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Junior high school mathematics. Thank you for your urgent solution.
Solution: extended ED=DH, connected CH,

∫BD = DC,ED=DH

A quadrilateral is a parallelogram.

∴BE∥CH,BE=CH

Point c is CF∨EH.

∫CH∨BE,CF∨EH

∴ quadrilateral EFCH is a parallelogram.

∴CH=EF,CF∥EH

∴BE=EF

∫CF∨EH

∴<; AFC =< doctorofaeronautics (abbreviation of doctor of aeronautics)

∵& lt; A =< doctorofaeronautics (abbreviation of Doctor of Aeronautics)

∴<; A =< automatic frequency control

△ AFC is an isosceles triangle.

Let cm?af be after point C.

∴FM=AM

Let BE=EF=a and AB=6.

∴AF=6-2a, then FM = af/2 = 3a.

Tan< AEC = cm/em = cm/(ef+FM) = cm/(a+3-a) = cm/3 = 4/3.

∴CM=4

∴s△abc= 1/2×ab×cm= 1/2×6×4= 12