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解:(1)過點P作PE⊥CD于E.則根據(jù)題意,得
EQ=16-2×3-2×2=6(cm),PE=AD=6cm;
在Rt△PEQ中,根據(jù)勾股定理,得
PE
2+EQ
2=PQ
2,即36+36=PQ
2,
∴PQ=6
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cm;
∴經(jīng)過2s時P、Q兩點之間的距離是6
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cm;
(2)設(shè)x秒后,點P和點Q的距離是10cm.
(16-2x-3x)
2+6
2=10
2,即(16-5x)
2=64,
∴16-5x=±8,
∴x1=
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,x2=
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;
∴經(jīng)過
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s或
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sP、Q兩點之間的距離是10cm;
(3)連接BQ.設(shè)經(jīng)過ys后△PBQ的面積為12cm
2.
①當(dāng)0≤y≤
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時,則PB=16-3y,
∴
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PB•BC=12,即
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×(16-3y)×6=12,
解得y=4;
②當(dāng)
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<x≤
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時,
BP=3y-AB=3y-16,QC=2y,則
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BP•CQ=
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(3y-16)×2y=12,
解得y
1=6,y2=-
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(舍去);
③
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<x≤8時,
QP=CQ-PQ=22-y,則
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QP•CB=
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(22-y)×6=12,
解得y=18(舍去).
綜上所述,經(jīng)過4秒或6秒△PBQ的面積為 12cm
2.
分析:(1)作PE⊥CD于E,表示出PQ的長度,利用PE
2+EQ
2=PQ
2列出方程求解即可;
(2)設(shè)x秒后,點P和點Q的距離是10cm.在Rt△PEQ中,根據(jù)勾股定理列出關(guān)于x的方程(16-5x)
2=64,通過解方程即可求得x的值;
(3)分類討論:①當(dāng)點P在AB上時;②當(dāng)點P在BC邊上;③當(dāng)點P在CD邊上時.
點評:本題考查了一元二次方程的應(yīng)用,另外還綜合考查了矩形的性質(zhì)、兩點間的距離、三角形的面積等知識點.解答(3)時,要分類討論,以防漏解.