Question

如图,平行四边形ABCD中,AB=8cm,BC=6cm,∠A=30°,点P从点A沿线段AB以1cm/s的速度向

如图，平行四边形ABCD中，AB=8cm，BC=6cm，∠A=30°，点P从点A沿线段AB以1cm/s的速度向点B移动.
1）当P点运动几秒时，△PBC为等腰三角形；（2）设S△PBC=y，请写出y（cm2）与点P的移动时间 t（s）之间的函数关系式，并写出t的取值范围；（3）是否存在否一点P，使S△PBC=1/3S平行四边形ABCD？若存在求AP的长，若不存在，请说明理由。

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Answer 1

(1) As the angle B is greater than 90 degrees, the only possibility will be BP=BC.
BC=6, AB=8, hence AP=2. t=2/1=2(second)
(2) We can see that AP=t*1=t, then BP=8-t,
(the height of PBC on the BP side)=BC*sinA=6*sin30=3
y=(8-t)*3/2=12-3/2t
(3) Yes, there is such a point P.
Area of ABCD=8*3=24
Area of PBC should be: 24/3=8
That is: 12-3/2t=8, t=8/3
AP=8/3
Sorry I can't type in Chinese in my school's computer. Don't forget my rewards!

追问

thank you very much....i good at English so i can know! Are you studying in school now? It's too late.....to pay attention to having a rest

追答

I'm very sorry, for my English skill is very limited, hence I can't understand your words.

追问

Now that it doesn't matter....please tell me the (2)..span

追答

好了我有中文输入法了。
第（2）问的话，
P点速度为1cm/s，则AP的长度可以看作是t乘速度，即为t，BP就是8减t了。
c点到AB的距离就是三角形PBC的高了，作C到AB的垂线，设它交AB延长线于E点，角CBE就等于角A为30度，
我们都知道直角三角形一个角为30度时的边长比为1：2：根号3
BC为6则CE，即高为3。
三角形面积y=BP*CE/2 对巴，剩下的很好算了。
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Answer 2

（1）△PBC为等边三角形那么BC=PB。
BC=6，那PB=6，AB=8，AB-PB=8-6=2CM
p点移动速度为1cm/秒，所以是2秒