重点中学——星华学校初三数学模拟试卷说明：全卷共4页，考试时间为100分钟，满分120分．请在答题卡上作答.（出卷者：倪迁华） 一、选择题（本大题10小题，每小题3分，共30分）在每小题列出的四个选项中，只有一个是正确的，请把答题卡上对应题目所选的选项涂黑． 1．有理数51-的绝对值为（ ） A ．51 B ．5- C ．51- D ．5 2．我们虽然把地球称为“水球”，但可利用淡水资源匮乏．我国淡水总量仅约为899000亿米3，用科学记数法表示这个数为（ ） A ．0.899×104亿米3B ．8.99×105亿米3C ．8.99×104亿米3D ．89.9×104亿米33．下列图形中对称轴只有两条的是（ ）A ．圆B ．等边三角形C ．矩形D ．等腰梯形 4．计算：322-=（ ）A ．3B ．22C ．2D ．425．已知等腰三角形的一个底角等于30°，则这个等腰三角形的顶角等于（ ） A 、150° B 、120° C 、75° D 、30° 6．如图所示的几何体的正视图是（ ）7．如图，直线AB ∥CD ，∠A ＝70︒，∠C ＝40︒，则∠E 等于（ ）A ．30° Ｂ．40° C ．60° Ｄ．70°8．袋子内有3个红球和2个蓝球，它们只有颜色上的区别，从袋子中随机地取出一个球，取出红球的概率是（ ） A ．52 B ．32 C ．53 D ．239．计算223)2(a a --的结果是（ ）A ．2a - B ．25a C ．25a - D ．2a 10．如图，在等腰梯形ABCD 中，BC ∥AD ，AD =5，DC =4，DE ∥AB 交BC 于点E ，且EC =3，则梯形ABCD 的周长是（ ） A ．21 B ．25 C ．26 D ．20ACBD E第7题图第10题图二、填空题（本大题6小题，每小题4分，共24分）请将下列各题的正确答案填写在答题卡相应的位置上．11．分解因式：m mn mn 962++＝___ ___．12．已知正比例函数)0(≠=k kx y ，点（2，﹣3）在函数上，则y 随x 的增大而 （增大或减小）． 13．如图，AB 为⊙O 的直径，CD 为⊙O 的一条弦， CD ⊥AB ，垂足为E ，已知CD =6，AE =1， 则⊙O 的半径为 ．14．在学校艺术节文艺汇演中，甲、乙两个舞蹈队队员的身高的方差分别是S 甲2=1.5，S 乙2=2.5，那么身高更整齐的是 队（填“甲”或“乙”）．15．不等式组：⎪⎩⎪⎨⎧->+≥-1230211x x 的解集是 ． 16．观察下列图形的排列规律（其中▲、■、★分别表示三角形、正方形、五角星），若第一个图形是三角形，则第18个图形是 ．（填图形名称）▲■★■▲★▲■★■▲★▲■★■▲★▲■★三、解答题（一）（本大题3小题，每小题6分，共18分）17．计算：︒-+-+-60sin 6272)12(118先化简，再求值：a －1a ＋2·a 2＋2a a 2－2a ＋1÷1a 2－1，其中a 为整数且－3＜a ＜2.19．如图，Rt △ABC 的斜边BC =8，AC =6（1）用尺规作图作AB 的垂直平分线l ，垂足为D ，（保留作图痕迹，不要求写作法、证明）；（2）连结D 、C 两点，求CD 的长度．四、解答题（二）（本大题3小题，每小题8分，共24分）20．如图，某同学在楼房的A 处测得荷塘的一端B 处的俯角为30︒，荷塘另一端D 处与C 、B 在同一条直线上，已知AC=32米，CD=16米，求荷塘宽BD 为多少米？(取3 1.73≈，结果保留整数)第13题图C BA第19题图第20题图21．已知一元二次方程012=+++q px x 的一根为 2． （1）求q 关于p 的关系式； （2）若q p 2=，求方程的另一根；（3）求证：抛物线q px x y ++=2与x 轴有两个交点．22．某中学为了解本校学生对球类运动的爱好情况，采用抽样的方法，从乒乓球、羽毛球、篮球和排球四个方面调查了若干名学生，在还没有绘制成功的“折线统计图”与“扇形统计图”中，请你根据已提供的部分信息解答下列问题。

（1）在这次调查活动中，一共调查了 名学生，并请补全统计图； （2）“羽毛球”所在的扇形的圆心角是 度；（3）若该校有学生1200名，估计爱好乒乓球运动的约有多少名学生？五、解答题（三）（本大题3小题，每小题9分，共27分）23．如图，在平面直角坐标系xOy 中，函数)0(4>=x xy 的图象与一次函数k kx y -=的图象交点为A （m ，2）. （1）求一次函数的解析式；（2）设一次函数k kx y -=的图象与y 轴交于点B ，若P 是x 轴上一点，且满足△PAB 的面积是4，直接写出P 的坐标．24．如图1，在△ABC 和△EDC 中，AC =CE =CB =CD ，∠ACB =∠ECD =90，AB 与CE 交于F ，ED 与AB 、BC 分别交于M 、H ．（1）求证:CF =CH ；（2）如图2，△ABC 不动，将△EDC 绕点C 旋转到∠BCE =45时，试判断四边形ACDM 是什么四边形？并证明你的结论．25． 已知，如图，在平面直角坐标系中，Rt △ABC 的斜边BC 在x 轴上，直角顶点A 在y 轴的正半轴上，A （0，2），B （－1，0）。

（1）求点C 的坐标；（2）求过A 、B 、C 三点的抛物线的解析式和对称轴；（3）设点P （m ，n ）是抛物线在第一象限部分上的点，△PAC 的面积为S ，求S 关于m 的函数关系式，并求使S 最大时点P 的坐标．A（图1） （图2）（24题图）九年级数学第一次模拟题参考答案和评分标准一、ABCBB DACDA二、11、2(3)m n + 12、减小 13、5， 14、甲 15、21≤<-x 16、五角星． 三、解答题（一）（本大题3小题，每小题5分，共15分） 17．解：原式=3333211-+-··································································· 4分 =21 ··························································································· 5分 18．解：设原计划平均每亩产量是x 万斤根据题意得：205.193636=+-xx ······························································· 2分 解得：3.0=x ····················································································· 4分 经检验：3.0=x 是原方程的根 45.05.1=x答：改良前亩产0.3万斤，改良后亩产0.45万斤． ······································· 5分 19．解：（1）作图正确（不保留痕痕迹的得1分）， ··········································· 3分（2）因为在Rt ABC △中，BC =8，AC =6∴ 1022=+=AC BC AB ， ························································ 4分 ∴521==AB CD ····································································· 5分 四、解答题（二）（本大题3小题，每小题8分，共24分）20．