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Question Number 194240 by cherokeesay last updated on 01/Jul/23

Answered by BaliramKumar last updated on 01/Jul/23

28 ✓

$$\mathrm{28}\:\checkmark \\ $$

Answered by a.lgnaoui last updated on 01/Jul/23

t=(3/(14)) ⇒cosα=(1/( (√(1+t^2 )))) =((14(√(205)))/( 205)) BI=BFcos α sin α=((3(√(205)))/(205)) sin α=(3/(2BF)) BF=(3/(2sin α))=((√(205))/2) ⇒BI=((√(205))/2)×((14(√(205)))/(205))=7 dinc MI=11−BI=4 Area=((ME×MI)/2)=((14×4)/2) Area(MEF)=28

$$\mathrm{t}=\frac{\mathrm{3}}{\mathrm{14}}\:\:\:\Rightarrow\mathrm{cos}\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:\:=\frac{\mathrm{14}\sqrt{\mathrm{205}}}{\:\mathrm{205}} \\ $$$$\mathrm{BI}=\mathrm{BFcos}\:\alpha\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{3}\sqrt{\mathrm{205}}}{\mathrm{205}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{\mathrm{2BF}}\:\:\:\:\mathrm{BF}=\frac{\mathrm{3}}{\mathrm{2sin}\:\alpha}=\frac{\sqrt{\mathrm{205}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{BI}=\frac{\sqrt{\mathrm{205}}}{\mathrm{2}}×\frac{\mathrm{14}\sqrt{\mathrm{205}}}{\mathrm{205}}=\mathrm{7} \\ $$$$\mathrm{dinc}\:\:\:\:\:\mathrm{MI}=\mathrm{11}−\mathrm{BI}=\mathrm{4} \\ $$$$\mathrm{Area}=\frac{\boldsymbol{\mathrm{ME}}×\boldsymbol{\mathrm{MI}}}{\mathrm{2}}=\frac{\mathrm{14}×\mathrm{4}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Area}}\left(\boldsymbol{\mathrm{MEF}}\right)=\mathrm{28} \\ $$

Commented by a.lgnaoui last updated on 01/Jul/23

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