辽宁省数学中考试题答案

辽宁省2003年中等学校招生考试
数学试题参考答案及评分标准
一、（选两个或两个以上答案不给分）
1．B 2．C 3．B 4．C 5．C 6．C 7．B 8．D 9．B 10．A 二、
11．x≥1且x≠2 12．3 13．8，7.5 14
．15．七
16．2 17．y2-8y-20=0（或写成y2-20=8y）18．3.6
19．
（1+20．15°或75°
（注：13题错1个扣1分，顺序错不给分；17题写成分式方程不给分；19题不写单位扣1分；20题只写对一解扣1分）
三、
21．
-=- ·································· 2分
x y
x y
+
=
-
···························································································· 4分当x=2，y=3时，原式
23
5
23
+
==-
-
························································· 6分（注：不化简，直接代数求值，按相应步骤给分）
22．如图
（O点找对） ·················································· 3分
（切线画对） ·················································· 6分
（注：不用尺规作图，不给分，没有保留作图痕迹不给分）
A B
M
O
C
23． （1）（频数）12，（频率）0.24 ······························································ 2分 （2）补全频率分布直方图 ··································································· 4分 （3）50···························································································· 6分 （4）80.5～90.5 ················································································· 8分 （5）216人 ····················································································· 10分
（注：（1）中每空1分，（2）中直方图1个1分，（3）中样本容量写单位的扣1分）
四、说明：本题给分点由两部分组成，一部分是图形设计（满分5分），按设计合理性和测量数据多少给分（5分、3分、1分、0分）；另一部分是依据图形计算（满分5分）。

