如圖,在等腰△ABC中,AB=AC=5cm,BC=6cm,點(diǎn)P從點(diǎn)B開始沿BC邊以每秒1cm的速度向點(diǎn)C運(yùn)動(dòng),點(diǎn)Q從點(diǎn)C開始沿CA邊以每秒2 cm的速度向點(diǎn)A運(yùn)動(dòng),DE保持垂直平分PQ,且交PQ于點(diǎn)D,交BC于點(diǎn)E.點(diǎn)P,Q分別從B,C兩點(diǎn)同時(shí)出發(fā),當(dāng)點(diǎn)Q
![](http://hiphotos.baidu.com/zhidao/pic/item/728da9773912b31bfa6544ad8518367adab4e12d.jpg)
運(yùn)動(dòng)到點(diǎn)A時(shí),點(diǎn)Q、p停止運(yùn)動(dòng),設(shè)它們運(yùn)動(dòng)的時(shí)間為x cm.
(1)當(dāng)x=______秒時(shí),射線DE經(jīng)過點(diǎn)C;
(2)當(dāng)點(diǎn)Q運(yùn)動(dòng)時(shí),設(shè)四邊形ABPQ的面積為ycm
2,求y與x的函數(shù)關(guān)系式(不用寫出自變量取值范圍);
(3)當(dāng)點(diǎn)Q運(yùn)動(dòng)時(shí),是否存在以P、Q、C為頂點(diǎn)的三角形與△PDE相似?若存在,求出x的值;若不存在,請(qǐng)說明理由.
![](http://hiphotos.baidu.com/zhidao/pic/item/b3b7d0a20cf431ad78a1cabd4836acaf2fdd98c7.jpg)
(1)x=2;
當(dāng)DE經(jīng)過點(diǎn)C時(shí),∵DE⊥PQ,PD=QD,
∴PC=CQ,PC=6-x,CQ=2x,
即6-x=2x,得x=2,
∴當(dāng)x=2時(shí),當(dāng)DE經(jīng)過點(diǎn)C;
(2)分別過點(diǎn)Q、A作QN⊥BC,AM⊥BC垂足為M、N.
∵AB=AC=5cm,BC=6cm,
∴
AM==4(cm),
∵QN∥AM,
∴△QNC∽△AMC,
∴
=,即
=,
∴
QN=x,
又PC=6-x,
∴S
△PCQ=
PC?QN=
(6?x)?x,
∴y=S
△ABC-S
△PCQ=
×6×4-
(6?x)?x,
即
y=x2?x+12;
![](http://hiphotos.baidu.com/zhidao/pic/item/c995d143ad4bd1135dbf3eb659afa40f4afb05c7.jpg)
(3)存在.
理由如下:
∵DE⊥PQ,
∴PQ⊥AC時(shí)△PQC∽△PDE
此時(shí),△PQC∽△AMC
∴
=即
=∴
x=.