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设正方形边长为a,延长AG、DC相交于H,分别由P、Q做AD的垂线交AD于M、N,∵∠AGB=∠CGH
∠BAG=∠CHG
∴△ABG∽△CHG
∴CH/AB=CG/BG
∴CH=1/2a
DH=3/2a
同理△AEP∽△DHP ∴ DP/EP=DH/AE= 9/2
∴DP/DE = 9/11
△AFQ∽△DHQ ∴ DQ/FQ=DH/AF= 9/4
∴DQ/DF = 9/13
△DMP∽△ADE ∴DP/DE = MP/AE = 9/11
MP=3/11a △DNQ∽△ADF ∴DQ/DF = NQ/AF = 9/13
NQ=6/13a
S△DPQ=S△ADQ-S△ADP= 1/2×6/13a×a-1/2×3/11a×a=27/286S希望对你有帮助!已赞过已踩过你对这个回答的评价是?评论
收起设正方形边长为a,延长AG、DC交于H,△ABG和△GCH,根据鸟头模型可知:△ABG/△GCH=2/1
CH=1/2a
DH=3/2a△AEP和△DPH, 同理:△AEP/△PDH=EP/DP=AE/DH=2/9△AFQ和△DQH ,同理:△AFQ/△DQH=FQ/QD=AF/DH=4/9易知:△DPQ/△DEF=81/143△DPQ=81/143x1/6S=27/286S学过奥数的都懂鸟头模型,相信你也懂的 ^_^.收起
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