题目内容
如图,直角△ABC中,∠C=90°,AB=2
,sinB=
,点P为边BC上一动点,PD∥AB,PD交AC于点D,连结AP.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230314186412655.jpg)
(1)求
、
的长;
(2)设
的长为
,
的面积为
.当
为何值时,
最大并求出最大值.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418625332.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418641399.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230314186412655.jpg)
(1)求
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418657407.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418672406.png)
(2)设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418688391.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418703280.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418781523.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418797313.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418703280.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418797313.png)
(1)2,4;(2)2,1.
试题分析:(1)在Rt△ABC中,根据∠B的正弦值及斜边AB的长,可求出AC的长,进而可由勾股定理求得BC的长;
(2)由于PD∥AB,易证得△CPD∽△CBA,根据相似三角形得出的成比例线段,可求出CD的表达式,也就求出AD的表达式,进而可以AD为底、PC为高得出△ADP的面积,即可求出关于y、x的函数关系式,根据所得函数的性质,可求出y的最大值及对应的x的值.
试题解析:(1)在Rt△ABC中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418859653.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418875600.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418891751.png)
∴AC=2,
根据勾股定理得:BC=4;
(2)∵PD∥AB,
∴△ABC∽△DPC,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418906878.png)
设PC=x,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418922652.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823031418937688.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230314189532113.png)
∴当x=2时,y的最大值是1.
![](http://thumb2018.1010pic.com/images/loading.gif)
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