题目内容
(1997•河北)如图,已知扇形AOB的半径为12,OA⊥OB,C为OB上一点,以OA为直径的半圆O1;和以BC为直径的半圆O2相切于点D,则图中阴影部分的面积是( )![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_ST/images0.png)
A.6π
B.10π
C.12π
D.20π
【答案】分析:要求阴影的面积,扇面AOB减去两半圆面积就是,半圆O1半径已知是6,只要求得半圆O2的半径即可,连接O1O2,因为OA⊥OB,所以由勾股定理OO12+OO22=O1O22可得r=4,所以阴影面积=
π122-
π62-
π42=10π.
解答:解:如图所示
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/images3.png)
连接O1O2,设BC=2r,AO=2R,
∵半圆O1,半圆O2相切,
∴O1O2过D点,O1O2=6+r,
∵OA⊥OB,
∴OO12+OO22=O1O22,
∴R2+(12-r)2=(6+r)2,
∴r=4,
所以阴影面积=
π×122-
π×62-
π×42=10π.
点评:本题考查了相切圆的性质,扇面面积的计算,以及勾股定理的运用,同学们应熟练掌握.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/2.png)
解答:解:如图所示
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/images3.png)
连接O1O2,设BC=2r,AO=2R,
∵半圆O1,半圆O2相切,
∴O1O2过D点,O1O2=6+r,
∵OA⊥OB,
∴OO12+OO22=O1O22,
∴R2+(12-r)2=(6+r)2,
∴r=4,
所以阴影面积=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232859209106020/SYS201310212328592091060007_DA/5.png)
点评:本题考查了相切圆的性质,扇面面积的计算,以及勾股定理的运用,同学们应熟练掌握.
![](http://thumb2018.1010pic.com/images/loading.gif)
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