2021－2022学年第二学期福州市九年级数学适应性练习答案(1)

1、2021 2022数学适应性练习答案及评分标准评分说明：1本解答给出了一种或几种解法供参考，如果学生的解法与本解答不同，可根据习题的主要考查内容比照评分参考制定相应的评分细则2对于计算题，当学生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分3解答右端所注分数，表示学生正确做到这一步应得的累加分数4只给整数分数选择题和填空题不给中间分一、选择题（共10 小题，每小题4 分，满分40 分；在每小题给出的四个选项中，只有一项是符合题目要求的，请在答题卡的相应位置

2、填涂）1C 2A 3B 4A 5C6C 7D 8D 9B 10D二、填空题（共6 小题，每小题4 分，满分24 分，请在答题卡的相应位置作答）11（2，3） 124 132143 153 216(1)(3)(4)三、解答题（共9 小题，满分86 分，请在答题卡的相应位置作答）17（本小题满分8 分）解：a =1，b = -4，c = -7，·····················

3、83;·················································

4、83;··············· 1 分D = b2 - 4ac= - 2 - ´ ´ - = > ，··························

5、83;·················································

6、83;·· 3 分( 4) 4 1 ( 7) 44 0 x-b ± b2 - 4ac= ········································&

7、#183;·················································&

8、#183;····· 4 分2a -(-4) ± 44= = 2 ± 11 2´1，····································&#

9、183;··············································· 6 分x1 = 2 + 1

10、1 ，x = - ················································

11、83;·································· 8 分2 2 1118（本小题满分8 分）解：连接OB，OC··········&#

12、183;·················································&#

13、183;········································· 1 分AB»C = B»C ，ÐA = 30° ，

14、8;BOC = 2ÐA = 60°·············································

15、83;·············································· 3 分 OB = OC ，ODBC，OO

16、D 平分BOC，ÐODB = 90° ，·············································

17、································ 4 分 1 30ÐBOD = ÐBOC = ° ···········&#

18、183;·················································&#

19、183;························· 5 分D2B C 在RtOBD 中，OB = 6 ， 1 3BD = OB = ，·················&#

20、183;·················································&#

21、183;······························ 6 分2OD = OB2 - BD2 = 3 3 ，···············

22、;··················································

23、;···················· 8 分OD 的长是3 3 九年级数学试题答案及评分参考 第1 页（共7 页）19（本小题满分8 分）解：（1）设黑棋为A1，A2，白棋为B1，B2，B3A1 A2 B1 B2 B3A1 （A1，A2） （A1，B1） （A1，B2） （A1，B3）A2 （A2，A1） （A2，B1） （A2，B2） （A2，B3） B1 （B1，A1） （B1，A2）

24、（B1，B2） （B1，B3） B2 （B2，A1） （B2，A2） （B2，B1） （B2，B3） B3 （B3，A1） （B3，A2） （B3，B1） （B3，B2）····································

25、;·························· 4 分 （2）由（1）得，共有20 种结果，并且它们发生的可能性相等，··················

26、;····················· 5 分 其中摸出2 枚白棋的情况共有6 种，·························

27、············································ 6 分 分别为（B1，B2），（B1，B3），（B2，B1），（B2，

28、B3），（B3，B1），（B3，B2） P（摸出的2 枚棋都是白棋） 6 3= = ···········································

29、··············· 8 分20 10 20（本小题满分8 分）解：（1）当 1t = 时，S = 6 ；当t =1时，S = 9 ，2ì1 + 1 = 6ï ，a b 4 2 ····················&

30、#183;·················································&

31、#183;························· 2 分íï + =îa b 9，ìa = -6，解得í = ···············

32、;··················································

33、;··································· 3 分îb 15，该函数的解析式是s = -6t2 +15t ········

34、3;·················································

35、3;·········· 4 分 （2）t 0 1 1 52 4S 0 6 9 758S（m）12 758 963012 15 32 5 t（s）4 2 2····························

36、3;··································· 7 分该函数的图象如图所示汽车刹车后到停下来前进了758米·········

37、3;·················································

38、3;·········· 8 分21（本小题满分8 分）解：（1）B CEOF···································&

39、#183;·························· 3 分九年级数学试题答案及评分参考 第2 页（共7 页） 如图所示，点O 即为所求作的旋转中心···············&

40、#183;·············································· 4 分（2）证明：连接OA，OB，O

41、C，OD，OE，OF线段BC 绕点O 旋转 得到线段EF，点B 的对应点是E，OB = OE ，OC = OF ，BC = EF ，ÐBOE = ÐCOF = a ，······························ 5 分OBCOEF，ÐOBC = 

