解:[(a+
![](http://thumb.1010pic.com/pic5/latex/53.png)
a)
2-2×
![](http://thumb.1010pic.com/pic5/latex/13.png)
a
2]÷[
![](http://thumb.1010pic.com/pic5/latex/13.png)
a
2+
![](http://thumb.1010pic.com/pic5/latex/253318.png)
-a
2],
=2(1+
![](http://thumb.1010pic.com/pic5/latex/53.png)
)÷(1+
![](http://thumb.1010pic.com/pic5/latex/53.png)
),
=2(倍),
答:正八边形的面积是阴影部分面积的2倍.
分析:如图:设正八边形ABCDEFGH的边长为a,DL的长度是
![](http://thumb.1010pic.com/pic5/latex/54.png)
a,则正方形KLMN的面积是
![](http://thumb.1010pic.com/pic5/latex/253319.png)
,正八边形ABCDEFGH的面积是
![](http://thumb.1010pic.com/pic5/latex/253319.png)
-2×
![](http://thumb.1010pic.com/pic5/latex/13.png)
a
2,里面空白的小正方形的面积是a
2,正方形ACEG的面积:2a
2+
![](http://thumb.1010pic.com/pic5/latex/253318.png)
,所以阴影部分的面积是:
![](http://thumb.1010pic.com/pic5/latex/13.png)
a
2+
![](http://thumb.1010pic.com/pic5/latex/253318.png)
-a
2,用正八边形的面积除以阴影部分面积即可.
![](http://thumb.1010pic.com/pic5/upload/201310/5286ce2e7e53e.png)
点评:关键是添加辅助线,运用勾股定理求出正八边形的面积与阴影部分的面积.