{square}}.\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{these}}\:\boldsymbol{\mathrm{squares}}\:\boldsymbol{\mathrm{is}}\:\mathrm{68}\boldsymbol{\mathrm{cm}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{lengths}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{two}} \\ $$$$\boldsymbol{\mathrm{pieces}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{wire}}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
x=4a_1 40−x=4a_2 a_1 ^2 +a_2 ^2 =68 4a_1 +4a_2 =40 a_1 ^2 +(10−a_1 )^2 =68 2a_1 ^2 −20a_1 +32=0 a_1 ^2 −10a_1 +16=0 (a_1 −2)(a_1 −8)=0 if a_1 =2 a_2 =8 if a_1 =8 a_2 =2 lentth of two piece are 4×2 and 4×8 8cm and 32 cm
$${x}=\mathrm{4}{a}_{\mathrm{1}} \:\:\:\:\mathrm{40}−{x}=\mathrm{4}{a}_{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{68} \\ $$$$\mathrm{4}{a}_{\mathrm{1}} +\mathrm{4}{a}_{\mathrm{2}} =\mathrm{40} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}} +\left(\mathrm{10}−{a}_{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{68} \\ $$$$\mathrm{2}{a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{20}{a}_{\mathrm{1}} +\mathrm{32}=\mathrm{0} \\ $$$${a}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{10}{a}_{\mathrm{1}} +\mathrm{16}=\mathrm{0} \\ $$$$\left({a}_{\mathrm{1}} −\mathrm{2}\right)\left({a}_{\mathrm{1}} −\mathrm{8}\right)=\mathrm{0} \\ $$$${if}\:{a}_{\mathrm{1}} =\mathrm{2}\:\:{a}_{\mathrm{2}} =\mathrm{8}\:\: \\ $$$${if}\:{a}_{\mathrm{1}} =\mathrm{8}\:\:{a}_{\mathrm{2}} =\mathrm{2} \\ $$$${lentth}\:{of}\:{two}\:{piece}\:\:{are}\:\mathrm{4}×\mathrm{2}\:\:{and}\:\mathrm{4}×\mathrm{8} \\ $$$$\mathrm{8}{cm}\:{and}\:\mathrm{32}\:{cm} \\ $$

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