已知如下图,四边形ABCD面积为52cm²,AD、BC交于点O,S△AOB=6cm²,S△BOD=7cm²,求阴影部分面积?
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分享交流一下我的解题思路:
S△AOC=S1 S△DOC=S2
S1/S2=AO/DO
S△AOB/S△DOB=AO/DO=6/7
S1/S2=6/7
S△ACD=52-6-7=39
S阴=S△DOC=7/13×39=21cm²
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