Question Number 194240 by cherokeesay last updated on 01/Jul/23
Answered by BaliramKumar last updated on 01/Jul/23
$$\mathrm{28}\:\checkmark \\ $$
Answered by a.lgnaoui last updated on 01/Jul/23
$$\mathrm{t}=\frac{\mathrm{3}}{\mathrm{14}}\:\:\:\Rightarrow\mathrm{cos}\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:\:=\frac{\mathrm{14}\sqrt{\mathrm{205}}}{\:\mathrm{205}} \\ $$$$\mathrm{BI}=\mathrm{BFcos}\:\alpha\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{3}\sqrt{\mathrm{205}}}{\mathrm{205}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{3}}{\mathrm{2BF}}\:\:\:\:\mathrm{BF}=\frac{\mathrm{3}}{\mathrm{2sin}\:\alpha}=\frac{\sqrt{\mathrm{205}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{BI}=\frac{\sqrt{\mathrm{205}}}{\mathrm{2}}×\frac{\mathrm{14}\sqrt{\mathrm{205}}}{\mathrm{205}}=\mathrm{7} \\ $$$$\mathrm{dinc}\:\:\:\:\:\mathrm{MI}=\mathrm{11}−\mathrm{BI}=\mathrm{4} \\ $$$$\mathrm{Area}=\frac{\boldsymbol{\mathrm{ME}}×\boldsymbol{\mathrm{MI}}}{\mathrm{2}}=\frac{\mathrm{14}×\mathrm{4}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Area}}\left(\boldsymbol{\mathrm{MEF}}\right)=\mathrm{28} \\ $$
Commented by a.lgnaoui last updated on 01/Jul/23
