解:S正方形=6×6=36(平方厘米),
![](http://thumb.1010pic.com/pic5/latex/352335.png)
=
![](http://thumb.1010pic.com/pic5/latex/355497.png)
=
![](http://thumb.1010pic.com/pic5/latex/241554.png)
=0.5,
![](http://thumb.1010pic.com/pic5/latex/415086.png)
=0.5,
因为S(OEF)+S(ODE)=S(DEF)=0.5S(CDF)=S(CEF)=4.5(平方厘米),
S(EOF)=
![](http://thumb.1010pic.com/pic5/latex/8.png)
S(DEF)=1.5(平方厘米),
S空白=S(DEF)+S(BCE)-S(EOF)-S(CEF)=9+9-1.5-4.5=12(平方厘米),
S阴影=S(ABCD)-S空白=36-12=24(平方厘米).
答:阴影部分的面积是24平方厘米.
分析:首先连接BD、EF,设DF与BE相交为O,因为E、F分别为CD、BC中点,则
![](http://thumb.1010pic.com/pic5/latex/352335.png)
=
![](http://thumb.1010pic.com/pic5/latex/355497.png)
=
![](http://thumb.1010pic.com/pic5/latex/241554.png)
=0.5,因为等高三角形面积比等于底边比,所以
![](http://thumb.1010pic.com/pic5/latex/415086.png)
=0.5,又因为S(OEF)+S(ODE)=S(DEF)=0.5S(CDF)=S(CEF)=4.5(平方厘米),所以S(EOF)=
![](http://thumb.1010pic.com/pic5/latex/8.png)
S(DEF)=1.5(平方厘米),S空白=S(DEF)+S(BCE)-S(EOF)-S(CEF)=9+9-1.5-4.5=12(平方厘米),所以S阴影=S(ABCD)-S空白=36-12=24(平方厘米)
点评:此题考查组合图形的面积,解决此题的关键是连接EF和BD,转化图形,进行计算.