【答案】
分析:如果求出了C點(diǎn)的坐標(biāo),那么只需將C點(diǎn)的坐標(biāo)代入反比例函數(shù)的解析式
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/0.png)
,即可求出k的值.由于AB=5,所以當(dāng)Rt△ABC的三邊長(zhǎng)均為整數(shù)時(shí),分AB為斜邊和AB為直角邊進(jìn)行討論:①如果AB=5為斜邊,那么兩條直角邊分別為3,4.當(dāng)AC=3時(shí),易求C(4,3);當(dāng)BC=3時(shí),根據(jù)直角三角形的性質(zhì)及三角函數(shù)的定義,可求出C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/2.png)
);②如果AB=5為直角邊,那么另外兩條邊分別為12,13.當(dāng)AC=12時(shí),根據(jù)相似三角形的判定與性質(zhì)可求C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/3.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/4.png)
);當(dāng)BC=12時(shí),根據(jù)相似三角形的判定與性質(zhì)可求C(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/6.png)
).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/images7.png)
解:∵A(4,0),B(0,3),
∴OA=4,OB=3,
∵∠AOB=90°,
∴AB=5.
分兩種情況:
①如果AB=5為斜邊,那么兩條直角邊分別為3,4.
當(dāng)AC=3時(shí),則BC=4,C
1點(diǎn)坐標(biāo)為(4,3),
所以k=4×3=12;
當(dāng)BC=3時(shí),設(shè)AC
2與BC
1交于點(diǎn)P,過(guò)點(diǎn)C
2作C
2D⊥BC
1于D.
由AAS易證△BPC
2≌△APC
1,則BP=AP,PC
2=PC
1.
設(shè)PC
1=x,則AP=BP=4-x,
在△APC
1中,由勾股定理,
得x
2+3
2=(4-x)
2,解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/7.png)
.
則AP=BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/8.png)
,
∴BD=BC
2•cos∠C
2BD=BC
2•cos∠C
1AP=3×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/10.png)
,C
2D=BC
2•sin∠C
2BD=BC
2•sin∠C
1AP=3×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/12.png)
,
∴OB+C
2D=3+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/images16.png)
∴C
2點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/16.png)
),
∴k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/17.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/19.png)
;
②如果AB=5為直角邊,那么另外兩條邊分別為12,13.
當(dāng)AC=12時(shí),∠BAC=90°.過(guò)點(diǎn)C
3作C
3D⊥x軸于D.
∵∠C
3DA=∠AOB=90°,∠C
3AD=∠ABO=90°-∠OAB,
∴△C
3DA∽△AOB,
∴C
3D:AO=DA:OB=C
3A:AB,
即C
3D:4=DA:3=12:5,
∴C
3D=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/20.png)
,DA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/21.png)
,
∴OD=OA+AD=4+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/images26.png)
∴C
3點(diǎn)坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/24.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/25.png)
),
∴k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/26.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/28.png)
;
當(dāng)BC=12時(shí),∠ABC=90°.過(guò)點(diǎn)C
4作C
4D⊥y軸于D.
∵∠C
4DB=∠BOA=90°,∠C
4BD=∠OAB=90°-∠ABO,
∴△C
4DB∽△BOA,
∴C
4D:BO=DB:OA=C
4B:BA,
即C
4D:3=DB:4=12:5,
∴C
4D=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/29.png)
,DB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/30.png)
,
∴OD=OB+BD=3+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/32.png)
,
∴C
4點(diǎn)坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/33.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/34.png)
),
∴k=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/35.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/37.png)
.
綜上可知,k的值為12,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/38.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/39.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/40.png)
.
故答案為:12,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/41.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/42.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202750461512093/SYS201311032027504615120015_DA/43.png)
.
點(diǎn)評(píng):本題考查了直角三角形的性質(zhì),相似三角形的判定與性質(zhì),勾股定理,解直角三角形,綜合性較強(qiáng),熟記常見(jiàn)的勾股數(shù)及將Rt△ABC分AB為斜邊和AB為直角邊進(jìn)行討論是解題的關(guān)鍵.