2007年福建省宁德市初中毕业、升学考试数 学 试 题（满分：150分，考试时间：120分钟）一、填空题（本大题有10小题，每小题3分，共30分） 1．2-= ．2．分解因式：22a a -= ． 3．若23a b =，则a bb += ． 4．如图，CD AB ⊥，垂足为1130C ∠=，，则2∠= 度．5．2007年4月27日，我国公布了第一批19座著名风景名胜山峰高程数据，其中“五岳”山峰高程数据分别是：泰山1532.7米，华山2154.9米，衡山1300.2米，恒山2016.1米，嵩山1491.7米．这五个数据的中位数是 米．6．如图，ABCD 的周长为20，对角线AC 的长为5，则ABC △的周长为 ．7．反比例函数(0)ky x x=>图象如图所示，则y 随x 的增 大而 ．8．如图，若把太阳看成一个圆，则太阳与地平线l 的位置关 系是 （填“相交”、“相切”、“相离”）．9．按下面程序计算，输入3x =-，则输出的答案是 ．10．一元二次方程210x x +-=的解是 ．DBAC1 2第4题图 A B C D 第6题图 第8题图输入x 平方 x + 2÷答案 xy第7题图O二、选择题（本大题有6小题，每小题4分，共24分．在每小题给出的四个选项中，只有一个选项是正确的，请把你认为正确选项的代号填写在题中的括号内） 11．下列计算错误的是（ ） A ．347x x x =B ．236()x x =C ．33x x x ÷=D ．4442x x x +=12．北京2008年第29届奥运会火炬接力活动历时130天，传递总里程约13.7万千米．传递总里程用科学记数法表示为（ ） A ．1.3710⨯千米 B ．51.3710⨯千米 C ．41.3710⨯千米D ．413.710⨯千米13．若如图所示的两个四边形相似，则α∠的度数是（ ） A ．87B ．60C ．75D ．12014．下列事件是必然事件的是（ ）A ．2008年奥运会刘翔能夺得男子110米栏冠军B ．这次数学考试李红会得满分C ．太阳每天从东方升起D ．李大爷买了一注“36选7的体育彩票”会中特等奖15．如图，AB 是O 的直径，20C ∠=，则BOC ∠的度数是（ ） A ．40B ．30C ．20D ．1016．如图1是一个小正方体的侧面展开图，小正方体从如图2所示的位置依次翻到第1格、第2格、第3格，这时小正方体朝上面的字是（ ） A ．和 B ．谐 C ．社 D ．会三、解答题（本大题有10小题，共96分） 17．（本题满分8分） 求值：2(2)(1)(5)x x x +++-，其中2x =．6075α60138第13题图ABC O第15题图 图1图2 第16题图解不等式153x x --≤，并把解集在数轴上表示出来． 19．（本题满分8分） 如图，已知AB CF DE CF ⊥，⊥，垂足分别为B E AB DE =，，．请添加一个．．适当条件，使ABC DEF △≌△，并予以证明．添加条件： ． 20．（本题满分8分） 已知：如图，直线l 是一次函数y kx b =+的图象． 求：（1）这个函数的解析式； （2）当4x =时，y 的值．4321---- 0 1 2 3 45A BCDE F3-2- 1- 12 3 4Oxy1- 2- 3- 4- 1 2 3 4l育才中学现有学生2870人，学校为了进一步丰富学生课余生活，拟调整兴趣活动小组，为此进行一次抽样调查．根据采集到的数据绘制的统计图（不完整）如下：请你根据图中提供的信息，完成下列问题：（1）图1中“电脑”部分所对应的圆心角为 度； （2）在图2中，将“体育”部分的图形补充完整；（3）爱好“书画”的人数占被调查人数的百分数是 ； （4）估计育才中学现有的学生中，有 人爱好“书画”． 22．（本题满分10分）图1是小明在健身器材上进行仰卧起坐锻炼时情景．图2是小明锻炼时上半身由EM 位置运动到与地面垂直的EN 位置时的示意图． 已知0.64BC =米，0.24AD =米， 1.30AB =米． （1）求AB 的倾斜角α的度数（精确到1）；（2）若测得0.85EN =米，试计算小明头顶由M 点运动到N 点的路径MN 的长度（精确到0.01米）书画 电脑 35%音乐 体育人数（人） 电脑体育 音乐 书画 兴趣小组 28242016 12 8 4 图1 图2 图1图2BCEDAMαN我国“八纵八横”铁路骨干网的第八纵通道——温（州）福（州）铁路全长298千米．将于2009年6月通车，通车后，预计从福州直达温州的火车行驶时间比目前高速公路上汽车的行驶时间缩短2小时．已知福州至温州的高速公路长331千米，火车的设计时速是现行高速公路上汽车行驶时速的2倍．求通车后火车从福州直达温州所用的时间（结果精确到0.01小时）． 24．（本题满分10分）汉字是世界上最古老的文字之一，字形结构体现人类追求均衡对称、和谐稳定的天性．如图，三个汉字可以看成是轴对称图形．（1）请在方框中再写出2个类似轴对称图形的汉字； （2）小敏和小慧利用“土”、“口”、“木”三个汉字设计一个游戏，规则如下：将这三个汉字分别写在背面都相同的三张卡片上，背面朝上洗匀后抽出一张，放回洗匀后再抽出一张，若两次抽出的汉字能构成上下结构的汉字（如“土”“土”构成“圭”）小敏获胜，否则小慧获胜．你认为这个游戏对谁有利？请用列表或画树状图的方法进行分析并写出构成的汉字进行说明． 解：（1） 25．（本题满分12分）如图，点O 是等边ABC △内一点，110AOB BOC α∠=∠=，．将BOC △绕点C 按顺时针方向旋转60得ADC △，连接OD ． （1）求证：COD △是等边三角形；（2）当150α=时，试判断AOD △的形状，并说明理由； （3）探究：当α为多少度时，AOD △是等腰三角形？土 口 木ABCDO 110α已知：矩形纸片ABCD 中，26AB =厘米，18.5BC =厘米，点E 在AD 上，且6AE =厘米，点P 是AB 边上一动点．．．按如下操作： 步骤一，折叠纸片，使点P 与点E 重合，展开纸片得折痕MN （如图1所示）； 步骤二，过点P 作PT AB ⊥，交MN 所在的直线于点Q ，连接QE （如图2所示） （1）无论点P 在AB 边上任何位置，都有PQ QE （填“>”、“=”、“<”号）； （2）如图3所示，将纸片ABCD 放在直角坐标系中，按上述步骤一、二进行操作： ①当点P 在A 点时，PT 与MN 交于点11Q Q ，点的坐标是（ ， ）； ②当6PA =厘米时，PT 与MN 交于点22Q Q ，点的坐标是（ ， ）； ③当12PA =厘米时，在图3中画出MN PT ，（不要求写画法），并求出MN 与PT 的交点3Q 的坐标；（3）点P 在运动过程，PT 与MN 形成一系列的交点123Q Q Q ，，，…观察、猜想：众多的交点形成的图象是什么？并直接写出该图象的函数表达式． CxAPBCMD(P )E B图10(A )BCDE6121824 y61218 1Q2Q图3ANP BCMDE QT 图2N2007年福建省宁德市初中毕业、升学考试数学试题参考答案及评分标准（1）本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可参照本答案的评分标准的精神进行评分．