对不同设计方案（如1、2、3），同一图形可能字母标记不同，但只要计算正确即可得5分。

根据不合理方案（如图d 、e ），计算正确给3分。

24．解：
方案1：（1）如图a （测三个数据） ··················· 5分
（2）解：设HG=x 在Rt △CHG 中 CG=x •cot β 在Rt △DHM 中 DM=（x －n ）•cot α
∴x •cot β=（x －n ）•cot α ··························· 8分
∴x=
β
αα
cot cot cot -•n ····································· 10分
方案1图a
方案2：（1）如图b （测四个数据） ··················· 3分 （2）解：设HG=x
在Rt △AHM 中 AM=(x －n) •cot γ 在Rt △DHM 中 DM=（x －n ）•cot α
∴(x －n) •cot γ=（x －n ）•cot α+m ················ 6分
∴x=
α
γα
γcot cot cot cot -•-•+n n m ························· 8分
方案3：（1）如图c （测五个数据） ··················· 1分
（2）参照方案1（2）或方案2（2）给分 ······ 6分
注：①如果在设计和计算中，考虑了测倾器高度，参照以
上标准给分.
②以下两种方案（图d 、e ）或其他与其相似的图形不给分，但如果计算正确给3分．
方案3图c
图d
图e
五、 25．解：
（1）设s 与t 的函数关系式为s=at 2+bt+c
由题意得 1.5422255 2.5a b c a b c a b c ++=-⎧⎪++=-⎨⎪++=⎩ （或 1.54220a b c a b c c ++=-⎧⎪
++=-⎨⎪=⎩
）
解得1220
a b c ⎧=⎪⎪⎨=-⎪⎪=⎩
∴s=21
22t t - ······················································································ 4分
（2）把s=30代入s=21
22t t -
得30=21
22t t -
解得t 1=10，t 2=-6（舍）
答：截止到10月末公司累积利润可达到30万元 ······································· 7分 （3）把t=7代入，得
s=2121
72710.522⨯-⨯== 把t=8代入，得
s=21
828162⨯-⨯= 16-10.5=5.5
答：第8个月公司获利润5.5万元． ······················································ 10分
六、
26．解：设每周参观人数与票价之间的一次函数关系式为y=kx+b
由题意得
107000
154500
k b
k b
+=
⎧
⎨
+=
⎩
····································································· 2分
解得
500
12000 k
b
=-
⎧
⎨
=
⎩
∴y=-500x+12000 ················································································ 4分根据题意，得xy=40000
即x(-500x+12000)=40000 ··································································· 6分x2-24x+80=0
解得x1=20x2=4 ············································································· 8分把x1=20，x2=4分别代入y=-500x+12000中
得y1=2000，y2=10000 ······································································10分因为控制参观人数，所以取x=20，y=2000 ··············································11分答：每周应限定参观人数是2000人，门票价格应是20元．·······················12分（注：其他方法按相应步骤给分）
七、 27．
（1）证明：
①连结BD ∵AB 是⊙O 的直径
∴∠ADB ＝90°
∴∠AGC ＝∠ADB ＝90° 又∵ACDB 是⊙O 内接四边形
∴∠ACG ＝∠B
∴∠BAD ＝∠CAG ······································ 3分
②连结CF
∵∠BAD ＝∠CAG ∠EAG ＝∠FAB ∴∠DAE ＝∠FAC 又∵∠ADC ＝∠F ∴△ADE ∽△AFC ·············································································· 5分
∴
AD AE
=
AF AC
∴AC ·AD ＝AE ·AF ·············································································· 6分
（其他方法相应给分） （2）①图形正确 ················································ 8分
②两个结论都成立，证明如下： ①连结BC ∵AB 是直径
∴∠ACB ＝90°
∴∠ACB ＝∠AGC ＝90°
∵GC 切⊙O 于C
图(a)
B
O A
F
D
C G
E l
·
B
O A ·
E
C (
D )
G F
∴∠GCA＝∠ABC
∴∠BAC＝∠CAG（即∠BAD＝∠CAG）·······10分
②连结CF
∵∠CAG＝∠BAC，∠GCF＝∠GAC
∴∠GCF＝∠CAE，∠ACF＝∠ACG－∠GFC，∠E＝∠ACG－∠CAE
∴∠ACF＝∠E
∴△ACF∽△AEC
∴AC AF
= AE AC
∴AC2＝AE·AF（即AC·AD＝AE·AF） ·········································12分（注：其他方法证明，按相应步骤给分）
图(b)
八、
x 轴、y 轴分别交于点C 、P
∴C （-0），P(0,-8)
∴cot ∠OCD=cot ∠OPC=∴∠OCD=∠OPC
∵∠OPC+∠PCO=90° ∴∠OCD+∠PCO=90°
∴PC 是⊙D 的切线 ······································ 5分 （2）设直线PC 上存在一点E(x,y)，使S △EOP ＝4S △CDO
11
84122
x ⨯⨯=⨯⨯⨯ 解得 x= 由8y =--可知：
当x 时，y =-12， 当x 时，y =-4·············································· 9分 ∴在直线PC 上存在点E ，-12）或（，-4）使S △EOP ＝4S △CDO ······ 10分 （注：只求出一个点，扣2分）
（3）解法一：
作直线PF 交劣弧AC 于F ，交⊙D 于Q ，连结DQ 由切割线定理得：PC 2＝PF ·PQ ····················· ① 在△CPD 和△OPC 中
∵∠PCD ＝∠POC ＝90° ∠CPD ＝∠OPC
∴△CPD ∽△OPC ∴
PC PD
PO PC
＝
⌒
即PC 2＝PO ·PD ········································· ② 由①、②得：PO ·PD ＝PF ·PQ ，又∵∠FPO ＝∠DPQ ∴△FPO ∽△DPQ
PF PD =
FO DQ ，即m 9
=n 3
∴m=3n ··························································································· 13分 （
2<n< ·················································································· 14分
PF 交劣弧AC 于F
F(x,y)，作FM ⊥y 轴，M 为垂足，连结DF ， m 2-(8+y )2=x 2 2-y 2=x 2
m 2-64-16y -y 2=n 2-y 2 m 2-64-16y =n 2 ·········① 32-(1-y )2=x 2 32-(1-y )2=n 2-y 2
y=28
2
n ··············②
②代入①，解得：m=3n ，m=-3n (舍)
∴m =3n ·················································· 13分 (2<n< ············································ 14分 （注：其他方法证明或求解，按相应步骤给分）
⌒。