9−t5
解得:
t=1.
答:
当t=1s,点P、Q、F三点在同一条直线上.
25.已知,正方形ABCD中,∠MAN=45°,∠MAN绕点A顺时针旋转,它的两边分别交
CB、DC(或它们的延长线)于点M、N,AH⊥MN于点H.
(1)如图①,当∠MAN绕点A旋转到BM=DN时,请你直接写出AH与AB的数量关系:
;
(2)如图②,当∠MAN绕点A旋转到BM≠DN时,
(1)中发现的AH与AB的数量关系还成立吗?
如果不成立请写出理由.如果成立请证明;
(3)如图③,已知∠MAN=45°,AH⊥MN于点H,且MH=2,NH=3,求AH的长.
(可利用
(2)得到的结论)
【关键词】正方形与旋转
【答案】解:
(1)如图①AH=AB………………………..1分
(2)数量关系成立.如图②,延长CB至E,使BE=DN
∵ABCD是正方形
∴AB=AD,∠D=∠ABE=90°
∴Rt△AEB≌Rt△AND………………………………3分图①
∴AE=AN,∠EAB=∠NAD
∴∠EAM=∠NAM=45°
∵AM=AM
∴△AEM≌△ANM………………………………….4分
∵AB、AH是△AEM和△ANM对应边上的高,
∴AB=AH……………………………………………...5分
(3)如图③分别沿AM、AN翻折△AMH和△ANH,得到△ABM和△AND
∴BM=2,DN=3,∠B=∠D=∠BAD=90°
分别延长BM和DN交于点C,得正方形ABCE.由
(2)可知,AH=AB=BC=CD=AD.
设AH=x,则MC=x−2,NC=x−3图②在Rt⊿MCN中,由勾股定理,得
MN2=MC2+NC2
∴52=(x−2)2+(x−3)2………………………6分
解得x1=6,x2
=−1.(不符合题意,舍去)
∴AH=6.……………………………………………7分
图③
1.如图,在△ABC中,AB=AC=2,∠BAC=20�.动点P,Q分别在直线BC上运动,且始终保持∠PAQ=100�.设BP=x,CQ=y,则y与x之间的函数关系用图象大致可以表示为()
A
BC
A.B.C.D.
【关键词】函数的图象
【答案】A
(12分)如图,已知抛物线y=1x2+bx+c与y轴相交于C,与x轴相交于A、B,点A
2
的坐标为(2,0),点C的坐标为(0,-1).
(1)求抛物线的解析式;
(2)点E是线段AC上一动点,过点E作DE⊥x轴于点D,连结DC,当△DCE的面积最大时,求点D的坐标;
(3)在直线BC上是否存在一点P,使△ACP为等腰三角形,若存在,求点P的坐标,若不存在,说明理由.
【关键词】二次函数及动点问题y
【答案】
D
BoAx
E
C
26题图
解:
(1)∵二次函数y=1x2+bx+c的图像经过点A(2,0)C(0,-1)
2
⎨
⎧2+2b+c=0
∴
⎩c=−1
解得:
b=-1
2
c=-1-------------------2分
∴二次函数的解析式为y=1x2−1x−1
--------3分
22
(2)设点D的坐标为(m,0)(0<m<2)
∴OD=m∴AD=2-m
由△ADE∽△AOC得,AD
AO
DE
--------------4分
OC
2−mDE
∴=
21
∴DE=
2−m
2
---------------------------5分
∴△CDE的面积=1
2−m
×
×m=−
m2+m
=−1
(m−1)2+1
224244
当m=1时,△CDE的面积最大
∴点D的坐标为(1,0)--------------8分
(3)存在由
(1)知:
二次函数的解析式为y=1x2−1x−1
22
设y=0则0=1x2−1x−1解得:
