北京市中考数学丰台二模测试卷

丰台区2021年初三学业水平考试统一练习（二）

初三数学评分标准及参考答案

一、选择题（本题共16分，每小题2分）

二、填空题（本题共16分，每小题2分）

三、解答题（本题共68分，第17 - 22题，每小题5分，第23 - 26题，每小题6分，

第27 - 28题，每小题7分）

17. 解：原式，···············································4分

.···············································5分

18. 解：

解不等式①得：x3，········································2分

解不等式②得：x8.·········································4分

∴原不等式组的解集为x3.·······································5分

19. 证明：

∵∠BAE∠DAC，

∴∠BAE+∠EAC∠DAC+∠EAC，

即：∠BAC∠DAE.············································1分

在△BAC和△DAE中，

∴△BAC≌△DAE. ············································4分

∴∠C∠E.··················································5分

20. 解：原式，···············································2分

.···············································3分

∵，

∴原式····················································5分

21. （1）如图所示，

·························································2分

（2）AQ, BQ.············································4分

经过半径的外端并且垂直于这条半径的直线是圆的切线. ···············5分

22. （1）证明：∵AE∥BC，CE∥AD，

∴四边形ADCE是平行四边形. ····································1分

∵∠BAC90°，AD是BC边上的中线，

∴ADBDCD.

∴四边形ADCE是菱形. ·········································2分

（2）解：过点E作EH⊥BA交BA的延长线于点H.

在Rt△ABC中，∠ABC30°，AC2，

∴BC4，AB.

∴ADBC2.

∵四边形ADCE是菱形，

∴AEAD2.

∵AE//BC，

∴∠EAH∠ABC30°.

在Rt△AEH中，EH1，AH.

∴HBAH+AB. ···············································4分

在Rt△BEH中，

BE. ·······················································5分

23. 解：（1）∵点，在反比例函数的图象上，

∴，解得：. ················································4分

（2） ·····················································6分

24. （1）证明：

∵D是AC中点，OP过⊙O的圆心，

∴OP⊥AC，ADCD.

∴PCPA.

∴∠PCA∠PAC. ·············································1分

∵PA是⊙O的切线，OA为半径，

∴OA⊥PA.·················································2分

∴∠PAC+∠BAC90°.

∵AB是直径，

∴∠ACB90°.

∴∠ABC+∠BAC90°.

∴∠PCA∠ABC. ··············································3分

（2）解：∵OP⊥AC，

∴∠PAC+∠APO90°.

∵∠PAB90°，

∴∠PAC+∠BAC90°.

∴∠APO∠BAC.  ·············································4分

∵BC4，tan∠APO，

∴AC8.

∴AB.

∴AO. ·····················································5分

∴AP.·····················································6分

25. （1）

m91.5；·················································2分

（2）解：（人）；···········································4分

（3）解：①；

·························································5分

②八年级的成绩较好，因为与九年级相比，八年级的平均成绩略高，且方差较小，成绩波动不大.( 用中位数也可以)

·························································6分

26. 解：

（1）∵抛物线的对称轴是直线，

∴.  ∴.····················································1分

（2）∵抛物线 的对称轴为直线，

∴.

∴抛物线的顶点坐标为（1，）. ···································3分

（3）答案不唯一：或满足方程 的一组值····························6分

27.（1）如图所示：···········································1分

（2）证明：连接CA.

∵OP平分∠MON，

∴∠AOC∠FOC.

在△AOC和△FOC中，

∴△AOC≌△FOC. ···········································2分

∴ACFC.

∵CE是线段AB的垂直平分线，

∴CBCA.

∴CBCF.···················································3分

（3）.·····················································4分

证明：∵CBCF，

∴∠CFB∠CBF.

∵△AOC≌△FOC，

∴∠CAO∠CFB.

∴∠CAO∠CBF.

∵∠CBO+∠CBF180°，

∴∠CAO+∠CBO180°.

∴∠AOB+∠ACB180°.

∵∠AOB90°，

∴∠ACB90°.

又∵CACB，

∴△ACB是等腰直角三角形.

∴.

∴.

 ··············································7分

28. 解：

（1）点B，点C；············································2分

（2）由题意可知，点P关于⊙O的旋转点形成的图形为以点G（0，2）为圆心，以2个单位长度为半径的⊙G.

当直线与⊙G相切时：

如图1，求得：，

如图2，求得：.

因为直线上存在点P关于⊙O的旋转点，所以，.

·························································5分

图1

图2

（3）.·····················································7分

若考生的解法与给出的解法不同，正确者可参照评分标准相应给分.