2026年河北省中考麒麟卷数学试题及答案（八）

一、选择题（本大题共12小题，每小题3分，共36分）1-5CBCDC6-10CAACB11-12DC二、填空题（本大题共4小题，每小题3分，共12分）【解析】延长BO交⊙O于点C，过点A作AD⊥BC于点D，过点O作OH⊥AB于点HOB42则AB＝2BH＝6+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）习题1第一步，习题2第一步·································································2分）－）＝ 7分18.解1）-2＋2＋0＝0，-1＋3＋0＝2，-1＋1－2＝-2每条边上的三个数的和不相等，所以小颖的尝试不正确 （2）3＋c＝3b（能转化为3＋c＝3b的式子均可） 5分（3）每条边上三个数之和为b，b最大，即3＋c最大即c最大时b最大c为三个顶点处三个数之和，-2，-1，0，1，2，3中三个数的和最大为1＋2＋3＝6∴c最大为6，此时b最大3∵平均得分不低于8分为优秀∴本学期班委工作优秀···········································································4分（2）360°-90°-90°-120°=60°=7.75（分）·································································································7分又AD＝BE，DC＝EC整理得AD2＝4×9，解得AD＝6∴BE＝AD＝6································································································8分21.解1）如下图，点O即为所求点（2）连接OB，如右图设圆O的半径为xcm，则OB＝xcm∵C为AB的中点OC＝OD－CDx－8）cm2∴圆O的半径为13cm·····················································································6分（3）如右图，连接OE，DB，设OE与DB交于点F∵E为的中点∴OE⊥DB于点F，DFDBcm∵圆O与MN相切于点E∴OE⊥MN又OE⊥DB∴BD∥MNst＝4.9t2··································································3分∴物体的运动时间为3s····················································································5分∴物体到达地面时的速度为49m/s······································································7分2∴物体掉落的初始高度为122.5m·······································································9分23.（1）4············································································································2分解2）由折叠可得∠NME=∠NMA∵ME平分∠DMN∴AM＝ME＝2MD∴MDcm，AM＝2MDcm·································································5分（3）连接NENF＝y，EF＝AB＝8cm，CE8－x）cm，NC8－y）cm＝NC22y228－y）28－x）216y8－x）2【解析】由AG⊥BM得∠AGB＝90°,点G在以AB为直径的圆上运动点E在点D时，点M为AD中点，AM＝4cm，BM＝45cm，AGcm设AB中点为点O，过点O作OH⊥AG于点H当E在点C时，MN与BD重合，点G为BD中点P点G运动路径所在弧所对圆心角为90°-53°=37°,半径为4cm点G运动路径的长度为cm24.解1）∵抛物线L1：y＝x2＋bx＋c经过点A（0，2B（3，-1）抛物线L1的解析式为y＝x2－4x＋2···························································2分=（x－2）2－2∴P（2，-2）······················································································3分理由：点D在抛物线L1：yx－2）2－2上7x－2）2－2D（-1，7）或（5，7）2：y＝a（x－2）2n＝3－a若L2过点（-1，7）即7＝a（-1－2）2＋3－a2（3）∵A（0，2B（3，-1）∴直线AB的解析式为y＝-x＋2当Δ=0时∵a时，抛物线L2与BA延长线有唯一交点，不符题意，舍去22①当时∴抛物线L2与线段AB没有交点，不符题意，舍去②当a时，抛物线L2过点C（1，3）且对称轴为直线x＝2∴a时抛物线L2与线段AB始终有两个交点，不符题意，舍去2∴抛物线L2与线段A