x1=2x2=-1
22
∴点B的坐标为(-1,0)C(0,-1)设直线BC的解析式为:
y=kx+b
⎧−k+b=0
⎨
∴
⎩b=−1
解得:
k=-1b=-1
∴直线BC的解析式为:
y=-x-1
在Rt△AOC中,∠AOC=900OA=2OC=1
由勾股定理得:
AC=
∵点B(-1,0)点C(0,-1)
∴OB=OC∠BCO=450
①当以点C为顶点且PC=AC=5时,设P(k,-k-1)
过点P作PH⊥y轴于H
∴∠HCP=∠BCO=450
CH=PH=∣k∣在Rt△PCH中
k2+k2=(
5)2
1010
解得k=,k=-
12
22
∴P1(,-
22
−1)P2(-
2
,−1)---10分
2
②以A为顶点,即AC=AP=
设P(k,-k-1)
过点P作PG⊥x轴于GAG=∣2-k∣GP=∣-k-1∣在Rt△APG中AG2+PG2=AP2
(2-k)2+(-k-1)2=5解得:
k1=1,k2=0(舍)
∴P3(1,-2)---------11分
③以P为顶点,PC=AP设P(k,-k-1)过点P作PQ⊥y轴于点Q
PL⊥x轴于点L
∴L(k,0)
∴△QPC为等腰直角三角形
PQ=CQ=k
由勾股定理知CP=PA=k
∴AL=∣k-2∣,PL=|-k-1|
在Rt△PLA中(
5
k)2=(k-2)2+(k+1)2
57
解得:
k=
∴P4(
2
-)-----------12分
22
综上所述:
存在四个点:
P1(
2
,-−1)
2
P2(-
2
,−1)P3(1,-2)P4(
2
57
-)
22
如图,四边形ABCD是边长为1的正方形,四边形EFGH是边长为2的正方形,点D与点
F重合,点B,D(F),H在同一条直线上,将正方形ABCD沿F→H方向平移至点B与点H重合时停止,设点D、F之间的距离为x,正方形ABCD与正方形EFGH重叠部分的面积为y,则能大致反映y与x之间函数关系的图象是()
EE
G10题图G
【关键词】A函数图像及动点问题BCD
【答案】B
1如图,正方形ABCD的边长是3cm,一个边长为1cm的小正方形沿
着正方形ABCD的边AB→BC→CD→DA→AB连续地翻转,那AB
么这个小正方形第一次回到起始位置时,它的方向是下图的
()
ABCD
【关键词】翻转,旋转
【答案】A
DC
第7题图
2.如图,已知直角梯形ABCD中,AD∥BC,AB⊥BC,AD=2,AB=8,CD=10.
(1)求梯形ABCD的周长;
(2)动点P从点B出发,以1cm/s的速度沿B→A→D→C方向向点C运动;动点Q从点
C出发,以1cm/s的速度沿C→D→A方向向点A运动;过点Q作QF⊥BC于点F.若P、
Q两点同时出发,当其中一点到达终点时整个运动随之结束,设运动时间为t秒.问:
①当点P在B→A上运动时,是否存在这样的t,使得直线PQ将梯形ABCD的周长平分?
若存在,请求出t的值;若不存在,请说明理由.
②在运动过程中,是否存在这样的t,使得以P、D、Q为顶点的三角形恰好是以DQ为一腰的等腰三角形?
若存在,请求出所有符合条件的t的值;若不存在,请说明理由.
【关键词】运动与等腰三角形
【答案】解:
(1)过点D作DE⊥BC于点E
∵四边形ABCD是直角梯形
∴四边形ABED是矩形
∴AD=BE=2,AB=DE=8
在Rt△DEC中,CE===6
∴梯形ABCD的周长=AB+BC+CD+DA=28.
(2)①∵梯形ABCD的周长为28,PQ平分梯形ABCD的周长
∴BP+BC+CQ=14
又∵BP=CQ=t
∴t+8+t=14
∴t=3
∴当t=3时,PQ平分梯形ABCD的周长.
