即S??25t2?65t.
(3)能.
①当DE∥QB时,如图4.
∵DE⊥PQ,∴PQ⊥QB,四边形QBED是直角梯形. 此时∠AQP=90°.
由△APQ ∽△ABC,得AQ?APACAB,
B 即t3?3?t5. 解得t?98.
②如图5,当PQ∥BC时,DE⊥BC,四边形QBED是直角梯形. 此时∠APQ =90°. Q 由△AQP ∽△ABC,得
AQAPD EAB?AC, A P
C 图5
即t5?3?t3. 解得t?158
. B
(4)t?5452或t?14. ①点P由C向A运动,DE经过点C. Q G 连接QC,作QG⊥BC于点G,如图6.
PC?t,QC2?QG2?CG2?[3(5?t)]24D 5?[4?5(5?t)]2.
A PC(E) 图6 由PC2?QC2,得t2B ?[35(5?t)]2?[4?455(5?t)]2,解得t?2.
②点P由A向C运动,DE经过点C,如图7.
Q G (6?t)2?[3(5?t)]2?[4?4(5?t)]255,t?4514】
D C(6.解(1)①30,1;②60,; ……………………A PE)
图7 4分 (2)当∠α=900时,四边形EDBC是菱形.
∵∠α=∠ACB=900,∴BC ∵CE ……………………6分 在Rt△ABC中,∠ACB=900,∠B=600,BC=2,
∴∠A=300.
∴AB=4,AC=23. ∴AO=
12AC=3 . ……………………8分 在Rt△AOD中,∠A=300,∴AD=2. ∴BD=2. ∴BD=BC.
又∵四边形EDBC是平行四边形,
∴四边形EDBC是菱形 ……………………10分
7.解:(1)如图①,过A、D分别作AK?BC于K,DH?BC于H,则四边形ADHK是矩形∴KH?AD?3. ·······················1分
在Rt△ABK中,AK?ABgsin45??42.22?4 BK?ABgcos45??42g22?4 ················2分 在Rt△CDH中,由勾股定理得,HC?52?42?3
∴BC?BK?KH?HC?4?3?3?10 ··············3分 A D A D
N
B K H C B C
G M (图①) (图②) 2)如图②,过D作DG∥AB交BC于G点,则四边形ADGB是平行四边形∵MN∥AB ∴MN∥DG ∴BG?AD?3
∴GC?10?3?7 ·····················4分 由题意知,当M、N运动到t秒时,CN?t,CM?10?2t. ∵DG∥MN
∴∠NMC?∠DGC 又∠C?∠C
∴△MNC∽△GDC ∴CNCD?CMCG ·······················5分 即t5?10?2t7 解得,t?5017 ·······················6分
(3)分三种情况讨论:
①当NC?MC时,如图③,即t?10?2t
∴t?103 ·························7分
A D
A D N N B M C B M H E C
(图③) (图④) ②当MN?NC时,如图④,过N作NE?MC于E 解法一:
(
由等腰三角形三线合一性质得EC?12MC?12?10?2t??5?t 在Rt△CEN中,cosc?EC5?tNC?t 又在Rt△DHC中,cosc?CHCD?35
∴5?t3t?5
解得t?258 ························ 8分
解法二:
∵∠C?∠C,?DHC??NEC?90? ∴△NEC∽△DHC ∴NCECDC?HC 即t5?5?t3 ∴t?258 ························· 8分
③当MN?MC时,如图⑤,过M作MF?CN于F点.FC?112NC?2t
解法一:(方法同②中解法一)
1cosC?FCt3A D MC?210?2t?5
解得t?60N 17
F 解法二:
B ∵∠C?∠C,?MFC??DHC?90? H M
C ∴△MFC∽△DHC (图⑤) ∴FCMCHC?DC 1t即210?2t3?5 ∴t?6017
综上所述,当t?10253、t?8或t?6017时,△MNC为等腰三角形 ·· 9分
8.解(1)如图1,过点E作EG?BC于点G. ··· 1分
∵E为AB的中点,
∴BE?1A D
2AB?2.
