2021年山东省临沂市中考数学真题及答案

2021年山东省临沂市中考数学真题及答案
注意事项：
1．本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题），共8页，满分120分，考试时间120分钟．答卷前，考生务必用0.5毫米黑色签字笔将自己的姓名、准考证号、座号填写在试卷和答题卡规定的位置．考试结束后，将本试卷和答题卡一并交回．
2．答题注意事项见答题卡，答在本试卷上不得分．
第Ⅰ卷（选择题
共42分）
一、选择题（本大题共14小题，每小题3分，共42分）在每小题所给出的四个选项中，只
有一项是符合题目要求的．
1．1
2
-的相反数是
（A）1
2
-.
（B）2-.
（C）2.
（D）
12
.2．2021年5月15日，天问一号探测器成功着陆火星，中国成为全世界第二个实现火星着
陆的国家.据测算，地球到火星的最近距离约为55000000km.将数据55000000用科学记数法表示应为（A）5.5×106
.（B）0.55×108
.（C）5.5×107.（D）55×106
.
3．计算3325a a ⋅的结果是
（A）610a .（B）910a .（C）37a .
（D）67a .
4．如图所示的几何体的主视图是
5．如图，AB ∥CD ，∠AEC =40°，CB 平分∠DCE ，则∠ABC 的度数为
（A）10°.
（B）20°.（C）30°.（D）40°.
6．方程256x x -=的根是
（A）17x =，28x =.（B）17x =，28x =-.（C）17x =-，28x =.（D）17x =-，28x =-.
7．不等式
1
13
x x -<+的解集在数轴上表示正确的是
－2
－2
8．计算11
(()a b b a
-÷-的结果是
（A）a
b -.
（B）a b .（C）b
a
-.
（D）
b a
.9．如图，点A ，B 都在格点上，若213
3
BC =
，则AC 的长为..（C）.（D）10．现有4盒同一品牌的牛奶，其中2盒已过期，随机抽取2盒，至少有一盒过期的概率是
（第5题图）
A
B
C
D
E
（A ）（B ）（C ）（D ）
00
（第9题图）B
A
C
（A）
12
.（B）
23
.（C）
34
.（D）
56
.11．如图，PA ，PB 分别与⊙O 相切于点A ，B ，∠P =70°，C 为⊙O 上一点，则∠ACB 的度数是
（A）110°.（B）120°.（C）125°.（D）130°.
12．某工厂生产A，B 两种型号的扫地机器人.B 型机器人比A 型机器人每小时的清扫面积多50%；清扫1002m 所用的时间，A 型机器人比B 型机器人多用40分钟.两种型号扫地机器人
每小时分别清扫多少面积？若设A 型扫地机器人每小时清扫2
m x ，根据题意可列方程为
（A）1001002
.0.53x x =+（B）1002100
.0.53x x +=（C）
1002100
.3 1.5x x
+=（D）
1001002
.1.53
x x =+13．已知a b >，下列结论：①2a ab >；②22a b >；③若0b <，则2a b b +<；④若0b >，
则
11
a b
<.其中正确的个数是（A）1.
（B）2.
（C）3.
（D）4.
14．实验证实，放射性物质放出射线后，质量将减少，减少的速度开始较快，后来较慢，物
质所剩的质量与时间成某种函数关系.下图为表示镭的放射规律的函数图象.
据此可计算32mg 镭缩减为1mg 所用的时间大约是（A）4860年.
（B）6480年.
（C）8100年.
（D）9720年.
（第14题图）
18m 时间/年
质量16204860
3240
012m O
014m 0
m
第Ⅱ卷（非选择题
共78分）
注意事项：
1．第Ⅱ卷分填空题和解答题．
2．第Ⅱ卷所有题目的答案，考生须用0.5毫米黑色签字笔答在答题卡规定的区域内，在试卷上答题不得分．
二、填空题（本大题共5小题，每小题3分，共15分）15．分解因式：328a a -
=.
16．比较大小：26
5（填“﹥”“﹤”或“=”）.
17．某学校八年级（2）班有20名学生参加学校举行的“学
党史、看红书”知识竞赛，成绩统计如图.这个班参赛学生的平均成绩是
.
