梯形ABCD中,AD//BC,AC与BD交与O点,若S△AOD=1,S△COD=4,求梯形ABCD的面积.

题目

梯形ABCD中,AD//BC,AC与BD交与O点,若S△AOD=1,S△COD=4,求梯形ABCD的面积.

答案

S△ADB=S△DAC(同底等高) 各减 S△AOD=1
=> S△COD=S△AOB=4
S△COD:S△AOD=CO:OA=S△BOC:S△AOB=4:1=S△BOC:4 =>S△BOC=16
so S[ABCD]=1+4+4+16=25