Question-181019

Question Number 181019 by cherokeesay last updated on 20/Nov/22

Answered by som(math1967) last updated on 20/Nov/22

cos120=((2^2 +2^2 −AE^2 )/(2.2.2)) −(1/2)×8=8−AE^2 ⇒AE=(√(12))=2(√3) ar△AEF=ar△DCE=(1/2)×2×2sin120 =(√3) sq unit ar △ALE=(1/2)×1×2(√3)=(√3) ar△BLC=(1/2)×1×2×sin120 =((√3)/2)squnit majent ar=(√3)+(√3)+(√3)+((√3)/2)=((7(√3))/2) ar o hexagon=6×((√3)/4)×4=6(√3)squ ((Green)/(Magenta))=((6(√3)−((7(√3))/2))/((7(√3))/2))=(5/7)

$$\:\:{cos}\mathrm{120}=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −{AE}^{\mathrm{2}} }{\mathrm{2}.\mathrm{2}.\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}=\mathrm{8}−{AE}^{\mathrm{2}} \\ $$$$\Rightarrow{AE}=\sqrt{\mathrm{12}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${ar}\bigtriangleup{AEF}={ar}\bigtriangleup{DCE}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{2}{sin}\mathrm{120} \\ $$$$\:\:=\sqrt{\mathrm{3}}\:{sq}\:{unit} \\ $$$${ar}\:\bigtriangleup{ALE}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{2}\sqrt{\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$$${ar}\bigtriangleup{BLC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\mathrm{2}×{sin}\mathrm{120} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{squnit} \\ $$$${majent}\:{ar}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${ar}\:{o}\:{hexagon}=\mathrm{6}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{4}=\mathrm{6}\sqrt{\mathrm{3}}{squ} \\ $$$$\:\frac{{Green}}{{Magenta}}=\frac{\mathrm{6}\sqrt{\mathrm{3}}−\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$ \\ $$

Commented by som(math1967) last updated on 20/Nov/22