苏科版数学九年级上册期中模拟试卷06（含答案）

这是一份苏科版数学九年级上册期中模拟试卷06（含答案），共11页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
苏科版数学九年级上册期中模拟试卷一、选择题1． 下列方程中，是一元二次方程的是···································【 ▲ 】A．x＋2y＝1 B．x2－2xy＝0 C．x2＋＝3 D．x2－2x＋3＝02． 下列图形中，不是中心对称图形的是·································【 ▲ 】A．正方形 B．正五边形 C．正六边形 D．正八边形3． 已知⊙O的半径为5cm，点A到圆心O的距离OA＝5cm，则点A与⊙O的位置关系为···【 ▲ 】A．点A在圆上 B．点A在圆内 C．点A在圆外 D．无法确定4． 已知直角三角形的两条直角边长分别为6和8，它的内切圆半径是············【 ▲ 】A．2 B．2.4 C．5 D．65． 已知关于x的一元二次方程＝0有一个解为0，则的值为 【 ▲ 】A． B． C． D．6． 如图，点A、B、C、D都在⊙O上，O点在∠D的内部，四边形OABC为平行四边形，则∠ADC的度数为····················································【 ▲ 】A．30° B．45° C．60° D．90°      二、填空题7． 一元二次方程x2＝2x的解为   ▲   ．8． 数据2，3，4，4，5的众数为   ▲   ．9． 圆内接正六边形的一条边所对的圆心角的度数为   ▲   ．10．一只自由飞行的小鸟，如果随意落在如图所示的方格地面上（每个小方格形状完全相同），那么小鸟落在阴影方格地面上的概率是   ▲   ．11．若a是方程x2－x－1＝0的一个根，则2a2－2a＋5＝   ▲   ．12．某药品原价为每盒25元，经过两次连续降价后，售价为每盒16元．若该药品平均每次降价的百分数是x，则可列方程为   ▲   ． 13．如图，正方形ABCD的边长为4，先以点A为圆心，AD的长为半径画弧，再以AB边的中点为圆心，AB长的一半为半径画弧，则两弧之间的阴影部分面积是   ▲   ．（结果保留）14．某种蔬菜按品质分成三个等级销售，销售情况如下表：等级单价（元/千克）销售量（千克）一等5.020二等4.540三等4.040则售出蔬菜的平均单价为   ▲   元/千克．15．如图，从⊙O外一点P引⊙O的两条切线PA、PB，切点分别是A、B，若PA＝8cm，C是上的一个动点（点C与A、B两点不重合），过点C作⊙O的切线，分别交PA、PB于点D、E，则△PED的周长是   ▲   cm．16．如图，四边形ABCD中，AB＝AD，连接对角线AC、BD，若AC＝AD，∠CAD＝76°，则∠CBD＝________°．三、解答题17．解方程：＝0．（用配方法）   18．某公司招聘一名部门经理，对A、B、C三位候选人进行了三项测试，成绩如下（单位：分）：候选人语言表达微机操作商品知识A608070B507080C608065如果语言表达、微机操作和商品知识的成绩按3∶3∶4计算，那么谁将会被录取？        19．如图，沿一条母线将圆锥侧面剪开并展平，得到一个扇形，若圆锥的底面圆的半径r＝2 cm，扇形的圆心角＝120°．（1）求该圆锥的母线长l；（2）求该圆锥的侧面积．      20．一个不透明的口袋中装有2个红球（记为红1、红2），1个白球、1个黑球，这些球除颜色外都相同，将球搅匀．（1）从中任意摸出1个球，恰好摸到红球的概率是   ▲   ；（2）先从中任意摸出一个球，再从余下的3个球中任意摸出1个球，请用画树状图或列表法求两次都摸到红球的概率．     21．甲、乙、丙三位运动员在相同条件下各射靶10次，每次射靶的成绩如下：甲：9，10，8，5，7，8，10，8，8，7；乙：5，7，8，7，8，9，7，9，10，10；丙：7，6，8，5，4，7，6，3，9，5．（1）根据以上数据完成下表： 平均数中位数方差甲88▲乙882.2丙6▲3（2）依据表中数据分析，哪位运动员的成绩最稳定，并简要说明理由．    22．已知△ABC中，∠A＝25°，∠B＝40°．（1）求作：⊙O，使⊙O经过A、C两点，且圆心落在AB边上；（要求：尺规作图，保留作图痕迹，不写作法．）（2）求证：BC是（1）中所作⊙O的切线．          23．已知关于x的一元二次方程x2－2x－m2＝0．（1）求证：该方程有两个不相等的实数根；（2）若该方程有两个实数根为x1，x2，且x1＝2x2＋5，求m的值．                 24．如图，△ABC中，AB＝AC，以AB为直径的⊙O分别与BC、AC交于点D、E，过点D作⊙O的切线DF，交AC于点F．（1）求证：DF⊥AC；（2）若⊙O的半径为4，∠CDF＝22.5°，求阴影部分的面积．        25．小颖妈妈的网店加盟了“小神龙”童装销售，有一款童装的进价为60元/件，售价为100元/件，因为刚加盟，为了增加销量，准备对大客户制定如下促销优惠方案：若一次购买数量超过10件，则每增加一件，所有这一款童装的售价降低1元/件．例如：一次购买11件时，这11件的售价都为99元/件．请解答下列问题：（1）一次购买20件这款童装的售价为   ▲   元/件，所获利润为   ▲   元；（2）促销优惠方案中，一次购买多少件这款童装，所获利润为625元？                