一。如图。设A(0,0),D(2a,0),B(3a,0),C(3b,3h),F(b,h).有(1/2)×3a×3h=S=S⊿ABC.ah=2S/9.FB方程:h/(b-3a)=y/(x-3a).即hx-3ah=by-3ay.CD构成:3h/(3b-2a)=y/(x-2a)。即3hx-6ah=3by-2ay.消去x,得y=3h/7.S⊿HDB=(1/2)×(3h/7)×a=3ha/14=(3/14)×(2S/9)=S/21.同理S⊿AGF=S⊿CKE=S/21.四边形EKHB面积=四边形GADH面积=四边...
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