初中毕业学业与升学考试数学试卷及答案(8.21)

九年级学业与升学考试（新课程）
数 学 试 卷
本试卷满分为150分，考试时间为120分钟．
一、选择题（每小题只有一个正确选项，将正确选项的序号填入题后的括号内．10小题，每小题3分，共30分） 1．2(2)3--的值是（ ）
Ａ．1 Ｂ．2 Ｃ．1- Ｄ．2- 2．下列运算中，不正确的是（ ） Ａ．23ab ab ab += Ｂ．2ab ab ab -= Ｃ．22ab ab ab ⨯=
Ｄ．
1
22
ab ab ÷=
3．如果一个四边形绕对角线的交点旋转90，所得的图形与原来的图形重合，那么这个四边形一定是（ ） Ａ．平行四边形 Ｂ．矩形 Ｃ．菱形 Ｄ．正方形 4．下列物体中，是同一物体的为（ ）
Ａ．（1）与（2） Ｂ．（1）与（3） Ｃ．（1）与（4） Ｄ（2）与（3） 5．“数轴上的点并不都表示有理数，如图中数轴上的点P 所表示的数是5”，这种利用图形直观说明问
题的方式体现的数学思想方法叫（ ） Ａ．代入法
Ｂ．换元法
Ｃ．数形结合的思想方法 Ｄ．分类讨论的思想方法
6．在一幅长60cm ，宽40cm 的矩形风景画的四周镶一条金色纸边，制成一幅矩形挂图，如图所示．如果要使整个挂图的面积是22816cm ，设金色纸边的宽为cm x ，那么x 满足的方程是（ ）
Ａ．(602)(402)2816x x ++= Ｂ．(60)(40)2816x x ++= Ｃ．(602)(40)2816x x ++= Ｄ．(60)(402)2816x x ++=
7．夏天，一杯开水放在桌子上，杯中水的温度()
T C随时间t变化的关系的大致图象是（）
8．如图，
ABC
△的边长都大于
2，分别以它的顶点为圆心，
1为半径画弧（弧的端点分别
在三角形的相邻两边上），则这三条弧的长的和是（）
Ａ．4πＢ．3πＣ．6πＤ．5π
9．一人乘雪橇沿如图所示的斜坡笔直滑下，滑下的距离S（米）与时间t（秒）间的关系式为2
10
S t t
=+，若滑到坡底的时间为2秒，则此人下滑的高度为（）
Ａ．24米Ｂ．12米Ｃ．123米Ｄ．6米
10．某超市（商场）失窃，大量的商品在夜间被罪犯用汽车运走．三个嫌疑犯被警察局传讯，警察局已经掌握了以下事实：（1）罪犯不在甲、乙、丙三人之外；（2）丙作案时总得有甲作从犯；（3）乙不会开车．在此案中，能肯定的作案对象是（）
Ａ．嫌疑犯乙Ｂ．嫌疑犯丙Ｃ．嫌疑犯甲Ｄ．嫌疑犯甲和丙
二、填空题（把答案填在题中的横线上．8小题，每小题4分，共32分）
11．10月12日9时15分，我国“神州六号”载人飞船发射成功．飞船在太空飞行了77圈，路程约3300000千米，用科学记数法表示这个路程为千米．
12．如图中两圆的位置关系是（相交，外切，外离）．
13．反比例函数
2
k
y
x
=的图象的两个分支分别位于象限．
14．下列是三种化合物的结构式及分子式，则按其规律第4个化合物的分子式为．
15．小明将一把钥匙放进自己家中的抽屉中，他记不清到底放进三个抽屉中的哪一个了，那么他一次选对抽屉的概率是．
16．如图，AB CD
∥，EG平分BEF
∠，若260
∠=，则1
∠=．
A．B．C．D．
17．如图，A D AC DF ==∠∠，
，则需要补充条件： （写出一个即可），才能使ABC DEF △
≌△．
18．二次函数2y ax bx c =++图象上部分点的对应值如下表：
x 3- 2- 1-
0 1 2 3 4 y
6 0 4-
6- 6- 4-
6
则使0y <的x 的取值范围为 ．
三、解答题（解答时，必须写出必要的解题步骤．5小题，共38分）
19．（6分）先化简，再求值：23111a a a a a a
-⎛⎫- ⎪-+⎝⎭·，其中22a =-． 20．（8分）有7名同学测得某楼房的高度如下（单位：米）： 29，28.5，30，30，32，31，33．
（1）求这组数据的中位数、众数、平均数；
（2）你认为此楼房大概有多高？
21．（8分）一次函数图象如图所示，求其解析式．
第17题图
第16题图 阅读材料 所谓中位数，就是一组数按从小到大（或从大到小）的顺序排列，位于中间的那个数（当数据个数为奇数时）或者位于中间的两个数的平均数（当数据个数为偶数时）． 所谓众数，就是一组数据中出现次数最多的那个数．
22．（8分）如图，已知AB DC AC DB ==，．求证：12∠=∠．
23．（8分）为节约用电，某学校在本学期初制定了详细的用电计划．如果实际每天比计划多用2度电，那么本学期的用电量将会超过2990度；如果实际每天比计划节约2度电，那么本学期的用电量将不超过2600度．若本学期的在校时间按130天计算，那么学校原计划每天用电量应控制在什么范围内？