（2）对解答题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的立意，可酌情给分；如果有较严重的错误，就不给分．（3）解答右端所注分数表示考生正确作完该步应得的累加分数． （4）评分只给整数分．一、填空题（本大题有10小题，每小题3分，共30分）1．2； 2．(2)a a -； 3．53； 4．40； 5．1532.7；6．15； 7．减少； 8．相离； 9．3； 10．1152x -+=，2152x --=；或10.618x =，2 1.618x =-．二、选择题（本大题有6小题，每小题4分，共24分）11．C ； 12．B ； 13．A ； 14．C ； 15．A ； 16．D 三、解答题（本大题有10小题，共96分）17．解：原式224445x x x x =+++-- ················································································· 4分 221x =-． ······························································································································· 6分 当2x =时， 原式22(2)1=⨯-3= ··········································································································································· 8分说明：若考生直接将2x =代入计算，且结果正确不扣分．18．解：1153x x --≤． ······································································································ 2分 416x ≤． ······························································································································· 4分 4x ≤． ·································································································································· 6分解集表示正确． ······················································································································ 8分 19．添加条件：C F ∠=∠（或AC DF =，或CE FB =等）． ·············································· 2分 证明：AB CF ⊥∵，DE CF ⊥， 90ABC DEF ∠=∠=∴°． ······································································································· 4分 又AB DE =∵，C F ∠=∠， ABC DEF ∴△≌△． ············································································································· 8分 20．解：（1）依题意，得 201k b b -+=⎧⎨=⎩，． ···························································································································· 4分4321---- 0 1 2 3 4 5解得112k b ==，． ……………………4分112y x =+∴． ······································· 6分 （2）当4x =时，3y =． ····················· 8分 21．（1）126； ······································· 2分 （2）画图，如图所示； ························ 4分 （3）10%； ············································ 6分 （4）287． ············································· 8分22．解：（1）过A 作AF DC ∥， 分别交BC NE ，延长线于F H ，．AD CD ⊥∵，BC CD ⊥，AD BC ∴∥．∴四边形AFCD 为矩形．0.4BF BC AD =-=∴． ······························································ 2分 在Rt ABF △中，0.40sin 1.30BF AB α==∵， 18α≈∴°．即AB 的倾斜角度数约为18°． ······························· 7分（2）NE AF ⊥∵， 901872AEH ∠=-=∴°°°． 180108MEN AEH ∠=-∠=∴°°． ·························· 8分 MN ∴的长108π0.851.60180⨯⨯=≈（米）． 答：小明头顶运动的路径MN 的长约为1.60米． ······························································ 10分23．解：设通车后火车从福州直达温州所用的时间为x 小时． ··········································· 1分 依题意，得29833122x x =⨯+．·································································································· 5分 解这个方程，得14991x =． ····································································································· 8分 经检验14991x =是原方程的解． ······························································································ 9分 1481.6491x =≈． 