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Question Number 11114 by ABD last updated on 12/Mar/17

Answered by ajfour last updated on 12/Mar/17

Commented by ajfour last updated on 12/Mar/17

((AB)/(BC))=((sin θ)/(cos θ)) ⇒BC=((4cos θ)/(sin θ)) AG=AB−FD=4−lsin θ GD=BF=7−FE=7−lcos θ Ar(ABCD)=Ar(trapeziumBCDG) +Ar(△AGD) =(1/2)(BC+GD)(FD)+(1/2)AG×GD =(1/2)(((4cos θ)/(sin θ))+7−lcos θ)(lsin θ) +(1/2)(4−lsin θ)(7−lcos θ) comes out =14 .

$$\frac{{AB}}{{BC}}=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:\Rightarrow{BC}=\frac{\mathrm{4cos}\:\theta}{\mathrm{sin}\:\theta} \\ $$$${AG}={AB}−{FD}=\mathrm{4}−{l}\mathrm{sin}\:\theta \\ $$$${GD}={BF}=\mathrm{7}−{FE}=\mathrm{7}−{l}\mathrm{cos}\:\theta \\ $$$${Ar}\left({ABCD}\right)={Ar}\left({trapeziumBCDG}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{Ar}\left(\bigtriangleup{AGD}\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({BC}+{GD}\right)\left({FD}\right)+\frac{\mathrm{1}}{\mathrm{2}}{AG}×{GD} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4cos}\:\theta}{\mathrm{sin}\:\theta}+\mathrm{7}−{l}\mathrm{cos}\:\theta\right)\left({l}\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}−{l}\mathrm{sin}\:\theta\right)\left(\mathrm{7}−{l}\mathrm{cos}\:\theta\right) \\ $$$${comes}\:{out}\:=\mathrm{14}\:. \\ $$

Answered by mrW1 last updated on 13/Mar/17

since DE//AC ⇒A_(ΔACD) =A_(ΔACE) A_(ABCD) =A_(ΔABC) +A_(ΔACD) =A_(ΔABC) +A_(ΔACE) =A_(ΔABE) =((7×4)/2)=14

$${since}\:{DE}//{AC} \\ $$$$\Rightarrow{A}_{\Delta{ACD}} ={A}_{\Delta{ACE}} \\ $$$${A}_{{ABCD}} ={A}_{\Delta{ABC}} +{A}_{\Delta{ACD}} ={A}_{\Delta{ABC}} +{A}_{\Delta{ACE}} \\ $$$$={A}_{\Delta{ABE}} =\frac{\mathrm{7}×\mathrm{4}}{\mathrm{2}}=\mathrm{14} \\ $$

Commented by mrW1 last updated on 12/Mar/17

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