26．如图，在扇形AOB中，OA、OB是半径，且OA＝4，∠AOB＝120°．点P是弧AB上的一个动点，连接AP、BP，分别作OC⊥PA，OD⊥PB，垂足分别为C、D，连接CD．（1）如图①，在点P的移动过程中，线段CD的长是否会发生变化？若不发生变化，请求出线段CD的长；若会发生变化，请说明理由；（2）如图②，若点M、N为的三等分点，点I为△DOC的外心．当点P从点M运动到N点时，点I所经过的路径长为__________．（直接写出结果）          27．如图，AB是⊙O的直径，点C，D分别在两个半圆上（不与点A、B重合），AD、BD的长分别是关于x的方程＝0的两个实数根．（1）求m的值；（2）连接CD，试探索：AC、BC、CD三者之间的等量关系，并说明理由；（3）若CD＝，求AC、BC的长．
参考答案一、选择题（每小题3分，共18分）题号123456答案DBAABC二、填空题（每小题3分，共30分）7． x1＝0，x2＝2． 8． 4． 9． 60°． 10．．11．7． 12．25(1－x)2＝16． 13．． 14．4.4．15．16． 16．38°．三、解答题17．（本题满分6分）解：＝．＝．······························································2分＝3．······························································3分＝7．······························································4分∴＝，＝．·························································6分（说明：根写对一个给1分）18．（本题满分7分）解：A的成绩＝＝70（分）；············································2分B的成绩＝＝68（分）；················································4分C的成绩＝＝68（分）．················································6分∵A的成绩最高，∴A将会被录取．·····················································7分19．（本题满分7分）解：（1）由题意，得＝．··············································3分∴＝＝6（cm）．·····················································4分（2）S侧＝＝（cm2）．················································7分20．（本题满分8分）解：（1）．························································3分（2）用表格列出所有可能出现的结果：···································6分 红1红2白球黑球红1 （红1，红球2）（红1，白球）（红1，黑球）红2（红2，红球1） （红2，白球）（红2，黑球）白球（白球，红1）（白球，红2） （白球，黑球）黑球（黑球，红1）（黑球，红2）（黑球，白球） 由表格可知，共有12种可能出现的结果，并且它们都是等可能的，其中“两次都摸到红球”有2种可能．              7分∴P（两次都摸到红球）＝＝．··········································8分21．（本题满分8分）（1）甲的方差为2；··················································3分丙的中位数为6．·····················································6分（2）∵甲的方差＜乙的方差＜丙的方差，而方差越小，数据波动越小，···········7分∴甲的成绩最稳定．··················································8分22．（本题满分8分）（1）解：如答图所示，⊙O就是所要求作的圆．·····························4分     （2）证明：连接OC．∵∠BOC＝2∠A＝50°，∠B＝40°，∴∠BOC＝90°．·····················································6分∴OC⊥BC．·························································7分∴BC是（1）中所作⊙O的切线．·········································8分23．（本题满分10分）（1）证明：∵b2－4ac＝(－2)2－4(－m2)＝4＋4m2．·························2分∵≥0，∴4＋4m2＞0．∴b2－4ac＞0．∴该方程有两个不相等的实数根．········································4分（2）解：由题意，得x1＋x2＝2，x1x2＝ －m2．