资料简介
钦州市 2006 年秋季学期期中考试
八年级数学
题 号
一 二 三
总 分
1~10 11~18 19 20 21 22 23 24 附加题
得 分
说明:1.可以使用计算器,但未注明精确度的计算问题不得采取近似计算,建议根据题型特
点把握好使用计算器的时机.
2.本试卷满分 120 分,在 120 分钟内完成.相信你一定会有出色的表现!
一、填空题:本大题共 10 小题;每小题 2 分,共 20 分.请将答案填
写在题中的横线上.
1.化简: 1 = , 4 = .
2.比较大小: 2 3
2
(用“>”或“<”填空).
3.对 398.15 取近似值,精确到十分位是 .
4.如图的格点三角形 ABC 是 三角形.
5.在△ABC 中,∠A=60°,∠B= 时,△ABC 是等边三角形.
6.在△ABC 中,∠C=90°,若 AB=13 cm,BC=5 cm,则 AC= cm.
7. 是从镜子中看到的一串数字,这串数字应为 .
8.如图,在△ABC 中,∠C=90°,AD 平分∠BAC,若
CD=6 cm,则点 D 到 AB 的距离是 cm.
9.如图,等腰梯形 ABCD 中,AD∥BC,AD=3,AB=6,BC=8.
且 DE∥AB,则△DEC 的周长是 .
10.如图,△ABC 是等腰直角三角形,D 是斜边 BC 上的中点,
△ABD 绕点 A 旋转到△ACE 的位置,恰好与△ACD 组成正方
形 ADCE,则△ABD 按 时针方向旋转了 .
得 分 评卷人
二、选择题:本大题共 8 小题;每小题 3 分,共 24 分.在每小题给出
的四个选项中,只有一项是正确的,请将正确答案前的字母填入题后
的括号内.每小题选对得 3 分,选错,不选或多选均得零分.
11.近似数 2.023 的有效数字是··································································· ( ).
(A)1 个 (B)2 个 (C)3 个 (D)4 个
12.······································································································· 144 的
平方根是···························································································( ).
(A)±12 (B)12 (C)-12 (D)± 12
13.······································································································· 下列各数
组中,不是勾股数的是········································································ ( ).
(A)5,12,13 (B)7,24,25
(C)8,12,15 (D)3k,4k,5k(k 为正整数)
14.······································································································· 用刻度尺
分别画下列图形的对称轴可以不用刻度尺上的刻度画的是···························( ).
(A)①④ (B)②③ (C)③④ (D)①②
15.······································································································· 在 4,3.5,
π, 7 , 9 五个数中,无理数有··························································( ).
(A)1 个 (B)2 个 (C)3 个 (D)4 个
16.某数的平方根等于它的立方根,则这个数················································( ).
(A)1 (B)-1 (C)0 (D)以上都不对
17.······································································································· 下列条件
中,能说明四边形 ABCD 是平行四边形的是············································ ( ).
(A)∠A=30°,∠B=150°,∠C=30°,∠D=150°
(B)∠A=60°,∠B=60°,∠C=120°,∠D=120°
(C)∠A=60°,∠B=90°,∠C=60°,∠D=150°
(D)∠A=60°,∠B=70°,∠C=110°,∠D=120°
得 分 评卷人
18.······································································································· 下列说法
中,正确的是·····················································································( ).
(A)在成中心时称的图形中,连结对称点的线段不一定都经过对称中心
(B)在成中心对称的图形中,连结对称点的线段都被对称中心平分
(C)若两个图形的对应点连成的线段都经过某一点.那么这两个图形一定关于这一点成中心对称
(D)以上说法都正确
三、解答题:本大题共 6 小题,共 76 分.解答应写出文字说明或演算步骤.
19.(本小题满分 16 分)
计算:(1)2+ 2 - 3 (精确到百分位);
(2) 5 +π(保留 2 个有效数字).
20.(本小题满分 12 分)
如图,一个长方形的运动场,有一个球落到了点 C,小明要从点 A 走到点 C 捡球,至少
要走多少米?
