2026年河北省中考麒麟卷数学试题及答案（六）

一、选择题（本大题共12小题，每小题3分，共36分）1-5DBCAC6-10BACBC11-12DC【解析】∵等腰直角三角形ABC的边BC∥x轴，BC＝2，B（2，2）∴C（4，2A（2，4）双曲线y＝k（k＞0，x＞0）经过点B时，k最小x∴k最小值为4∵等腰直角三角形ABC，y＝k（k＞0，x＞0）都是关于直线y＝x对称x∴当y＝k（k＞0，x＞0）经过AC中点E时，k最大x∵C（4，2A（2，4）∴E（3，3）此时k＝9∴双曲线y＝k（k＞0，x＞0）与△ABC有交点时，4≤k≤9x故选：D【解析】延长AB，DC交于点M∵正六边形ABCDEF∴△BCM是等边三角形∵点G是2过点G作GH⊥DM，垂足为H∴结论①正确过点C作CN⊥AM于点N∵△BCM为等边三角形，CM＝a2∴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） 1【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∴△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）∴不等式的正整数解为：1，2···········································································7分-2x2－2x＋4＝-8（2）嘉嘉的说法正确·····························································································4分理由如下：由题意得（-x2＋2x＋3M＝-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 19.（1）证明：由题意可知，∠ABE＝90°∴△BCD≌△EBF∴BD＝EF······················································································4分（2）解：在Rt△BCD中360∴机械臂从C旋转到E处时，扫过的面积为90πx22=π（m2）························8分36020.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）∴m＝12·····························································································4分（2）众数所在的组别为C组····················································································5分A组所对应的圆心角为×360°=18°··························································6分（3）属于C组的大约有8800×=5280（万人）＝5.28×107（人）·························8分答：估计属于C组的大约有5.28×107人∴由勾股定理可得BD···············································3分（2）∵点P为AB的中点9∴当AQ＝或AQ＝4时，以A，P，Q为顶点的三角形与△BCD相似·····················6分（3）∵P，C都在以Q为圆心的圆上设AQ的长为x，则DQ＝8－x则32＋x28－x）2＋62∴AQ=····································································································9分22.（1）1············································································································1分解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，用时小时如下图所示：5分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y1＝kx＋90（k≠0）∴30＝k＋901＝-60x＋90······························································································7分设甲从A地到B地的函数关系式为y2＝mx（m≠0）甲从A地到B地的函数关系式为y2＝90x·····························································8分∴F（3，54）······························································································9分523.（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∵C所在圆的半径为4∴O'即为C所在圆的圆心∴嘉嘉的说法正确······················································································6分又AB是⊙O的直径解得AM····························································································9分（4）O'O的最大值为42·····················································································10分 ∴当OE＝OB＝4时，OE最大，此时点B，E，D重合（如图1OO'最大，OO'＝42当OE＝0时，OE最小，此时点E，O重合（如图2OO'最小，OO'＝4∴OO'的最大值为42，最小值为424.解1）将C（0，3）代入y＝a（x＋1x－3得a＝-1∴抛物线G1的解析式为y＝-x2＋2x＋3 3分（2）由平移性质及题中图象可知抛物线G2过C（0，3）设抛物线G2的解析式为y＝-x2＋bx＋3把（-1，6）代入y＝-x2＋bx＋3∴抛物线G2的解析式为y＝-x2－4x＋3＝-（x＋2）2＋7∴抛物线G2的顶点D的坐标为（-2，7）抛物线G1的解析式为y＝-x2＋2x＋3＝-（x－1抛物线G1的顶点E的坐标为（1，4）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∴抛物线G1平移的最短距离为32···································································7分（3）①∵直线l：y＝kx＋b（k≠0）过点C（0，3）∵抛物线G1的解析式为y＝-（x＋1x－3）∴A（-1，0B（3，0）∵直线l：y＝kx＋3过点B（3，0）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）＋3-x＋3∴当x＝时，S取得最大值··································