白银等九市州试题友情提示:抛物线2y ax bx c =++的顶点坐标是2424b ac b a a ⎛⎫-- ⎪⎝⎭，．一、选择题:本大题共10小题,每小题3分,共30分．每小题给出的四个选项中,只有一项是符合题目要求的,将此选项的代号填入题后的括号内． 1．化简:4=( )A ．2B ．-2C ．4D ．-42． 如图1,一个碗摆放在桌面上,则它的俯视图是( )3． 2008年在北京举办的第29届奥运会的火炬传递在各方面都是创记录的:火炬境外传递城市19个,境内传递城市和地区116个,传递距离为137万公里,火炬手的总数达到21780人．用科学记数法表示21780为( ) A ．2.178×105 B ．2.178×104 C ．21.78×103 D ．217.8×102 4． 如图2,小红和小丽在操场上做游戏,她们先在地上画出一个圆圈,然后蒙上眼在一定距离外向圆圈内投小石子,则投一次就正好投到圆圈内是( ) A ．必然事件(必然发生的事件) B ．不可能事件(不可能发生的事件) C ．确定事件(必然发生或不可能发生的事件)D ．不确定事件(随机事件) 5． 把不等式组110x x +⎧⎨-⎩≤的解集表示在数轴上,正确的为图3中的( )A ．B ．C ．D ．图3 6． 张颖同学把自己一周的支出情况,用如图4所示的统计图来表示．则从图中可以看出( ) A ．一周支出的总金额图1图2图8 图6 B ．一周各项支出的金额C ．一周内各项支出金额占总支出的百分比D ．各项支出金额在一周中的变化情况 7． 如图5①～④是四种正多边形的瓷砖图案．其中,是轴对称图形但不是中心对称的图形为( )A ．①③B ． ①④C ．②③D ．②④图58．中央电视台2套“开心辞典”栏目中,有一期的题目如图6所示,两个天平都平衡,则与2个球体相等质量的正方体的个数为( ) A ．5 B ．4 C ．3 D ．29． 高速公路的隧道和桥梁最多．图7是一个隧道的横截面,若它的形状是以O 为圆心的圆的一部分,路面AB =10米,净高CD =7米,则此圆的半径OA =( ) A ．5B ．7C．375 D ．37710．如图8,把矩形ABCD 沿EF 对折后使两部分重合,若150∠=,则AEF ∠=( ) A ．110° B ．115° C ．120° D ．130°二、填空题:本大题共8小题,每小题4分,共32分．把答案填在题中的横线上．11． 若向南走2m 记作2m -,则向北走3m 记作 m ． 12．点P(-2,3)关于x 轴的对称点的坐标是________．13． 已知等腰三角形的一条腰长是5,底边长是6,则它底边上的高为 ． 14． 抛物线 y=x 2+x-4与y 轴的交点坐标为 ．15． 如图9,将左边的矩形绕点B 旋转一定角度后,位置如右边的矩形,则∠ABC=___ ___ ．① ② ③ ④ 图7OD ABC16．某商店销售一批服装,每件售价150元,打8折出售后,仍可获利20元,设这种服装的成本价为每件x元,则x 满足的方程是．17．一个函数具有下列性质:①它的图像经过点(-1,1)；②它的图像在二、四象限内；③在每个象限内,函数值y随自变量x的增大而增大．则这个函数的解析式可以为．18．如图10(1)是一个等腰梯形,由6个这样的等腰梯形恰好可以拼出如图10(2)所示的一个菱形．对于图10(1)中的等腰梯形,请写出它的内角的度数或腰与底边长度之间关系的一个正确结论: ．三、解答题(一):本大题共5小题,共38分．解答时,应写出必要的文字说明、证明过程或演算步骤．19．(6分) 化简:24()22a a aa a a---+．20．(6分)请你类比一条直线和一个圆的三种位置关系,在图11①、②、③中,分别各画出一条直线,使它与两个圆都相离、都相切、都相交,并在图11④中也画上一条直线,使它与两个圆具有不同于前面3种情况的位置关系．21．(8分)图12是某种蜡烛在燃烧过程中高度与时间之间关系的图像,由图像解答下列问题:(1)此蜡烛燃烧1小时后,高度为cm；经过小时燃烧完毕；(2)22．(8分)如图13,在ABCD中,点E是CD的中点,AE的延长线与BC的延长线相交于点F．（1）（2）图10图1171Oy(cm)x(小时)15图12(1)求证:△ADE ≌△FCE ；(2)连结AC 、DF,则四边形ACFD 是下列选项中的( )． A ．梯形 B ．菱形 C ．正方形 D ．平行四边形23．(10分) 某校八年级320名学生在电脑培训前后各参加了一次水平相同的考试,考试成绩都以同一标准划分成“不及格”、“及格”和“优秀”三个等级．为了了解电脑培训的效果,用抽签方式得到其中32名学生培训前后两次考试成绩的等级,并绘制成如图14的统计图,试结合图形信息回答下列问题:(1) 这32名学生培训前后考试成绩的中位数所在的等级分别是 、 ； (2)估计该校整个八年级学生中,培训后考试成绩的等级为“及格”与“优秀”的学生共有多少名？四、解答题(二):本大题共5小题,共50分．解答时,应写出必要的文字说明、证明过程或演算步骤．24．(8分))图15是一盒刚打开的“兰州”牌香烟,图16(1)是它的横截面(矩形ABCD),已知每支香烟底面圆的直径是8mm ．(1) 矩形ABCD 的长AB= mm ； (2)利用图15(2)求矩形ABCD 的宽AD ．(3≈1.73,结果精确到0.1mm)25．(10分)如图17①,在一幅矩形地毯的四周镶有宽度相同的花边． 如图17②,地毯中央的矩形图案长6米、宽3米,整个地毯的面积是40平方分米．求花边的宽．优秀及格不及格11678824人数培训后培训前图14图15 (1)O 1 O 2 O 3 图16 （2）26．