DA广东省中山市中考真题
2008 年广东省中山市初中毕业生学业考试 数 学 参 考 答 案
一、选择题（每小题 3 分） 1．B； 2．C； 3．D； 4．C； 5．B． 二、填空题（每小题 4 分） 6．2； 7． y
2 ； 8． 9 3 3 ； 9．60； 10．30． x
三、解答题（一） （每小题 6 分） 11．解： 原式
1 1 1 ··········· ··········· ·········· · 3 分 ··········· ·········· ··········· · ·········· ··········· ··········· · 2 2 2 ··········· ··········· ·········· ····· 6 分 ··········· ·········· ··········· ····· ·········· ··········· ··········· ·····
y x 1
2 2 x y 5
12．（本题满分 6 分）解方程
（1） （2）
解：把（1）代入（2）得， x 2 ( x 1) 2 5 ， ······················ 2 分 ··········· ·········· · ·········· ··········· · — 2x 1 5 x 3 ··········· ··········· ···· 分 ··········· ·········· ····· ·········· ··········· ···· 4 把 x 3 代入（1）得， y 2
所以方程组的解为
x 3 ··········· ··········· ···· 分 ··········· ·········· ···· 6 ·········· ··········· ···· y 2
13．解： （1）作图正确得 2 分（不保留痕迹的得 1 分）·················· 2 分 ··········· ······· ·········· ········ （2）在△ABC 中，AB=AC，AD 是△ABC 的中线， ∴ AD⊥ BC， ······································ 3 分 ··········· ·········· ··········· ······ ·········· ··········· ··········· ······
BD CD
1 1 BC 8 4 ． ·························· 分 ··········· ·········· ····· ·········· ··········· ···· 4 2 2
2 2 2
在 Rt△ABD 中，AB＝10，BD＝4， AD BD AB ， ··············· 分 ··········· ··· 5 ·········· ···· ··········· ·········· · ·········· ··········· · AD AB2 BD2 102 42 2 21 ． ······················ 6 分 A
B 图3
C
14．解：由题意得，
y 4 x 5 , 1 y 2 x 4.
解得，
··········· ··········· ·········· ··· 分 ··········· ·········· ··········· ··· ·········· ···········
··········· ·· 1
x 2, y 3.
··········· ··········· ·········· ····· 分 ··········· ·········· ··········· ····· ·········· ··········· ··········· ···· 3
∴ 直线 l1 和直线 l2 的交点坐标是（2，－3） ······················· 分 ． ··········· ·········· · 4 ·········· ··········· · 交点（2，－3）落在平面直角坐标系的第四象限上．·················· 6 分 ··········· ······· ·········· ········ 15．解：设小正方形的边长为 xcm ． ·························· 分 ··········· ·········· ····· ·········· ··········· ···· 1 由题意得， 10 8 4 x 80% 10 8 ．····················· 分 ··········· ·········· ·········· ·········· 3
2
解得， x1 2, x2 2 ． ····························· 4 分 ··········· ·········· ········ ·········· ··········· ········ 经检验， x1 2 符合题意， x2 2 不符合题意舍去． ∴ x 2 ． ··········· ··········· ·········· ······· 分 ··········· ·········· ··········· ······ 5 ·········· ··········· ··········· ······ 答：截去的小正方形的边长为 2cm ． ························ 分 ··········· ·········· ··· ·········· ··········· ·· 6 四、解答题（二） （每小题 7 分） 16．解：设抢修车的速度为 x 千米/时，则吉普车的速度为 1.5x 千米/时． ······· 分 ······· ······ 1 由题意得，
