【答案】
分析:求出x、y的值,設內切圓O的半徑是R,與AC、BC、AB分別切于F、D、E,連接OA、OB、OC、OD、OE、OF,①AC=2,BC=3時,由勾股定理求出AB,由三角形的面積公式推出AC×BC=AC×OF+BC×OD+AB×OE,代入求出R即可;②AC=2,AB=3時,由勾股定理求出BC,同樣由三角形的面積公式求出R即可.
解答:解:∵|x
2-4|+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/0.png)
=0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/images1.png)
∴x
2-4=0,y
2-6y+9=0,
解得:x=±2,y=3,
∵x、y表示直角三角形的兩邊長,
∴x=2,y=3,
設內切圓O的半徑是R,與AC、BC、AB分別切于F、D、E,連接OA、OB、OC、OD、OE、OF,
①AC=2,BC=3時,由勾股定理得:AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/2.png)
,
由三角形的面積公式得:S
△ABC=S
△ACO+S
△BCO+S
△ABO,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/3.png)
AC×BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/4.png)
AC×OF+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/5.png)
BC×OD+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/6.png)
AB×OE,
即2×3=2R+3R+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/7.png)
R,
解得:R=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/8.png)
,
②AC=2,AB=3時,由勾股定理得:BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/10.png)
由三角形的面積公式得:S
△ABC=S
△ACO+S
△BCO+S
△ABO,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/11.png)
AC×BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/12.png)
AC×OF+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/13.png)
BC×OD+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/14.png)
AB×OE,
即2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/15.png)
=2R+3R+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/16.png)
R,
解得:R=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192853769371010/SYS201311011928537693710008_DA/17.png)
.
故選C.
點評:本題考查了非負數(shù)性質,勾股定理,三角形的面積,三角形的內切圓等知識點的應用,關鍵是能根據(jù)題意求出x、y的值和求出符合條件的所有情況,題型較好,但有一定的難度.