2026年河北省中考麒麟卷数学试题及答案（四）

一、选择题（本大题共12小题，每小题3分，共36分）1-5CDCBC6-10BDDAA11-12DC【解析】①当AB'与AB的夹角为20°时②当AB'与AD的夹角为20°时综上，∠APB的度数是80°或55°或35°故选：D【解析】∵每个大圆圈上的四个数字的和都等于21∴3个大圆圈上的数字和为21×3＝63∵各个小圆圈上数字和为1＋2＋3＋4＋5＋6＋7＋8＋9＝45∴45＋x＋y＋x＋y＝632＋92＝285，每个大圆圈上的四个数字的平方和分别记为A，＋y2x＋y）2＝4112＋y2x＋y）2＝1262故结论Ⅱ正确故选：C二、填空题（本大题共4小题，每小题3分，共12分）13.（x－3x＋3） x-7三、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） （2）当Q＝53时，53＝5a－7b即b5a－53）75∴a的最小整数值为8······················································································7分2＝2a2＋2······················································4分2223∴a＝5········································································································8分3∴△BDE≌△ABC（SAS）（2）解：∵△BDE≌△ABC20.解1）a＝100，b＝145，c＝0.29·······························································（2）根据以上抽样调查结果，游客最满意的主题板块是A板块······································4分答：当本届灯会实际接待游客达150000人时，估计最满意A板块的人数是54000人···6分（3）画树状图如图：共有12种等可能结果，其中“两名女生”的结果有2种答：恰好抽到两名女生的概率是·····································································8分21.解1）连接OO'，AO'，记OO'与AB交于点C由折叠的性质知，AB垂直平分OO'∴△OO'A是等边三角形在Rt△OAC中，AO＝3cm，∠AOC＝60°解得ACcm（2）设圆锥的底面半径为r连接O'D在Rt△ODO'中，OD＝O'O2_O'D2＝22cm即圆锥的高OD为22cm················································································6分（3）∵圆锥侧面展开图为优弧A所成的扇形AM为⊙F的直径∴点M为优弧A的中点∴GC＝3cm由（1）可知ACcm在Rt△AGC中，由勾股定理可得AG2＝AC2+GC2∴所走的最短距离为cm·············································································9分22.解1）①（-2－6）÷2＝-45＋∴B（-4，3）设光线AB所在直线的解析式为y＝kx＋b（k，b为常数，且k≠0）将A（5，0）和B（-4，3）分别代入y＝kx＋b得理由如下：将M（m，m＋2）代入y得m444∴光线AB穿过点M，此时点M的坐标为····························7分（2）符合条件的m的整数值为-1或0·······································································9分【解析】当光线经过点P时，设光线AP所在直线的解析式为y＝k1x＋b1（k1，b1为常数，且k1≠0）将A（5，0）和P（-2，5）分别代入y＝k1x＋b1得∴光线AP所在直线的解析式为y当光线经过点Q时，设光线AQ所在直线的解析式为y＝k2x＋b2（k2，b2为常数，且k2≠0）将A（5，0）和Q（-6，1）分别代入y＝k2x＋b2得∴光线AQ所在直线的解析式为y∴符合条件的m的整数值为-1或0 4分（2）①证明：由旋转可知PC＝PE∵CM⊥AB，EN⊥AB∴△PCM≌△EPN··············································································7分②解：由得CM＝8，BM则AM＝AB－BM＝6由①知△PCM≌△EPN∴EN＝PM＝6－x，AN＝PN－PA＝812－xx－4∵EN∥CM7（3）解：由旋转得PE＝PC，PF＝PD，∠DPF=∠CPE＝90°∴△PEF可看作△PCD绕点P逆时针旋转90°当∠PEF＝90°时，∴∠PCD＝90°可知点E在直线AB上，如右图在Rt△BCP中，得BP＝6在Rt△AEF中，由勾股定理可得24.（10，3）····································································································3分∴抛物线L1的顶点坐标为························································6分∴抛物线L2的解析式为y＝x2＋bx＋3∵新抛物线与x轴只有一个交点2∵两条抛物线能形成封闭图形，且有整点2的顶点坐标为（3，0）联立∴两条抛物线的交点为（0，3由①可知：L1的顶点坐标为∴图象G中的整点有（1，11，21，31，42，12，22，3（2，43，2）共9个∴直线l上方的整点有（1，31，42，4共3个直线l下方的整点有（1，12，12，23，2）共4个∴抛物线L3的解析式为y＝ax2－4ax＋31，L3与y轴的交点均为（0，3）3与L1只有1个交点时