已知⊙Ο1、⊙Ο2相交于點A、B,公共弦與連心線O1O2交于點G,若AB=48,⊙Ο1、⊙Ο2的半徑分別是30、40,則△AO1O2的面積是 .
【答案】
分析:由題意,可知:△AO
1O
2的面積=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/0.png)
O
1O
2×AG,AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/1.png)
,O
1O
2=O
1G+O
2G,在△AO
1G和△AO
2G中,兩次利用勾股定理,分別求得O
1G的長和O
2G的長,故△AO
1O
2的面積可求.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/images2.png)
解:∵AB=48,∴AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/2.png)
=24,
又∵O
2A=30,O
1A=40,
∴O
1G=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/3.png)
=32,∴O
2G=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/4.png)
=18;
∵O
1O
2=O
1G+O
2G,∴O
1O
2=50,
∴△AO
1O
2的面積=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/5.png)
O
1O
2×AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/6.png)
×50×24=600.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/images8.png)
當(dāng)∵O
1O
2=O
1G-O
2G,∴O
1O
2=14,
∴△AO
1O
2的面積=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/7.png)
O
1O
2×AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103001308953922726/SYS201311030013089539227015_DA/8.png)
×14×24=168.
故答案為:600或168.
點評:此題綜合運用了直角三角形的勾股定理、垂徑定理.