九年级数学（五）参考答案_2026河北中考麒麟卷数学

2026 年初中毕业班（九年级）练习 数学（五） 参考答案 一、选择题（本大题共12小题，每小题3分，共36分） 1-5ACCBD 6-10ADCCB 11-12DA 10.B 【解析】由条件可知a≠1，且a为正整数 2a1 2（a1）3 3  ＝2＋ a1 a1 a1 ∴a＝2，4 ∴2＋4＝6 故选：B 11.D 【解析】设∠AME的度数为2x，则∠AEM＝90°－2x ∵2∠DEF＋∠AEM＝180°，∠EFB＝∠DEF＝3x ∴2×3x＋（90°－2x）＝180° ∴x＝22.5° ∴∠EFB＝67.5° 故选：D 12.A 【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 ∴∠BAD＝90°，AB＝AD ∴∠BAG＝90°－∠DAO＝∠ADO ∵∠BGA＝∠AOD＝90° ∴△BGA≌△AOD（AAS） ∵B（-8，6） ∴D（0，2） 把y＝2代入y＝-x－2，解得x＝-4 ∴m＝4 故选：A 二、填空题（本大题共4小题，每小题3分，共12分） 数学练习（五） 参考答案 第 1 页 共 7 页13.a（a＋6） 14.675 15.13 【解析】如下图，延长AE，BC交于点G ∵四边形ABCD是平行四边形 ∴AD＝BC＝BF＋CF＝7＋3＝10，AD∥BC ∴∠DAE＝∠G ∵AE平分∠DAF ∴∠DAE＝∠FAE ∴∠FAE＝∠G ∴AF＝FG＝FC＋CG ∵E是CD中点，AD∥CG ∴∠D＝∠ECG，DE＝CE，且∠DEA＝∠CEG 在△ADE和△GCE中 DECG  DE CE  DEACEG ∴△ADE≌△GCE（ASA） ∴GC＝AD＝10 ∴FG＝CF＋CG＝3＋10＝13＝AF 故答案为：13 16. 21 【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 5 （2）解：原式＝（-20）÷（- ）×3 3 3 ＝（-20）×（- ）×3 5 ＝12×3 ＝36································································································ 7分 18. 解：（1）解不等式①，得x≤m＋3 数学练习（五） 参考答案 第 2 页 共 7 页由数轴可知，不等式①的解集为x≤2 ∴m＋3＝2 ∴m＝-1······························································································3分 7 （2）解不等式②，得x＞- 2 7 综合①②得- ＜x≤2····················································································· 6分 2 其解集在数轴上表示如下图 ······················································ 8分 19. （1）证明：∵△ABC和△DEF都是等边三角形，且AC＝DF ∴AC＝BC＝DF＝DE＝EF，∠ACB＝∠DEF＝60° ∵∠ACB＋∠ACF＝∠DEF＋∠DEB＝180° ∴∠ACF＝∠DEB ∵BC＝EF ∴BC－EC＝EF－EC 即BE＝FC 在△AFC和△DBE中 AC DE  ACF DEB  FC BE ∴△AFC≌△DBE（SAS）·································································4分 （2）解：∵△AFC≌△DBE ∴AF＝DB 又AB＝DF ∴四边形ABDF为平行四边形 当AB⊥AF时，四边形ABDF为矩形 ∴AD＝BF 在Rt△ABF中，∠ABF＝60°，AB＝4 ∴BF＝8 ∴AD＝8································································································ 8分 20. （1）9，8·······································································································2分 （2）解：∵6×10×20%＋8×10×40%＋9×10×10%＋10×10×30%＝83 ∴C款机器人的运动能力测试成绩p为83分················································· 4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 ∴s2＜1.85 由表知s2 ＜s2 A C ∴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） 数学练习（五） 参考答案 第 3 页 共 7 页C款机器人的综合成绩为90×40%＋83×60%＝85.8（分） ∵86.2＞85.8 ∴综合成绩最高的是B款机器人·································································8分 21. 解：（1）连接CD，过点D作DE⊥AB于点E，如下图 ∵∠COD＝60°，OC＝OD ∴△OCD为等边三角形 ∴∠OCD＝60° ∵优弧CD与直线AB相切于点C ∴OC⊥AB ∴∠OCB＝90° ∴∠DCE＝30° 1 ∴DE＝ CD＝5 2 ∴OC＝CD＝10···················································································· 3分 （2）①6或18······································································································5分 ②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 ∵t＝3 ∴∠MOC＝45° ∵优弧CD与直线AB相切于点C ∴OC⊥AB ∵直线l∥OC ∴直线l⊥AB ∵OF⊥MN ∴四边形OFEC为矩形 ∴∠FOC＝90° ∴∠FOM＝45° ∴∠MON＝2∠FOM＝90° 90π102 1 ∴阴影部分的面积为S ＝S －S ＝  ×10×10＝25π－50········9分 阴影 扇形OMN △OMN 360 2 22. 