Tricky Geometry Problem: can you figure this out?

[Guest]

Hello everyone,,

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Thanks for the help....

[Guest]

Let EH be the height of the triangle ABE . Let EM bethe height of the triangle ECD.
[tex]S_{ABCD }[/tex]=?

[tex]\triangle[/tex]ABE[tex]\approx[/tex][tex]\triangle[/tex]ECD (1 sign) [tex]\Rightarrow[/tex] [tex]\frac{S_{ABE }}{S_{ECD }}[/tex]=[tex]\frac{EH^{2}}{EM^{2}}[/tex]
[tex]\frac{18}{8}[/tex]=([tex]\frac{EH}{EM})^{2}[/tex] ; [tex]\sqrt{\frac{9}{4}}[/tex]=[tex]\frac{MH-EM}{EM}[/tex] ; [tex]\frac{3}{2}[/tex]=[tex]\frac{MH}{EM}[/tex]-1 ; [tex]\frac{MH}{EM}[/tex]=[tex]\frac{5}{2}[/tex] [tex]\Rightarrow[/tex]

MH=5x ;EM=2x ; EH=5x-2x=3x (1)
[tex]S_{ABE }[/tex]=18
[tex]\frac{AB.EH}{2}[/tex]=18
AB.3x=36
AB=[tex]\frac{12}{x}[/tex] (2)

[tex]S_{ECD }[/tex]=8
[tex]\frac{CD.EM}{2}[/tex]=8
CD.2x=16
CD=[tex]\frac{8}{x}[/tex] (3)

[tex]S_{ABCD }[/tex]=[tex]\frac{1}{2}[/tex](AB+CD)MH=[tex]\frac{1}{2}[/tex]([tex]\frac{12}{x}[/tex]+[tex]\frac{8}{x}[/tex])5x=[tex]\frac{1}{2}[/tex].[tex]\frac{20}{x}[/tex].5x=50
B) 50