②(i)当0≤t≤8时,过点Q作QG⊥AB于点G
∵AP=8-t,DQ=10-t,AD=2,sinC=4,cosC=3
55
∴CF=3t,QF=4t,PG=t−4t=1t,QG=8-3t
55555
PD2=AP2+AD2=(8-t)2+22=t2+16t+68,
PQ2=QG2+PG2=(8-3t)2+(1t)2=2t2−48t+64
5555
若DQ=PD,则(10-t)2=t2+16t+68,解得:
t=8;
若DQ=PQ,则(10-t)2=2t2−48t+64,
55
解得:
t1=26−234
3
,t2=26+2
3
34>8(舍去),此时t=26−2
3
34;
(ii)当8<t<10时,PD=DQ=10-t,
∴此时以DQ为一腰的等腰△DPQ恒成立;
而当t=10时,点P、D、Q三点重合,无法构成三角形;
(iii)当10<t≤12时,PD=DQ=t-10,
∴此时以DQ为一腰的等腰△DPQ恒成立;
综上所述,当t=26−2
3
34或8≤t<10或10<t≤12时,以P、D、Q为顶点的三角形恰好是
以DQ为一腰的等腰三角形.
25.如图,已知等边三角形ABC中,点D,E,F分别为边AB,AC,BC的中点,M为直线BC上一动点,△DMN为等边三角形(点M的位置改变时,△DMN也随之整体移动).
(1)如图①,当点M在点B左侧时,请你判断EN与MF有怎样的数量关系?
点F是否在直线NE上?
都.请.直.接.写出结论,不必证明或说明理由;
(2)如图②,当点M在BC上时,其它条件不变,
(1)的结论中EN与MF的数量关系是否仍然成立?
若成立,请利用图②证明;若不成立,请说明理由;
(3)若点M在点C右侧时,请你在图③中画出相应的图形,并判断
(1)的结论中EN
与MF的数量关系是否仍然成立?
若成立?
请直接写出结论,不必证明或说明理由.
AAA
DE
D··E
BB
FC·
B·
MMFCFC
N
图①图②图③
第25题图
【关键词】等边三角形
【答案】
25.
(1)判断:
EN与MF相等(或EN=MF),点F在直线NE上,····························3分
(说明:
答对一个给2分)
(2)成立.··································································································································4分证明:
法一:
连结DE,DF.·······································································································5分
∵△ABC是等边三角形,∴AB=AC=BC.又∵D,E,F是三边的中点,
∴DE,DF,EF为三角形的中位线.∴DE=DF=EF,∠FDE=60°.
又∠MDF+∠FDN=60°,∠NDE+∠FDN=60°,
∴∠MDF=∠NDE.················································································································7分在△DMF和△DNE中,DF=DE,DM=DN,∠MDF=∠NDE,
∴△DMF≌△DNE.··············································································································8分
∴MF=NE.··············································································································9分
AA
BMF
CBMFC
法二:
延长EN,则EN过点F.··································································································5分
∵△ABC是等边三角形,∴AB=AC=BC.
又∵D,E,F是三边的中点,∴EF=DF=BF.
∵∠BDM+∠MDF=60°,∠FDN+∠MDF=60°,
∴∠BDM=∠FDN.····················································································································7分又∵DM=DN,∠ABM=∠DFN=60°,
∴△DBM≌△DFN.··················································································································8分
∴BM=FN.
∵BF=EF,∴MF=EN.··········································································································9分法三:
连结DF,NF.··························································································································5分
∵△ABC是等边三角形,
∴AC=BC=AC.
又∵D,E,F是三边的中点,
∴DF为三角形的中位线,∴DF=1
2
AC=
1AB=DB.
2
又∠BDM+∠MDF=60°,∠NDF+∠MDF=60°,
∴∠BDM=∠FDN.················································································································7分在△DBM和△DFN中,DF=DB,
DM=DN,∠BDM=∠NDF,∴△DBM≌△DFN.
∴∠B=∠DFN=60°.···················································································································8分又∵△DEF是△ABC各边中点所构成的三角形,
∴∠DFE=60°.
∴可得点N在EF上,
∴MF=EN.··············································································································9分
(3)画出图形(连出线段NE),··························································································11分
MF与EN相等的结论仍然成立(或MF=NE成立).······················································12分
N
BFCM
1.如图,在等边∆ABC中,线段AM为BC边上的中线.动点D在直.线.AM上时,以CD为一边且在CD的下方作等边∆CDE,连结BE.
(1)填空:
∠ACB=______度;
(2)当点D在线.段.AM上(点D不运动到点A)时,试求出AD的值;
BE
(3)若AB=8,以点C为圆心,以5为半径作⊙