【答案】
分析:(1)過點(diǎn)P,作PD⊥BC于D,利用30度的銳角所對(duì)的直角邊等于斜邊的一半,即可求得PD的長(zhǎng),然后利用三角形的面積公式即可求解;
(2)分PC=QC和PC=QC兩種情況進(jìn)行討論,求解;
(3)PA為半徑的圓與以Q為圓心,QC為半徑的圓相切時(shí),分為兩圓外切和內(nèi)切兩種情況進(jìn)行討論.在直角△PFQ中利用勾股定理即可得到關(guān)于t的方程,從而求解.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/images0.png)
解:在Rt△ABC中,AB=6米,BC=8米,∴AC=10米
由題意得:AP=2t,則CQ=t,則PC=10-2t
(1)①過點(diǎn)P,作PD⊥BC于D,
∵t=2.5秒時(shí),AP=2×2.5=5米,QC=2.5米
∴PD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/0.png)
AB=3米,∴S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/1.png)
•QC•PD=3.75平方米;
②過點(diǎn)Q,作QE⊥PC于點(diǎn)E,
易知Rt△QEC∽R(shí)t△ABC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/3.png)
,
解得:QE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/4.png)
,
∴S=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/5.png)
•PC•QE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/6.png)
•(10-2t)•
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/7.png)
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/8.png)
+3t(0<t≤5)
(2)當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/9.png)
秒(此時(shí)PC=QC),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/10.png)
秒(此時(shí)PQ=QC),或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/11.png)
秒(此時(shí)PC=PQ)時(shí),△CPQ為等腰三角形;
∵△ABC中,∠B=90°,AB=6米,BC=8米,
∴AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/13.png)
=10,
當(dāng)PC=QC時(shí),PC=10-2t,QC=t,即10-2t=t,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/14.png)
秒;
當(dāng)PQ=CQ時(shí),如圖1,過點(diǎn)Q作QE⊥AC,則CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/15.png)
,CQ=t,可證△CEQ∽△CBA,故
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/17.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/19.png)
,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/20.png)
秒;
當(dāng)PC=PQ時(shí),如圖2,過點(diǎn)P作PE⊥BC,則CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/21.png)
,PC=10-2t,可證△PCE∽△ACB,故
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/23.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/24.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/25.png)
,解得t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/26.png)
秒.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/images28.png)
(3)如圖3,過點(diǎn)P作PF⊥BC于點(diǎn)F.
則△PCF∽△ACB
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/29.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/32.png)
∴PF=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/33.png)
,F(xiàn)C=8-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/34.png)
則在直角△PFQ中,PQ
2=PF
2+FQ
2=(6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/35.png)
)
2+(8-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/36.png)
-t)
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/37.png)
t
2-56t+100
如圖4,當(dāng)⊙P與⊙Q外切時(shí),有PQ=PA+QC=3t,此時(shí)PQ
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/38.png)
t
2-56t+100=9t
2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/images41.png)
整理得:t
2+70t-125=0
解得:t
1=15
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/39.png)
-35,t
2=-15
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/40.png)
-35<0(舍去)
故當(dāng)⊙P與⊙Q外切時(shí),t=(15
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/41.png)
-35)秒;
當(dāng)⊙P與⊙Q內(nèi)切時(shí),PQ=PA-QC=t,此時(shí),
∵PQ
2=PF
2+FQ
2,
∴PQ
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/42.png)
t
2-56t+100=t
2整理得:9t
2-70t+125=0,解得:t
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/43.png)
,t
2=5
故當(dāng)⊙P與⊙Q內(nèi)切時(shí),t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192936848956939/SYS201311011929368489569021_DA/44.png)
秒或5秒.
點(diǎn)評(píng):本題主要考查了相似三角形的性質(zhì),以及圓和圓的位置關(guān)系,正確把圖形之間的位置關(guān)系轉(zhuǎn)化為線段之間的相等關(guān)系是解題的關(guān)鍵.