(10分)如图18,在梯形ABCD 中,AD ∥BC,∠B=90°,AD=2,BC=5,tanC=34． (1)求点D 到BC 边的距离； (2)求点B 到CD 边的距离．27．(10分)小明和小慧玩纸牌游戏． 图19是同一副扑克中的4张扑克牌的正面,将它们正面朝下洗匀后放在桌上,小明先从中抽出一张,小慧从剩余的3张牌中也抽出一张．小慧说:若抽出的两张牌的数字都是偶数,你获胜；否则,我获胜． (1)请用树状图表示出两人抽牌可能出现的所有结果；(2)若按小慧说规则进行游戏,这个游戏公平吗？请说明理由．28．(12分)如图20,在平面直角坐标系中,四边形OABC 是矩形,点B 的坐标为(4,3)．平行于对角线AC 的直线m 从原点O 出发,沿x 轴正方向以每秒1个单位长度的速度运动,设直图19①②图17图18BDAαβ C 图22 (2)线m 与矩形OABC 的两边．．分别交于点M 、N,直线m 运动的时间为t(秒)． (1) 点A 的坐标是__________,点C 的坐标是__________； (2) 当t= 秒或 秒时,MN=21AC ； (3) 设△OMN 的面积为S,求S 与t 的函数关系式；(4) 探求(3)中得到的函数S 有没有最大值？若有,求出最大值；若没有,要说明理由．附加题 (12分)1．(5分)如图21,网格小正方形的边长都为1．在⊿ABC 中,试画出三边的中线(顶点与对边中点连结的线段),然后探究三条中线位置及其有关线段之间的关系,你发现了什么有趣的结论？请说明理由．2．(7分)如图22(1),由直角三角形边角关系,可将三角形面积公式变形, 得 ABC S △=12bc·sin ∠A ． ① 即 三角形的面积等于两边之长与夹角正弦之积的一半． 如图22(2),在⊿ABC 中,CD ⊥AB 于D,∠ACD=α, ∠DCB=β． ∵ ABC ADC BDC S S S =+△△△, 由公式①,得12AC·BC·sin(α+β)= 12AC·CD·sin α+12BC·CD·sin β, 即 AC·BC·sin(α+β)= AC·CD·sin α+BC·CD·sin β． ②你能利用直角三角形边角关系,消去②中的AC 、BC 、CD 吗？不能, 说明理由；能,写出解决过程．白银等九市试题答案b BAC c图22 (1)图20 B AC图21一、选择题:本大题共10小题,每小题3分,共30分．1．A 2．C 3．B 4．D 5．B 6．C 7．A 8．A 9．D 10．B 二、填空题:本大题共8小题,每小题4分,共32分．11．3 12．(-2,-3) 13．4 14．(0,-4)15．90o16．150×80%－x＝20 17．y=-x118．答案不唯一．可供参考的有:①它内角的度数为60°、60°、120°、120°；②它的腰长等于上底长；③它的上底等于下底长的一半．三、解答题(一):本大题共5小题,共38分．19．本小题满分6分解法1:原式=(a+2)-(a-2) ·····································································4分=4． ···············································································6分解法2:原式=22(2)(2)44a a a a aa a+---⋅-·······································2分=(2)(2)a a+--·······················································4分=4．··············································································6分20．本小题满分6分答案不唯一．可供参考的有:相离:·······························1分相切: ·································3分相交: ·································5分其它:························································ 6分21． 本小题满分8分解:(1)7,158． ················································································· 4分 (2)设所求的解析式为y kx b =+, ································································ 5分 ∵ 点(0,15)、(1,7)在图像上,∴ 15,7.b k b =⎧⎨=+⎩ ……………………………………………………………………… 6分 解得 8k =-,15b =．∴ 所求的解析式为815y x =-+． (0≤x ≤158) …………………………… 8分 说明:只要求对8k =-、15b =,不写最后一步,或者未注明x 的取值范围,都不扣分． 22． 本小题满分8分证明:(1)∵ 四边形ABCD 是平行四边形,∴ AD ∥BF ,∴ ∠D =∠ECF ． ······························································ 3分∵ E 是CD 的中点,∴ DE = CE ．