2026年河北省中考麒麟卷数学试题及答案（七）

一、选择题（本大题共12小题，每小题3分，共36分）1-5ACBAC6-10BDACA11-12CD【解析】设桌面的宽为x，则2c＝4x，即c＝2x由题意，得a＝4x＝2c，故①正确故选：C【解析】甲：设点P表示x则P1表示的数为x＋2，P2表示的数为x＋2－4＝x－21，P2表示的数互为相反数∴点P表示0，故甲说法正确乙：∵点P表示-11表示的数为-1＋2＝1P2表示的数为1－4＝-3P3表示的数为-3＋6＝3P4表示的数为3－8＝-5P5表示的数为-5＋10＝5P6表示的数为5－12＝-7∵点Pn到原点的距离为9∴n＝9或n＝8，故乙说法错误丙：设点P表示x1表示的数为x＋2P2表示的数为x＋2－4＝x－2P3表示的数为x－2＋6＝x＋4P4表示的数为x＋4－8＝x－4P5表示的数为x－4＋10＝x＋6P6表示的数为x＋6－12＝x－6n综上可知：甲对，乙、丙不对故选：D二、填空题（本大题共4小题，每小题3分，共12分）13.2（答案不唯一）【解析】第1个化学式中有1个C和4个H第2个化学式中有2个C和6个H第3个化学式中有3个C和8个H……以此类推，第n个化学式中有n个C和（2n＋2）个H∵烷烃CnHm化学式中的m，n满足此规律故答案为：5【解析】如下图∵点C为⊙B上一点，BC＝4在x轴上取OD＝OA＝8，连接BD，交⊙B于点E∴OM是△ACD2当OM最小时，即CD最小而D，B，C三点共线时，即当C在BD与⊙B的交点处时，OM最小，此时点E与点C重合即OM的最小值为42_2故答案为：42_2三、解答题（本大题共8小题，共72分.解答应写出文字说明、证明过程或演算步骤）在数轴上表示如下图·············································································2分2x＜3·············································································································4分此解集在数轴上表示如上图所示········································································5分（3）-1，0，1，2 7分 2分=x2+1_x2+2x+2＋1·············································································3分检验：当x＝-时，x＋1≠0······································································7分∴x＝-是原方程的解··············································································8分∵E是菱形ABCD的边CD的中点∴△ADE≌△FCE（AAS）·································································4分（2）解：∵E为菱形ABCD的边CD的中点∴由勾股定理可得EA＝23······································································6分∵△ADE≌△FCE∴AF＝2EA＝43····················································································8分20.解1）如下图，直线l即为所求（2）由（1）可知，EF为OC的垂直平分线由题意可知∠DOC＝30°∴CG＝23··································································································6分（3）由题意可知∠AOC=∠DOC＝30°∴DG＝6-23∴DE＝33_3······························································································8分21.解1）由图可知，A（优秀，分数范围x≥26）有4名，该公司共有员工4＋7＋6＋3＝20（名）∴m=×100%＝20%·········································································3分（2）该公司共有20名员工，则成绩中位数23是第10名与第11名员工成绩的平均数∵该公司成绩排名（从高到低）第10名员工的成绩为24分设排名为第11名员工的成绩为n分则＝23，解得n＝22···············································································6分答：排名为第11名员工的成绩为22分（3）∵20＜21∴员工进修情况要发生变化，即成绩平均分需不低于21分，则总分需要增加20分∵该公司部分员工有科研技术奖励分值，且每项科研技术奖励分值为10分∴员工至少有2项科研技术··············································································9分22.（1）解：设直线MN的解析式为：y＝kx＋b∵每个台阶宽、高均分别为30cm和20cm∴M（0，160N（240，0）将M（0，160）和N（240，0）代入解析式得：k+b∴y＝-x＋160··················································································3分由题意得，点B1的坐标为（210，20）∴点B1（210，20）在直线MN上 4分（2）在 5分22（3）解：把N（240，0）代入y＝mx－260m＋180（m≠0）解得m＝9把M（0，160）代入y＝mx－260m＋180（m≠0）∴≤m≤9···························································································9分23.解1）依题意，设嘉嘉第一个蛙跳的路线抛物线L1的解析式为y＝a（x－1）2＋0.4（2）①∵以第一个蛙跳起跳点为原点，落在点A处，对称轴为x＝1∴A（2，0）∵第二个蛙跳路线为抛物线L2：y＝a（x－h）2＋k（a≠0其开口大小和方向均与第一个蛙跳的路线抛物线L1相同，第二个蛙跳路线最高点为m又L2过点A（2，0）第二个蛙跳落地点与点A的距离为2.6－2）×2＝1.2（m）······························5分∴第二个蛙跳落地点距离第一个蛙跳的起跳点的距离为3.2m 6分②嘉嘉在第二个蛙跳中不会越过可调节支撑杆 7分理由如下：∴嘉嘉在第二个蛙跳中不会越过可调节支撑杆···················································9分（3）2≤h≤·····································································································11分【解析】∵点P是抛物线L1与直线y＝mx的交点解得m2或由题意可知L2的解析式为y＝-0.4（x－h）2＋0.4整理得（2－2.5m－h）2＝6.25m2-5m+1解得h＝1（舍去）或h＝3－5m∴当且抛物线L2与抛物线L1的顶点的纵坐标恰好相等时，2≤h≤24.（1）①证明：∵四边形ABCD是正方形∵△AEF是含有45°的直角三角尺∴△AEF是等腰直角三角形即∠BAE=∠DAF∴△ABE≌△ADF（SAS）∴∠ABG=∠ADF·········································································3分②BE＝DF，BE⊥DF··················································································4分证明：由①可知△ABE≌△ADF即BE⊥DF∴BE＝DF，BE⊥DF······································································7分（2）解：∵△BAD是直角三角形，O是BD的中点2∴△BGD是直角三角形 12 12（3）点G经过路线的长度为22π············································