第二十七章相似 达标测试卷(word版含答案)

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名称 第二十七章相似 达标测试卷(word版含答案)
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更新时间 2021-10-01 09:30:58

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第二十七章达标测试卷
1.在下列各组线段中,不成比例的是(  )
A.a=3,b=6,c=2,d=4
B.a=1,b=2,c=2,d=4
C.a=4,b=6,c=5,d=10
D.a=1,b=,c=,d=
2.如图,四边形ABCD和四边形A′B′C′D′是以点O为位似中心的位似图形,若OA?OA′=2?3,则四边形ABCD与四边形A′B′C′D′的面积比为(  )
A.4∶9
B.2∶5
C.2∶3
D.3∶2
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(第2题)   
(第3题)  
 
(第4题)   
(第5题)
3.如图,l1∥l2∥l3,直线a,b与l1,l2,l3分别相交于点A,B,C和点D,E,F,若=,DE=6,则EF的长是(  )
A.8
B.9
C.10
D.12
4.如图,在△ABC中,点D,E分别在边AB,AC上,下列条件中不能判定△ABC∽△AED的是(  )
A.∠AED=∠B
B.∠ADE=∠C
C.=
D.=
5.如图,在平行四边形ABCD中,
EF∥AB交AD于点E,交DB于点F,DE?EA=3?4,EF=3,则CD的长为(  )
A.4
B.7
C.3
D.12
6.两个相似三角形的最短边长分别是5
cm和3
cm,它们的周长之差为12
cm,那么较小三角形的周长为(  )
A.14
cm
B.16
cm
C.18
cm
D.30
cm
7.【教材P42习题T3(1)变式】下列选项中的四个三角形,与如图中的三角形相似的是(  )
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8.《孙子算经》是中国古代重要的数学著作,成书于约一千五百年前,其中有首歌谣:今有竿不知其长,量得影长一丈五尺,立一标杆,长一尺五寸,影长五寸,问竿长几何?意即:有一根竹竿不知道有多长,量出它在太阳下的影子长一丈五尺.同时立一根一尺五寸的标杆,它的影长五寸(提示:1丈=10尺,1尺=10寸),则竹竿的长为(  )
A.五丈
B.四丈五尺
C.一丈
D.五尺
9.【教材P43习题T10变式】为了测量校园水平地面上一棵不可攀登的树的高度,学校数学兴趣小组做了如下探索:根据光的反射定律,利用一面镜子和一根皮尺,设计如图所示的测量方案:把一面很小的镜子水平放置在离树8.4
m远的点E处,然后沿着直线BE走到点D,这时恰好在镜子里看到树梢顶点A,再用皮尺量得DE=3.2
m,观察者眼高CD=1.6
m,则树AB的高度为(  )
A.4.2
m
B.4.8
m
C.6.4
m
D.16.8
m
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(第9题)     
(第10题)
10.如图,半圆O的直径BC=7,延长CB到A,割线AED交半圆于点E,D,且AE=ED=3,则AB的长为(  )
A.
B.2
C.
D.9
二、填空题(每题3分,共24分)
11.如果=,那么=________.
12.如果两个相似三角形的面积之比是9?25,其中小三角形一个角的平分线长是12
cm,那么大三角形对应角的平分线的长是________cm.
13.【教材P41练习T2改编】如图是测量河宽的示意图,AE与BC相交于点D,∠B=∠C=90°,测得BD=150
m,DC=75
m,EC=62.5
m,则河宽AB=________m.
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(第13题)
      
(第14题)
      
