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Question Number 173321 by pete last updated on 09/Jul/22

The diagonal of a rectangualar field is 169m.  If the ratio of the lengh to the width is  12:5, find its dimensions.

$$\mathrm{The}\:\mathrm{diagonal}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangualar}\:\mathrm{field}\:\mathrm{is}\:\mathrm{169m}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lengh}\:\mathrm{to}\:\mathrm{the}\:\mathrm{width}\:\mathrm{is} \\ $$$$\mathrm{12}:\mathrm{5},\:\mathrm{find}\:\mathrm{its}\:\mathrm{dimensions}. \\ $$

Answered by Rasheed.Sindhi last updated on 09/Jul/22

Length: 12k , Width: 5k  D=(√((12k)^2 +(5k)^2 )) =169            144k^2 +25k^2 =(169)^2               169k^2 =169^2                      k^2 =169                    k=13  Length: 12k=12×13=156  Width:   5k=5×13=65

$${Length}:\:\mathrm{12}{k}\:,\:{Width}:\:\mathrm{5}{k} \\ $$$${D}=\sqrt{\left(\mathrm{12}{k}\right)^{\mathrm{2}} +\left(\mathrm{5}{k}\right)^{\mathrm{2}} }\:=\mathrm{169} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{144}{k}^{\mathrm{2}} +\mathrm{25}{k}^{\mathrm{2}} =\left(\mathrm{169}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{169}{k}^{\mathrm{2}} =\mathrm{169}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}^{\mathrm{2}} =\mathrm{169} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}=\mathrm{13} \\ $$$${Length}:\:\mathrm{12}{k}=\mathrm{12}×\mathrm{13}=\mathrm{156} \\ $$$${Width}:\:\:\:\mathrm{5}{k}=\mathrm{5}×\mathrm{13}=\mathrm{65} \\ $$

Commented by pete last updated on 09/Jul/22

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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