15 15 15 ． ··········· ··········· ······ 3 分 ··········· ·········· ······· ·········· ··········· ······· x 1.5 x 60 解得， x 20 ． ···································· 分 ··········· ·········· ··········· ··· 5 ·········· ··········· ··········· ···
经检验， x 20 是原方程的解，并且 x 20, 1.5 x 30 都符合题意．········ 分 ······· 6 ······· 答：抢修车的的速度为 20 千米/时，吉普车的速度为 30 千米/时． ·········· 分 ·········· ········· 7 17．解： （1）设红球的个数为 x ， ···························· 1 分 ··········· ·········· ··
····· ·········· ··········· ·······
2 0.5 ····························· 2 分 ··········· ·········· ········ ·········· ··········· ········ 2 1 x 解得， x 1 ．
由题意得， 答：口袋中红球的个数是 1． ····························· 分 ··········· ·········· ······· 3 ·········· ··········· ······· （2）小明的认为不对． ································ 分 ··········· ·········· ··········· ·········· ··········· ·········· 4 树状图如下：
开始
白1
白2
黄
红
··········· ··········· ·· 分 ··········· ·········· ·· 6 ·········· ··········· ·· ∴ P (白)
2 1 1 1 ， P (黄) ， P (红) ． 4 2 4 4
∴ 小明的认为不对． ································ 7 分 ··········· ·········· ··········· ·········· ··········· ··········· 18． （1）证明：
CF 平分ACB ，
∴ 1 2 ． ················ 1 分 ··········· ····· ·········· ······ 又∵ DC AC ， ∴CF 是△ACD 的中线， ∴ 点 F 是 AD 的中点． ··········· 2 分 ··········· ·········· · ∵ 点 E 是 AB 的中点， B ∴EF∥ BD， 即 EF∥ BC． ·················3 分 ··········· ······ ·········· ······ （2）解：由（1）知，EF∥ BD， ∴△AEF∽ ABD ， △ ∴
A
E
F
1 2
D
C
SAEF AE 2 ··········· ·········· ·········· ·········· ··········· ·········· ( ) ． ··········· ··········· ········· 4 分 SABD AB
1 AB ， 2
又∵ AE
··········· ········· ·········· ·········· SAEF SABD S四边形BDFE SABD 6 ···················· 5 分 ∴
SABD 6 1 2 ··········· ·········· ········· 6 ·········· ··········· ········· ( ) ， ··········· ··········· ········· 分 SABD 2
∴ SABD 8 ， ∴ ABD 的面积为 8． ······························· 分 ··········· ·········· ········· 7 ·········· ··········· ········· 19．解：过点 A 作 AF⊥ BC，垂足为点 D A F． 在 Rt△ABF 中，∠ B=60° ，AB=6， AF AB sin B ∴ i 1: 3
6 sin 60
3 3．
B C BF AB c
os B F E 6 cos 60 3 ． ··········· ··········· ·········· ······ 分 ··········· ·········· ··········· ····· 2 ·········· ··········· ··········· ·····
∵AD∥ BC，AF⊥ BC，DE⊥ BC， ∴ 四边形 AFED 是矩形， ∴ DE AF 3 3 ， FE AD 4 ． ······················· 分 ··········· ·········· · 3 ·········· ··········· · 在 Rt△CDE 中， i
ED 1 ， EC 3
∴ EC 3ED 3 3 3 9 ，
∴ BC BF FE EC 3 4 9 16 ． ····················· 5 分 ··········· ·········· ·········· ··········· ∴ S梯形ABCD
1 ( AD BC )DE 2 1 (4 16) 3 3 2 ≈ 52.0 ．
