九年级数学（八）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（八） 参考答案 一、选择题（本大题共12小题，每小题3分，共36分） 1-5CBCDC 6-10CAACB 11-12DC 二、填空题（本大题共4小题，每小题3分，共12分） 13.1 14.12（或13、或14） 15.30 16. 6 2 【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H 1 1 ∠AOC＝30°，AD＝ OA＝ OB 2 2 1 1 1 S ＝ ·OB·AD＝ ·OB· ·OB＝1 △AOB 2 2 2 OB＝2 BH 6  2 6  2 sin∠BOH＝ ，则BH＝OB·sin∠BOH＝2× ＝ OB 4 2 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 x1 3x （2）  1 2 3 x1 3x ×6－ ×6＝1×6 2 3 3（x－1）－2（3－x）＝6 3x－3－6＋2x＝6 5x＝15 x＝3·············································································································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································3分 （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 数学练习（八） 参考答案 第 1 页 共 5 页3c ∴c最大为6，此时b ＝ ＝3 最大 3 ∴每条边上三个数之和的最大值为3·································································· 8分 1 1 19. 解：（1）平均得分x ×（9＋8＋6＋9）＝ ×32＝8（分） 4 4 ∵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° 60 90 120 90 平均得分为9× ＋8× ＋6× ＋9× 360 360 360 360 ＝1.5＋2＋2＋2.25 ＝7.75（分）·································································································7分 7.75＜8 ∴本学期班委工作不优秀，（1）的评价结果改变了··············································· 8分 20. 解：（1）∵∠ADE＝∠ACD ∴∠ADE＋∠CDE＝∠ACD＋∠CDE ∴∠ADC＝∠BEC 又AD＝BE，DC＝EC ∴△ADC≌△BEC（SAS）·····································································5分 （2）∵∠ADE＝∠ACD，∠DAE＝∠CAD ∴△ADE∽△ACD AE AD 4 AD ∴  ，即  AD AC AD 45 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 ············································································3分 （2）连接OB，如右图 设圆O的半径为xcm，则OB＝xcm ∵C为AB的中点 ∴OD⊥AB于点C OC＝OD－CD＝（x－8）cm 1 CB＝ AB＝12cm 2 ∵OC2＋CB2＝OB2 ∴（x－8）2＋122＝x2 解得x＝13 ∴圆O的半径为13cm·····················································································6分 数学练习（八） 参考答案 第 2 页 共 5 页（3）如右图，连接OE，DB，设OE与DB交于点F ∵OD⊥AB ∴DB＝ CD2 CB2  82 122 4 13cm ⌒ ∵E为DB的中点 1 ∴OE⊥DB于点F，DF＝ DB＝2 13cm 2 ∴OF＝ OD2 DF2  132 （2 13）2 3 13cm ∴FE＝OE－OF＝（13－3 13）cm ∵圆O与MN相切于点E ∴OE⊥MN 又OE⊥DB ∴BD∥MN ∴点B到地面MN的距离为（13－3 13）cm······················································9分 22. 