题目内容
已知直线l1∥l2∥l3∥l4,正方形ABCD的四个顶点分别在四条直线上,正方形ABCD的面积为S.(1)如图1,已知平行线间的距离均为m,求S.(用含有m的式子表示)
(2)如图2,改变平行线之间的距离,但仍使四边形ABCD为正方形,
①求证:h1=h3.
②求证:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_ST/0.png)
③若
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_ST/images2.png)
【答案】分析:(1)根据过D点作EF⊥l1于E交l4于F,首先得出△ADE≌△DCF,再利用勾股定理得出S;
(2)①首先过A点作AP⊥l2于P,过C点作CQ⊥l3于Q,得出△ABP≌△CDQ,即可得出AP=CQ,即h1=h3,
②首先过D点作EF⊥l1于E交l4于F,则ED=h1+h2,DF=h3,进而得出△ADE≌△DCF,则AE=DF=h3,再利用勾股定理AD2=AE2+DE2,求出即可;
③利用
,以及②中所求得出S的值即可.
解答:
解:
(1)如图1,过D点作EF⊥l1于E交l4于F,
则ED=2m,DF=m,
∵∠ADC=90°,
∴∠ADE+∠CDF=90°,
∵∠FCD+∠CDF=90°,
∴∠ADE=∠DCF,
在△ADE和△DCF中,
,
∴△ADE≌△DCF(AAS),
∴AE=DF=m,
在Rt△ADE中由勾股定理可得:
AD2=AE2+DE2=m2+(2m)2=5m2,
∴S=AD2=5m2,
(2)如图2所示:
①过A点作AP⊥l2于P,过C点作CQ⊥l3于Q,
∵∠EAD+∠DAP=90°,
∠EAD=∠ADQ,
∴∠DAP+∠ADQ=90°,
∵∠CDQ+∠ADQ=90°,
∴∠DAP=∠DQC,
∵∠ABP+∠BAP=90°,∠BAP+∠DAP=90°,
∴∠ABP=∠QDC,
在△ABP和△CDQ中,
,
∴△ABP≌△CDQ(AAS),
∴AP=CQ,即h1=h3,
②过D点作EF⊥l1于E交l4于F,则ED=h1+h2,DF=h3,
∵∠ADC=90°,
∴∠ADE+∠CDF=90°,
∵∠FCD+∠CDF=90°,
∴∠ADE=∠DCF,
在△ADE和△DCF中,
,
∴△ADE≌△DCF(AAS),
则AE=DF=h3,
在Rt△ADE中由勾股定理可得:AD2=AE2+DE2=
+(h1+h2) 2,
又∵h1=h3,
∴S=AD2=(h1+h2) 2+
,
③∵
,
∴
,
∴S=(h1+1-
h1)2+
,
=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/10.png)
-h1+1,
=
(h1-
)2+
,
又∵
,
解得0<h1<
,
∴当0<h1<
时,S随h1的增大而减小;
当h1=
时,S取得最小值
;
当
<h1<
时,S随h1的增大而增大.
点评:此题主要考查了二次函数的综合应用以及全等三角形的判定与性质和勾股定理等知识,利用二次函数的增减性得出S随h1的关系是解题关键.
(2)①首先过A点作AP⊥l2于P,过C点作CQ⊥l3于Q,得出△ABP≌△CDQ,即可得出AP=CQ,即h1=h3,
②首先过D点作EF⊥l1于E交l4于F,则ED=h1+h2,DF=h3,进而得出△ADE≌△DCF,则AE=DF=h3,再利用勾股定理AD2=AE2+DE2,求出即可;
③利用
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/0.png)
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/images1.png)
(1)如图1,过D点作EF⊥l1于E交l4于F,
则ED=2m,DF=m,
∵∠ADC=90°,
∴∠ADE+∠CDF=90°,
∵∠FCD+∠CDF=90°,
∴∠ADE=∠DCF,
在△ADE和△DCF中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/1.png)
∴△ADE≌△DCF(AAS),
∴AE=DF=m,
在Rt△ADE中由勾股定理可得:
AD2=AE2+DE2=m2+(2m)2=5m2,
∴S=AD2=5m2,
(2)如图2所示:
①过A点作AP⊥l2于P,过C点作CQ⊥l3于Q,
∵∠EAD+∠DAP=90°,
∠EAD=∠ADQ,
∴∠DAP+∠ADQ=90°,
∵∠CDQ+∠ADQ=90°,
∴∠DAP=∠DQC,
∵∠ABP+∠BAP=90°,∠BAP+∠DAP=90°,
∴∠ABP=∠QDC,
在△ABP和△CDQ中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/2.png)
∴△ABP≌△CDQ(AAS),
∴AP=CQ,即h1=h3,
②过D点作EF⊥l1于E交l4于F,则ED=h1+h2,DF=h3,
∵∠ADC=90°,
∴∠ADE+∠CDF=90°,
∵∠FCD+∠CDF=90°,
∴∠ADE=∠DCF,
在△ADE和△DCF中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/3.png)
∴△ADE≌△DCF(AAS),
则AE=DF=h3,
在Rt△ADE中由勾股定理可得:AD2=AE2+DE2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/4.png)
又∵h1=h3,
∴S=AD2=(h1+h2) 2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/5.png)
③∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/6.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/7.png)
∴S=(h1+1-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/9.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/11.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/14.png)
又∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/15.png)
解得0<h1<
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/16.png)
∴当0<h1<
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/17.png)
当h1=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/18.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/19.png)
当
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101185859315514379/SYS201311011858593155143023_DA/21.png)
点评:此题主要考查了二次函数的综合应用以及全等三角形的判定与性质和勾股定理等知识,利用二次函数的增减性得出S随h1的关系是解题关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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