····························5分又∵x1＝2x2＋5，∴x1＝3，x2＝－1．··················································7分∴－m2＝－3，即m2＝3．解得m＝．··························································8分24．（本题满分10分）（1）证明：连结OD．∵OB＝OD，∴∠ABC＝∠ODB．∵AB＝AC，∴∠ABC＝∠ACB．∴∠ODB＝∠ACB．∴OD∥AC．·························································3分∵DF是⊙O的切线，∴DF⊥OD．∴DF⊥AC．·························································5分（2）连结OE．∵DF⊥AC，∠CDF＝22.5°，∴∠ABC＝∠ACB＝67.5°．∴∠BAC＝45°．·····················································7分∵OA＝OE，∴∠AOE＝90°．∴⊙O的半径为4，∴S扇形AOE＝，S△AOE＝8．············································9分∴S阴影＝S扇形AOE－S△AOE＝－8．·····································10分25．（本题满分12分）解：（1）售价为90；·················································3分利润为600．························································6分（2）设一次购买x件这款童装，所获利润为625元．根据题意，得＝625．····························································9分解得x1＝x2＝25．…………………………………………………………………………11分答：一次购买25件这款童装，所获利润为625元．····························12分26．（本题满分12分）解：（1）线段CD的长不会发生变化．·····································2分连接AB，过O作OH⊥AB于H．∵OC⊥PA，OD⊥PB，∴AC＝PC，BD＝PD．∴CD＝AB． 4分∵OA＝OB，OH⊥AB，∴AH＝BH＝AB，∠AOH＝∠AOB＝60°．····································5分在Rt△AOH中，∵∠OAH＝30°，∴OH＝＝2．························································6分∴在Rt△AOH，由勾股定理得AH＝＝．·····································8分∴AB＝．∴CD＝．···························································9分（2）．···························································12分27．（本题满分14分）解：（1）由题意，得　b2－4ac≥0．∴≥0．化简整理，得　≥0．·················································2分∴≤0，即≤0．······················································3分又∵≥0，∴＝5．····························································4分（2）AC＋BC＝CD．···················································6分理由是：如图，由（1），得　当m＝5时，b2－4ac．∴ AD＝BD．························································7分∵AB是⊙O的直径，∴∠ACB＝∠ADB＝90°．将△ADC绕点D逆时针旋转90°后，得△BDE．∴△ADC≌△BDE．∴∠DAC＝∠DBE．∵∠DAC＋∠DBC＝180°，∴∠DBE＋∠DBC＝180°．∴点C、B、E三点共线．∴△CDE为等腰直角三角形．············································9分∴CE＝CD．即AC＋BC＝CD．·····················································10分（3）由（1），得　当m＝5时，b2－4ac．∴ AD＝BD＝5．∵∠ACB＝∠ADB＝90°，∴AB＝10． 11分∴AC2＋BC2＝102＝100．　①··········································11分由（2）得，AC＋BC＝CD＝7＝14．　②···································12分由① ②解得AC＝6，BC＝8或AC＝8，BC＝6．·······························14分