【答案】
分析:(1)過點(diǎn)D作DE⊥BC于點(diǎn)E,由已知得AD=BE,DE=AB=20cm.在Rt△DEC中,根據(jù)勾股定理得EC=15cm.由題意得
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=
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,由此可以求出AD的長,然后可以求出梯形的面積;
(2)設(shè)P、Q兩點(diǎn)運(yùn)動(dòng)的時(shí)間為t,則點(diǎn)P運(yùn)動(dòng)的路程為3t(cm),點(diǎn)Q運(yùn)動(dòng)的路程為4t(cm).
①當(dāng)0<t≤
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時(shí),P在AD上運(yùn)動(dòng),Q在AB上運(yùn)動(dòng),此時(shí)四邊形APCQ的面積S=S
梯形ABCD-S
△BCQ-S
△CDP=70t;
②當(dāng)
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<t≤5時(shí),P在DC上運(yùn)動(dòng),Q在AB上運(yùn)動(dòng),此時(shí)四邊形APCQ的面積S=S
梯形ABCD-S
△BCQ-S
△ADP=34t+60;
③當(dāng)5<t<10時(shí),P在DC上運(yùn)動(dòng),Q在BC上運(yùn)動(dòng),此時(shí)四邊形APCQ的面積S=S
梯形ABCD-S
△ABQ-S
△ADP=-46t+460.
(3)根據(jù)(2)的函數(shù)關(guān)系式,分別把已知梯形面積的
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代入其中就可以求出相應(yīng)的t,然后結(jié)合已知條件進(jìn)行取舍
最后得到t的取值.
解答:解:(1)過點(diǎn)D作DE⊥BC于點(diǎn)E,由已知得AD=BE,DE=AB=20cm.
在Rt△DEC中,根據(jù)勾股定理得EC=15cm.由題意得
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=
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,
∴
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=
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.解得AD=5.
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∴梯形ABCD的面積=
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=
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=250(cm
2).
(2)當(dāng)P、Q兩點(diǎn)運(yùn)動(dòng)的時(shí)間為t(秒)時(shí),點(diǎn)P運(yùn)動(dòng)的路程為3t(cm),點(diǎn)Q運(yùn)動(dòng)的路程為4t(cm).
①當(dāng)0<t≤
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時(shí),P在AD上運(yùn)動(dòng),Q在AB上運(yùn)動(dòng).
此時(shí)四邊形APCQ的面積S=S
梯形ABCD-S
△BCQ-S
△CDP=70t.
②當(dāng)
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<t≤5時(shí),P在DC上運(yùn)動(dòng),Q在AB上運(yùn)動(dòng).
此時(shí)四邊形APCQ的面積S=S
梯形ABCD-S
△BCQ-S
△ADP=34t+60.
③當(dāng)5<t<10時(shí),P在DC上運(yùn)動(dòng),Q在BC上運(yùn)動(dòng).
此時(shí)四邊形APCQ的面積S=S
梯形ABCD-S
△ABQ-S
△ADP=-46t+460.
(3)①當(dāng)0<t≤
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時(shí),由S=70t=250×
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,解得t=
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.
②當(dāng)
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<t≤5時(shí),由S=34t+60=250×
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,解得t=
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.
又∵
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<t≤5,
∴t=
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不合題意,舍去.
③當(dāng)5<t<10時(shí),由S=-46t+460=250×
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,
解得t=
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.
∴當(dāng)t=
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或t=
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時(shí),四邊形APCQ的面積恰為梯形ABCD的面積的
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.
點(diǎn)評(píng):在有關(guān)動(dòng)點(diǎn)的幾何問題中,由于圖形的不確定性,我們常常需要針對(duì)各種可能出現(xiàn)的圖形對(duì)每一種可能的情形都分別進(jìn)行研究和求解.換句話說,分類思想在動(dòng)態(tài)問題中運(yùn)用最為廣泛.