答：通车后火车从福州直达温州所用的时间约为1.64小时． ··········································· 10分24．解：（1）如： 等 ································································································· 2分 （2）这个游戏对小慧有利． ································································································· 3分 每次游戏时，所有可能出现的结果如下：（列表） 一回 人数（人）电脑 体育 音乐 书画 兴趣小组 28 24 20 16 12 8 4B C ED A M α NF H土 口 木土 （土，土） （土，口） （土，木） 口 （口，土） （口，口） （口，木） 木（木，土） （木，口） （木，木）（树状图）················································································································································· 7分 总共有9种结果，每种结果出现的可能性相同， 其中能组成上下结构的汉字的结果有4种：（土，土）“圭”，（口，口）“吕”，（木，口）“杏”或“呆”，（口，木）“呆”或“杏”．()49P =小敏获胜∴，()59P =小慧获胜． ····························································································· 9分 ()P <小敏获胜()P 小慧获胜．∴游戏对小慧有利．············································································································· 10分 说明：若组成汉字错误，而不影响数学知识的考查且结论正确，本题只扣1分 25．（1）证明：CO CD =∵，60OCD ∠=°， COD ∴△是等边三角形． ······································································································ 3分 （2）解：当150α=°，即150BOC ∠=°时，AOD △是直角三角形． ································ 5分 BOC ADC ∵△≌△，150ADC BOC ∠=∠=∴°． 又COD ∵△是等边三角形， 60ODC ∠=∴°． 90ADO ∠=∴°．即AOD △是直角三角形． ····································································································· 7分 （3）解：①要使AO AD =，需AOD ADO ∠=∠． 190AOD α∠=-∵°，60ADO α∠=-°，土口 木开始 土（土，土） 口（土，口） 木（土，木） 土（口，土）口（口，口） 木（口，木） 土（木，土）口（木，口） 木（木，木）19060αα-=-∴°°． 125α=∴°．②要使OA OD =，需OAD ADO ∠=∠． 180()50OAD AOD ADO ∠=-∠+∠=∵°°， 6050α-=∴°°． 110α=∴°．③要使OD AD =，需OAD AOD ∠=∠． 19050α-=∴°°． 140α=∴°．综上所述：当α的度数为125°，或110°，或140°时，ABC △是等腰三角形． ·············· 12分 说明：第（3）小题考生答对1种得2分，答对2种得4分． 26．（1）PQ QE =． ·············································································································· 2分（2）①(03)，；②(66)，． ······································································································· 6分 ③画图，如图所示． ··············································································································· 8分解：方法一：设MN 与EP 交于点F ． 在Rt APE △中，2265PE AE AP =+=∵，1352PF PE ==∴．390Q PF EPA ∠+∠=∵°，90AEP EPA ∠+∠=°， 3Q PF AEP ∠=∠∴．又390EAP Q FP ∠=∠=∵°， 3Q PF PEA ∴△∽△． 3Q P PFPE EA=∴． 315PE PFQ P EA==·∴． 3(1215)Q ∴，． ························································································································ 11分 方法二：过点E 作3EG Q P ⊥，垂足为G ，则四边形APGE 是矩形． 6GP =∴，12EG =．设3Q G x =，则336Q E Q P x ==+．在3Rt Q EG △中，22233EQ EG Q G =+∵． 222(6)12x x +=+∴． 9x =∴． 3125Q P =∴．3(1215)Q ∴，． ························································································································ 11分（3）这些点形成的图象是一段抛物线． ············································································ 12分 函数关系式：213(026)12y x x =+≤≤．··········································································· 14分 说明：若考生的解答：图象是抛物线，函数关系式：21312y x =+均不扣分．0(A )B CDE6121824 xy 61218 1Q2Q3QFM G P。