在Rt△EBG中,∠B?60?,∴∠BEG?30?. ·· 2分
E
F ∴BG?12BE?1,EG?22?12?3.
B G
C 即点E到BC的距离为3. ········· 3分
图1
(2)①当点N在线段AD上运动时,△PMN的形状不发生改变.
∵PM?EF,EG?EF,∴PM∥EG. ∵EF∥BC,∴EP?GM,PM?EG?3.
同理MN?AB?4. ·······················4分
如图2,过点P作PH?MN于H,∵MN∥AB, ∴∠NMC?∠B?60?,∠PMH?30?. A
N
D
∴PH?12PM?32. E
P F H
∴MH?PMgcos30??32.
B
G M
C
则NH?MN?MH?4?352?2.
图2
22在Rt△PNH中,PN?NH2?PH2???5??3??2??????2???7. ?∴△PMN的周长=PM?PN?MN?3?7?4. ··········6分
②当点N在线段DC上运动时,△PMN的形状发生改变,但△MNC恒为等边三角形. 当PM?PN时,如图3,作PR?MN于R,则MR?NR.
类似①,MR?32.
∴MN?2MR?3. ·······················7分 ∵△MNC是等边三角形,∴MC?MN?3.
此时,x?EP?GM?BC?BG?MC?6?1?3?2. ·········8分
A D A
D A D N
E P
F
E
P
F E F(P) R
N N B
G
M
C
B G
M C
B G
M
C
图3
图4
图5
当MP?MN时,
如图4,这时MC?MN?MP?3.
此时,x?EP?GM?6?1?3?5?3.
当NP?NM时,如图5,∠NPM?∠PMN?30?.
则∠PMN?120?,又∠MNC?60?, ∴∠PNM?∠MNC?180?.
因此点P与F重合,△PMC为直角三角形. ∴MC?PMgtan30??1. 此时,x?EP?GM?6?1?1?4.
综上所述,当x?2或4或?5?3?时,△PMN为等腰三角形. ··· 10分 9解:(1)Q(1,0) ·······················1分 点P运动速度每秒钟1个单位长度. ················2分 (2) 过点B作BF⊥y轴于点F,BE⊥x轴于点E,则BF=8,OF?BE?4.
∴AF?10?4?6.
y 在Rt△AFB中,AB?82?62?10 3分 D 过点C作CG⊥x轴于点G,与FB的延长线交于点H. C∵?ABC?90?,AB?BC ∴△ABF≌△BCH. AMP ∴BH?AF?6,CH?BF?8. FBH∴ONQOG?FH?8?6?14,CG?8?4?12.
EGx∴所求C点的坐标为(14,12). 4分 (3) 过点P作PM⊥y轴于点M,PN⊥x轴于点N, 则△APM∽△ABF.
∴
APAMMPtAMMPAB?AF?BF. ?10?6?8. ∴AM?3t,PM?4t. ∴PN?OM?10?3t,ON?PM?45555t. 设△OPQ的面积为S(平方单位) ∴S?1?(10?3t)(1?t)?5?4710t?32510t2(0≤t≤10)
············ 5分 说明:未注明自变量的取值范围不扣分.
47 ∵a??310<0 ∴当t??10473?6时, △OPQ的面积最大. ····· 6分
2?(?10) 此时P的坐标为(
9415,5310) . ·················· 7分 (4) 当 t?53或t?29513时, OP与PQ相等. ············ 9分
10.解:(1)正确. ············ (1分)
证明:在AB上取一点M,使AM?EC,连接ME.(2分)?BM?BE.??BME?45°,??AME?135°.
A D QCF是外角平分线,
M F ??DCF?45°,
??ECF?135°.
B E C G ??AME??ECF.
Q?AEB??BAE?90°,?AEB??CEF?90°, ??BAE??CEF.
?△AME≌△BCF(ASA)
. ·················· (5分) ?AE?EF.
