2026年河北省中考麒麟卷数学试题及答案（五）

一、选择题（本大题共12小题，每小题3分，共36分）1-5ACCBD6-10ADCCB11-12DA【解析】由条件可知a≠1，且a为正整数故选：B【解析】设∠AME的度数为2x，则∠AEM＝90°-2x∴2×3x90°-2x180°故选：D【解析】过点B作BG⊥x轴于点G，如下图把B（-8，6）代入y＝kx－2，得6＝-8k－2解得k＝-1∴直线l的解析式为y＝-x－2∵四边形ABCD是正方形∴△BGA≌△AOD（AAS）∵B（-8，6）∴D（0，2）把y＝2代入y＝-x－2，解得x＝-4故选：A二、填空题（本大题共4小题，每小题3分，共12分）【解析】如下图，延长AE，BC交于点G∵四边形ABCD是平行四边形∵AE平分∠DAF∵E是CD中点，AD∥CG在△ADE和△GCE中∴△ADE≌△GCE（ASA）故答案为：13【解析】由题意知HK＝5，过G作GN⊥HK，垂足为N，由勾股定理可得GK＝21三、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.（1）②,③····································································································2分由数轴可知，不等式①的解集为x≤2∴m＝-1······························································································3分727277其解集在数轴上表示如下图19.（1）证明：∵△ABC和△DEF都是等边三角形，且AC＝DF∴AC＝BC＝DF＝DE＝EF，∠ACB=∠DEF＝∴BC－EC＝EF－EC即BE＝FC在△AFC和△DBE中∴△AFC≌△DBE（SAS）·································································4分（2）解：∵△AFC≌△DBE又AB＝DF∴四边形ABDF为平行四边形当AB⊥AF时，四边形ABDF为矩形20.（1）9，8·······································································································2分（3）B·················································································································5分【解析】由折线统计图可判断B款机器人的得分波动比A款机器人的得分波动小2由表知s2A＜s2C∴测试员对B款机器人运动能力测试表现评价的一致性程度更高；故答案为：B（4）解：A款机器人的综合成绩为87×40%＋85×60%＝85.8（分）B款机器人的综合成绩为85×40%＋87×60%＝86.2（分）C款机器人的综合成绩为90×40%＋83×60%＝85.8（分）21.解1）连接CD，过点D作DE⊥AB于点E，如下图∴△OCD为等边三角形∵优弧CD与直线AB相切于点C②设直线l与优弧的另一个交点为N，连接ON，过点O作OF⊥MN于点F，如下图∵优弧CD与直线AB相切于点C∵直线l∥OC∵OF⊥MN∴四边形OFEC为矩形∴∠MON＝2∠FOM＝90°∴阴影部分的面积为S阴影＝S扇形OMN－S△OMN分22.解1）an＝21＋n····························································································1分n（2）如下图4分4分解得[bn]＝5an＋50（方法不唯一，合理即可）·······················································7分（4）鞋号为42的鞋适合的脚长范围是258mm～262mm·················································8分【解析】根据[bn]＝5n＋155可知[bn]能被5整除，而270－2≤bn≤270＋2若脚长为268mm故应购买44号的鞋故答案为：4423.（1）解：嘉嘉发现的结论是正确的 理由：由折叠知BA'＝BA＝20cm由勾股定理得CA''＝DA''2-CD2＝20cm∴嘉嘉的结论是正确的································································3分（2）①证明：如下图，连接AC，AC与BD相交于点O，设CC'与BD相交于点P∵四边形ABCD是矩形∴OP∥AC'②解：在矩形ABCD中，BD＝AC在Rt△BCD中，由面积公式得BC·CDBD·CP24.（1）x＝2a······································································································2分（2）解：若a＜0∵当-2＜x＜1时，函数值y随着x的增大而减小∵当-2＜x＜1时，函数值y随着x的增大而减小2综上所述，a的取值范围为a≤-1或a≥····················································5分（3）①解：a＝1时，抛物线为y＝x2－4x－5x－2）2－9将抛物线向左平移m个单位长度后，抛物线变为yx－2＋m）2－9情况1：对称轴在区间左侧：2－m≤-3，即m≥5当-3≤x≤0时，函数值y随着x的增大而增大最小值（x＝-3y小m－5）2－9最大值（x＝0y大m－2）2－9由最值差为6，列方程（m－2）2m－5）2＝6情况2：对称轴在区间内：-3＜2－m＜0函数在顶点处取最小值y小＝-9，最大值为x＝0时的函数值：y大m－2）2－9由最值差为6，列方程[（m－2）2－9]-96解得m＝2＋6（满足3.5≤m＜5）或m＝2－6（舍去）函数在顶点处取最小值y小＝-9，最大值为x＝-3时，y大m－5）2－9由最值差为6，列方程（m－5）2－9-96解得m＝5＋6（舍去）或m＝5－6（满足2＜m＜3.5）情况3：对称轴在区间右侧：2－m≥0，即m≤2当-3≤x≤0时，函数值y随着x的增大而减小最大值（x＝-3y大m－5）2－9最小值（x＝0y小m－2）2－9由最值差为6，列方程（m－5）2m－2）2＝6【解析】当a＝1时，y＝x2－4x－5x－2）2－9如下图，原抛物线交y轴于点C（0，