42、8;OEF AABC，DEF 都是等边三角形，AB = BC ，DE = EF ，ÐABC = ÐDEF = 60° ， AB = DE ，ÐABC + ÐOBC = ÐDEF + ÐOEF ，BC即ÐABO = ÐDEO ，·······················

43、3;·················································

44、3;······· 6 分EABODEO，OA = OD ，·······································&

45、#183;·················································&

46、#183; 7 分ÐAOB = ÐDOE ，DOÐAOB + ÐAOE = ÐDOE + ÐAOE ， 即ÐAOD = ÐBOE = a ，F 点A 绕点O 顺时针旋转 到点D，DEF 可由ABC 绕点O 顺时针旋转 得到·························

47、;·················· 8 分22（本小题满分10 分）解：（1）将x = 6 代入y = x - 5，得y =1，该交点坐标是（6，1） ······················

48、··················································

49、·········· 2 分将（6，1）代入y k= ，得1 = ，····································

50、···································· 3 分k x 6k = 6···········

51、3;·················································

52、3;·········································· 4 分（2）设A（m， 6），其中m > 0m过点A 作ACy 轴于点C，过点B 作

53、BDy 轴于点D，yÐACO = ÐODB = 90°，AC = m，CO 6= ，mÐCAO + ÐCOA = 90° ···································

54、;················································ 5 分AC 线段OA

55、 绕点O 顺时针旋转90°得到线段OB，OB = OA，ÐAOB = 90°，点B 在第四象限，·······································

56、83;················ 6 分O xÐCOA+ ÐDOB = 90° ， ÐCAO = ÐDOB ，DBOACBOD，·····················&#

57、183;·················································&#

58、183;·················· 7 分OD = AC = m ，DB CO 6= = ，mB（ 6，-m ）·························

59、;··················································

60、;·················· 8 分m点B 在一次函数y = x - 5的图象上， m 6 5- = - ，·························&

61、#183;·················································&

62、#183;···················· 9 分m解得 1 2m = ， 2 3m = ，点A 的坐标是（2，3）或（3，2）······················

63、············································10 分23（本小题满分10 分）（1）证明：连接ODOA = OD ，

64、ÐOAD = ÐODA··············································

65、83;······································· 1 分 AD 平分BAC，EÐEAD = ÐOAD ，····&

66、#183;·················································&

67、#183;······························· 2 分C DÐEAD = ÐODA，FODAE，············&#

68、183;·················································&#

69、183;································· 3 分ÐAED + ÐODE =180° A O B DEAC 于点E，ÐAED = 90°，九年级数学试题答案及评分参考 第3 页（共7 页）

70、ÐODE = 90° ，ODDE·············································

71、3;················································· 4 分

72、点D 是半径OD 的外端点，DE 是半圆O 的切线··············································

73、;·································· 5 分 （2）解：连接BD，AB 是直径， ÐADB = 90° ， ÐAED = ÐADB ÐEAD = ÐOAD ，AEDA

74、DB，·················································

75、83;······································· 6 分 AE AD= EAD AB半圆O 的半径为4，AE = 6 ，C DFAB = 8 ，G 6 AD= ，AD 8A

76、O BAD = 4 3 ················································

77、················································ 7 分ODAE，OBG

78、ABE，ÐAEF = ÐDGF ，ÐEAF = ÐGDF ，OG OB 1 AE AB 2= = ，······································&#

79、183;·················································&#

80、183;·· 8 分AEFDGF， OG = 3，DF DG= ，···········································

81、;········································ 9 分AF AE DG = OD - OG =1，DF 1 AF 6= ，即 1 4 3DF = AD = 

82、3;·················································

83、3;··································10 分7 724（本小题满分12 分）D 解：（1）ABDCBE，ÐBAD = ÐBCE ······&#

84、183;·················································&#

85、183;·································· 2 分 ÐABC + ÐADC = 90° ，AÐBAD + ÐBCD = 270° ，···

86、;··················································

87、;····························· 3 分 ÐBCE + ÐBCD = 270° ，··············&

88、#183;·················································&

89、#183;················· 4 分BCÐDCE = 360° -ÐBCE - ÐBCD = 90° ，ECCD·····················

90、··················································

91、····························· 5 分（2）ABDCBE， 5AB= ，BC 8E 5DB AD AB= = = ，ÐABD = ÐCBE BE CE BC 8 BD = 20，BE = 32······

92、;··················································

93、;············································ 6 分过点A 作AFCD 于点F，D1CD × AFS A

94、F2ÐAFD = 90°， ACD = = ············································

95、3;·············· 7 分GS 1 CD CE CE×CDE2A 5S = S ，FACD CDE B16C AF 5 = ，··························

96、;··········································H·······

97、3;······················· 8 分CE 16又 AD 5 CE 8= ，EAD = 2AF ·····················&

98、#183;·················································&

99、#183;························· 9 分取AD 中点G，连接FG，九年级数学试题答案及评分参考 第4 页（共7 页） 1FG = AD = AG = DG ，2FG = AF = AG ，AFG 是等边三角形， ÐFAD = 60°， ÐADC = 30°ÐABC + &

