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湖北省部分县市2 026年 4月 中考模拟
数学试题
《2026 年九年级期中考试》参考答案及评分细则
题号 1 2 3 4 5 6 7 8 9 10
答案 B A A D C C C B A D
11． 1 12． 2（答案不唯一，正数即可）
1
13． 24 14．
3
15． （1）4 ， （2） 2 5 2
（说明：解答题学生用非参考答案方法解答，只要正确，均按评分标准对应给分）
16．解： | 2 | 9 ( 3)0
=2+3-1·····································································································4分
=4···········································································································6分
17．证明：∵C为 BD的中点
∴CB CD ······························································································ 1分
∵ BCE ACD
∴ ACB ECD ····················································································· 3分
又∵CA CE
∴△ABC ≌△ABC （SAS)········································································· 5分
∴ AB ED．··························································································· 6分
18．解：在 Rt△ADC中，∵∠ADC=60°，CD=5米．
AC
∵tan∠ADC= ，∴AC=5tan60°=5 3．···················································· 3分DC
在 Rt△BDC中，∵∠BDC=45°，∴BC=CD=5，············································· 4分
∴AB=AC﹣BC=5（ 3﹣1）米．·································································· 5分
答：广告牌的高度 AB为 5（ 3﹣1）米．······················································6分
19．解：（1） a 95 ； b 90 ； m 20 ；··············· 3分（每空一分）
（2）5000 30% 1500（台），···································································4分
答：估计该月生产的 B型扫地机器人“优秀”等级的台数有 1500台；················· 5分
（3）会建议他们选购 A型号，·····································································6分
理由：在平均数均为 90的情况下，A型号的除尘量“优秀”等级所占百分比高于 B型号的．
（说其他理由，只要正确一样评分）····························································· 8分
第 1 页 共 8 页
20．解：（1）答案为： x 8， x 7， x 7， x 8；（错一个扣一分扣完 3分为止）3分
依题意，位置C的数字为 x，
∴ 结合日历的特征，位置 A的数字为 x 8，位置 B的数字为 x 7，位置D的数字为 x 7，
位置 E的数字为 x 8．
（2）结合日历的特征，规律为： (x 7)(x 7) (x 8)(x 8) 15；·························· 4分
（3）证明： (x 7)(x 7) (x 8)(x 8)
(x2 49) (x2 64)
x2 49 x2 64
15；·····································································································6分
（4）答案为：11．···················································································· 8分
∵ 最小的数和最大的数的乘积为 57，
∴ (x 8)(x 8) 57，
∵为正整数， x 8为正整数，且3 19 57，
则 x 8 3，x 8 19，
∴ x 11，
即中间C位置上的数为 11．
21．（1）证明：连接OD，
在△AOC和△AOD中，
OC OD

