解:过D作DE⊥AB于E,分为三种情况:
①如图1,当P在AD上时,此时0≤x≤6,
![](http://thumb.1010pic.com/pic5/upload/201304/51d660d9e3b07.png)
∵D(6,3),
∴OE=6,DE=3,
∵MN⊥AB.DE⊥AB,
∴PQ∥DE,
∴△AQP∽△AED,
∴
![](http://thumb.1010pic.com/pic5/latex/262361.png)
=
![](http://thumb.1010pic.com/pic5/latex/515975.png)
,
∴
![](http://thumb.1010pic.com/pic5/latex/28230.png)
=
![](http://thumb.1010pic.com/pic5/latex/515976.png)
,
PQ=
![](http://thumb.1010pic.com/pic5/latex/13.png)
x,
∴S=S
△APQ=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×AQ×PQ=
![](http://thumb.1010pic.com/pic5/latex/13.png)
•x•
![](http://thumb.1010pic.com/pic5/latex/13.png)
x=
![](http://thumb.1010pic.com/pic5/latex/96.png)
x
2;
![](http://thumb.1010pic.com/pic5/upload/201304/51d660da1dfd2.png)
②如图2,P在DC上,此时6<x≤12,
DP=EQ=x-6,PQ=DE=3,AQ=x,
S=S
四边形ADPQ=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×(DP+AQ)×PQ=
![](http://thumb.1010pic.com/pic5/latex/13.png)
•(x-6+x)•3=3x-9;
③如图3,P在BC上,此时12<x<15,
过C作CF⊥AB于F
则PQ∥CF,
∵B(15,0),C(12.3),D(6,3),
∴CF=3,BA=15,BQ=15-x,BF=15-12,DC=12-6=6,
∵CF∥PQ,
![](http://thumb.1010pic.com/pic5/upload/201304/51d660da4bb60.png)
∴△PQB∽△CFB,
∴
![](http://thumb.1010pic.com/pic5/latex/515977.png)
=
![](http://thumb.1010pic.com/pic5/latex/515978.png)
,
∴
![](http://thumb.1010pic.com/pic5/latex/515976.png)
=
![](http://thumb.1010pic.com/pic5/latex/515979.png)
,
PQ=15-x,
∴S=S
五边形ADCPQ=S
梯形ABCD-S
△BPQ=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×(DC+AB)×CF-
![](http://thumb.1010pic.com/pic5/latex/13.png)
×BQ×PQ
=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×(6+15)×3-
![](http://thumb.1010pic.com/pic5/latex/13.png)
•(15-x)•(15-x)
=-
![](http://thumb.1010pic.com/pic5/latex/13.png)
x
2+15x-81,
④当x≥15时,S=S
梯形ABCD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×(6+15)×3=31.5;
综合上述,S=
![](http://thumb.1010pic.com/pic5/latex/515980.png)
.
分析:过D作DE⊥AB于E,画出符合的四种情况,根据A、B、C、D的坐标求出PQ的值,根据面积公式求出即可.
点评:本题考查了相似三角形的性质和判定和分段函数,关键是求出符合条件的所有情况,用了分类讨论思想.