Question

如图,梯形ABCD中,AD//BC(AD<BC),AC、BD相交于点O。若S△OAB=6/25S梯形ABCD,则△AO

Answer 1

设S△OAB=S1，S△OBC=S2，春亮S△OCD=S3，S△ODA=S4，S梯形ABCD=S，OC/OA=k
因为AD//BC，AD<BC, 所以
OC/OA=k=BC/AD>喊森孝郑稿1, S1+S2=S2+S3, S2/S1=OC/OA, S2/S4=(OC/OA)^2
S1=S3=6/25 * S, S2=kS1=6/25*kS, S4=S1/k=6/(25k)*S
S1+S2+S3+S4=S, 6k^2-13k+6=0, k=3/2 or k=2/3(舍去，因为k>1)
S△ODA=S4=6/(25k)*S=4/25*S, S△OBC=S2=9/25*S

Answer 2

图呢