AD AC

AO AO
∴△AOC≌△AOD（SSS)，······································································· 2分
∴ ACO ADO，
又 ∵AD是的⊙O切线，点D是切点，
∴OD AD，即 ADO 90 ACO，
∵OC是半径，
∴AC是⊙O的切线；·················································································4分
（2）由（1）可知 ACB 90 ，
在Rt△ABC中， AB 15，BC 9
∴ AC AB2 BC2 12，则DB AB AD AB AC 3 ································ 6分
第 2 页 共 8 页
设⊙O的半径为 x，则OC OD x，OB 9 x，
在Rt△ODB中，OB2 OD2 DB2，····························································7分
∴ (9 x)2 x2 32 ，解得 x 4，
∴⊙O的半径为 4．··················································································· 8分
22．解：（1）设购买 1颗 A型芯片需要 m元，购买 1颗 B型芯片需要 n元．
2 + = 900
根据题意，得 3 + 2 = 1450，··································································· 1分
= 350
解得 = 200 ．························································································· 2分
答：购买 1颗 A型芯片需要 350元，购买 1颗 B型芯片需要 200元．··················3分
（2）设购买 A型芯片 a颗，则购买 B型芯片（10000﹣a）颗．
根据题意，得10000 a 1 a，
3
解得 a 7500，·························································································4分
设所需资金W元，则W＝350a+200（10000﹣a）＝150a+2000000，
∵150＞0，
∴W随 a的增大而增大，············································································5分
∵ a 7500，
∴当 a＝7500时W值最小，W最小＝150×7500+2000000＝3125000（元）．
答：当购买 A型芯片 7500颗时，所需资金最少，最少资金是 3125000元．···········6分
（3）① 80 ························································································7分
② 1.5或 4.5或 6.5 ·····················································（对一个得一分）10分
具体思路：
①乙车的速度为（480﹣60）÷7＝60（km/h），
当 x＝3时，y 乙＝60+60×3＝240，
则甲车的速度为 240÷3＝80（km/h）．
故答案为：80．
②y 甲＝80x，
当 80x＝480时，解得 x＝6，
∴y 甲与 x之间的函数关系式为 y 甲＝80x（0≤x≤6），
第 3 页 共 8 页
y 乙与 x之间的函数关系式为 y 乙＝60x+60（0≤x≤7），
当 0≤x≤6时，当甲、乙两车相距 30km时，得|y 乙﹣y 甲|＝30，即|60x+60﹣80x|＝30，
解得 x＝1.5或 4.5，
当 6＜x≤7时，当甲、乙两车相距 30km时，得 480﹣y 乙＝30，即 480﹣（60x+60）＝30，
解得 x＝6.5，
∴当甲、乙两车相距 30km时，x的值为 1.5或 4.5或 6.5．
故答案为：1.5或 4.5或 6.5．
23．解：（1）由旋转可得： AE AB， AF AC， BAE CAF，··················1分
AB AE
∴ ，························································································· 2分
AC AF
∴△ABE∽△ACF；··················································································· 3分
（2）如图 1，过点 G作GM AD，垂足为 M，则 GMD 90 ，
∵四边形 ABCD是矩形， AB 6， AD 8，
∴ AF AC AB2 BC 2 10， ADC DCB 90
图 1
则 GMD MDC DCG 90 ，
∴四边形 GMDC是矩形，··········································································· 4分
∴MG DC ，
由旋转得： AB AE， BC EF ， ABC AEF， ACB AFE，
∴MG EA， GMF AEF 90 ，
又∵ F F ，
∴△GMF ≌△AEF （AAS)········································································ 5分
∴GF AF AC，MF EF BC AD，
∴GF EF AF MF AC AD 10 8 2，
即GE 2；······························································································6分
（3）①如图 2，延长 EH交 BC于点 N，连接 AN，DF，设 EF交 AD于点 S，
第 4 页 共 8 页
图 2
由旋转得 AB AE， ABC AEF 90 ， ACB AFE，
∴ ABE AEB， ABC AEN 90
∴ NBE ABC ABE AEN AEB BEN ，
∴ NB NE， HNC 2 EBN ，································································7分
∵ AHF CHN， ACB AFE，
∴180 AHF AFE 180 CHN ACB，即 FAH HNC，
∴ FAH 2 EBN ，················································································ 8分
∵四边形 ABCD是矩形，
∴ EBC ACB DAC，
∴ FAH 2 EBN 2 DAC DAC DAF ，
∴ DAC DAF，
∴ AD平分 FAC ；·················································································· 9分
AH 25
②答案： ．················································································11分
AC 39
∵ AB AE， NB NE，
∴AN垂直平分 BE，
∴ BAN ABE ABE EBN 90 ，
∴ BAN EBN ACB，
又∵ ABN CBA 90
∴△ABN∽△CBA，
BN BA 6 BN 6
∴ ，即 ，
BA BC 8 6 8
∴ BN 9 ， NC BC 7 BN ,
2 2
∵ DAC DAF， AD AD， AC AF ，
第 5 页 共 8 页
∴△ACD ≌△AFD （SAS)，
∴CD FD， ADC ADF 90 ，
∴D为 FC的中点，而 SD // NC，
∴ SD 1 NC 7 7 25 ， AS AD SD 8 ，
2 4 4 4
∵ AS // NC，
∴△ASH∽△CNH，
AH AS 25 / 4 25 25
∴ ， AH HC，
HC CN 7 / 2 14 14
AH AH 25
∴ ．
AC AH HC 39
24．解：（1）∵OB OC 3，
∴ B 3,0 ,C 0,3 ，····················································································1分
把 B 3,0 ,C 0,3 代入 y x 2 bx c ，得：
c 3 b 2

9 3b c 0
，解得：
c 3
∴抛物线的解析式为 y x2 2x 3，··························································2分
设直线 BC的解析式为 y k1x b1，
将 B 3,0 ,C 0,3 代入 y k1x b1，得：
3k1 b1 0 k1 1

b1 3
，解得：
b1 3
∴直线 BC的解析式为 y x 3；······························································· 4分
（2）①依题 P(m, m2 2m 3)（ 0 m 3），则G(m, m 3)，
∴ d m2 2m 3 ( m 3) m2 3m，··················································· 6分
②由题设易知 PGC BGH HBG 45 ，
∴若△PCG为直角三角形，则：
当 CPG 90 ，则 PC PG，即m m2 3m，
解得：m 2，m 0（舍去），····································································7分
第 6 页 共 8 页
当 PCG 90 ，过 C作CF PG ,则 PG 2CF ，即 m2 3m 2m，
解得：m 1，m 0（舍去），
综上，m 1，或m 2；············································································ 8分
③m，n的数量关系为：mn 3 m n 11，·················································· 10分
定点坐标为 3, 2 ．················································································· 12分
详解如下：
依题： P m, m 2 2m 3 ,Q n, n 2 2n 3 （ n 3），
由 y x2 2x 3 0，解得： xA 1，
∴ A 1,0 ，
设直线 AP的解析式为： y ax p，
a p 0 a 3 m
则：
ma p m
2 2m 3，解得： p 3 m，
∴ y 3 m x 3 m，
当 x 0时， y 3 m，
∴D 0,3 m ，
同法可得： E(0,3- n)，
∴OD OE 3 m n 3 2，
∴3 m n mn 9 2 ，
∴mn 3 m n 11，
设直线 PQ的解析式为： y tx q，
mt q m2 2m 3 t m n 2
则： ，解得： ，
nt

q n2 2n 3 q mn 3
∴ y m n 2 x mn 3
m n 2 x 3 m n 8
x m n 2x 3 m n 8，
第 7 页 共 8 页
3 x m n 2x 8，
∴当 x 3时， y 2 3 8 2，
∴定点坐标为 3, 2 ．
第 8 页 共 8 页

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