题目内容
(2009•门头沟区一模)已知:如图,在梯形ABCD中,AD∥BC,∠D=90°,∠B=60°,CD=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_ST/images2.png)
【答案】分析:如图,过点A作AF⊥BC于点F,这样把梯形分割成直角三角形和矩形,然后在Rt△AFB中解直角三角形求出BF、AD,最后在Rt△ADE解直角三角形就可以求出AE.
解答:
解:如图,过点A作AF⊥BC于点F.(1分)
∵∠D=90°,
∴AF∥DC.
又∵AD∥BC,
∴四边形AFCD是矩形.
∴FA=CD=
.(2分)
在Rt△AFB中,∠B=60°,
∴BF=AF÷tan60°=
÷
=4.(3分)
∴AD=FC=BC-BF=9-4=5.(4分)
在Rt△ADE中,∠D=90°,∵
,
∴
.
∴
.(5分)
点评:此题考查了梯形的一种常用辅助线-作梯形的高,把梯形分割成直角三角形和矩形,然后解直角三角形就可以解题.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/images0.png)
∵∠D=90°,
∴AF∥DC.
又∵AD∥BC,
∴四边形AFCD是矩形.
∴FA=CD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/0.png)
在Rt△AFB中,∠B=60°,
∴BF=AF÷tan60°=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/2.png)
∴AD=FC=BC-BF=9-4=5.(4分)
在Rt△ADE中,∠D=90°,∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/3.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/4.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205382528046497/SYS201310201205382528046017_DA/5.png)
点评:此题考查了梯形的一种常用辅助线-作梯形的高,把梯形分割成直角三角形和矩形,然后解直角三角形就可以解题.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目