18．在平面直角坐标系中，□ABCD 的对称中心是坐标原点，顶点A ，B 的坐标分别是(1-，
1)，(2，1).将□ABCD 沿x 轴向右平移3个单位长度，则顶点C 的对应点C 1的坐标是
.
19．数学知识在生产和生活中被广泛应用.下列实例所应用的最主要的几何知识，说法正确
的是
（只填写序号）.
①射击时，瞄准具的缺口、准星和射击目标在同一直线上，应用了“两点确定一条直线”；②车轮做成圆形，应用了“圆是中心对称图形”；
③学校门口的伸缩门由菱形而不是其它四边形组成，应用了“菱形对角线互相垂直且平分”；
④地板砖可以做成矩形，应用了“矩形对边相等”.
（第19题图）
准星
缺口
三、解答题（本大题共7小题，共63分）20.（本小题满分7分）
计算：2
2
11|2|2222⎫⎫+--+⎪⎪⎭⎭.
10290成绩
95100人数
（第17题图）
35850
21．（本小题满分7分）
实施乡村振兴计划以来，我市农村经济发展进入了快车道.为了解梁家岭村今年一季度经济发展状况，小玉同学的课题研究小组从该村300户家庭中随机抽取了20户，收集到他们一季度家庭人均收入的数据如下（单位：万元）：
0.690.730.740.800.810.980.930.810.890.690.74
0.99
0.98
0.78
0.80
0.89
0.83
0.89
0.94
0.89
研究小组的同学对以上数据进行了整理分析，得到下表：
（1）表格中：a =，b =，c =，d =；
（2）试估计今年一季度梁家岭村家庭人均收入不低于0.8万元的户数；
（3）该村梁飞家今年一季度人均收入为0.83万元，能否超过村里一半以上的家庭？请说明理由.
22．（本小题满分7分）
如图，在某小区内拐角处的一段道路上，有一儿童在C 处玩耍，一辆汽车从被楼房遮挡的拐角另一侧的A 处驶来.已知CM =3m，CO =5m，DO =3m，∠AOD =70°，汽车从A 处前行多少米，才能发现C 处的儿童（结果保留整数）？
（参考数据：sin37°≈0.60，cos37°≈0.80，tan37°≈0.75；sin70°≈0.94，cos70°≈0.34，tan70°≈2.75.）
分组频数0.65≤x <0.7020.70≤x <0.7530.75≤x <0.801
0.80≤x <0.85a 0.85≤x <0.9040.90≤x <0.9520.95≤x <1.00
b
统计量平均数中位数众数数值
0.84c
d
A
B
C
D O
M （第22题图）
2号楼
23．（本小题满分9分）
已知函数3
, 1,3, 11,
3
, 1.x x y x x x x
⎧±-⎪⎪
=-±±⎨⎪⎪±⎩（1）画出函数图象；
列表：
x ……y
…
…
描点，连线.得到函数图象.
（2）该函数是否有最大或最小值？若有，求出其值，若没有，简述理由；（3）设（x 1，y 1），（x 2，y 2）是函数图象上的点，若x 1+x 2=0，证明：y 1+y 2=0.24．（本小题满分9分）
如图，已知在⊙O 中， AB = BC =
CD ，OC 与AD 相交于点E .求证：（1）AD //BC ；
（2）四边形BCDE 为菱形.
≤
≥（第24题图）
C
D
O
E
A
B
＜＜y O x
4
2
62
4-2
-4
-6
-2-4
25．（本小题满分11分）
公路上正在行驶的甲车，发现前方20m 处沿同一方向行驶的乙车后，开始减速.减速后甲车行驶的路程s （单位：m）、速度v （单位：m/s）与时间t （单位：s）的关系分别可以用二次函数和一次函数表示，其图象如图所示.
（1）当甲车减速至9m/s 时，它行驶的路程是多少？
（2）若乙车以10m/s 的速度匀速行驶，两车何时相距最近，最近距离是多少？
26．（本小题满分13分）
如图，已知正方形ABCD ，点E 是BC 边上一点，将△ABE 沿直线AE 折叠，点B 落在点F 处，连接BF 并延长，与∠DAF 的平分线相交于点H ，与AE ，CD 分别相交于点G ，M ，连接HC .
