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共回答了18个问题采纳率:94.4% 举报
S△ABC=AC.BC.sin(∠ACB)/2=1;
S△CEF=CE.CF.sin(∠ECF)/2; (1)
CE=2.BC, CF=5.AC, sin(∠ECF)=sin(π-∠ACB)=sin(∠ACB);代入(1)得:
S△CEF=10.AC.BC.sin(∠ACB)/2=10.S△ABC=10;
同理,S△ABC=AB.AC.sin(∠BAC)/2=1;
S△ADF=AD.AF.sin(∠DAF)/2; (2)
AD=2.AB, AF=4.AC, sin(∠DAF)=sin(π-∠BAC)=sin(∠BAC); 代入(2)得:
S△ADF=8.AB.AC.sin(∠BAC)/2=8S△ABC=8;
同理,S△ABC=AB.BC.sin(∠ABC)/2=1;
S△BDE=BD.BE.sin(∠DBE)/2; (3)
BD=AB, BE=3BC, sin(∠DBE)=sin(π-∠ABC)=sin(∠ABC); 代入(3)得:
S△BDE=3.AB.BC.sin(∠ABC)/2=3.S△ABC=3;
所以:
S△DEF=S△ABC+S△CEF+S△ADF+S△BDE=1+10+8+3=22;为所求.
1年前
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