又 ∠AED =∠FEC , ····································· 4分∴ △ADE ≌△FCE ． ·································· 5分 (2) D ．或填“平行四边形”． ······························ 8分 23． 本小题满分10分解；(1)不及格,及格； ············································································· 4分 (2)抽到的考生培训后的及格与优秀率为(16+8)÷32=75%, ································· 6分 由此,可以估计八年级320名学生培训后的及格与优秀率为75%． ··················· 8分 所以,八年级320名学生培训后的及格与优秀人数为75%×320=240． ··············· 10分 四、解答题(二):本大题共5小题,共50分． 24． 本小题满分8分解:(1)56； ····························································································· 3分 (2)如图,△O 1 O 2 O 3是边长为8mm 的正三角形,作底边O 2O 3上的高O 1 D ． ······························· 4分 则 O 1D =O 1O 3·sin60°=43≈6.92． ···················· 6分 ∴ AD =2(O 1D +4)=2×10.92≈21.8(mm)． ·················· 8分 说明:(1)用勾股定理求O 1D ,参考本标准评分； (2)在如图大正三角形中求高后再求AD ,也参考本标准评分．25． 本小题满分10分解:设花边的宽为x 分米, ···································································· 1分 根据题意,得40)32)(62(=++x x ． ················································ 5分 解得121114x x ==-，． ··························································· 8分 x 2=114-不合题意,舍去． ··································································· 9分 答: 花边的宽为1米． ····························································· 10分 说明:不答不扣分． 26． 本小题满分10分解:(1)如图①,作DE ⊥BC 于E , ···························· 1分 ∵ AD ∥BC ,∠B =90°, ∴ ∠A =90°．又∠DEB =90°,∴ 四边形ABED 是矩形． ·································································· 2分 ∴ BE =AD =2, ∴ EC =BC -BE =3． ·························································· 3分 在Rt △DEC 中,DE = EC ·t a n C =433⨯=4． ············································ 5分 (2)如图②,作BF ⊥CD 于F ． ································································· 6分 方法一:在Rt △DEC 中,∵ CD =5, ······································ 7分 ∴ BC =DC ,又∠C =∠C , ·········································· 8分 ∴ Rt △BFC ≌Rt △DEC ． ······································ 9分 ∴ BF = DE =4． ················································ 10分 方法二:在Rt △DEC 中,∵ CD =5, ········································································· 7分O 1O 2O 3D图①图②∴ sin C =54． ······················································································ 8分 在Rt △BFC 中,BF =BC ·sin C =455⨯=4． ····················································· 10分27． 本小题满分10分解:(1) 树状图为:共有12种可能结果． ············································································ 4分 说明:无最后一步不扣分．