解： 如图，依题意得： ∠BAC =60°， ························································ 2分在Rt △ABC 中，∵tan ∠BAC =ACBC, ················································· 3分 ∴BC =tan 32⨯60°332= ····································································· 6分∴荷塘宽3916332≈-=-=CD BC BD （米） ····································· 7分答：约荷塘宽BD 约为39米 ·································································· 8分 21．解：（1）∵A （m ，2）在函数)0(4>=x xy 的图象上 ∴m42=， 2=m ······································································ 2分 ∴A （2，2）∵A （2，2）一次函数k kx y -=的图象上∴k k -=22，2=k ···································································· 3分 ∴一次函数的解析式为：22-=x y ························································· 4分 （2）1(1,0)P -，2(3,0)P ·········································································· 8分 22．解：（1）200 ························································································ 2分∵喜欢篮球的人数：200×20%=40（人） 喜欢羽毛球的人数：200-80-20-40=60（人）喜欢排球的20人，应占10020020⨯℅=10℅ 喜欢羽毛球的应占统计图的1－20%－40%－10%=30%∴根据以上数据补全统计图： ·································································· 4分（2）108° ···························································································· 6分 （3）该校1200名学生中估计爱好乒乓球运动的约有：40%×1200=480（人） ····································································· 8分 五、解答题（三）（本大题3小题，每小题9分，共27分） 23．解:（1）2=x 代入方程012=+++q px x 得：01222=+++q p ·············································································· 1分52--=p q ························································································ 2分 （2）若q p 2=，则⎩⎨⎧--==522p q q p ，∴⎩⎨⎧-=-=12q p ········································· 3分原方程变为：022=-x x ······································································ 4分 ∴01=x ，22=x方程的另一根为0 ·················································································· 5分 （3）∵208)52(44222++=---=-=∆p p p p q p ······························ 6分 04)4(2>++=p ··································································· 7分 ∴方程02=++q px x 有两个不等的实根 ·················································· 8分 ∴抛物线q px x y ++=2与x 轴有两个交点． ············································ 9分24．解: （1）证明:在△ACB 和△ECD 中 ∵∠ACB =∠ECD = 90∴∠1+∠ECB =∠2+∠ECB ,∴∠1=∠2 ········································································ 1分又∵AC =CE =CB =CD ,∴∠A =∠D = 45 ····································································· 2分∴△ACF ≌△DCH , ···························································· 3分∴CF =CH ············································································· 4分（2）答: 四边形ACDM 是菱形 ······························································· 5分 证明: ∵∠ACB =∠ECD = 90, ∠BCE =45∴∠1=45, ∠2=45 ···················································· 6分又∵∠E =∠B = 45,∴∠1=∠E , ∠2=∠B ······················································· 7分 ∴AC ∥MD , CD ∥AM ,∴ACDM 是平行四边形 ···················································· 8分 又∵AC =CD , ∴ACDM 是菱形 ······································· 9分25．解：（1）∵A （0，2），B （－1，0），∴OA=2，OB=1。