九年级数学（一）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（一） 参考答案 一、选择题（本大题共12小题，每小题3分，共36分） 1-5ADBCD 6-10BDCAB 11-12CC 二、填空题（本大题共4小题，每小题3分，共12分） 13.2 14.22.5° 15.55 2 85 16. 85 【解析】连接AC，交BD于点O，过点E作EG⊥OD于点G ∵四边形ABCD是菱形 1 ∴AD＝AB＝4 5 ，BO＝OD＝ BD＝8，AO⊥BD 2 ∴AO＝ AD2 OD2 ＝4，EG∥AO DE EG DG ∴△DEG∽△DAO，∴ ＝ ＝ DA AO DO ∵E是AD的中点 DE EG DG 1 ∴ ＝ ＝ ＝ DA 4 8 2 ∴EG＝2，DG＝4 ∴FG＝BD－BF－DG＝16－3－4＝9 ∴EF＝ FG2 EG2 ＝ 92 22 ＝ 85 EG 2 2 85 ∴sin∠EFD＝ ＝ ＝ EF 85 85 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）6－（-6）×2－5＝6＋12－5＝13···································································3分 （2）设□为x，则：6－2x－5＜-13 6－2x－5＜-13 -2x＜-14 x＞7 ∴指针应指向8所在区域·······················································································7分 18. 解：（1）二············································································································2分 正确解答：①×2，得：2x－4y＝8③ ③－②，得：-y＝7 解得：y＝-7······························································································ 3分 把y＝-7代入①，得：x－2×（-7）＝4 解得：x＝-10·····························································································4分 x10 ∴原方程组的解为 ···········································································5分 y7 数学练习（一） 参考答案 第 1 页 共 4 页y3x1 （2）联立得： ······························································································6分 y x5 x3 解得： ·······································································································7分 y8 ∴P（3，8）位于第一象限·····················································································8分 19. 证明：（1）∵∠CAD＝∠EAB ∴∠CAD－∠BAD＝∠EAB－∠BAD ∴∠CAB＝∠EAD ∵∠C＝∠E，AC＝AE ∴△ABC≌△ADE（ASA）········································································5分 （2）∵△ABC≌△ADE ∴AB＝AD ∵F为BD的中点 ∴AF⊥BD···········································································································8分 20. （1）50 36····································································································2分 56101815162010 解：（2）x ＝13（元）························································4分 50 众数为10元···································································································5分 学生人数为50人，从小到大排列后位于第25第26名捐款为15元 所以中位数是15元··························································································6分 （3）设李老师的捐款金额为x，目前捐款总人数为51人，中位数位于第26位，捐款数仍是15元 650x 则： ＝15··································································································7分 501 x＝115 李老师的捐款金额为115元····················································································8分 21. 解：（1）30°·········································································································2分 （2）方法一：如下图，点M即为所求 ·······························································································5分 提示：∵∠CMD＝2∠CED ∴∠MED＝∠MDE ∴作出DE的垂直平分线交CE于点M即可 方法二：如下图，点M即为所求 ····················································································5分 数学练习（一） 参考答案 第 2 页 共 4 页提示：∵∠CMD＝2∠CED＝60° ∴∠CDM＝90° ∴过点D作CD的垂线交CE于点M即可 （答案不唯一，作法正确即可） （3）如图1，若⊙O与CD，DE两边相切，则点O在∠CDE的平分线上 又CD＝DE ∴OD⊥CE ∴OE＝DEcos30°＝3 3 如图2，若⊙O与BC，DE两边相切 ∴∠BCD＝120°，∠ECD＝30° ∴∠BCE＝90° ∴C是⊙O与BC边的切点 连接OD，则OD＝OC ∴∠ODC＝∠OCD＝30° ∴∠EOD＝60° ∴∠ODE＝90° ∴D是⊙O与DE边的切点 DE ∴OE＝ ＝4 3 cos30 综上，OE的长为3 3或4 3··················································································9分 22. 解：（1）3.5×103×0.12×[15－（-10）]＝1.05×104J······················································3分 （2）①0.4c×[20－（-5）]＝3×104 2 c＝3×103J/（kg·℃）·······················································································5分 2 2.7104 ②温度降低量为 ＝30℃·········································································· 7分 0.33103 （3）由题意得：3.5×103×25×m－3×103×25×m＝2.1×104，解得m＝1.68 ∴冷冻的猪肉的质量为1.68kg·················································································9分 23. 解：（1）90°·········································································································3分 （2）由题知点A'在以B为圆心，以BA长为半径的圆上，当B，A'，D共线时，DA'的值最小···· 4分 BD＝ 62 82 ＝10 ∴DA'的值最小为10－6＝4·····················································································5分 此时，FA'⊥BD，∠FA'D＝∠BAD＝90°，∠FDA'＝∠BDA ∴△FA'D∽△BAD································································································6分 DF DA 1 ∴ ＝ ＝ DB DA 2 ∴DF＝5·············································································································8分 1 （3）点A'到直线BC的距离为3时，sin∠A'BC＝ 2 ∴∠A'BC＝30° 6062π 若A'在直线BC上方，α＝90°－30°＝60°，则边BA扫过区域的面积为 ＝6π 360 12062π 若A'在直线BC下方，α＝90°＋30°＝120°，则边BA扫过区域的面积为 ＝12π 360 数学练习（一） 参考答案 第 3 页 共 4 页∴边BA扫过区域的面积为6π或12π····································································10分 （4） 73－3············································································································11分 24. 解：（1）将A（3，1），B（0，-2），代入y＝x2＋bx＋c 93bc1 b2 得 ，解得 c2 c2 ∴y＝x2－2x－2··························································································· 3分 ∵y＝x2－2x－2＝（x－1）2－3 ∴顶点G的坐标为（1，-3）·········································································4分 （2）①把x＝5代入y＝x2－2x－2中，y＝136 ∴点P（5，6）不在图象C 上············································································· 6分 1 ②根据平移规律可得新抛物线解析式为：y＝（x－1－n）2－3 当C 经过点P（5，6）时，则有6＝（5－1－n）2－3 2 解得：n＝1或n＝7····························································································8分 （3）①设直线AB的解析式为y＝kx＋a 3ka1 将A（3，1），B（0，-2）代入得 a2 k 1 解得 a2 ∴直线AB的解析式为y＝x－2 设直线GG'的解析式为y＝x＋d，过G（1，-3） -3＝1＋d，d＝-4 ∴直线GG'的解析式为y＝x－4，G'在x轴上，G'的坐标为（4，0） ∴G移动的距离GG'为 32 32 3 2 ··································································10分 ②G'（ 5 ， 5 －4）··························································································12分 【解析】图象沿射线BA方向平移时，上下与左右平移的距离相等 设向上，向右平移m个单位长度 ∴A'（3＋m，1＋m），G'（1＋m，-3＋m） 由平移得AA'＝GG'，AA'∥GG' ∴四边形A'AGG'是平行四边形 ∵线段AG'与A'G交于点M m4 m2 ∴M（ ， ） 2 2 ∵点M落在图象C 上 1 m2 m4 ∴ ＝（ －1）2－3 2 2 解得m＝-1± 5 ∵沿射线BA方向平移 ∴m＝-1－ 5 （舍去） ∴G'（1＋ 5 －1，-3＋ 5 －1） 即G'（ 5 ， 5 －4） 数学练习（一） 参考答案 第 4 页 共 4 页