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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 查看更多

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