2021丰台初三数学二模(答案).pdf

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）题号12345678答案DAABDCBC二、填空题（本题共16分，每小题2分）9.x110.511.答案不唯一，如：5440050012.13.14.3xx3015.0m116.（1）是；（2）2025三、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）2xyx2y17.解：原式22312，4分20.解：原式，·····2分2xyxy2分x22.··················5.···················3分xy∵x2y，2x3x6,①2y18.解：2x5∴原式2.······················5分x1.②2yy3解不等式①得：x3，··············2分解不等式②得：x8.················4分（）如图所示，∴原不等式组的解集为x3.···········5分21.1P19.证明：∵∠BAE∠DAC，ACBl∴∠BAE+∠EAC∠DAC+∠EAC，Q即：∠BAC∠DAE.·····················1分在△BAC和△DAE中，················································2分（2）AQ,BQ.······························4分ABAD经过半径的外端并且垂直于这条半径BACDAE的直线是圆的切线.·······················5分ACAE∴△BAC≌△DAE.······················4分22.（1）证明：∵AE∥BC，CE∥AD，∴∠C∠E.································5分∴四边形ADCE是平行四边形.·······1分1∵∠BAC90°，AD是BC边上的中∴OP⊥AC，ADCD.线，∴PCPA.∴ADBDCD.∴∠PCA∠PAC.·······················1分∴四边形ADCE是菱形.···············2分∵PA是⊙O的切线，OA为半径，（2）解：过点E作EH⊥BA交BA的延∴OA⊥PA.··································2分长线于点H.∴∠PAC+∠BAC90°.在Rt△ABC中，∠ABC30°，AC2，∵AB是直径，∴∠ACB90°.∴BC4，AB422223.∴∠ABC+∠BAC90°.1∴ADBC2.∴∠PCA∠ABC.························3分2∵四边形ADCE是菱形，P∴AEAD2.∵AE//BC，C∴∠EAH∠ABC30°.D在Rt△AEH中，EH1，AH3.AOB∴HBAH+AB33.·················4分在Rt△BEH中，（2）解：∵OP⊥AC，∴∠PAC+∠APO90°.BE12(33)227.··············5分∵∠PAB90°，∴∠PAC+∠BAC90°.C∴∠APO∠BAC.······················4分D1E∵BC4，tan∠APO，2∴AC8.HAB∴AB45.∴AO25.······························5分23.解：（1）∵点A（-1，n），B（2，-1）m∴AP45.································6分在反比例函数y(m0)的图象上，xm12m2∴，解得：.········4分mn2n1（2）a2.·······························6分24.（1）证明：25.（1）∵D是AC中点，OP过⊙O的圆心，2（2）证明：连接CA.频数（学生人数）12∵OP平分∠MON，1110∴∠AOC∠FOC.98在△AOC和△FOC中，76OAOF54AOCFOC32OCOC105060708090100成绩/分∴△AOC≌△FOC.······················2分m91.5；······························2分∴ACFC.11∵CE是线段AB的垂直平分线，（2）解：300165（人）；·····4分20∴CBCA.85.2408520∴···································