题目内容
(2004•上海模拟)如图,在梯形ABCD中,AD∥BC,∠A=90°,BC=DC,sinC=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_ST/images1.png)
【答案】分析:作直角梯形的另一高.构造了一个直角三角形和矩形.根据锐角三角函数的概念和勾股定理得到直角三角形的比,结合BC=10,最后求得梯形的高.
解答:解:如图,过点D作DE⊥BC,垂足为E.
∵BC=DC,BC=10,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_DA/images0.png)
∴CD=10,
∵AD∥BC,∠A=90°,
∴∠A+∠ABC=180°,
得∠ABC=90°,
又∵AD∥BC,DE⊥BC,
∴AB=DE
在Rt△DCE中,∠DEC=90°,
∴sinC=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_DA/0.png)
又∵sinC=
,BC=10,
∴DE=6,即AB=6.
点评:作梯形的另一高是直角梯形中常见的辅助线.
解答:解:如图,过点D作DE⊥BC,垂足为E.
∵BC=DC,BC=10,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_DA/images0.png)
∴CD=10,
∵AD∥BC,∠A=90°,
∴∠A+∠ABC=180°,
得∠ABC=90°,
又∵AD∥BC,DE⊥BC,
∴AB=DE
在Rt△DCE中,∠DEC=90°,
∴sinC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_DA/0.png)
又∵sinC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232956465359594/SYS201310212329564653595020_DA/1.png)
∴DE=6,即AB=6.
点评:作梯形的另一高是直角梯形中常见的辅助线.
![](http://thumb2018.1010pic.com/images/loading.gif)
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