已知直角梯形的一腰与下底的夹角为60°,下底与其中的一腰都等于6,则梯形的中位线的长为 .
【答案】
分析:由已知条件易求得上底的长,再根据梯形中位线性质:中位线的长等于
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/0.png)
(上底+下底),即可求得中位线的长.
解答:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/images1.png)
解:根据题意可作出如图:DE⊥BC,
①当DC=BC=6,∠C=60°,则EC=6×cos60°=3,
∵∠A=∠B=90°,AD∥BC,DE⊥BC,
∴四边形ABED中矩形,
∴AD=BE=BC-EC=6-3=3,
∴梯形ABCD的中位线长=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/1.png)
(AD+BC)=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/2.png)
(6+3)=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/3.png)
.
②当BC=AB=6时,DE=AB=6,则EC=DE÷tan60°=6÷
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/4.png)
=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/5.png)
,
∴AD=BE=6-2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/6.png)
,
∴梯形ABCD的中位线长=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/7.png)
(AD+BC)=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/8.png)
(6+6-2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/9.png)
)=6-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/10.png)
,
故答案为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/11.png)
或6-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131022161254842910785/SYS201310221612548429107013_DA/12.png)
.
点评:本题考查了直角梯形的性质、矩形的性质、梯形中位线性质等知识点,解直角梯形一般是通过作高线构造矩形和直角三角形的方式来解决.本题考查了分类讨论的思想.