发布时间 : 星期日 文章(云南名校试卷合集)2019届九年级初三数学期中考试卷16份word文档合集更新完毕开始阅读b100fd6c5627a5e9856a561252d380eb629423f1
25．（本题满分12分）
小颖妈妈的店加盟了“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．（本题满分12分）
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AB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．（本题满分14分）
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
NBDDIPBPMCO图①
（第26题图）
CAAO图②
1的长分别是关于x的方程x2?102x?(m2?10m?225)＝0的两个实数根．
4（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＝72，求AC、BC的长．
DAOC（第27题图）
B九年级上学期数学期中考试试题答案
一、选择题（每小题3分，共18分）
题号 答案 1 D 2 B 3 A 4 A 5 B 6 C 二、填空题（每小题3分，共30分） 7． x1＝0，x2＝2． 8． 4． 11．7． 15．16． 三、解答题
17．（本题满分6分）
解：x2?4x＝?1．
x2?4x?4＝?1?4． ···············································································2分
9． 60°． 10．
1． 412．25(1－x)2＝16． 13．2?． 16．38°．
14．4.4．
·························································································3分 (x?2)2＝3． ·
x?2＝7． ·····························································································4分
∴x1＝2?3，x2＝2?3． ····································································6分 （说明：根写对一个给1分） 18．（本题满分7分）
解：A的成绩＝B的成绩＝C的成绩＝
60?3?80?3?70?4＝70（分）； ··········································2分
3?3?450?3?70?3?80?4＝68（分）； ················································4分
3?3?460?3?80?3?65?4＝68（分）． ················································6分
3?3?4∵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2πr＝
120πl． ····························································3分 180∴l＝3r＝6（cm）． ················································································4分 120π?62（2）S侧＝＝12π（cm2）． ·····························································7分
36020．（本题满分8分）
解：（1）
1． ·························································································3分 2（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2 白球 黑球 （红1，黑球） （红2，黑球） （白球，黑球） （红1，红球2） （红1，白球） （红2，白球） （红2，红球1） （白球，红1） （黑球，红1） （白球，红2） （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）＝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 ∵m2≥0， ∴4＋4m2＞0．
（第22题答图）
21＝． ···························································8分 126CAOB