四、解答题（解答时，必须写出必要的解题步骤．5小题，共50分） 24．（8分）如图是两个半圆，点O 为大半圆的圆心，AB 是大半圆的弦关与小半圆相切，且24AB =．问：能求出阴影部分的面积吗？若能，求出此面积；若不能，试说明理由． 25．（10分）一架长5米的梯子AB ，斜立在一竖直的墙上，这时梯子底端距墙底3米．如果梯子的顶端沿墙下滑1米，梯子的底端在水平方向沿一条直线也将滑动1米吗？用所学知识，论证你的结论．
26．（10分）某公司现有甲、乙两种品牌的计算器，甲品牌计算器有A B C ，，三种不同的型号，乙品牌计算器有D E ，
两种不同的型号，新华中学要从甲、乙两种品牌的计算器中各选
购一种型号的计算器．
（1）写出所有的选购方案（利用树状图或列表方法表示）；
（2）如果（1）中各种选购方案被选中的可能性相同，那么A型号计算器被选中的概率是多少？
（3）现知新华中学购买甲、乙两种品牌计算器共40个（价格
如图所示），恰好用了1000元人民币，其中甲品牌计算器为A
型号计算器，求购买的A型号计算器有多少个？
27．（12分）如图，在M中，AB所对的圆心角为120，已知圆的半径为2cm，并建立如
图所示的直角坐标系．
（1）求圆心M的坐标；
（2）求经过A B C
，，三点的抛物线的解析式；
（3）点D是弦AB所对的优弧上一动点，求四边形ACBD的最大面积；
（4）在（2）中的抛物线上是否存在一点P，使PAB
△
和ABC
△相似？若存在，求出点P 的坐标；若不存在，
请说明理由．
公司
计算器单价（单位：元）型：60
型：40
型：25
型：50
型：20
张掖市初中毕业学业与升学考试（新课程）
数学试题参考答案与评分标准
一、选择题：每小题3分，共30分
ＡＣＤＢＣＡＢＤＢＣ
二、填空题：每小题4分，共32分
11．6
3.310
⨯12．外离13．一，三14．
410
C H
15．1
3
16．6017．答案不唯一；如：C F
=
∠∠，或AB DE
=，等
18．23
x
-<<
三、解答题：共38分
19．本小题满分6分
解：原式24
a
=+，·······················································································4分
当2 a=
时，原式=．········································································6分
说明：会运用分配律且对者给2分，会用公式21(1)(1)
x x x
-=-+且对者给1分，若先通分且对者给2分．
20．本小题满分8分
解：（1）在这组数据中，中位数是30，·····························································2分众数是30，·································································································4分平均数是30.5；····························································································6分（2）该楼房大概高30米（未写单位不扣分） ····················································8分21．本小题满分8分
解：设一次函数解析式为y kx b
=+， ·······························································1分
则
01
20.
k b
k b
=⨯+
⎧
⎨
-=⨯+
⎩
，
··························································································5分
解得
2
2. k
b
=
⎧
⎨
=-⎩
，
所以，一次函数解析式为22
y x
=-． ·······························································8分
说明：只要求出
2
2.
