2026年河北省中考麒麟卷数学试题及答案（二）

一、选择题（本大题共12小题，每小题3分，共36分）1-5BDACC6-10DACBC11-12AD【解析】数轴上AC＝8-1018，直尺测量AC＝5.4cm，5.4÷18＝0.3（cm）数轴上一个单位长度的长是0.3cm，直尺测量AB＝1.8cm，1.8÷0.3＝6数轴上AB＝6，-10＋6＝-4∴点B对应的数是-4.故选：A.【解析】作AG⊥BC，EH⊥BC，垂足分别为G，H∴AG∥EH∵在平行四边形纸片ABCD中，AB＝CD＝10cm，AD∥BC∴AE∥GH，AG＝EH，∴四边形AGHE是矩形∴AG＝EH＝8cm在Rt△ABG中，BG∵AD∥BC设AE＝GH＝xcm，则BE＝DE16－x）cm，BH＝BG＋GH6＋x）cm在Rt△BEH中，BH2＋EH2＝BE2，即（6＋x）2＋8216－x）2二、填空题（本大题共4小题，每小题3分，共12分）【解析】如右图，连接O2C∵正六边形的边长为11A1B2C故答案为：【解析】过点A作AD⊥x轴于点D，过点B作BE⊥x轴于点E，则∠ADO=∠BEC＝90。∵直线y＝x与反比例函数y＝9（x＞0）交于点Ax由平移知，BC∥OA∴BE＝9，∴点B的纵坐标为9x∵将直线y＝x沿y轴向上平移b个单位长度得到直线y＝x＋b∴把B的坐标代入得1＋b＝9，解得b＝8故答案为：8三、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）17.解1）嘉嘉的错误在第一步，琪琪的错误在第三步···············································4分（2）原式＝16×5-8）÷4＝80＋2＝82································································7分 即被遮挡部分的代数式为a＋4································································3分（2）①∵a－4≠0∴小颖认为“原算式的值不可能为9”···················································②小颖的说法不全面·······················································································6分理由如下：∵a－4≠0且a－5≠0∴a≠4且a≠5∴原式的值不可能为9，10∴小颖的说法不全面 19.（1）84，100···································································································4分理由：从平均数的角度看，七年级（2）班的平均数87.2高于七年级（1）班的平均数87，所以从平均数的角度七年级（2）班学生计算能力较好······································6分从中位数的角度看，七年级（2）班的中位数86高于七年级（1）班的中位数84，所以从中位数的角度七年级（2）班学生计算能力较好······································6分从众数的角度看，七年级（2）班的众数100高于七年级（1）班的众数98，所以从众数的角度七年级（2）班学生计算能力较好·········································6分从方差的角度看，七年级（2）班的平均数高于七年级（1）班的平均数，且同时方差小于七年级（1）班，所以从方差的角度七年级（2）班学生计算能力较好·····················6分答：估计参加此次数学计算比赛活动成绩达到90分及以上的学生约有1200名···········8分∴△ABC≌△ECD（SAS）···········································································4分（2）解：∵△ABC≌△ECD∴BD＝DE2+BE2＝82+42＝45·························································8分（2）解：设线段AB对应的函数关系式为F＝kh＋c（k，c为常数，且k≠0）将A（4，12）和B（10，6）分别代入得解得∴线段AB对应的函数关系式为F＝-h＋16（4≤h≤10）此时弹簧测力计显示的读数是10.8N·····························································9分22.解1）如右图，过点O作OD⊥AC于点D，连接AO∴AD＝AC＝8cm又AO＝10cm即圆心O到线段AC的距离为6cm···························································3分（2）如上图，过点O作OE⊥AB于点E，连接OA，OB依题意，OE＝8cm∴在Rt△AOE中，AE＝AO2-OE2＝6cm的长为cm··········································································7分（3）如上图，过点O作OF∥AM交CN于点F又CN是⊙O的切线∴CN⊥OC23.（1）90°·······································································································2分解2）∵点E和点F分别是AC边，BC边的中点∴EFABcm）··································································4分（3）如下图，点P1，P2即为所求（作法提示：以EF为直径作圆，与AB交于点P）当点P位于点P1时，连接P1E，P1F，四边形CEP1F是矩形当点P位于点P2时，则EF垂直平分CP2综上，AP＝2.5cm或16cm················································································9分5【解析】由对称可知FC＝FN＝1.5cm，FA＝FM，MN＝AC＝4cm∴点N在以F为圆心，CF长为半径的圆上运动点M在以F为圆心，AF长为半径的圆上运动如下图，当点E，F，N三点依次共线时，△EMN的面积最大此时，EN＝EF＋FN＝4cm（提示：因为AC＝MN，当EN取得最大值且MN⊥EN时面积最大）24.解1）将点A（-1，0点B（2，3）代入y＝ax2＋bx＋3，得解得∴二次函数的解析式为y＝-x2＋2x＋3····························································4分∵抛物线开口向下，对称轴为直线x＝1∵3－1＜1-5）∴当x＝-5时，y取最小值，最小值＝-（-5）2＋2×（-53＝-32∴当-5≤x≤3时，二次函数y＝-x2＋2x＋3的最大值为4，最小值为-32····················8分3当-2m＋3＞0时，即3当-2m＋3＜0时，即mMN＝2m－3，MN的长度随m增大而增大，不符合题意②线段MN与二次函数y＝ax2＋bx＋3（-1≤x的图象只有1个交点时，1≤m＜或-1≤m≤；线段MN与二次函数y＝ax2＋bx＋3（-1≤x的图象有2个交点时m＜1·················································································································12分【解析】∵0＜MN≤5解得：-1≤m<如图①,当m＝1时，点M在最高点，MN与图象有1个交点如图②,m增大过程中，当1＜m＜