【答案】
分析:(1)設(shè)經(jīng)過x秒后,根據(jù)△PBQ的面積等于8cm
2.得出方程
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/0.png)
×(6-x)×2x=8,求出方程的解即可;
(2)設(shè)經(jīng)過y秒后,△PCQ的面積等于12.6cm
2.那么可分以下情況討論設(shè)經(jīng)過y秒后,△PCQ的面積等于12.6cm
2.
(1)0<y≤4(Q在BC上,P在AB上)時(shí),連接PC,求出CQ=8-2y,PB=6-y,根據(jù)三角形的面積公式得出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/1.png)
×(8-2y)×(6-y)=12.6,求出方程的解即可;(2)4<y≤6(Q在CA上,P在AB上),過點(diǎn)P作PM⊥AC,交AC于點(diǎn)M,求出CQ=2y-8,AP=y,根據(jù)sinA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/3.png)
,推出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/5.png)
,求出PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/6.png)
y,根據(jù)三角形的面積公式求出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/7.png)
×(2y-8)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/8.png)
y=12.6,求出方程的解即可;(3)6<y≤9(Q在CA上,P在BC上),過點(diǎn)Q作QD⊥BC,交BC于點(diǎn)D,根據(jù)QD∥AB得出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/9.png)
,代入求出QD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/10.png)
,根據(jù)三角形的面積公式得出方程
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/11.png)
×(14-y)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/12.png)
=12.6,求出方程的解即可.
解答:解:(1)設(shè)經(jīng)過x秒后,△PBQ的面積等于8cm
2.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/13.png)
×(6-x)×2x=8,
解得x
1=2 x
2=4,
答:經(jīng)過2或4秒后,△PBQ的面積等于8cm
2.
(2)設(shè)經(jīng)過y秒后,△PCQ的面積等于12.6cm
2.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/images14.png)
①0<y≤4(Q在BC上,P在AB上)時(shí),如圖:(1)連接PC,
則CQ=8-2y,PB=6-y,
∵S
△PQC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/14.png)
CQ×PB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/15.png)
×(8-2y)×(6-y)=12.6,
解得y
1=5+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/16.png)
>4(不合題意,舍去),y
2=5-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/17.png)
;
②4<y≤6(Q在CA上,P在AB上),如圖(2)
過點(diǎn)P作PM⊥AC,交AC于點(diǎn)M,
由題意可知CQ=2y-8,AP=y,
在直角三角形ABC中,sinA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/19.png)
,
在直角三角形APM中,sinA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/20.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/22.png)
,
∴PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/23.png)
y,
∵S
△PCQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/24.png)
CQ×PM,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/25.png)
×(2y-8)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/26.png)
y=12.6,
解得y
1=2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/27.png)
>6(舍去),y
2=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/28.png)
<0(負(fù)值舍去);
③6<y≤9(Q在CA上,P在BC上),如圖(3),
過點(diǎn)Q作QD⊥BC,交BC于點(diǎn)D,
∵∠B=90°,
∴QD∥AB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/29.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/30.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/31.png)
,
∴QD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/32.png)
,
∵S
△CQP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/33.png)
×CP×QD,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/34.png)
×(14-y)×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/35.png)
=12.6
解得:y
1=7,y
2=11(不合題意,舍去)
答:當(dāng)(5-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192056564238879/SYS201311011920565642388028_DA/36.png)
)秒或7秒后,△PCQ的面積等于12.6cm
2點(diǎn)評:應(yīng)注意應(yīng)先表示出兩直角三角形的面積所需要的邊和高,然后分情況進(jìn)行討論.