(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

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

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