九年级数学（三）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（三） 参考答案 一、选择题（本大题共12小题，每小题3分，共36分） 1-5CBBCB 6-10ADCCD 11-12CA 二、填空题（本大题共4小题，每小题3分，共12分） 13.2 2（答案不唯一） 14.6 15.4＜k≤6 π 16. （1）15°（1分） （2） （2分） 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）（-3）2＋2-2×4÷（-1）2026 1 ＝9＋ ×4÷1 4 ＝9＋1 ＝10·····································································································3分 （2）原式＝x2－4x＋4－（x2＋x－2） ＝x2－4x＋4－x2－x＋2 ＝-5x＋6································································································6分 当x＝5时，原式＝-25＋6＝-19········································································7分 18. 解：【规律运用】（1）12×18＝1×（1＋1）×100＋2×8＝216···································1分 （2）24×26＝2×（2＋1）×100＋4×6＝624···································2分 （3）33×37＝3×（3＋1）×100＋3×7＝1221································· 3分 【规律证明】规律表示：（10a＋b）（10a＋c）＝100a（a＋1）＋bc＝100a2＋100a＋bc····5分 证明：（10a＋b）（10a＋c） ＝100a2＋10ac＋10ab＋bc ＝100a2＋10a（c＋b）＋bc ＝100a2＋10a×10＋bc ＝100a2＋100a＋bc····································································8分 19. 解：（1）中位数在400≤x＜450这组····································································2分 （2）续航里程不低于450km的频数：18＋7＝25 25 1 恰好是“优秀续航”的概率：P＝ ＝ ··························································5分 100 4 （3）10＋n≤（100＋n）×12％ 3 解得n≤2 11 ∵n为整数 ∴n的最大值为2····························································································8分 数学练习（三） 参考答案 第 1 页 共 6 页20. 解：（1）延长AB交DE的延长线于点F 由题意可知：AC＝12米，CD＝30米，∠BDF＝30° 在Rt△BDF中，DF＝AC＝12米 3 3 ∴BF＝ DF＝12× ＝4 3 （米） 3 3 ∵AF＝CD＝30米 ∴AB＝30－4 3 ≈30－4×1.73≈23.1（米） ∴教学楼AB的高度约为23.1米······························································4分 BF 4 （2）在Rt△BEF中 tanα＝ ＝ EF 5 4 3 4 ∴ ＝ EF 5 ∴EF＝5 3 米 ∴DE＝DF－EF＝（12－5 3 ）米 ∴无人机全程的路程为：30＋12－5 3 ≈42－5×1.73≈33.35（米） 平均速度为：33.35÷3≈11.12（米/秒） 11.12米/秒在10-20米/秒之间 ∴无人机能在3秒内回到点C的位置·································································8分 21. 解：（1）甲步行的速度：1500÷30＝50（米/分钟）·················································1分 乙骑车的速度：2000÷（25－5）＝100（米/分钟）·····································2分 学校门口和操场的距离：2000－1500＝500（米）·······································3分 （2）设y 与x的函数关系式为y ＝kx 甲 甲 将（30，1500）代入，得30k＝1500 ∴k＝50 ∴y ＝50x 甲 设y 与x的函数关系式为y ＝ax＋b 乙 乙 代入（5，0）和（25，2000） 5ab0 a100 得 ，解得 25ab2000 b500 ∴y ＝100x－500 乙 当乙追上甲时y －y ＝500 乙 甲 ∴100x－500－50x＝500，解得x＝20··································································6分 （3）y＝2000＋（x－30）×100 ＝100x－1000·······························································································7分 x的值为45···································································································9分 【解析】当乙到达学校门口时，乙一共行驶的路程y＝2000＋1500＝3500（米） 把y＝3500代入y＝100x－1000 得100x－1000＝3500 数学练习（三） 参考答案 第 2 页 共 6 页解得x＝45 22. （1）证明：∵EF⊥DE，DG⊥DE ∴∠DEF＝∠EDG＝90° ∵GF∥DE ∴∠DGF＝180°－∠EDG＝180°－90°＝90° ∴四边形DEFG为矩形··············································································2分 （2）证明：∵四边形ABCD为正方形 ∴AD＝CD＝BC＝AB＝12cm，∠CDA＝90° ∴∠CDE＋∠ADE＝90° ∵∠ADE＋∠GDA＝90° ∴∠GDA＝∠CDE 又∠G＝∠C＝90° ∴△DGA∽△DCE DG DA ∴ ＝ DC DE ∵DA＝DC＝12cm ∴DG·DE＝DA·DC＝12×12＝144（cm2），即矩形DEFG的面积为144cm2 ∵正方形ABCD的面积为12×12＝144（cm2） ∴矩形DEFG的面积和正方形ABCD的面积相等·········································5分 （3）证明：∵KJ⊥DE ∴∠DJK＝90° ∵∠C＝90° ∴∠CDE＋∠CED＝90° ∵四边形DEFG为矩形 ∴∠F＝∠DEF＝90°＝∠DJK ∴∠CED＋∠BEM＝90° ∴∠CDE＝∠BEM ∵∠BEM＋∠BME＝90°，∠AMF＋∠FAM＝90°，∠BME＝∠AMF ∴∠BEM＝∠FAM＝∠CDE 又DK＝AM ∴△DKJ≌△AMF·················································································7分 （4）解：方法一：∵DE＞CD，AD＞DG ∴DE＞DG ∴只存在DE∶DG＝3∶2的情况 144 设CE＝acm，DE＝ a2 144 cm，DG＝ cm a2 144 144 ∴ a2 144 ∶ ＝3∶2 a2 144 解得a＝6 2cm 即CE＝6 2cm··········································································· 9分 数学练习（三） 参考答案 第 3 页 共 6 页方法二：∵DE＞CD，AD＞DG ∴DE＞DG ∴只存在DE∶DG＝3∶2的情况 设DG＝2xcm，则DE＝3xcm ∵矩形DEFG的面积为144cm2 ∴2x·3x＝144 解得x＝2 6 ∴DE＝3×2 6＝6 6 （cm） ∴CE＝ （6 6）2 122 ＝6 2（cm）················································9分 23. 解：（1）把x＝0代入y＝-2x＋4 得y＝4 ∴点A的坐标为（0，4）·······································································1分 把y＝0代入y＝-2x＋4 得-2x＋4＝0 ∴x＝2 ∴点B的坐标为（2，0）·······································································2分 点C的坐标为（1，2）········································································· 3分 （2）∵抛物线W的顶点为C（1，2） ∴其解析式可设为：y＝a（x－1）2＋2 代入点A（0，4） a＋2＝4 ∴a＝2 ∴y＝2（x－1）2＋2（或y＝2x2－4x＋4）····························································6分 （3）点A不是线段BD的中点·················································································7分 ∵抛物线W 和W关于y轴对称 1 ∴抛物线W 的顶点为（-1，2） 1 ∴抛物线W 的解析式为y＝2（x＋1）2＋2 1 令2（x＋1）2＋2＝-2x＋4 解得x＝0，x＝-3 1 2 ∴点D的横坐标为-3，当x＝-3时，y＝-2×（-3）＋4＝10 ∴点D的坐标为（-3，10） 已知点B（2，0） 1 ∴线段BD的中点坐标为（- ，5） 2 ∴点A不是线段BD的中点··············································································9分 （4）1＜m＜1.5···································································································11分 13 【解析】当y＝y 时，对称轴为x＝ ＝2 2 1 2 ∵y＜y，抛物线开口向上 2 1 ∴抛物线W 的对称轴离着x＝3比x＝1近 2 数学练习（三） 参考答案 第 4 页 共 6 页∴抛物线W 的对称轴在x＝2的右侧，故m＞2-1，即m＞1 2 14 当y＝y 时，对称轴为x＝ ＝2.5 1 3 2 ∵y＜y，抛物线开口向上 1 3 ∴抛物线W 的对称轴离着x＝1比x＝4近 2 ∴抛物线W 的对称轴在x＝2.5的左侧，故m＜2.5－1，即m＜1.5 2 综上m的取值范围是：1＜m＜1.5 24. 解：（1）4 3 ，8······························································································2分 过点O作OP⊥CD于点P，连接OD 1 在Rt△CPO中，∠OCP＝90°－60°＝30°，OC＝ BC＝2 3 cm 2 1 ∴OP＝ OC＝ 3 cm，CP＝ 3 ·OP＝3cm 2 ∴CD＝2CP＝6cm，∠COD＝180°－30°－30°＝120° 120π（2 3）2 1 ∴阴影部分的面积是： － ×6× 3 ＝（4π－3 3 ）cm2·········4分 360 2 （2）设半圆O 和直线AC相切的切点为点E，连接OE，连接OO并延长交AC于点F 1 1 1 ∴∠OEF＝90° 1 ∵OF∥AB 1 ∴∠EFO＝∠A＝60° 1 在Rt△OEF中，∠EOF＝30°，OE＝2 3 cm 1 1 1 OE 3 ∴cos∠EOF＝ 1  1 OF 2 1 2 3 3 ∴  OF 2 1 ∴OF＝4cm 1 1 ∵OF∥AB，OC＝ BC 2 1 ∴OF＝ AB＝2cm 2 ∴OO＝4－2＝2（cm） 1 ∴半圆O平移的距离为2cm··············································································7分 （3）①在旋转过程中点C 到直线AC的距离先越来越小，再越来越大（当BC⊥AC时，点C 到AC 2 1 2 2 的距离最大），再越来越小 当α＝0°时，过点C 作CG⊥AC于点G，连接CC 1 1 1 ∵CC＝2cm，∠GCC＝∠A＝60° 1 1 ∴CG＝1cm，GC＝ 3 cm··································8分 1 当BC⊥AC时，设垂足为H 1 2 AB＝4＋2＝6（cm），∠A＝60° 1 1 ∴AH＝ AB＝3cm，BH＝ 3 AH＝3 3 cm 2 1 1 数学练习（三） 参考答案 第 5 页 共 6 页∴CH＝BC－BH＝4 3 －3 3 ＝ 3 （cm） 2 1 2 1 此时∠ABC＝30° 1 2 ∴α＝90°－30°＝60° ∵GC＝CH 1 2 ∴当点C 到直线AC的距离最大时，α的值为0°或60°··································10分 2 ②10°＜α＜49°·························································································12分 【解析】当半圆O 经过点M时，过点M作MN⊥AB于点N 2 在Rt△AMN中，AB＝4cm，∠A＝60° ∴AN＝2cm，MN＝2 3 cm 在Rt△BMN中，BN＝6－2＝4（cm） 1 1 ∴BM＝ 42 （2 3）2 ＝2 7（cm） 1 MN 2 3 3 tan∠ABM＝   1 BN 4 2 1 ∴∠ABM≈41° 1 ∵BC 为直径 1 2 ∴∠BMC＝90° 1 2 2 7 21 ∴cos∠MBC＝  1 2 4 3 6 ∴∠MBC≈39° 1 2 ∴α＝90°－41°－39°＝10° 当直径BC 过点M时 1 2 α＝90°－41°＝49° ∴10°＜α＜49° 数学练习（三） 参考答案 第 6 页 共 6 页