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Question Number 48976 by peter frank last updated on 30/Nov/18

Answered by $@ty@m last updated on 01/Dec/18

L=3B−4 ...(1) b=B−2 ...(2) l=2b+2...(3) L.B−lb=44 ...(4) ⇒(3B−4).B−(2b+2).b=44 ⇒(3B−4).B−{2(B−2)+2).(B−2)=44 ⇒3B^2 −4B−(2B−2).(B−2)=44 ⇒3B^2 −4B−(2B^2 −6B+4)=44 ⇒B^2 +2B−48=0 ⇒B=((−2±(√(4+192)))/2) ⇒B=6 (B=−8 is inadmissible) ⇒L=3×6−4=14 ∴Area=84m^2

$${L}=\mathrm{3}{B}−\mathrm{4}\:...\left(\mathrm{1}\right) \\ $$$${b}={B}−\mathrm{2}\:...\left(\mathrm{2}\right) \\ $$$${l}=\mathrm{2}{b}+\mathrm{2}...\left(\mathrm{3}\right) \\ $$$${L}.{B}−{lb}=\mathrm{44}\:...\left(\mathrm{4}\right) \\ $$$$\Rightarrow\left(\mathrm{3}{B}−\mathrm{4}\right).{B}−\left(\mathrm{2}{b}+\mathrm{2}\right).{b}=\mathrm{44} \\ $$$$\Rightarrow\left(\mathrm{3}{B}−\mathrm{4}\right).{B}−\left\{\mathrm{2}\left({B}−\mathrm{2}\right)+\mathrm{2}\right).\left({B}−\mathrm{2}\right)=\mathrm{44} \\ $$$$\Rightarrow\mathrm{3}{B}^{\mathrm{2}} −\mathrm{4}{B}−\left(\mathrm{2}{B}−\mathrm{2}\right).\left({B}−\mathrm{2}\right)=\mathrm{44} \\ $$$$\Rightarrow\mathrm{3}{B}^{\mathrm{2}} −\mathrm{4}{B}−\left(\mathrm{2}{B}^{\mathrm{2}} −\mathrm{6}{B}+\mathrm{4}\right)=\mathrm{44} \\ $$$$\Rightarrow{B}^{\mathrm{2}} +\mathrm{2}{B}−\mathrm{48}=\mathrm{0} \\ $$$$\Rightarrow{B}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{192}}}{\mathrm{2}} \\ $$$$\Rightarrow{B}=\mathrm{6}\:\left({B}=−\mathrm{8}\:{is}\:{inadmissible}\right) \\ $$$$\Rightarrow{L}=\mathrm{3}×\mathrm{6}−\mathrm{4}=\mathrm{14} \\ $$$$\therefore{Area}=\mathrm{84}{m}^{\mathrm{2}} \\ $$

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