试题分析:(1)作CM⊥OA于点M,知CM
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529848326.png)
,由∠AOC=60°易求BM=1,求出C点坐标;由B点坐标可求BC的长,从而梯形面积可求;
(2)用含有t的代数式分别表示△OPQ的高和底,求出△OPQ的的面积即可表示出S与运动时间t的函数关系式;
(3)分点Q分别在边BC、OC、OA上运动时进行讨论,即可求出t的值.
试题解析:(1)作CM⊥OA于点M,
∵∠AOC=60°,∴∠OCM=30°,
∵B(3,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529863313.png)
),BC∥AO,∴CM
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529848326.png)
,
设OM=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529879266.png)
,则OC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529895349.png)
,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529910727.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529926329.png)
,∴OM=1,OC=2,
∴C(1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529863313.png)
),
∵B(3,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529863313.png)
),∴BC=2,
∵A(6,0),∴OA=6,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230325299731306.png)
,
(2)如图1,当动点Q运动到OC边时,OQ=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529988331.png)
,
作QG⊥OP,∴∠OQG=30°,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230325300192265.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530019871.png)
,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530035789.png)
,
又∵OP=2t,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530051884.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530066723.png)
(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032529832463.png)
);
(3)根据题意得出:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530082458.png)
,
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530097466.png)
时,Q在BC边上运动,延长BC交y轴于点D,
此时OP=2t,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530113827.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230325301291004.png)
,
∵∠POQ<∠POC=60°,
∴若△OPQ为直角三角形,只能是∠OPQ=90°或∠OQP=90°,
若∠OPQ=90°,如图2,则∠PQD=90°,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230325301442194.png)
∴四边形PQDO为矩形,
∴OP=QD,∴2t=3-t,
解得t=1,
若∠OQP=90°,如图3,则OQ
2+PQ
2=PO
2,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230325301602088.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530175984.png)
,
解得:t
1=t
2=2,
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530175466.png)
时,Q在OC边上运动,
若∠OQP=90°,
∵∠POQ=60°,∴∠OPQ=30°,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530191608.png)
,
若∠OPQ=90°,同理:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530207602.png)
,
而此时OP=2t>4,OQ<OC=2,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530222641.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823032530238629.png)
,
故当Q在OC边上运动时,△OPQ不可能为直角三角形,
综上所述,当t=1或t=2时,△OPQ为直角三角形。
考点: 1.二次函数;2.直角三角形的判定.