解：（1）a＝21＋n····························································································1分 n [b]＝160＋5（n－1）＝5n＋155······························································ 2分 n 数学练习（五） 参考答案 第 4 页 共 7 页（2）如下图 ·····································································4分 一次函数······································································································5分 （3）由a＝21＋n与[b]＝5n＋155 n n 解得[b]＝5a＋50（方法不唯一，合理即可）·······················································7分 n n （4）鞋号为42的鞋适合的脚长范围是258mm～262mm·················································8分 （5）44··············································································································· 9分 【解析】根据[b]＝5n＋155可知[b]能被5整除，而270－2≤b≤270＋2 n n n 若脚长为268mm 所以[b]＝270 n 将[b]＝270代入[b]＝5a＋50，得a＝44 n n n n 故应购买44号的鞋 故答案为：44 23. （1）解：嘉嘉发现的结论是正确的······································································1分 理由：由折叠知BA'＝BA＝20cm ∴A'C＝（20 2－20）cm 在图17-2中，∠C＝90°，DA''＝DA＝20 2cm 由勾股定理得CA''＝ DA''2CD2 ＝20cm ∴BA''＝（20 2－20）cm ∴A'C＝BA'' ∴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 ∴OA＝OC 由翻折可得BD⊥CC'，CP＝C'P ∴OP∥AC' ∴∠AC'C＝∠OPC＝90° ∴∠AC'C＝90°················································································ 6分 数学练习（五） 参考答案 第 5 页 共 7 页②解：在矩形ABCD中，BD＝AC＝ （20 2）2 202 ＝20 3 （cm） 1 1 在Rt△BCD中，由面积公式得 BC·CD＝ BD·CP 2 2 BCCD 20 220 20 6 ∴CP＝   （cm） BD 20 3 3 40 6 ∴CC'＝ cm 3 40 6 20 3 在Rt△ACC'中，根据勾股定理：AC'＝ AC2 CC'2 ＝ （20 3）2 （ ）2 ＝ （cm） 3 3 ··········································································································9分 （3）如下图（答案不唯一）·················································································· 11分 24. （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 ∴x＝2a≤-2 ∴a≤-1 若a＞0 ∵当-2＜x＜1时，函数值y随着x的增大而减小 ∴x＝2a≥1 1 ∴a≥ 2 1 综上所述，a的取值范围为a≤-1或a≥ ····················································5分 2 （3）①解：a＝1时，抛物线为y＝x2－4x－5＝（x－2）2－9 将抛物线向左平移m个单位长度后，抛物线变为y＝（x－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＝-3）：y ＝（m－5）2－9 小 最大值（x＝0）：y ＝（m－2）2－9 大 由最值差为6，列方程 （m－2）2－（m－5）2＝6 解得m＝4.5 m＝4.5与m≥5矛盾，舍去 情况2：对称轴在区间内：-3＜2－m＜0 当-3＜2－m≤-1.5时，即3.5≤m＜5 函数在顶点处取最小值y ＝-9，最大值为x＝0时的函数值：y ＝（m－2）2－9 小 大 由最值差为6，列方程 [（m－2）2－9]－（-9）＝6 数学练习（五） 参考答案 第 6 页 共 7 页解得m＝2＋ 6（满足3.5≤m＜5）或m＝2－ 6 （舍去） 当-1.5＜2－m＜0时，即2＜m＜3.5 函数在顶点处取最小值y ＝-9，最大值为x＝-3时，y ＝（m－5）2－9 小 大 由最值差为6，列方程（m－5）2－9－（-9）＝6 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＝-3）：y ＝（m－5）2－9 大 最小值（x＝0）：y ＝（m－2）2－9 小 由最值差为6，列方程 （m－5）2－（m－2）2＝6 解得m＝2.5 m＝2.5与m≤2矛盾，舍去 综上m＝2＋ 6 或m＝5－ 6································································10分 ②-4.5≤n＜-2.5··························································································· 12分 【解析】当a＝1时，y＝x2－4x－5＝（x－2）2－9 如下图，原抛物线交y轴于点C（0，-5），抛物线的顶点为D（2，-9），设直线EF交y轴 于点N（0，n），根据图象折叠的对称性，则点N在CC'和DD'中垂线上 由中点坐标公式得，点C'（0，2n＋5），点D'（2，2n＋9） 若翻折后所得部分与x轴有交点，且交点都位于x轴的正半轴 则点C'在y轴的负半轴，点D'在x轴上或x轴的上方 即2n＋5＜0且2n＋9≥0 解得：-4.5≤n＜-2.5 数学练习（五） 参考答案 第 7 页 共 7 页