解：（1）v＝v＋at＝0＋9.8t＝9.8t·········································································1分 0 v v 9.8t s＝vt＝ 0 t＝ t＝4.9t2··································································3分 2 2 （2）v＝9.8t＝29.4，解得t＝3 ∴物体的运动时间为3s····················································································5分 （3）v＝9.8t＝9.8×5＝49m/s ∴物体到达地面时的速度为49m/s······································································7分 s＝4.9t2＝4.9×52＝122.5m ∴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 ∴∠NME＝∠DME ∴∠NMA＝∠NME＝∠DME＝60° ∴AM＝ME＝2MD 8 16 ∴MD＝ cm，AM＝2MD＝ cm·································································5分 3 3 （3）连接NE NF＝y，EF＝AB＝8cm，CE＝（8－x）cm，NC＝（8－y）cm ∵NF2＋EF2＝NC2＋CE2 y2＋82＝（8－y）2＋（8－x）2 16y＝（8－x）2 1 1 y＝ （8－x）2＝ x2－x＋4·········································································· 9分 16 16 37 （4） πcm······································································································11分 45 【解析】由AG⊥BM得∠AGB＝90°，点G在以AB为直径的圆上运动 48 8 5 点E在点D时，点M为AD中点，AM＝4cm，BM＝4 5cm，AG＝  cm 4 5 5 数学练习（八） 参考答案 第 3 页 共 5 页设AB中点为点O，过点O作OH⊥AG于点H 1 4 5 HG 5 ∴HG＝ AG cm，sin∠HOG＝  2 5 OG 5 ∴∠HOG＝26.5°，∠AOG＝53° 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 37π4 37π 点G运动路径的长度为  cm 180 45 24. 解：（1）∵抛物线L：y＝x2＋bx＋c经过点A（0，2），B（3，-1） 1 ∴c＝2 -1＝32＋3b＋2 b＝-4 抛物线L 的解析式为y＝x2－4x＋2···························································2分 1 y＝x2－4x＋2 ＝（x－2）2－2 ∴P（2，-2）······················································································3分 （2）L 能经过点D································································································ 4分 2 理由：点D在抛物线L：y＝（x－2）2－2上 1 ∴y＝7 7＝（x－2）2－2 解得x＝-1，x＝5 1 2 D（-1，7）或（5，7） ∵L：y＝a（x－2）2＋n过点C（1，3） 2 ∴3＝a（1－2）2＋n n＝3－a y＝a（x－2）2＋3－a 若L 过点（-1，7） 2 即7＝a（-1－2）2＋3－a 1 解得a＝ 2 1 若L 过点（5，7），同理可得a＝ ···································································6分 2 2 （3）∵A（0，2），B（3，-1） ∴直线AB的解析式为y＝-x＋2 令a（x－2）2＋3－a＝-x＋2 ax2－4ax＋4a＋3－a＝-x＋2 ax2＋（1－4a）x＋（3a＋1）＝0 Δ＝（1－4a）2－4a（3a＋1）＝4a2－12a＋1 当Δ＝0时 12 122 16 128 2 32 2 a＝  ＝ 8 8 2 数学练习（八） 参考答案 第 4 页 共 5 页32 2 ∵a＝ 时，抛物线L 与BA延长线有唯一交点，不符题意，舍去 2 2 32 2 ∴a＝ 2 当Δ＞0时 32 2 32 2 即0＜a＜ 或a＞ 时 2 2 32 2 ①当0＜a＜ 时 2 4a1 1 x＋x＝ ＝4－ ＜0 1 2 a a 3a1 1 x·x＝ ＝3＋ ＞0 1 2 a a ∴x＜0，x＜0 1 2 ∴抛物线L 与线段AB没有交点，不符题意，舍去 2 32 2 ②当a＞ 时，抛物线L 过点C（1，3）且对称轴为直线x＝2 2 2 32 2 ∴a＞ 时抛物线L 与线段AB始终有两个交点，不符题意，舍去 2 2 32 2 ∴抛物线L 与线段AB有唯一交点时a＝ ················································10分 2 2 m2 4m2 （4）a ·························································································12分 m2 10m24 【解析】∵点M的横坐标为m，∴M（m，m2－4m＋2） ∵四边形ABMN为平行四边形 ∴AB＝MN，AB∥MN ∵A（0，2），B（3，-1） ∴点N为点M左移3个单位长度，上移3个单位长度得到的 ∴N（m－3，m2－4m＋5） ∵点N在L 上 2 ∴m2－4m＋5＝a（m－3－2）2＋3－a m2－4m＋5＝a（m2－10m＋25）＋3－a m2－4m＋5＝am2－10am＋25a＋3－a am2－10am＋24a＝m2－4m＋2 m2 4m2 a m2 10m24 数学练习（八） 参考答案 第 5 页 共 5 页