作AN⊥EF于N,交BC于M,
∵BC ∥ EF,
∴AM⊥BC于M,
∴△ABC ∽ △AEF,
∴
BC
EF =
AM
AN ,
∵AM=0.6,AN=30,BC=0.12,
∴EF=
BC?AN
AM =
0.12×30
0.6 =6m.
故选D.
作AN⊥EF于N,交BC于M,
∵BC ∥ EF,
∴AM⊥BC于M,
∴△ABC ∽ △AEF,
∴
BC
EF =
AM
AN ,
∵AM=0.6,AN=30,BC=0.12,
∴EF=
BC?AN
AM =
0.12×30
0.6 =6m.
故选D.