我们知道,三条边都相等的三角形叫等边三角形.类似地,我们把弧长等于半径的扇形称为“等边扇形”.小明准备将一根长为120cm的铁丝剪成两段,并把每一段铁丝围成一个“等边扇形”.
(1)小明想使这两个“等边扇形”的面积之和等于625cm2,他该怎么剪?
(2)这两个“等边扇形”的面积之和能否取得最小值?若能,请求出这个最小值;若不能,请说明理由.
解:(1)假设一段为x,则另一段为120-x,
∴扇形的半径分别为:
![](http://thumb.1010pic.com/pic5/latex/2987.png)
,40-
![](http://thumb.1010pic.com/pic5/latex/2987.png)
,
∴扇形面积分别为:S
1=
![](http://thumb.1010pic.com/pic5/latex/13.png)
lr=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×
![](http://thumb.1010pic.com/pic5/latex/2987.png)
×
![](http://thumb.1010pic.com/pic5/latex/2987.png)
=
![](http://thumb.1010pic.com/pic5/latex/246416.png)
,
S
2=
![](http://thumb.1010pic.com/pic5/latex/13.png)
LR=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×(40-
![](http://thumb.1010pic.com/pic5/latex/2987.png)
)
2=
![](http://thumb.1010pic.com/pic5/latex/246417.png)
-
![](http://thumb.1010pic.com/pic5/latex/11801.png)
x+800,
∴S
1+S
2=
![](http://thumb.1010pic.com/pic5/latex/246416.png)
+
![](http://thumb.1010pic.com/pic5/latex/246417.png)
-
![](http://thumb.1010pic.com/pic5/latex/11801.png)
x+800,
=
![](http://thumb.1010pic.com/pic5/latex/664.png)
-
![](http://thumb.1010pic.com/pic5/latex/11801.png)
x+800,
当625=
![](http://thumb.1010pic.com/pic5/latex/664.png)
-
![](http://thumb.1010pic.com/pic5/latex/11801.png)
x+800,
∴x
2-120x+1575=0,
解得:x
1=15,x
2=105,
∴一段为15,则另一段为105.
(2)假设两个“等边扇形”的面积之和为y,
∴y=
![](http://thumb.1010pic.com/pic5/latex/664.png)
-
![](http://thumb.1010pic.com/pic5/latex/11801.png)
x+800,
当x=-
![](http://thumb.1010pic.com/pic5/latex/2530.png)
=60时,
y取到最小值为:
![](http://thumb.1010pic.com/pic5/latex/8487.png)
=400.
分析:(1)根据弧长等于半径的扇形称为“等边扇形”,得出扇形半径长,再利用扇形面积公式S=
![](http://thumb.1010pic.com/pic5/latex/13.png)
lr求出;
(2)根据二次函数的最值公式求出即可.
点评:此题主要考查了扇形的面积公式以及二次函数的综合应用,得出二次函数的解析式再求出最值是解决问题的关键.