Question

长方形abcd的长为a宽为b,ef为bc、cd中点,de、bf交于点a,求abgd面积?

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Answer 1

作gm⊥cd交cd于m，作gn⊥bc交bc于n.
gm/(bc/2)=dm/cd
gm=[bc*(cd-gn)]/(2cd)
=bc/2-(bc*gn)/(2cd)
=a/2-(a*gn)(2b) -----(1)
gn/(cd/2)=bn/bc
gn=[cd*(bc-gm)]/(2bc)
=b/2-(b*gm)/(2a)
2a*gn=ab-b*gm ------(2）
(1)代入(2)
2a*gn=ab-b*[a/2-(a*gn)(2b)]
=ab/2+a*gn/2
gn=b/3
gm=a/3
S△bcg=bc*gn/2=ab/6
S△cdg=cd*gm/2=ab/6
Sabgd=Sabcd-S△bcg-S△cdg
=ab-ab/6-ab/6
=(2ab)/3
答：abgd面积(2ab)/3。

Answer 2

kao