da2007年泰安市中考數(shù)學(xué)試題

1、泰安市二七年中等學(xué)校招生數(shù)學(xué)試題（）（課改區(qū)用）參考一、選擇題（每小題 3 分，共 36 分）及評分標(biāo)準(zhǔn)二、填空題（本大題共 7 小題，每小題 3 分，共 21 分）13 x1 = 2 ， x2 = -715 a > 414 6016 (5，4)17 5n + 218 (300 +1002) m19 3，2，9 三、解答題（本大題共 7 小題，滿分 63 分）20（本小題滿分 6 分）解：（1）20，16如右圖 ···············

2、;·················（2） 800´ 573 （或 574） ····························

3、83;···········································4 分43·····

4、3;··························6 分60頻數(shù)20161176時間/天3.55.57.59.5（第 20 題）11.513.521（本小題滿分 8 分）（1）證明： AD BCÐDBC = ÐADB又 ÐABD = ÐDBCÐABD = ÐADB AB =

5、 AD ································又 AF = 1 AB ， AG = 1 AD·············

6、83;···2 分（第 21 題）2 AF = AG ································2···········

7、;················································3 分又 Ð

8、BAE = ÐDAE ， AE = AEAFE AGE EF = EG ································（2）當(dāng) AB = 2EC 時， E AB = 2EC AD = 2ECGD =AD = EC ···&

9、#183;····························2又 GD EC····················

10、;············D ·····································

11、83;·················································

12、83;··5 分6 分1··············································&#

13、183;··7 分學(xué)數(shù)學(xué)用專頁搜上第 1 頁 共 5 頁題號123456789101112DCBCBA彰顯數(shù)學(xué)！演繹！四邊形GECD 是平行四邊形 E····································

14、;··················································

15、;·····8 分22（本小題滿分 9 分）解： 設(shè)第一次購書的進價為 x 元， 則第二次購書的進價為 (x +1) 元 根據(jù)題意得：1200 +10 = 1500································&

16、#183;·················································&

17、#183;·4 分x解得： x = 51.2x經(jīng)檢驗 x = 5 是原方程的解··········································

18、3;·······························6 分1200= 240 （本）所以第一次購書為5第二次購書為240 +10 = 250 （本）第一次賺錢為240´(7 - 5) = 480 （元）第二次賺錢為200´(7 - 5´1.2)

19、+ 50´(7 ´ 0.4 - 5´1.2) = 40 （元）所以兩次共賺錢480 + 40 = 520 （元） ······································

20、···················8 分···· 9 分答：該兩次售書總體上是賺錢了，共賺了 520 元 ·····················

21、3;··········23（本小題滿分 9 分）（1）證明：連結(jié) AD，ODAB 是 O 的直徑AG AD BC································&

22、#183;···············2 分ABC 是等腰三角形 BD = DC又 AO = BOOD ACEOFCBDF ACOF OD DF OD DF 是 O 的切線 ·······················

23、;·········（2） AB 是 O 的直徑 BG ACABC 是等邊三角形 BG 是 AC 的垂直平分線GA = GC ································

24、又 AG BC ， ÐACB = 60··············································

25、··4 分（第 23 題）··············································&

26、#183;··5 分··············································

27、3;·············7 分ÐCAG = ÐACB = 60ACG 是等邊三角形ÐAGC = 60 ···························

28、3;····24（本小題滿分 9 分）解：（1） y = 80x +100(900 - x)·······································

29、················9 分學(xué)數(shù)學(xué)用專頁搜上第 2 頁 共 5 頁彰顯數(shù)學(xué)！演繹！= -20x + 90000 ··························

30、83;·····（2）由題意得：-20x + 90000 82000-x + 4500 4100x 400即購 A 種樹不少于 400 棵 ································（3） 92%x + 98

31、%(900 - x) 94%´ 90092x + 98´900 - 98x 94´900-6x -4´900·····································

32、3;·3 分···········································5 分x 600 ···

33、3;·················································

34、3;··········································7 分y = -20x + 90000 隨 x 的增大而減小當(dāng) x = 600 時，購樹費用

35、最低為 y = -20´ 600 + 90000 = 78000 （元）當(dāng) x = 600 時， 900 - x = 300此時應(yīng)購 A 種樹 600 棵， B 種樹 300 棵 ································25（本小題滿分 10 分

36、）解：（1）過點 A¢作 A¢D 垂直于 x 軸，垂足為 D 則四邊形OB¢A¢D 為矩形在A¢DO 中，y·····················9 分A¢D = OA¢sin ÐA¢OD = 4 ´sin 60OD = A¢B¢ = AB = 2點 A¢

37、的坐標(biāo)為(2，2 3) ································= 23CB¢A¢·······3 分B（2） C(0，4) 在拋物線上，c = 4OD（第 25 題

38、）Ax y = ax2 + bx + 4A(4，0) ， A¢(2，2 3) ，在拋物線 y = ax2 + bx + 4 上 ìï16a + 4b + 4 = 0， ································í4a

39、+ 2b + 4 = 2 3···············································5 分&

40、#239;îì1- 3a =ï解之得í2ïb = 2 3 - 3î所求式為 y = 1-3 x2 + (23 - 3)x + 4 ··································&

41、#183;········7 分2（3）若以點O 為直角頂點，由于OC = OA = 4 ，點C 在拋物線上，則點C(0，4) 為滿足學(xué)數(shù)學(xué)用專頁搜上第 3 頁 共 5 頁彰顯數(shù)學(xué)！演繹！條件的點若以點 A 為直角頂點，則使PAO 為等腰直角三角形的點 P 的坐標(biāo)應(yīng)為 (4，4) 或(4，- 4) ，經(jīng)計算知；此兩點不在拋物線上若以點 P 為直角頂點，則使PAO 為等腰直角三角形的點 P 的坐標(biāo)應(yīng)為 (2，2) 或(2，- 2) ，經(jīng)過算知；此兩點也不在拋物線上綜上述在拋物線上只有一點 P(0，4) 使OAP 為等

42、腰直角三角形 ···················10 分26（本小題滿分 12 分）（1）證明：在ADC 和EGC 中，ÐADC = ÐEGC = Rt Ð， ÐC = ÐCADC EGC=··············

43、;··················AFGEGCG·······················3 分ADCD（2） FD 與 DG 垂直 ····

44、····························證明如下：在四邊形 AFEG 中，ÐFAG = ÐAFE = ÐAGE = 90四邊形 AFEG 為矩形 AF = EGBC·········&

45、#183;··4 分DE（第 26 題）EGCG由（1）知=ADCDAFCG=································ADCDABC 為直角三角形， AD BCÐFAD = ÐCAFD CGDÐADF =

46、ÐCDG ································又ÐCDG + ÐADG = 90················································