如图,在矩形ABCD中,AB=3,AD=4,点P在AD上,PE⊥AC于E,PF⊥BD于F,则 PE+PF等于(  ) A.75 B.125 C.135 D.145

问题描述:

如图,在矩形ABCD中,AB=3,AD=4,点P在AD上,PE⊥AC于E,PF⊥BD于F,则
PE+PF等于(  )
A.

7
5

B.
12
5

C.
13
5

D.
14
5

连接OP,过D作DM⊥AC于M,
∵四边形ABCD是矩形,
∴AO=OC=

1
2
AC,OD=OB=
1
2
BD,AC=BD,∠ADC=90°
∴OA=OD,
由勾股定理得:AC=
32+42
=5,
∵S△ADC=
1
2
×3×4=
1
2
×5×DM,
∴DM=
12
5


∵S△AOD=S△APO+S△DPO
1
2
(AO×DM)=
1
2
(AO×PE)+
1
2
(DO×PF),
即PE+PF=DM=
12
5

故选B.