九年级数学（六）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（六） 参考答案 一、选择题（本大题共12小题，每小题3分，共36分） 1-5DBCAC 6-10BACBC 11-12DC 11.D 【解析】∵等腰直角三角形ABC的边BC∥x轴，BC＝2，B（2，2） ∴C（4，2），A（2，4） k 双曲线y＝ （k＞0，x＞0）经过点B时，k最小 x ∴k最小值为4 k ∵等腰直角三角形ABC，y＝ （k＞0，x＞0）都是关于直线y＝x对称 x k ∴当y＝ （k＞0，x＞0）经过AC中点E时，k最大 x ∵C（4，2），A（2，4） ∴E（3，3） 此时k＝9 k ∴双曲线y＝ （k＞0，x＞0）与△ABC有交点时，4≤k≤9 x 故选：D 12.C 【解析】延长AB，DC交于点M ∵正六边形ABCDEF ∴∠CBM＝∠BCM＝60° ∴△BCM是等边三角形 ∴BM＝CM＝BC＝a ∴DM＝2a ∵点G是AB的中点 3 ∴GM＝ a 2 过点G作GH⊥DM，垂足为H 3 3 3 3 在Rt△GMH中，GH＝GM·sin60°＝ a× ＝ a 2 2 4 3 1 3 MH＝GM·cos60°＝ a× ＝ a 2 2 4 3 5 ∴DH＝DM－MH＝2a－ a＝ a 4 4 3 3 a ∴tan∠GDC＝ GH  4  3 3 DH 5 5 a 4 数学练习（六） 参考答案 第 1 页 共 7 页∴结论①正确 过点C作CN⊥AM于点N ∵△BCM为等边三角形，CM＝a 3a ∴CN＝CM·sin60°＝ 2 1 3a2 ∴△GBC的面积为 ·GB·CN＝ 2 8 ∴结论②正确 故选：C 二、填空题（本大题共4小题，每小题3分，共12分） 13.5 2 1 14. 2 15.8 16.2 【解析】∵△BEF是直角三角形 ∴△BEF的外心为BF中点，设为点M ∵点F在直线CD上，点M为BF中点 ∴点M始终在BC的中垂线上移动 设BC的中垂线为ON，ON交BC于点K 当点E与点A重合时 点M与点O重合 当点E与点O重合时 点M与点K重合 ∵OK为BC的中垂线 1 ∴OK＝ AB＝2 2 ∴△BEF外心移动的距离为2 故答案为：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） 17. 解：（1）32＋（-1）×4＋3 27＋|-2| ＝9－4－3＋2 ＝4·······································································································3分 （2）7－3x＞2x－8 ∴-3x－2x＞-8－7 ∴-5x＞-15 ∴x＜3 ∴不等式的正整数解为：1，2···········································································7分 18. 解：（1）由题意，当x＝2时 -x2＋2x＋3＝3 -2x2－2x＋4＝-8 ∴M＝-8－3＝-11················································································3分 数学练习（六） 参考答案 第 2 页 共 7 页（2）嘉嘉的说法正确·····························································································4分 理由如下： 由题意得 （-x2＋2x＋3）＋M＝-2x2－2x＋4 ∴M＝-2x2－2x＋4－（-x2＋2x＋3）＝-x2－4x＋1················································· 6分 ＝-（x＋2）2＋5≤5 无论x取何值，M都不可能大于5 ∴嘉嘉的说法正确··························································································8分 19. （1）证明：由题意可知，∠ABE＝90° ∴∠CBD＋∠EBF＝90° ∵∠CBD＋∠BCD＝90° ∴∠BCD＝∠EBF ∵∠D＝∠F＝90°，CB＝BE ∴△BCD≌△EBF ∴BD＝EF······················································································4分 （2）解：在Rt△BCD中 ∵CD＝1.8m，α＝26° CD 1.8 ∴BC＝ ≈ ＝2（m） cos26 0.90 90π22 ∴机械臂从C旋转到E处时，扫过的面积为 ＝π（m2）························8分 360 20. 解：（1）由图可知，B组和D组人数占总人数35%，人数为84人 ∴总人数为：84÷35%＝240（人） ∴A组和C组总人数为：240－84＝156（人） ∴13m＝156 ∴m＝12·····························································································4分 （2）众数所在的组别为C组····················································································5分 12 A组所对应的圆心角为 ×360°＝18°··························································6分 240 1212 （3）属于C组的大约有8800× ＝5280（万人）＝5.28×107（人）·························8分 240 答：估计属于C组的大约有5.28×107人 21. 解：（1）∵四边形ABCD为矩形 ∴∠BAD＝90° ∵AB＝6，AD＝8 ∴由勾股定理可得BD＝ AB2  AD2 ＝10·················································3分 （2）∵点P为AB的中点 ∴AP＝3 当Rt△PAQ∽Rt△BCD时 PA AQ ＝ BC CD 数学练习（六） 参考答案 第 3 页 共 7 页3 AQ ∴ ＝ 8 6 9 ∴AQ＝ 4 当Rt△PAQ∽Rt△DCB时 PA AQ ＝ DC CB 3 AQ ∴ ＝ 6 8 ∴AQ＝4 9 ∴当AQ＝ 或AQ＝4时，以A，P，Q为顶点的三角形与△BCD相似·····················6分 4 （3）∵P，C都在以Q为圆心的圆上 ∴PQ＝CQ 设AQ的长为x，则DQ＝8－x 则32＋x2＝（8－x）2＋62 91 解得x＝ 16 91 ∴AQ＝ ····································································································9分 16 