2021年中考数学全真模拟预测试卷

友提有答案必须填答题卡应位上在本卷一无；2．物

ybx

的点标（

a

，

4ac4

2

一选题本题10小题每题4共分每题只有个确选，在题的应置涂1的倒是A．-2B．2C．2

D

122．图a则列选项，能直利“直平，错相”定1=是1aa1

1

a

a2

b

2

b

2

b

1

2

bA．B．D3．列算确的是

．A．

a

B．

aa

C．

a

D

a4．下调中，适采用普的是A．解某校班生力况1/19

B查2021

2拟央春的视2C．测批电灯的使用D了我市中生课余网时间5．图下几何体，左视不是矩的是A．B．D6．简x结是xx

．A．

x

B．

C．

x

Dx7．商利摸奖开促销活，中奖1是A．摸奖三，则至中奖一B．连摸两，不都奖C．只奖一次则也有能中奖

，下说正的ADOD若续奖次不奖则三次一

第题2/19

8．图四形的角相于且则列条件判四形矩的A

一

二

三

四CD．AB∥CD9．图在正形格，有个

1234小方被黑若图其小方任涂一，整图构一轴称形则小方的置以ABC）D）

第题某市需要设一条米管，了量减少工对城交造的响施时铺管道的度比原划增加％，结果提6完成．实际每铺设管的长度实际工数小同根题列方660x方中知示量A．际每天设管道长度．际施工天数3/19

．

C．计每天铺管道的度

D工天二填题本题题每题，分．将案填答卡相位）．计算：

1)2

＝________．分解因：

2

＝________十五累新城就人口147000科学记法表示_如图，甲，乙个可以由动转盘转止指都落在影域的率_

90°甲

120°乙第题图15如，离面5处拉线固电线杆拉线和面成50°，拉线的为米（确到米PD米A

50°

D

B

5

4/19A

B第题图

第16题图

＜x＜x71．如图，知形，是CD直径半上一动，接则BP的最值是_三解题本题9小，分请答卡相应位作）17分）简：

(a22)

.18分

分求等3

2

的数解.A

1

D19形ABCD

M

2

3边AB一BC延线点N.求：

BN208分九级有个班数例如图所．一数考中四1008060405/1920

67

740

1

2

3

4班级图

1

2a%b%4c%

3c%图四班均绩中数_；下说：3班以上人最；1，3班平分距小③本考年段成最的生4班其正的_（序若公

m2

（m

分表各平成）别算1班和班的平成绩，两班的算果会实平成相，说理.6/19

21）如图已中以B圆，BC为径弧别点，E连BD，ED．写图所的腰角；

A若AED=114°求∠度．

EDB

22分）图1在形中动P从A出，A路径运．设P动路程为，的面积为．映是点P运动过中y与x的数关系请根据象回答下问题矩边AD=________，AB写点P在运动程yx的函关式并y图2补函数图．PD

54321

0345x图7/19A

B图1

⌒⌒23分分如，已知以AB为径交AC点，

．求线

A若EA点

sinBED

35

，BE长O

ED

B24分线

1

与x交点（m交点，物

y

经AP线一点，点PPQ交物于．当时①求物的系；8/19

②设横标，含代式表示PQ的长并求x

为值＝8；5若PQ长的最值16，讨关于x一二方

2

ax

的的数的取范的系QO9/19

2514）们有组边等一对边平但不相的边称“菱证“菱”质角平一内角（求根图1出知求，明已：求：证：

B

AD图1

已，，∠A=90°若D分在BC上且边形为准形下给的满足条的所，并出应DE长ABC一都，够添CA________DE=10/19

DE=________

11数试参答及分准⑴解给了种几解供考如考的法与本答不，参本案评标的神行分⑵解题当生解在一出错时如后续部的解未变题立，酌给．⑶答端注数示生确完步得累分数．⑷分给数，择和空均给间．一选题有小题，每小，分分）1．D2345678910二填题有题每题分，满分）11(

．1.47

5

2

151613三解题本题题共86．在答题的相应置作）17分）解原=

aa

，················································=

a

．

························································7分18分）11/19

①＜②≤2.解解等①得

x

···················································

2解等②得

x

．················································4在一轴表不式②解，图--4-3--2∴不式的集

x

．····················································6∴不式的数-4，-3，-2，0．················19分）

A

1

D证：四形正形，∠A=∠ADC=∠BCD=90°．

MB

2

3N································································∴．∴∠A··························································∵，12/19

∴∠36∴eq\o\ac(△,≌)△DCN．

······················································································································20分））69；·············································································2分②··············································································用式

m2

计两的均绩，结会与实平成相，为3，4班权重人数或例）相．·······························································································8分21分10分）答等三形：△BCDeq\o\ac(△,,)；···············3解∵∠AED=114°∴∠°－∠

AAED=66°分

ED∵BD=BEB

∴∠BED=66°．∴∠°66°×2=48°·······················································分13/19

解一设，∴∠ABC=．∴－2x°∵BC=BD∴又△的角∴∠A+．··········································－2x+48，解得．∴分解二设，∴．∴－48°．∵BC=BD∴分又∠DBC+∠BCD+∠BDC=180°，∴x解得：x=76．∴分22分10分）(1)2，4··························································4y14/19

54321

(2)点P过中－x，∴

1)2

，即

y

分正作图············(提：生对数关系化简，写出取范围不)23分10分）解是径，

A∴∠ADB=90°．························1

O∴．

D又∠A=∠CBD∴∠CBD+．．∴AB⊥BC．································4又∵ABO的直，⊙线···············5连AE⊙径15/19

B

⌒△AEB∴∠AEB=∠ADB=90°．⌒△AEB∵∠BAD=∠BED∴

BED

35

．·····················································∴eq\o\ac(△,Rt)ABD中，

sin

BD35

．∵BD，．············································································∵E为AB点∴AE=BE．∴是腰直角角形．∴．∴

sin

22

分24分12分）解∴标，0

QOAx将x=0入

kx1

，y=2∴坐标为（，2将）入

yax

，

a20a2.

······························································16/19

xx2)

2a5c∴物的达为

285

．··························4②A(5入

，解：

25

．∴次数表为

·····································5分∴

的标

2(xx2)5

．又∵PQ∥y轴，∴标为

(,

5

．∴225

，25

2

．··········································································

7∵

，∴85

．得

，x

．∴x=1x=4时

85

．·········································9分设

y

ax2)ax

kx

．∴Sx次函数∵PQ的大为16∴S大为16，17/19

∴二函的象质知当，一元次方程

ax

有个；当，一元次方程当，一元次方程

axkxaxkx

无；有个12分(提：生对种情况得分未明由扣)25分14分）解：图中AB=AD∥(

BC

)．·······································