k
b
=
⎧
⎨
=-
⎩
，
无最后一步不扣分．
22．本小题满分8分
说明：
AB DC
AC DB
BC BC
=
⎧
⎪
=
⎨
⎪=
⎩
，
，
，
······················································································2分
ABC DCB
∴△≌△． ·····················································································4分
A D
∴∠=∠． ······························································································6分
又AOB DOC ∠=∠，
12∴∠=∠． ·
······························································································· 8分 23．本小题满分8分
解：设学校原计划每天用电量为x 度， ······························································ 1分 依题意得130(2)2990130(2)2600.x x +>⎧⎨-⎩
，≤ ··········································································· 5分
解得2122x <≤．
即学校每天的用电量，应控制在21～22度（不包括21度）范围内． ······················ 8分 说明：只要求出2122x <≤，无最后一步不扣分． 四、解答题：共50分 24．本小题满分8分 解法1：
能（或能求出阴影部分的面积）． ····································································· 1分 设大圆与小圆的半径分别为R r ，，··································································· 2分 作辅助线如图所示（作对）， ··········································································· 4分 可得22212R r -=， ······················································································· 6分
221
(ππ)72π2
S R r ∴=-=阴影． ·
···································································· 8分
解法2：
能（或能求出阴影部分的面积）． ····································································· 1分 设大圆与小圆的半径分别为R r ，······································································ 2分 平移小半圆使它的圆心与大半圆的圆心O 重合（如图）． ······································· 3分 作OH AB ⊥于H ，则OH r =，12AH BH ==． ················································· 5分 22212R r ∴-=，
··························································································· 6分 221
π()72π2
S S R r ∴==
-=阴影半圆环． ·
·························································· 8分
25．本小题满分10分 是． ··········································································································· 2分 证明1：
在Rt ACB △中，22354BC AB AC AB BC =
==
-=
，
，米．
·································· 4分 413DC =-=米．·
························································································ 6分 在Rt DCE △中，22354DC DE CE DE DC ===-=，，米． ································· 8分 1BE CE CB =-=．即梯子底端也滑动了1米． ·················································· 10分 证明2：
在Rt ACB △中，22354BC AB AC AB BC ===-=，，米． ·································· 4分 413DC =-=米．·
························································································ 6分 可证Rt Rt ECD ACB △≌△． ··········································································· 8分 4CE AC ∴==米．
1BE CE CB =-=．即梯子底端也滑动了1米． ·
················································· 10分 26．本小题满分10分 解：（1）树状图表示如下：
················································································································· 2分 列表表示如下：
················································································································· 2分 有6种可能结果：()()()()()()A D A E B D B E C D C E ，，，，，，，，，，，． ···························· 3分 说明：用其它方式表达选购方案且正确者，只给1分． （2）因为选中A 型号计算器有2种方案，即()()A D A E ，，，，所以A 型号计算器被选中的概
率是
21
63
=． ·
······························································································· 5分 （3）由（2）可知，当选用方案()A D ，时，设购买A 型号，D 型号计算器分别为x y ，个， 根据题意，得4060501000.x y x y +=⎧⎨+=⎩，解得100140.x y =-⎧⎨=⎩
，
经检验不符合题意，舍去； ············································································ 7分
当选用方案()A E ，时，设购买A 型号、E 型号计算器分别为x y ，个， 根据题意，得4060201000.x y x y +=⎧⎨+=⎩，解得535.x y =⎧⎨=⎩
， ····················································· 9分
甲
A B
C D ()D A ， ()D B ， ()D C ， E ()E A ， ()E B ， ()E C ， 乙 甲品牌 乙品牌
所以新华中学购买了5个A 型号计算器． ························································· 10分 说明：设购买A 型号计算器x 台，D （或E ）型号计算器为(40)x -个，用一元一次方程解答，同样给分．
27．本小题满分12分 解：（1）如图（1），连结MA MB ，． 则120AMB ∠=，
60CMB ∴∠=，30OBM ∠=． ·
······································································ 2分
1
12
OM MB ∴==，(01)
M ∴，． ········································································ 3分 （2）由A B C ，，三点的特殊性与对称性， 知经过A B C ，，三点的抛物线的解析式为2y ax c =+． ·
········································ 4分 1OC MC MO =-=，223OB MB OM =-=，
(01)(30)C B ∴-，，，． ···················································································· 5分
113
c a ∴=-=，．
21
13y x ∴=-．
····························································································· 6分 说明：只要求出1
13
c a =-=，，无最后一步不扣分． （3）
ABC ABD ACBD S S S =+△△四边形，又ABC S △与AB 均为定值， ···························· 7分
∴当ABD △边AB 上的高最大时，ABD S △最大，此时点D 为
M 与y 轴的交点，如图（1）
． ················································································································· 8分
2111
43cm 222
ABC ABD ACBD S S S AB OC AB OD AB CD ∴=+=
+==△△四边形···． ·
···· 9分 （4）方法1：
图（1）
如图（2），ABC △为等腰三角形，
303AB
ABC BC
∠==，，
ABC PAB
∴
△∽△
等价于
302336PAB PB AB PA PB ∠=====，，． ················· 10分 设()P x y ，且0x >，则cos3033323x PA
AO =-=-=·， sin303y PA ==·． ······················································································ 11分
又
(233)P ，的坐标满足2
113
y x =
-， ∴在抛物线2
113
y x =
-上，存在点(233)P ，
， 使ABC PAB △∽△． 由抛物线的对称性，知点(233)-，
也符合题意． ∴存在点P ，它的坐标为(233)，或(233)-，． ··················································· 12分 说明：只要求出(233)，
，(233)-，，无最后一步不扣分．下面的方法相同． 方法2：
如图（3），当ABC PAB △∽△时，30PAB BAC ∠=∠=，又由（1）知30MAB ∠=，
∴点P 在直线AM 上．
设直线AM 的解析式为y kx b =+， 将(30)(01)A M -，，，代入，解得3
1.k b ⎧=
⎪⎨⎪=⎩，
∴直线AM 的解析式为3
1y x =
+． ·
······························································ 10分
图（2）。