题目内容
阅读下列材料:问题:如图1,在正方形ABCD内有一点P,PA=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_ST/1.png)
小明同学的想法是:已知条件比较分散,可以通过旋转变换将分散的已知条件集中在一起,于是他将△BPC绕点B逆时针旋转90°,得到了△BP′A(如图2),然后连接PP′.
请你参考小明同学的思路,解决下列问题:
(1)图2中∠BPC的度数为______;
(2)如图3,若在正六边形ABCDEF内有一点P,且PA=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_ST/2.png)
【答案】分析:(1)根据旋转的性质得到∠P′BP=90°,BP′=BP=
,P′A=PC=1,∠BP′A=∠BPC,则△BPP′为等腰直角三角形,根据等腰直角三角形的性质得PP′=
PB=2,∠BP′P=45°,利用勾股定理的逆定理可得到△APP′为直角三角形,且∠AP′P=90°,则∠BPC=∠BP′A=45°+90°=135°;
(2)把△BPC绕点B逆时针旋转120°,得到了△BP′A,根据旋转的性质得到∠P′BP=120°,BP′=BP=4,P′A=PC=2,∠BP′A=∠BPC,则∠BP′P=∠BPP′=30°,得到P′H=PH,利用含30°的直角三角形三边的关系得到BH=
BP′=2,P′H=
BH=2
,得到P′P=2P′H=4
,再利用勾股定理的逆定理可得到△APP′为直角三角形,且∠AP′P=90°,于是有∠BPC=∠BP′A=30°+90°=120°;过A作AG⊥BP′于G点,利用含30°的直角三角形三边的关系得到GP′=
AP′=1,AG=
GP′=
,然后在Rt△AGB中利用勾股定理即可计算出AB长.
解答:解:(1)如图2.
∵△BPC绕点B逆时针旋转90°,得到了△BP′A,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/images9.png)
∴∠P′BP=90°,BP′=BP=
,P′A=PC=1,∠BP′A=∠BPC,
∴△BPP′为等腰直角三角形,
∴PP′=
PB=2,∠BP′P=45°,
在△APP′中,AP=
,PP′=2,AP′=1,
∵(
)2=22+12,
∴AP2=PP′2+AP′2,
∴△APP′为直角三角形,且∠AP′P=90°
∴∠BP′A=45°+90°=135°,
∴∠BPC=∠BP′A=135°;
(2)如图3.
∵六边形ABCDEF为正六边形,
∴∠ABC=120°,
把△BPC绕点B逆时针旋转120°,得到了△BP′A,
∴∠P′BP=120°,BP′=BP=4,P′A=PC=2,∠BP′A=∠BPC,
∴∠BP′P=∠BPP′=30°,
过B作BH⊥PP′于H,
∵BP′=BP,
∴P′H=PH,
在Rt△BP′H中,∠BP′H=30°,BP′=4,
∴BH=
BP′=2,P′H=
BH=2
,
∴P′P=2P′H=4
,
在△APP′中,AP=2
,PP′=4
,AP′=2,
∵(2
)2=(4
)2+22,
∴AP2=PP′2+AP′2,
∴△APP′为直角三角形,且∠AP′P=90°,
∴∠BP′A=30°+90°=120°,
∴∠BPC=120°,
过A作AG⊥BP′于G点,
∴∠AP′G=60°,
在Rt△AGP′中,AP′=2,∠GAP′=30°,
∴GP′=
AP′=1,AG=
GP′=
,
在Rt△AGB中,GB=GP′+P′B=1+4=5,
AB=
=
=2
,
即正六边形ABCDEF的边长为2
.
故答案为135°;120°,
.
点评:本题考查了旋转的性质:旋转前后两图形全等,即对应角相等,对应线段相等;对应点与旋转中心的连线段的夹角等于旋转角.也考查了正方形的性质、等腰直角三角形的判定与性质、勾股定理与逆定理以及含30°的直角三角形三边的关系.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/1.png)
(2)把△BPC绕点B逆时针旋转120°,得到了△BP′A,根据旋转的性质得到∠P′BP=120°,BP′=BP=4,P′A=PC=2,∠BP′A=∠BPC,则∠BP′P=∠BPP′=30°,得到P′H=PH,利用含30°的直角三角形三边的关系得到BH=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/8.png)
解答:解:(1)如图2.
∵△BPC绕点B逆时针旋转90°,得到了△BP′A,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/images9.png)
∴∠P′BP=90°,BP′=BP=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/9.png)
∴△BPP′为等腰直角三角形,
∴PP′=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/10.png)
在△APP′中,AP=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/11.png)
∵(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/12.png)
∴AP2=PP′2+AP′2,
∴△APP′为直角三角形,且∠AP′P=90°
∴∠BP′A=45°+90°=135°,
∴∠BPC=∠BP′A=135°;
(2)如图3.
∵六边形ABCDEF为正六边形,
∴∠ABC=120°,
把△BPC绕点B逆时针旋转120°,得到了△BP′A,
∴∠P′BP=120°,BP′=BP=4,P′A=PC=2,∠BP′A=∠BPC,
∴∠BP′P=∠BPP′=30°,
过B作BH⊥PP′于H,
∵BP′=BP,
∴P′H=PH,
在Rt△BP′H中,∠BP′H=30°,BP′=4,
∴BH=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/15.png)
∴P′P=2P′H=4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/16.png)
在△APP′中,AP=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/17.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/18.png)
∵(2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/19.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/20.png)
∴AP2=PP′2+AP′2,
∴△APP′为直角三角形,且∠AP′P=90°,
∴∠BP′A=30°+90°=120°,
∴∠BPC=120°,
过A作AG⊥BP′于G点,
∴∠AP′G=60°,
在Rt△AGP′中,AP′=2,∠GAP′=30°,
∴GP′=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/21.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/22.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/23.png)
在Rt△AGB中,GB=GP′+P′B=1+4=5,
AB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/24.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/25.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/26.png)
即正六边形ABCDEF的边长为2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/27.png)
故答案为135°;120°,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101190625419382012/SYS201311011906254193820021_DA/28.png)
点评:本题考查了旋转的性质:旋转前后两图形全等,即对应角相等,对应线段相等;对应点与旋转中心的连线段的夹角等于旋转角.也考查了正方形的性质、等腰直角三角形的判定与性质、勾股定理与逆定理以及含30°的直角三角形三边的关系.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目