10-正文-大兴区2020-2021九上期末数学-含答案练习题

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大兴区2020~2021学年度第一学期期末检测初三数学参考答案及评分标准一、选择题（本题共24分，每小题3分）题号12345678答案C ABD BCDB 二、填空题（本题共24分，每小题3分）；     10.  =；          11. （2，-4）； 12.  2；13. ；  14. 答案不唯一，例如：； 15.  ①②③； 16.  1<m<5.三、解答题（本题共52分，第17-21题，每小题5分，第22题6分，第23-25题,每小题7分）解答应写出文字说明、演算步骤或证明过程. 17.解：··········································································4分···········································································································5分 18.解：（1）∵抛物线经过点（1，-4），（0，-3），∴.······················································2分解得.  ∴． ····················································3分（2）令y=0,∴.解得: .∴ 抛物线与x轴的交点坐标是（-1,0），（3,0）. ··················5分     19. 解:（1）补全的图形如图所示：   ···········································2分     （2）60. ·········································································3分 一条弧所对的圆周角等于它所对的圆心角的一半.······················5分 20.解:作DE⊥AC，垂足为E，在Rt△CED中，sinC＝，∵∠C=30º,CD=20,∴DE＝10.        ····················································1分∵cosC＝，∴.∴CE＝10.    ··················································2分∵∠ADB是△ACD的外角，∠ADB=75º,∠C=30º，∴∠CAD＝45°.∴在Rt△ADE中，tan∠EAD＝，∴AE＝10.     ·······················································3分∴AC＝AE+CE＝10+10.   ········································4分∴在Rt△ABC中，sin∠C＝，∴AB＝5+5.答:这棵树AB的高度是(5+5)米. ··································5分  21.解：（1）将点A（1，2）代入中得k=1. ···················································1分将点A（1，2）代入中得m=2.  ·············································2分（2）①当点P在点A下方时,过点A作AG⊥x轴，交直线PQ于点H，∵PQ平行于x轴，∴∆APQ ∽∆ACB∴        ∴∵点A（1，2），∴点P纵坐标为1.，.∴P点坐标为（2，1）. ····································4分②当点P在点A上方时，过点A作AG⊥x轴，交直线PQ于点H.∵PQ平行于x轴，∴∆APQ ∽∆ACB.∴       ∴∵点A（1，2），∴P点纵坐标为3.代入得，∴P点坐标为. ···································5分∴P点坐标为（2，1）或（，3）.  22.(1)证明：连接OF.∵OC=OF，∴∠OCF＝∠OFC.∵四边形ABCD是矩形，∴∠B＝∠D=∠DCB=90°.又∵∠DAF＝∠BAC，∴∠AFD=∠ACB.   ···········································1分∵∠ACB＋∠ACD＝90°，∴∠AFD＋∠OFC＝90°.∴∠AFO＝90°.∴OF⊥AF于F.∴直线AF与⊙O相切．·······································2分 (2)解：∵tan∠DAF＝，∠DAF＝∠BAC，∴tan∠BAC＝.∵∠B =90°，∴tan∠BAC＝＝.∵AB＝4，∴BC＝2.························································3分∴.又∵四边形ABCD是矩形，∴BC=AD=2.又∵∠D＝90°，tan∠DAF＝，∴DF＝AD·tan∠DAF＝2×＝2. ·······················4分∴AF＝2.设⊙O的半径为r，在Rt△AFO中，∠AFO＝90°.∴OA2＝OF2＋AF2.即(2－r)2＝r2＋12. ········································5分解得r＝.∴⊙O的半径为.········································6分23.解：（1）∵，∴.  ·····································1分（2）把b=-2a代入y=ax2+bx+a+1得：y=ax2-2ax+a+1. 　　配方得：y=a（x-1）2+1.　 ∴顶点M（1，1）. ································2分（3）①1个.  ············································3分　　②　由①得，时，区域W内有1个整点.（Ⅰ）当抛物线过（-1，0）时，区域W内恰有3个整点.将（-1，0）代入y=ax2-2ax+a+1,得. ············································4分结合图象可得-.··························5分（Ⅱ）当抛物线过（0，-2）时，区域W内恰有3个整点.将（0，-2）代入y=ax2-2ax+a+1,得.综上所述，a的值范围是或.···············7分    24.（1）AE = BE.  ·························································1分（2）依题意补全图形：   ················································2分  ①AE = BE.        ·························································3分如图，作EM⊥AB于M.∵∠DBC = ∠ABC + ∠ABD = 60°+ ∠ABD，∠EBM = ∠EBD + ∠ABD = 60°+ ∠ABD，∴∠DBC = ∠EBM.在△DBC与△EBM中，∴△DBC≌ △EBM.∴ BC = BM.在△ABC中，∠C = 90°，∠BAC = 30°，∴.∴.∴EM垂直平分AB.∴AE = BE.∴AE = BD.···································································5分② .··················································7分   25.解：（1）①C(4，0)，E(－1，5). ··················································2分②（Ⅰ）当点（4，0）在直线y=kx+3上时，4k+3=0，k=.（Ⅱ）当点（3，1）在直线y=kx+3上时，3k+3=1，k=.（Ⅲ）当点（2，2）在直线y=kx+3上时，2k+3=2，k=.（Ⅰ）                        （Ⅱ）                         （Ⅲ）结合图象可得且.    ························5分（2）·······················································7分