若一个正六边形的周长为24,则该六边形的面积为 .
【答案】
分析:首先根据题意画出图形,即可得△OBC是等边三角形,又由正六边形ABCDEF的周长为24,即可求得BC的长,继而求得△OBC的面积,则可求得该六边形的面积.
解答:![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/images0.png)
解:如图,连接OB,OC,过O作OM⊥BC于M,
∴∠BOC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/0.png)
×360°=60°,
∵OB=OC,
∴△OBC是等边三角形,
∵正六边形ABCDEF的周长为24,
∴BC=24÷6=4,
∴OB=BC=4,
∴BM=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/1.png)
BC=2,
∴OM=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/2.png)
=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/3.png)
,
∴S
△OBC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/4.png)
×BC×OM=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/5.png)
×4×2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/6.png)
=4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/7.png)
,
∴该六边形的面积为:4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/8.png)
×6=24
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/9.png)
.
故答案为:24
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103202416562876758/SYS201311032024165628767010_DA/10.png)
.
点评:此题考查了圆的内接六边形的性质与等边三角形的判定与性质.此题难度不大,注意掌握数形结合思想的应用.