答：拦水坝的横断面 ABCD 的面积约为 52.0 面积单位．··············· 7 分 ··········· ···· ·········· ····· 五、解答题（三） （每小题 9 分） 20． （1）证明:因为△= (m 2) 2 4(2m 1) ······················ 分 ··········· ·········· 1 ·········· ··········· = (m 2) 2 4 ··························· 3 分 ··········· ·········· ······ ·········· ··········· ······ 所以无论 m 取何值时， △>0，所以方程有两个不相等的实数根． （2）解：因为方程的两根互为相反数，所以 x1 x2 0 ， ·············· 分 ··········· ·· 5 ·········· ··· 根据方程的根与系数的关系得 m 2 0 ，解得 m 2 ， ·············· 7 分 ··········· ··· ·········· ···· 所以原方程可化为 x 5 0 ，解得 x1 5 ， x2 5 ············· 9 分 ··········· ·· ·········· ···
2
21．解： （1）如图 7． C ∵△BOC 和△ABO 都是等边三角形， B 且点 O 是线段 AD 的中点， 5 E ∴OD=OC=OB=OA，∠ 1=∠ 2=60° ···· 分 ， ···· ··· 1 ∴∠ 4=∠ 5． 3 又∵ 4+∠ ∠ 5=∠ 2=60° ， 1 2 6 4 D ∴∠ 4=30° ················· 2 分 ． ··········· ······ ·········· ······· A O 同理，∠ 6=30° ················ 分 ． ··········· ····· ·········· ····· 3 图7 B ∵∠ AEB=∠ 4+∠ 6， ∴∠ AEB=60° ················ 分 ． ··········· ···· 4 ·········· ····· 5 （2）如图 8． C ∵△BOC 和△ABO 都是等边三角形， E 7 8 ∴OD=OC， OB=OA，∠ 1=∠ 2=60° ······ 5 分 ， ······ ······ 3 2 6 1 又∵ OD=OA， O ∴OD＝OB，OA＝OC， 图
8 4 ∴∠ 4=∠ 5，∠ 6=∠ 7． ············· 6 分 ··········· ·· ·········· ··· ∵∠ DOB=∠ 1+∠ 3， D ∠ AOC=∠ 2+∠ 3， ∴ DOB=∠ ∠ AOC． ····································· 分 ··········· ·········· ··········· ···· 7 ·········· ··········· ··········· ···· ∵∠ 4+∠ 5+∠ DOB=180° ∠ ， 6+∠ 7+∠ AOC=180° ， ∴2∠ 5=2∠ 6， ∴∠ 5=∠ ········································· 分 6． ········································· ·········· ··········· ··········· ········ 8 又∵∠ AEB=∠ 5， ∠ 8-∠ 8=∠ 2+∠ 6， ∴∠ AEB＝∠ 2＋∠ 5－∠ 5＝∠ 2， ∴∠ AEB＝60° ······································ 9 分 ． ··········· ·········· ··········· ······ ·········· ··········· ··········· ······
A
22．解： （1） 4 3 ， 4 3 ， ································· 分 ··········· ·········· ··········· 1 ·········· ··········· ··········· 等腰； ·································2 分 ··········· ·········· ··········· · ·········· ··········· ··········· · （2）共有 9 对相似三角形． （写对 3－5 对得 1 分，写对 6－8 对得 2 分，写对 9 对得 3 分） ① DCE、 △ △ABE 与△ACD 或△BDC 两两相似， 分别是： △DCE∽ ABE， △ △DCE∽ ACD， △ △DCE∽ BDC，△ABE∽ ACD，△ABE∽ BDC； △ △ △ （有 5 对） ② ABD∽ EAD，△ABD∽ EBC； △ △ △ （有 2 对） ③ BAC∽ EAD，△BAC∽ EBC； △ △ △ （有 2 对） 所以，一共有 9 对相似三角形．······························· 分 ··········· ·········· ·········· ·········· ··········· ········· 5 （3）由题意知，FP∥ AE， y ∴∠ 1＝∠ PFB， 又∵∠ 1＝∠ 2＝30° ， ∴∠ PFB＝∠ 2＝30° ， ∴FP＝BP．············· 6 分 ··········· ·· ·········· ··· D C H 过点 P 作 PK⊥ 于点 K， FB 则 FK BK
1 FB ． 2 1 (8 t ) ． 2
1 A F
E P 2
∵AF＝t，AB＝8， ∴FB＝8－t， BK
图10
K
B
G
x
在 Rt△BPK 中， PK BK tan 2
1 3 ·········· 7 ·········· (8 t ) tan 30 (8 t ) ． ··········· 分 2 6
∴△FBP 的面积 S
1 1 3 FB PK (8 t ) (8 t ) ， 2 2 6
∴S 与 t 之间的函数关系式为：
S
3 3 2 4
16 (t 8 2 ，或 S ) t t 3 ． ··········· ········ 分 ··········· ······· 8 ·········· ········ 12 12 3 3
t 的取值范围为： 0 ≤ t 8 ． ····························· 9 分 ··········· ·········· ········ ·········· ··········· ········