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中考数学得分专练24一、填空题(本大题每个空格1分,共18分.把答案填在题中横线上)1.2的相反数是,13的绝对值是,立方等于64的数是.2.点(12)A,关于x轴对称的点的坐标是;点A关于原点对称的点的坐标是.3.若30∠,则∠的余角是°,cos.4.在校园歌手大赛中,七位评委对某位歌手的打分如下:9.8,9.5,9.7,9.6,9.5,9.5,9.6,则这组数据的平均数是,极差是.5.已知扇形的半径为2cm,面积是24cm3,则扇形的弧长是cm,扇形的圆心角为°.6.已知一次函数ykxb的图象经过点(02)A,,(10)B,,则b,k.7.如图,已知DEBC∥,5AD,3DB,9.9BC,50B∠,则ADE∠°,DE,ADEABCSS△△.8.二次函数2yaxbxc的部分对应值如下表:x…320135…y…708957…二次函数2yaxbxc图象的对称轴为x,2x对应的函数值y.二、选择题(下列各题都给出代号为A,B,C,D的四个答案,其中有且只有一个是正确的,把正确答案的代号填在题后()内,每小题2分,共18分)9.在下列实数中,无理数是()A.13B.C.16D.22710.在函数12yx中,自变量x的取值范围是()A.2xB.2x≤C.2xD.2x≥11.下列轴对称图形中,对称轴的条数最少的图形是()A.圆B.正六边形C.正方形D.等边三角形12.袋中有3个红球,2个白球,若从袋中任意摸出1个球,则摸出白球的概率是()(第7题)ABCDEA.15B.25C.23D.1313.如图,图象(折线OEFPMN)描述了某汽车在行驶过程中速度与时间的函数关系,下列说法中错误..的是()A.第3分时汽车的速度是40千米/时B.第12分时汽车的速度是0千米/时C.从第3分到第6分,汽车行驶了120千米D.从第9分到第12分,汽车的速度从60千米/时减少到0千米/时14.下面各个图形是由6个大小相同的正方形组成的,其中能沿正方形的边折叠成一个正方体的是()15.小明和小莉出生于1998年12月份,他们的出生日不是同一天,但都是星期五,且小明比小莉出生早,两人出生日期之和是22,那么小莉的出生日期是()A.15号B.16号C.17号D.18号16.若二次函数222yaxbxa(ab,为常数)的图象如下,则a的值为()A.2B.2C.1D.217.如图,在ABC△中,10AB,8AC,6BC,经过点C且与边AB相切的动圆与CACB,分别相交于点PQ,,则线段PQ长度的最小值是()A.4.75B.4.8C.5D.42三、解答题(本大题共2小题,共18分.解答应写出演算步骤)18.(本小题满分10分)化简:(1)02229;(2)24142xx.(第13题)速度/(千米/时)时间/分604020O36912EFPMNA.B.C.D.(第16题)yOAx(第17题)ABCQP19.(本小题满分8分)解方程:(1)341xx;(2)2220xx.20.(本小题满分5分)已知,如图,在ABCD中,BAD∠的平分线交BC边于点E.求证:BECD.21.(本小题满分7分)已知,如图,延长ABC△的各边,使得BFAC,AECDAB,顺次连接DEF,,,得到DEF△为等边三角形.求证:(1)AEFCDE△≌△;(2)ABC△为等边三角形.(第20题)ABCDE(第21题)ABCDEF一、填空题(每个空格1分,共18分)1.2,13,4;2.(12),,(12),;3.60,32;4.9.6,0.3;5.43,120;6.2,2;7.50,6.6,49;8.1,8.二、选择题(本大题共9小题,每小题2分,共18分)题号91011121314151617答案BCDBCCDDB三、解答题(本大题共2题,第18题10分,第19题8分,共18分.解答应写出演算步骤)18.解:(1)原式1134············································································3分74.···············································································5分(2)原式42(2)(2)(2)(2)xxxxx··························································2分42(2)(2)xxx···············································································3分(2)(2)(2)xxx···············································································4分12x.·····················································································5分19.解:(1)去分母,得344xx.······························································1分解得,4x.·······························································································2分经检验,4x是原方程的根.原方程的根是4x.··················································································4分(2)2(1)3x,························································································2分13x.·······························································································3分113x,213x.······································································4分21.证明:四边形ABCD是平行四边形,ADBC∥,ABCD.DAEBEA∠∠.·····················································································1分AE平分BAD∠,BAEDAE∠∠.························································2分BAEBEA∠∠.······················································································3分ABBE.·······························································································4分又ABCD,BECD.········································································5分21.证明:(1)BFAC,ABAE,FAEC.·····································1分DEF△是等边三角形,EFDE.····························································2分又AECD,AEFCDE△≌△.·····························································4分(2)由AEFCDE△≌△,得FEAEDC∠∠,BCAEDCDECFEADECDEF∠∠∠∠∠∠,DEF△是等边三角形,60DEF∠,60BCA∠,同理可得60BAC∠.·························································5分ABC△中,ABBC.···············································································6分ABC△是等边三角形.················································································7分
本文标题:中考数学得分专练24
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