初三数学试题答案

初三数学试题答案

一. 选择题(每小题3分,共36分)

4

二. 填空题(每小题3分,共18分)

13. 1.3⨯1014. 三. 解答题(第19~21题各6分,22、23题8分,24、25题10分,,26题12分,共66分) 19.原式=1-1) +2⨯

2

3分

=11 ········································································· 5分 =2. ························································································ 6分 20. 方程两边同乘(x -2) ,得1=-(1-x ) -3(x -2) . 2分

解这个方程,得x =2. ································································ 4分 检验:当x =2时,x -2=0,所以x =2是增根,原方程无解. ·········· 6分

21. (1)特征1:都是轴对称图形;特征2:都是中心对称图形;特征3:这些图形的面积都等于4个单位面积;等 ············································································································· 4分 (2)满足条件的图形有很多,只要画正确一个,都可以得满分. ······················· 6分

22. 所有可能出现的结果如下:

总共有6种结果,(1)所有的结果中,满足A 在甲组的结果有3种,所以A 在甲组的概率是都在甲组的结果有1种,A ,B 都在甲组的概率是

1

,(4分) (2)满足A ,B 2

1

.(8分) 6

3, ∴BH =3. ∴OH =4. ∴5

H O ⊥A 23. 解:(1)如图2,作B

点B 的坐标为(4,3) .(4分)

BO =5,sin ∠BOA =,垂足为H ,在Rt △OHB 中,

AH ∴AH =6.

BH =3(2) 在Rt △AHB 中,BAO =(8= OA =10,OH =4,

AB 5分)

90⨯1

24. (1)180

+

90⨯2180

+

90⨯3180

=

2π(1+2+3) =3π (5分)

F

x

(2)易证 BCG ≅ DCF 可得∠F =∠G

∠F +∠FDC =900 ∴∠G +∠FDC =900 ∴BG ⊥DF (10分)

D

25. (1)设文化衫和相册的价格分别为x 元和y 元,则 ········································· 1分

A

E

⎧x -y =9

····························································································· 3分 ⎨

2x +5y =200⎩

解得⎨

⎧x =35

⎩y =26

答:文化衫和相册的价格分别为35元和26元. ··················································· 5分 (2)设购买文化衫t 件,则购买相册(50-t ) 本,则

1500≤35t +26(50-t ) ≤1530 ······································································· 7分

200230

≤t ≤ 99

·········································· 8分 t 为正整数,∴t =23,24,25,即有三种方案. ·

解得

第一种方案:购文化衫23件,相册27本,此时余下资金293元;

第二种方案:购文化衫24件,相册26本,此时余下资金284元; 第三种方案:购文化衫25件,相册25本,此时余下资金275元; ··························· 9分 所以第一种方案用于购买教师纪念品的资金更充足. ··········································· 10分

26. (1)根据题意,得CD =CB =OA =5,OD =3,

COD =90 ,∴OC =4.

4) ;(4分) ∴点C 的坐标是(0,

(2) AB =OC =4,设AE =x ,

则DE =BE =4-x ,

AD =OA -OD =5-3=2,

在Rt △DEA 中,DE =AD +AE .

2

2

2

∴(4-x ) 2=22+x 2.

解之,得x =

3, 2

即点E 的坐标是 5⎪.

设DE 所在直线的解析式为y =kx +b ,

⎛⎝3⎫2⎭

⎧3k +b =0,⎪∴⎨3

5k +b =,⎪⎩2

3⎧k =,⎪⎪4

解之,得⎨

9⎪b =-.⎪⎩4∴DE 所在直线的解析式为y =

39

x -;(8分) 44

(3) 点C (0,

4) 在抛物线y =2x 2+c 上,∴c =4.

即抛物线为y =2x 2+4.

假设在抛物线y =2x 2+4上存在点G ,使得△CMG 为等边三角形, 根据抛物线的对称性及等边三角形的性质,得点G 一定在该抛物线的顶点上. 设点G 的坐标为(m ,n ) ,

32-3b 2,n =, =∴m ==8

⎛32-3b 2⎫

即点G

的坐标为 ⎪ 8⎪.

⎝⎭

设对称轴x =与直线CB 交于点F ,与x 轴交于点H . 4⎛⎝

⎫,4⎪. ⎪4⎭

则点F

的坐标为 -

b 0,点G 在y 轴的右侧,

32-3b 23b 2=,FH =4,FG =4-.

CF =m =88 CM =CG =2CF =

⎛⎫⎛⎫⎛3b 2⎫222

∴在Rt △CGF 中,CG =CF +

FG , 2⎪⎪= 4⎪⎪+ 8⎪.

⎭⎝⎭⎝⎭⎝

解之,得b =-2.( b

22

2

32-3b 25=. ,n =∴m ==

825⎫

∴点G

的坐标为⎪⎪. 2⎝⎭

5⎫

G ,使得△CMG 为等边三角形.(12分) ∴

在抛物线y =2x +4(b

⎪⎪2⎭⎝

2


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