（1）求证：AG =GH ；
（2）若AB =3，BE =1，求点D 到直线BH 的距离；
（3）当点E 在BC 边上（端点除外）运动时，∠BHC 的大小是否变化？为什么？
试卷类型：A
参考答案及评分标准
说明：解答题给出了部分解答方法，考生若有其它解法，应参照本评分标准给分.
15.5
3043.5568t (s)
1456
16
8t (s)
23（第25题图）
s (m)v (m/s)
ＯＯ
（第26题图）
D
B
M
A
H
G E
F
C
一、选择题（每小题3分，共42分）题号1234567891011121314答案
D
C
A
B
B
C
B
A
B
D
C
D
A
C
二、填空题（每小题3分，共15分）15．2(2)(2)a a a +-；16．＜；17．95.5；18．(4,1)-；19．①③．
三、解答题
20．解：2
2
11||22⎫⎫+--⎪⎪
⎭⎭1111
)
2222
=+-+-················································3分
(1)=+-···············································································
5分=·····························································································
7分
21．解：（1）a =5，b =3，c =0.82，d =0.89.··················································4分
（2）14
30021020
⨯
=（户），因此，家庭人均收入不低于0.8万元的大约有210户.··························6分（3）因为样本的中位数是0.82，0.83＞0.82，
所以可以估计梁飞家的人均收入超过村里一半以上的家庭.····················7分
22．解：由题意，3
sin 0.65
COM ∠=
=，因此∠COM ≈37°.∵∠DOB 与∠COM 为对顶角，
∴∠DOB ≈37°.·······················································································2分在Rt△DOB 中，∵tan∠DOB =
BD
DO
，∴BD=DO ·tan∠DOB ≈3×0.75=2.25.·····························································4分在Rt△DOA 中，∵tan∠DOA =
AD DO
∴AD=DO ·tan∠DOA ≈3×2.75=8.25.·····························································6分∴AB=AD －BD =8.25－2.25=6（m）.
因此，汽车前行约6米，才能发现儿童.························································7分23．解：（1）列表：
x …4-3-2-1-1234…y
…
0.75
-1
- 1.5
-3
-3
1.5
1
0.75
…
·························2分
描点，画出如图所示图象.·····································································5分
y 11.5
O x
-1
-4-3-22
3
4
-3
3
-1.5（2）由图象可知，当1x =-时，函数取得最小值，=3y -最小；当1x =时，函数取
得最大值，=3y 最大.·························································································7分
（3）方法一：∵x 1+x 2=0，∴x 2=－x 1.当x 1≤－1时，x 2≥1，y 1=13x ，y 2=2
3x ，12121212
3()
330x x y y x x x x ++=
+==.当x 1≥1时，x 2≤－1，同理可知y 1+y 2=0····················································8分当－1＜x 1＜1时，－1＜x 2＜1，y 1=3x 1，y 2=3x 2，
121212333()0y y x x x x +=+=+=.
综上所述，若x 1+x 2=0，则y 1+y 2=0.···························································9分方法二：
因为该函数的图象关于原点中心对称，由x 1+x 2=0，得12x x =-，可知点（x 1，y 1）与
点（x 2，y 2）关于原点对称，所以y 1=-y 2，即y 1+y 2=0.············································9分24．证明：（1）如图24－1，连接BD .
∵ AB CD =，
∴ADB DBC ∠=∠···················2分∴AD ∥BC.······························3分（2）如图24－2，连接OB ，OD ，∵ BC
CD =，∴BC CD =.·······················································································4分
图24－1
B
D
O
E
A
C
方法一：
∵OB =OD ，BC =CD ，∴OC 垂直平分BD .
∴EB =ED .································5分∴BEC DEC ∠=∠.又∵AD ∥BC ,∴BCE DEC ∠=∠.∴BEC BCE ∠=∠.
∴BC =EB .·····························································································7分∴BC =CD =ED =EB .····················································································8分∴四边形BCDE 是菱形.···········································································9分方法二：
∵OB =OD ，BC =DC ，OC =OC ，∴△OBC ≌△ODC .