22. （1）1············································································································1分 1 1 解：（2）乙行驶1小时，休息了30分钟，即 小时，以90千米/小时的速度到达A地，用时 小时 2 3 1 1 11 ∴到达A地共用时1＋ ＋ ＝ （小时）···················································3分 2 3 6 如下图所示： ········································································5分 （3）设乙休息前距A地的距离y（千米）和经过的时间x（小时）之间的函数关系为y＝kx＋90（k≠0） 1 ∵E（1，30） ∴30＝k＋90 ∴k＝-60 ∴y＝-60x＋90······························································································7分 1 设甲从A地到B地的函数关系式为y＝mx（m≠0） 2 ∵C（1，90） ∴m＝90 甲从A地到B地的函数关系式为y＝90x·····························································8分 2  3 y 60x90 x  1 ，解得 5 y 2 90x  y54 数学练习（六） 参考答案 第 4 页 共 7 页3 ∴F（ ，54）······························································································9分 5 23. （1）4，4·······································································································2分 解：（2）连接OC，OD，O'C，O'D ∵点O关于CD的对称点为点O' 即CD是OO'的垂直平分线 ∴OC＝O'C，OD＝O'D ∵OC＝OD＝4 ∴O'C＝O'D＝4 ⌒ ∵CED所在圆的半径为4 ⌒ ∴O'即为CED所在圆的圆心 ∴嘉嘉的说法正确······················································································6分 （3）连接O'E，则O'E＝4，O'E⊥AB ∵E是OB的中点，OB＝4 1 ∴OE＝BE＝ OB＝2 2 ∴O'B＝ O'E2 BE2 ＝2 5 又AB是⊙O的直径 ∴∠AMB＝90°＝∠O'EB 又∠ABM＝∠O'BE ∴△AMB∽△O'EB AM AB ∴ ＝ O'E O'B AM 8 即 ＝ 4 2 5 16 5 解得AM＝ ····························································································9分 5 （4）O'O的最大值为4 2 ·····················································································10分 O'O的最小值为4··························································································11分 【解析】∵O'E＝4，OO'＝ OE2 OE2 ＝ OE2 16 ∴OO'的大小取决于OE的大小 ∴当OE＝OB＝4时，OE最大，此时点B，E，D重合（如图1），OO'最大，OO'＝4 2 当OE＝0时，OE最小，此时点E，O重合（如图2），OO'最小，OO'＝4 ∴OO'的最大值为4 2，最小值为4 数学练习（六） 参考答案 第 5 页 共 7 页24. 解：（1）将C（0，3）代入y＝a（x＋1）（x－3），得a＝-1 ∴抛物线G 的解析式为y＝-x2＋2x＋3······················································3分 1 （2）由平移性质及题中图象可知抛物线G 过C（0，3） 2 设抛物线G 的解析式为y＝-x2＋bx＋3 2 把（-1，6）代入y＝-x2＋bx＋3 得6＝-1－b＋3 解得b＝-4 ∴抛物线G 的解析式为y＝-x2－4x＋3＝-（x＋2）2＋7 2 ∴抛物线G 的顶点D的坐标为（-2，7） 2 抛物线G 的解析式为y＝-x2＋2x＋3＝-（x－1）2＋4 1 抛物线G 的顶点E的坐标为（1，4） 1 过点D作DF平行于y轴，过点E作EF平行于x轴，两线交于点F ∴EF＝3，DF＝3 ∵∠DFE＝90° ∴DE＝3 2 ∴抛物线G 平移的最短距离为3 2 ···································································7分 1 （3）①∵直线l：y＝kx＋b（k≠0）过点C（0，3） ∴b＝3 ∴直线l：y＝kx＋3 ∵抛物线G 的解析式为y＝-（x＋1）（x－3） 1 ∴A（-1，0），B（3，0） ∵直线l：y＝kx＋3过点B（3，0） ∴0＝3k＋3 ∴k＝-1 直线l：y＝-x＋3 设P（x，-x2＋2x＋3） 过点P作y轴平行线，交BC于点Q，交x轴于点M ∴Q（x，-x＋3） ∴PQ＝-x2＋2x＋3－（-x＋3）＝-x2＋3x 1 1 ∴△PBC的面积S＝ ·PQ·OM＋ ·PQ·MB 2 2 1 ＝ ·OB·PQ 2 1 ＝ ×3×（-x2＋3x） 2 3 3 27 ＝- （x－ ）2＋ 2 2 8 由题意可知0＜x＜3 3 27 ∴当x＝ 时，S取得最大值 ·····································································9分 2 8 3 15 此时点P（ ， ）················································································· 10分 2 4 ②0＜k＜2或-4＜k＜0··················································································· 12分 数学练习（六） 参考答案 第 6 页 共 7 页【解析】当直线l与图象G 只有一个交点时 1 ykx3  y（x1）（x3） ∴x2＋（k－2）x＝0有唯一解 ∴（k－2）2＝0 ∴k＝2 ∴当0＜k＜2时，直线l与图象G有三个交点 当直线l与图象G 只有一个交点时 2 ykx3  yx2 4x3 ∴x2＋（k＋4）x＝0有唯一解 ∴（k＋4）2＝0 ∴k＝-4 ∴当-4＜k＜0时，直线l与图象G有三个交点 综上所述：当0＜k＜2或-4＜k＜0时，直线l与图象G有三个交点 数学练习（六） 参考答案 第 7 页 共 7 页