(2)游戏公平． ·················································································· 6分 ∵ 两张牌的数字都是偶数有6种结果: (6,10),(6,12),(10,6),(10,12),(12,6),(12,10)．∴ 小明获胜的概率P =126=21． ···························································· 8分 小慧获胜的概率也为21．∴ 游戏公平． ··············································································· 10分 28． 本小题满分12分解:(1)(4,0),(0,3)； ··················································································· 2分 (2) 2,6； ····························································································· 4分 (3) 当0＜t ≤4时,OM =t ． 由△OMN ∽△OAC ,得OCONOA OM =, ∴ ON =t 43,S=283t ． ································· 6分 当4＜t ＜8时,如图,∵ OD =t,∴ AD = t-4． 方法一:由△DAM ∽△AOC ,可得AM =)4(43-t ,∴ BM =6-t 43． ···························· 7分 由△BMN ∽△BAC ,可得BN =BM 34=8-t,∴ CN =t-4． ·································· 8分S=矩形OABC 的面积-Rt △OAM 的面积- Rt △MBN 的面积- Rt △NCO 的面积=12-)4(23-t -21(8-t)(6-t 43)-)4(23-t =t t 3832+-． ·············································································· 10分 方法二:易知四边形ADNC 是平行四边形,∴ CN =AD =t-4,BN =8-t ． ·································· 7分 由△BMN ∽△BAC ,可得BM =BN 43=6-t 43,∴ AM =)4(43-t ． ··················· 8分 以下同方法一．(4) 有最大值．方法一:当0＜t ≤4时, ∵ 抛物线S=283t 的开口向上,在对称轴t=0的右边, S 随t 的增大而增大, ∴ 当t=4时,S 可取到最大值2483⨯=6； ·················································· 11分 当4＜t ＜8时,∵ 抛物线S=t t 3832+-的开口向下,它的顶点是(4,6),∴ S ＜6． 综上,当t=4时,S 有最大值6． ································································· 12分 方法二:∵ S=22304833488t t t t t ⎧<⎪⎪⎨⎪-+<<⎪⎩，≤， ∴ 当0＜t ＜8时,画出S 与t 的函数关系图像,如图所示． ······························ 11分 显然,当t=4时,S 有最大值6． ······························································ 12分 说明:只有当第(3)问解答正确时,第(4)问只回答“有最大值”无其它步骤,可给1分；否则,不给分．附加题 (12分)1． (1)三条中线交于一点； ···························2分(2)在同一条中线上,这个点到边中点的距离等于它到顶点距离的一半． ·······················5分2． 能消去AC 、BC 、CD ,得到sin(α+β)= sinα·cosβ+cosα·sinβ． ····························· 2分解:给AC ·BC ·sin(α+β)= AC ·CD ·sinα+BC ·CD ·sinβ两边同除以AC ·BC ,得BA Csin(α+β)= CDBC·sinα+CDAC·sinβ, ······················································4分∵CDBC=cosβ,CDAC=cosα． ························································6分∴ sin(α+β)= sinα·cosβ+cosα·sinβ． ·····················································7分说明:如果上边解法没有第1个步骤的采分点,则后边三个采分点得分分别改为2分、6分、7分．全卷说明:对于以上各解答题学生试卷中出现的不同解法,请参考本标准给分。