∴∠BCO ＝∠DCO .··················································································5分又∵AD ∥BC ，∴∠DEC ＝∠BCE .∴∠DCO ＝∠DEC .
∴DE =DC .·····························································································7分∴DE =BC .
∴四边形BCDE 是平行四边形.·································································8分∴四边形BCDE 是菱形.···········································································9分
C
D
O
E
A
B 图24－2
25．解：（1）由题意，设2s at bt =+，v =mt +16.
∵抛物线过点（2，30），（4，56），
∴4230,
16456.
a b a b +=⎧⎨
+=⎩··················································································1分解得1,216.
a b ⎧
=-⎪⎨⎪=⎩·····················································································2分
∴s 与t 的函数关系式为21
162
s t t =-+.·····················································3分
∵直线过点（8，8），∴8m +16=8.∴m =－1.
∴16v t =-+.·······················································································4分当9v =时，169t -+=，∴7t =，··························································5分
当7t =时，21
716787.52
s =-⨯+⨯=.
因此，当甲车减速至9m/s 时，行驶的路程是87.5m.·································7分（2）方法一：设两车距离为w ，则
21
1020(16)2
w t t =+--+.
整理得：
()2
1622
y t =
-+.·················································································8分∵
1
2
＞0，∴w 有最小值.当6t =时，·························································································9分w 最小= 2.
所以当甲车减速行驶6秒时，两车距离最近，最近距离为2米.····················11分方法二：
由题意知，当甲车速度减至10m/s 时，两车距离最近.
当10
v=时，1610
t-+=，∴6
t=.·························································8分此时，甲车行驶的路程为：
2
1616678(m)
2
s=-⨯+⨯=，····································································9分乙车行驶的路程为10660(m)
⨯=，
∴两车相距6020782(m)
+-=.
因此，当甲车减速行驶6秒时，两车相距最近，最近距离是2米.·················11分26．证明：（1）如图26－1.
∵点B，F关于AE对称，
∴AE⊥BF，BAG FAG
∠=∠.···································································1分∵AH平分∠DAF，
∴DAH FAH
∠=∠.
∴
1145
22
GAH GAF FAH BAF DAF
∠=∠+∠=∠+∠=︒.································2分
∴AG HG
=.·······················································································3分
（2）连接DH .
由题意可知AD =AB =AF ，
在△ADH 和△AFH 中，
,,,AD AF DAH FAH AH AH =⎧⎪
∠=∠⎨⎪=⎩
∴△ADH ≌△AFH .
∴DH=FH ,45AHD AHG ∠=∠=︒.∴90DHB ∠=︒.
∴DH 的长即为点D 到直线BH 的距离.······················································4分∵AB =3，BE =1，
∴AE =
==.·······················································5分
∵BG ⊥AE ，
∴AB BE BG AE ⋅=
==
·······························································6分
∴AG ===.···········································7分方法一：
连接BD
，BD =∵GH =AG ，
∴BH =BG+GH =BG+AG .∴
+=
.····································································8分
∴DH ==.∴点D 到直线BH
····························································9分方法二：
∵GF =BG ，AG =GH ，
图26－1
D
A
B
G E
F
H
M
C
∴DH =FH=GH -GF=AG -BG .···································································8分∴
=
∴点D 到直线BH
····························································9分（3）不变.
如图26－2，过点C 作CN ⊥BH ，垂足为N .∵90BAG ABG ∠+∠=︒，90CBN ABG ∠+∠=︒，
∴=BAG CBN ∠∠.
在△ABG 和△BCN 中，BAG CBN AGB BNC AB BC ∠=∠⎧⎪
∠=∠⎨⎪=⎩
，，
，∴△ABG ≌△BCN .
∴AG =BN ，BG =CN .·················································································10分又∵AG =GH ，
∴BN =GH .···························································································11分∴BG +GN =GN +NH .∴BG =NH .
∴CN =NH .···························································································12分∴45BHC ∠=︒.
所以，当点E 在BC 边上运动时，∠BHC 的大小不变.··································13分
图26－2
A
B
G H
E
F
M C
D
N。