得分 评卷人
得分 评卷人
A
BC
D
40m
30m
21.(本小题满分 12 分)
在 5×7 的方格纸上,任意选出 5 个小方块,涂上颜色,使着色图形为轴对称图形,并画
出其对称轴(要求:着色出一个图形,并画出一条对称轴即可).
22.(本小题满分 12 分)
△ABC 的边 AB 绕点 B 旋转到图中 BA′的位置,点 A′是 A 的对应点,画出△ABC 绕
点 B 旋转后的图形.
得分 评卷人
得分 评卷人
A
B
C
A
23.(本小题满分 12 分)
如图,在△ABC 中,AB=AC,∠ABD=∠ACE,BD,CE 相交于点 O,猜想:BO=CO
成立吗?并说明理由.
24.(本小题满分 12 分)
现有一块等腰直角三角形木板,你能通过剪切,使它成为一个含有 45°角的平行四边形
吗?请你设计一个最简的方案,并说明你的方案的正确性.
得分 评卷人
得分 评卷人
A
B C
DE
O
A
BC
附加题.(本小题满分 10 分)
请你把上面的解答再认真地检查一遍,别留下什么遗憾,并估算一下成绩是否达到了 80
分,如果你的全卷得分低于 80 分,则本题的得分将计入全卷总分,但计入后全卷总分最多不
超过 80 分;如果你全卷得分已经达到或超过 80 分,则本题的得分不计入全卷总分.
求下列各式中 x 的值:
(1)x2= 9
25
;
(2)x3=-0.125.
得分 评卷人
钦州市 2006 年春季学期期中考试
八年级数学参考答案及评分标准
一、填空题:(每小题 2 分,共 20 分)
1.1,2;2.<;3.398.2;4.等腰;5.60°;6.12; 7.81007;8.6;
9.17;10.逆,90°或填顺,270°.
二、选择题:(每小题 3 分,共 24 分)
题号 11 12 13 14 15 16 17 18
答案 D D C A B C A B
三、解答题:
19.解:(1)2+ 2 - 3 =2+1.414-1.732······················································4 分
≈1.68;····································································8 分
(2) 5 +π=2.24+3.14······································································· 4 分
≈5.4.··············································································8 分
20.解:连结 AC,在 Rt △ABC 中,由勾股定理得
AC2=AB2+BC2·················································································· 4 分
=302+402·····················································································6 分
=2500.·······················································································8 分
AC= 2500 =50.············································································ 10 分
答:小明至少要走 50 米.··································································· 12 分
21.解:图形有多种,如:
(每个图各 4 分)
22.解:如右图.···························12 分
A
B
C
A
C
23.解:成立.
在△ABC 中,因为 AB=AC,所以∠ACB=∠ABC.································· 4 分
理由是:等边对等角.········································································· 6 分
因为∠ABD=∠ACE,所以∠OBC=∠OCB.·········································· 8 分
在△OBC 中,因为∠OBC=∠OCB,所以 OB=OC.·······························10 分
理由是:等角对等边.········································································12 分
24.解:作等腰直角三角形底边上的中线 CD,交 AB 于点 D,将等腰 Rt △ABC 分成两个全
等的等腰直角三角形,如图翻折其中一个三角形使得 DC 与 CD 重叠就可得到一个
含有 45°角的.··················································································4 分
因为 CD 是等腰 Rt △ABC 底边上的中线,
所以 AD=BD,即 AD=CB′.
又因为 Rt △ABC 是等腰 Rt ,
所以∠A=45°,AC=CB,即 AC=DB′.
所以四边形 ACB′D,是一个含有 45°角的平行四边形.
理由是:有两组对边相等的四边形是平行四边形.··································· 12 分
附加题.(1)因为
23
5
= 9
25
,······································································· 3 分
所以 x=± 3
5
;·············································································5 分
(2)因为(-0.5)3=-0.125,····························································· 3 分
所以 x=-0.5.·············································································5 分
D
A
BC
B
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