························· (6分) (2)正确. ············· (7分) 证明:在BA的延长线上取一点N. 使AN?CE,连接NE. ········ (8分) ?BN?BE. N
F ??N??PCE?45A D °.
Q四边形ABCD是正方形, ?AD∥BE.
B C E G
??DAE??BEA. ??NAE??CEF.
?△ANE≌△ECF(ASA)
. ·················· (10分) ?AE?EF. (11分)
11.解(Ⅰ)如图①,折叠后点B与点A重合, 则△ACD≌△BCD.
设点C的坐标为?0,m??m?0?. 则BC?OB?OC?4?m. 于是AC?BC?4?m.
在Rt△AOC中,由勾股定理,得AC2?OC2?OA2, 即?4?m?2?m2?22,解得m?32. ?点C的坐标为???0,3?2??. ·······················4分
(Ⅱ)如图②,折叠后点B落在OA边上的点为B?, 则△B?CD≌△BCD. 由题设OB??x,OC?y,
则B?C?BC?OB?OC?4?y,
在Rt△B?OC中,由勾股定理,得B?C2?OC2?OB?2.
??4?y?2?y2?x2,
即y??18x2?2 ··························6分
由点B?在边OA上,有0≤x≤2,
? 解析式y??18x2?2?0≤x≤2?为所求.
? Q当0≤x≤2时,y随x的增大而减小,
?y的取值范围为32≤y≤2.···················7分
(Ⅲ)如图③,折叠后点B落在OA边上的点为B??,且B??D∥OB. 则?OCB????CB??D. 又Q?CBD??CB??D,??OCB????CBD,有CB??∥BA. ?Rt△COB??∽Rt△BOA. 有OB??OA?OCOB,得OC?2OB??. ··················9分 在Rt△B??OC中,
设OB???x0?x?0?,则OC?2x0.
由(Ⅱ)的结论,得2x120??8x0?2,
解得x0??8?45.Qx0?0,?x0??8?45. ?点C的坐标为0,85?16. ················· 10分
?? ∵MN?BE, ??EBC??BNM?90°. QNG?BC, ??MNG??BNM?90°,??EBC??MNG. 在△BCE与△NGM中
??EBC??MNG,? ?BC?NG,∴△BCE≌△NGM,EC?MG. ······ 5分
??C??NGM?90°.?
12解:方法一:如图(1-1),连接BM,EM,BE.
F M
A
D
E
51∵AM?AG?MG,AM=?1?. ·············· 6分
44AM1∴ ······················· 7分 ?.BN5
B N
C
图(1-1) 由题设,得四边形ABNM和四边形FENM关于直线MN对称. ∴MN垂直平分BE.∴BM?EM,BN?EN. ········ 1分
∵四边形ABCD是正方形,∴?A??D??C?90°,AB?BC?CD?DA?2. ∵CECD?12,?CE?DE?1.设BN?x,则NE?x,NC?2?x. 在Rt△CNE中,NE2?CN2?CE2.
∴x2??2?x?2?12.解得x?54,即BN?54. ·········· 3分 在Rt△ABM和在Rt△DEM中,
AM2?AB2?BM2, DM2?DE2?EM2,
?AM2?AB2?DM2?DE2. ················· 5分
设AM?y,则DM?2?y,∴y2?22??2?y?2?12.
解得y?14,即AM?14. ·················· 6分
∴AMBN?15. ······················· 7分 方法二:同方法一,BN?54. ················ 3分
如图(1-2),过点N做NG∥CD,交AD于点G,连接BE.
F
A M G D
E B
N C
图(1-2)
∵AD∥BC,∴四边形GDCN是平行四边形. ∴NG?CD?BC.
同理,四边形ABNG也是平行四边形.∴AG?BN?54.
类比归纳
245(或10);联系拓广
n2m2?2n?1n2m2?1
9217; ?n?1?n2?1 ··················10分 ························12分
初三数学几何的动点问题专题练习及答案
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