(第15题)
14.如图,锐角三角形ABC的边AB,AC上的高线CE,BF相交于点D,请写出图中的两对相似三角形:______________________________(用相似符号连接).
15.如图,请添加一个条件,使△ADB∽△ABC,你添加的条件是______________.
16.如图,在平行四边形ABCD中,点E在BC边上,且CE∶BC=2∶3,AC与DE相交于点F.若S△AFD=9,则S△EFC=________.
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(第16题)       
(第18题)
17.在平面直角坐标系中,点C,D的坐标分别为(2,3),(1,0),现以原点为位似中心,将线段CD放大得到线段AB.若点D的对应点B在x轴上且OB=2,则点C的对应点A的坐标为__________.
18.如图,将边长为6
cm的正方形ABCD折叠,使点D落在AB边的中点E处,折痕为FH,点C落在点Q处,EQ与BC交于点G,则△EBG的周长是________cm.
三、解答题(19题8分,22题10分,其余每题12分,共66分)
19.如图,DE∥BC,EC=AD,AE=2
cm,AB=7.5
cm,求DB的长.
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20.如图,△ABC在方格纸(小正方形的边长均为1)中.
(1)请在方格纸上建立平面直角坐标系,使点A的坐标为(3,4),点C的坐标为(7,3),并求出点B的坐标;
(2)以原点O为位似中心,相似比为2?1,在第一象限内将△ABC放大,画出放大后的位似图形△A′B′C′;
(3)计算△A′B′C′的面积S.
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21.如图,在Rt△ABC中,∠BAC=90°,AB=AC,E,D分别是BC,AC上的点,且∠AED=45°.
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(1)求证△ABE∽△ECD;
(2)若AB=4,BE=,求CD的长.
22.【教材P43习题T9变式】如图,九(1)班课外活动小组利用标杆测量学校旗杆的高度,已知标杆高度CD=3
m,标杆与旗杆的水平距离BD=15
m,人的眼睛与地面的高度EF=1.6
m,人与标杆CD的水平距离DF=2
m,求旗杆AB的高度.
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23.如图,⊙O是△ABC的外接圆,O点在BC边上,∠BAC的平分线交⊙O于点D,连接BD,CD,过点D作BC的平行线,与AB的延长线相交于点P.
(1)求证:PD是⊙O的切线;
(2)求证△PBD∽△DCA;
(3)当AB=6,AC=8时,求线段PB的长.
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24.如图①,在Rt△ABC中,∠B=90°,BC=2AB=8,点D,E分别是边BC,AC的中点,连接DE.将△EDC绕点C按顺时针方向旋转,记旋转角为α.
(1)问题发现
①当α=0°时,=________;②当α=180°时,=________.
(2)拓展研究
试判断:当0°≤α<360°时,的大小有无变化?请仅就图②的情况给出证明.
(3)问题解决
当△EDC旋转至A,D,E三点共线时,直接写出线段BD的长.
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答案
一、1.C 2.A 3.B 4.D 5.B 6.C
7.B 8.B 9.A
10.B 点拨:连接BE,CD.由圆内接四边形性质知∠ABE=∠ADC.
∵∠A=∠A,∴△ABE∽△ADC,从而有=,
∴AB·AC=AE·AD,即AB·(AB+7)=3×6,解得AB=2或AB=-9(舍去).
二、11. 12.20 13.125
14.△ABF∽△ACE,△BDE∽△CDF(答案不唯一)
15.∠ABD=∠C(答案不唯一) 16.4
17.(4,6)或(-4,-6)
18.12 点拨:由折叠的性质,得DF=EF,设EF=x
cm,则AF=(6-x)cm.
∵点E是AB的中点,
∴AE=BE=×6=3(cm).
在Rt△AEF中,由勾股定理,得AE2+AF2=EF2,即32+(6-x)2=x2,解得x=.
∴AF=6-=(cm).
∵∠FEG=∠D=90°,
∴∠AEF+∠BEG=90°.
∵∠AEF+∠AFE=90°,
∴∠AFE=∠BEG.
又∵∠A=∠B=90°,
∴△AEF∽△BGE.
∴==,
即==.
解得BG=4
cm
,EG=5
cm
.
∴△EBG的周长为3+4+5=12(cm).
三、19.解:∵DE∥BC,
∴=.
∵EC=AD,AE=2
cm,AB=7.5
cm,
∴=,
解得BD=4.5
cm(BD=12.5
cm舍去).
20.解:(1)建立平面直角坐标系如图所示.点B的坐标为(3,2).
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(2)如图所示.
(3)△A′B′C′的面积S为×4×8=16.
21.(1)证明:在Rt△ABC中,∠BAC=90°,AB=AC,∴∠B=∠C=45°.
∵∠AEC=∠B+∠BAE=∠AED+∠CED,∠AED=45°,
∴∠BAE=∠CED.
∴△ABE∽△ECD.
(2)解:在Rt△ABC中,∠BAC=90°,AB=AC=4,∴BC=4.
∵BE=,∴EC=3.
∵△ABE∽△ECD,
∴=,即=,
解得CD=.
22.解:作EH⊥AB于点H,交CD于点G.
∵CD⊥FB,AB⊥FB,
∴CD∥AB.
∴△CGE∽△AHE.
∴=,即=.
∴=,
解得AH=11.9
m.
∴AB=AH+HB=AH+EF=11.9+1.6=13.5(m).
答:旗杆AB的高度为13.5
m.
23.(1)证明:∵圆心O在BC上,
∴BC是⊙O的直径.
∴∠BAC=90°.
连接OD.
∵AD平分∠BAC,
∴∠BAC=2∠DAC.
∵∠DOC=2∠DAC,
∴∠DOC=∠BAC=90°,
即OD⊥BC.
∵PD∥BC,
∴OD⊥PD.
∵OD为⊙O的半径,
∴PD是⊙O的切线.
(2)证明:∵PD∥BC,
∴∠P=∠ABC.
∵∠ABC=∠ADC,
∴∠P=∠ADC.
∵∠PBD+∠ABD=180°,∠ACD+∠ABD=180°,
∴∠PBD=∠ACD.
∴△PBD∽△DCA.
(3)解:∵△ABC为直角三角形,
∴BC===10.
∵OD垂直平分BC,∴DB=DC.
∵BC为⊙O的直径,∴∠BDC=90°.
在Rt△DBC中,DB2+DC2=BC2,
即2DC2=BC2=100,
∴DC=DB=5.
由(2)知△PBD∽△DCA,
∴=,
则PB===.
24.解:(1)① ② (2)无变化.
证明:在题图①中,∵DE是△ABC的中位线,
∴DE∥AB.
∴=,∠EDC=∠B=90°.
在题图②中,∵△EDC在旋转过程中形状、大小不变,
∴=仍然成立.
又∵∠ACE=∠BCD=α,
∴△CEA∽△CDB.
∴=.
在Rt△ABC中,AC===4,
∴==.
∴=,即的大小不变.
(3)线段BD的长为4或.