内容发布更新时间 : 2024/5/5 0:14:02星期一 下面是文章的全部内容请认真阅读。

24．如图，AB是⊙O的直径，AC切⊙O于点A，连接BC交⊙O于点D，点E是BD的中点，连接AE交BC于点F． （1）求证：AC=CF；

（2）若AB=4，AC=3，求∠BAE的正切值．

25．如图，点P是AB 所对弦AB上一动点，点Q是AB与弦AB

CDFAEOB所围成的图形的内部的一定点，作射线PQ交AB于点C，连接BC．已知AB=6cm，设A，P两点间的距离为xcm，P，C两点间的距离为y1cm，B，C两点间的距离为y2cm．（当点P与点A重合时，x的值为0）．

小平根据学习函数的经验，分别对函数y1，y2随自变量x的变化而变化的规律进行了探究． 下面是小平的探究过程，请补充完整：

（1）按照下表中自变量x的值进行取点、画图、测量，分别得到了y与x的几组对应值；

x/cm 0 1 2 3 4 5 6 y1/cm 5.37 4.06 2.83 m 3.86 4.83 5.82 y2/cm 2.68 3.57 4.90 5.54 5.72 5.79 5.82 经测量m的值是 （保留一位小数）． （2）在同一平面直角坐标系xOy中，描出补全后的表中各组数值所对应的点(x，y1)， (x，y2)，并画出函数y1，y2的图象；

（3）结合函数图象，解决问题：当△BCP为等腰三角形时，AP的长度约为 cm．

26．平面直角坐标系xOy中，抛物线y?x2?2mx?m2?3与y轴

y654321交于点A，过A作AB∥x轴与直线x=4交于B点．

（1）抛物线的对称轴为x= （用含m的代数式表示）； （2）当抛物线经过点A，B时，求此时抛物线的表达式； （3）记抛物线在线段AB下方的部分图象为G（包含A，B两点），点P（m,0）是x轴上一动点，过P作PD⊥x轴于P，交–6–5–4–3–2–1O123–1图象G于点D，交AB于点C，若CD≤1，求m的取值范围． –2–3

–4

–5 –6

27．在△ABC中，∠ABC=120°，线段AC绕点A逆时针旋转60°得到线段AD，连接CD，BD交AC于P．

（1）若∠BAC=α，直接写出∠BCD的度数 （用含α的代数式表示）； （2）求AB，BC，BD之间的数量关系；

（3）当α=30°时，直接写出AC，BD的关系．

456xDPABC

28．对于平面直角坐标系xoy中的图形P，Q，给出如下定义：M为图形P上任意一点，N为图形Q上任意一点，如果M，N两点间的距离有最小值，那么称这个最小值为图形P，Q间的“非常距离”，记作d（P,Q）．已知点A（4,0），B（0,4），连接AB． （1）d（点O，AB）=

（2）⊙O半径为r，若d（⊙O，AB）=0，求r的取值范围； （3）点C（－3，－2），连接AC，BC，⊙T的圆心为T（t，0），半径为2，d（⊙T，△ABC），且0

北京市平谷区2019年中考一模数学试卷参考答案及评分标准 2019.4

一、选择题（本题共16分，每小题2分）

题号 答案 1 C 2 C 3 B 4 D 5 B 6 A 7 A 8 B 二、填空题（本题共16分，每小题2分）

9．正方； 10．x>-1； 11．甲； 12．答案不唯一，如BD=DC； 13．

?2.5x?2y?2022； 14．a?b??a?b??a?b?； 15．23； 16．(4,0)．

x?y?11?20三、解答题(本题共68分，第17-21题，每小题5分，第22-27题，每小题6分，第28题7分)解答应写出文字说明、演算步骤或证明过程. 17．（1）如图；····························································································· 1

ACPOBD （2）同位角相等，两直线平行； ································································ 3

等边对等角． ·················································································· 5 18．解：原式=2?3·························································· 4 ?1?23?3?1 ·

2 =0． ························································································· 5 19．解：由①得x<3 ······················································································· 1 由①得x+1>2， ················································································· 2 x>1． ················································································ 3

∴1

2?k?3?2?0，

······························································· 3 ∴方程总有两个实数根． ·(2) ∵x???k?1???k?3?22∴x1??1，x2?2?k． ······························································· 4

，

∵方程有一个根为正数， ∴2?k?0 k?2．··············································································· 5

21．（1）k=4；····························································································· 1

（2）①1个； ························································································· 2

②当直线AB经过点A（2，﹣2），（0,1）时区域W内恰有1个整点，

∴a?1． 2当直线AB经过点A（2，﹣2），（1,1）时区域W内没有整点， ∴a=1． ······················································································ 3 ∴当

1?a?1时区域W内恰有1个整点． ········································· 5 222．（1）证明：∵AB=AC，点D是BC边的中点，

∴AD⊥BC于点D． ··································································· 1 ∵AE∥BC，CE∥AD，

∴四边形ADCE是平行四边形． ··················································· 2 ∴平行四边形ADCE是矩形． ······················································ 3

（2）解： 过点E作EF⊥AC于F．

A∵AB=10， ∴AC=10．

∵对角线AC，DE交于点O， O∴DE=AC=10． F∴OE=5． ···················· 4 BD∵sin∠COE=

E4， 5C∴EF=4 ··················································································· 5 ∴OF=3．

∵OE=OC=5， ∴CF=2．

∴CE=25． ········································································· 6

23．（1）如图； ··························································································· 1

（2）31≤x＜34这组的圆心角度数是 78 度， ·················································· 2

如图（画图1分，数据1分）； ································································· 4 （3）统计表中中位数m的值是 36 ； ··························································· 5 （4）答案不唯一，如：费尔兹奖得主获奖时年龄集中在37岁至40岁． ··················· 6