【答案】
分析:(1)求面積要先求梯形的高,可根據(jù)兩底的差和CD的長,在直角三角形中用勾股定理進(jìn)行求解,得出高后即可求出梯形的面積.
(2)①PQ平分梯形的周長,那么AD+DQ+AP=BC+CQ+BP,已知了AD,BC的長,可以用t來表示出AP,BP,CQ,QD的長,那么可根據(jù)上面的等量關(guān)系求出t的值.
②本題要分三種情況進(jìn)行討論:
一,當(dāng)P在AB上時(shí),即0<t≤8,如果兩三角形相似,那么∠C=∠ADP,或∠C=∠APD,那么在△ADP中根據(jù)∠C的正切值,求出t的值.
二,當(dāng)P在AD上時(shí),即8<t≤10,由于P,A,D在一條直線上,因此構(gòu)不成三角形.
三,當(dāng)P在CD上時(shí),即10<t≤12,由于∠ADC是個(gè)鈍角,因此△ADP是個(gè)鈍角三角形因此不可能和直角△CQE相似.
綜合三種情況即可得出符合條件的t的值.
(3)和(2)相同也要分三種情況進(jìn)行討論:
一,當(dāng)P在AB上時(shí),即0<t≤8,等腰△PDQ以DQ為腰,因此DQ=DP或DQ=PQ,可以通過構(gòu)建直角三角形來表示出DP,PQ的長,然后根據(jù)得出的等量關(guān)系來求t的值.
二,當(dāng)P在AD上時(shí),即8<t≤10,由于BA+AD=CD=10,因此DP=DQ=10-t,因此DP,DQ恒相等.
三,當(dāng)P在CD上時(shí),即10<t≤12,情況同二.
綜合三種情況可得出等腰三角形以DQ為腰時(shí),t的取值.
解答:
解:(1)過D作DH∥AB交BC于H點(diǎn),
∵AD∥BH,DH∥AB,
∴四邊形ABHD是平行四邊形.
∴DH=AB=8;BH=AD=2.
∴CH=8-2=6.
∵CD=10,
∴DH
2+CH
2=CD
2∴∠DHC=90°.
∠B=∠DHC=90°.
∴梯形ABCD是直角梯形.
∴S
ABCD=
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(AD+BC)AB=
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×(2+8)×8=40.
(2)①∵BP=CQ=t,
∴AP=8-t,DQ=10-t,
∵AP+AD+DQ=PB+BC+CQ,
∴8-t+2+10-t=t+8+t.
∴t=3<8.
∴當(dāng)t=3秒時(shí),PQ將梯形ABCD周長平分.
②第一種情況:0<t≤8若△PAD∽△QEC則∠ADP=∠C
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∴tan∠ADP=tan∠C=
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=
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∴
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=
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,∴t=
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若△PAD∽△CEQ則∠APD=∠C
∴tan∠APD=tan∠C=
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=
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,∴
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=
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∴t=
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第二種情況:8<t≤10,P、A、D三點(diǎn)不能組成三角形;
第三種情況:10<t≤12△ADP為鈍角三角形與Rt△CQE不相似;
∴t=
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或t=
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時(shí),△PAD與△CQE相似.
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③第一種情況:當(dāng)0≤t≤8時(shí).過Q點(diǎn)作QE⊥BC,QH⊥AB,垂足為E、H.
∵AP=8-t,AD=2,
∴PD=
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=
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.
∵CE=
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t,QE=
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t,
∴QH=BE=8-
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t,BH=QE=
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t.
∴PH=t-
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t=
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t.
∴PQ=
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=
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,DQ=10-t.
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Ⅰ:DQ=DP,10-t=
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,
解得t=8秒.
Ⅱ:DQ=PQ,10-t=
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,
化簡得:3t
2-52t+180=0
解得:t=
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,t=
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>8(不合題意舍去)
∴t=
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第二種情況:8≤t≤10時(shí).DP=DQ=10-t.
∴當(dāng)8≤t<10時(shí),以DQ為腰的等腰△DPQ恒成立.
第三種情況:10<t≤12時(shí).DP=DQ=t-10.
∴當(dāng)10<t≤12時(shí),以DQ為腰的等腰△DPQ恒成立.
綜上所述,t=
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或8≤t<10或10<t≤12時(shí),以DQ為腰的等腰△DPQ成立.
點(diǎn)評(píng):本題主要考查了梯形的性質(zhì)以及相似三角形的判定和性質(zhì)等知識(shí)點(diǎn),要注意(2)中要根據(jù)P,Q的不同位置,進(jìn)行分類討論,不要漏解.