100、#208;ADC = 90° ，ÐABC = 60°，···········································

101、··················································

102、·10 分ÐDBE = ÐDBC + ÐCBE = ÐDBC + ÐABD = ÐABC = 60° 过点D 作DHBE 于点H，ÐDHB = 90°， ÐBDH = 30° ， 1 10BH = BD = ······················

103、3;·················································

104、3;················11 分2在RtBHD 中，DH = BD2 - BH 2 =10 3 ， 1 1 32 10 3 160 3S = BE × DH = ´ ´ = ··················

105、3;·····································12 分BDE2 225（本小题满分14 分）解：（1）抛物线y = mx2 - (1- 4m)x + c 过点（0，-1），c = -1·&

106、#183;·················································&

107、#183;·················································&

108、#183;· 1 分抛物线y = mx2 - (1- 4m)x + c 过点（1，a），（-1，a），抛物线的对称轴是y 轴，······································

109、3;·········································· 2 分 即1- 4 = 0 ，m 2m 1m = ，···&

110、#183;·················································&

111、#183;·················································

112、3 分4该抛物线的解析式是 1 2 1y = x - ·············································

113、83;······················· 4 分4（2）将x = 0 代入 1 2 1y = x - ，得y = -1，4顶点C 的坐标是（0，-1），OC =1 将x = 4 代入 1 2 1y = x - ，得y = 3 ，4点A 的坐标是（4，3），直线OA 的解析式是 3y = x ·····

114、83;·················································

115、83;··················· 5 分4解法一：将y = 3 x 代入y = 1 x2 -1 ，得1 2 1 3x - = x ，4 4 4 4解得x1 = 4 ，x2 = -1，点B 的横坐标为-1， 1 4 1 1 5 5S = S + S = ´ ´OC + ´ ´OC = OC = ·····&#

116、183;················ 6 分ABC AOC BOC2 2 2 2 连接AD，BD，过点D 作DQ 垂直x 轴交AB 于点Q，y分别过点A，B 作DQ 的垂线，垂足为M，N，1 1设D（t， t2 - ），MA4 QQ（t， 3t ）， D4Ox 1 1 BS = S + S = DQ × AM + DQ × BNN ABD ADQ BDQ2 2C4 5- t DQ t - (-1) DQ DQ=

117、+ = ················································ 7 分2 2

118、2 S = S ，ABD ABC5 5DQ = 2 2点D 是点A，C 之间的抛物线上的点（不与点A，C 重合），九年级数学试题答案及评分参考 第5 页（共7 页）点Q 在点D 上方，3 (1 2 1) 1t - t - = ，4 4即1 (3 ) 0t - t = ，4解得 1 0 2 3t = （舍去），t = ，点D 的坐标是（3， 5）·····················&

119、#183;··········································· 8 分4解法二：点D 是点A，C 之间的抛物线上的点（不与点A，C 重合

120、），点C，D 在AB 的同侧y分别过点C，D 作AB 的垂线，垂足为S，T， ÐCSB = ÐDTB = 90° ACSDTT ABC 的面积与ABD 的面积相等，DSO CS = DT xB四边形CSTD 是平行四边形，·····························

121、······························· 6 分C直线CD 可由直线AB 向下平移1 个单位得到，直线CD 的解析式是 3 1y = x - ··········

122、83;············································· 7 分4将y = 3 x -1代入 1 2 1y =

123、 x - ，4 4得1 2 1 3 1x - = x - ，4 4解得 1 0 2 3x = （舍去），x = ，点D 的坐标是（3， 5）······································&#

124、183;·························· 8 分4解法一：设点D 的坐标是（2n ，n2 -1）由待定系数法可得直线OD 的解析式是n2 -1y = x 2nn2 -1 4n将y = -2 代入 = ，得 = 1- ny x x ，2n 24n点F 的横坐标是 ·····

125、;··················································

126、;··········· 9 分1- n2yn2 -1将y x y = x - ，= 代入 1 2 12n 4得1 12 n -12x - = x ，4 2nE整理得nx2 - 2(n2 -1)x - 4n = 0 ，Ox2D解得 1 2x = n ， = - ，x2nlI G H F点E 的横坐标是 2- ···············

127、··················································

128、···10 分n过点F 作y 轴的垂线l，垂足为G；分别过点D，E 作直线l 的垂线，垂足为H，I，ÐFGO = ÐFHD = ÐFIE = 90° ，点G，H，I 的横坐标分别为0，2n， 2- ，n OGDHIE，OE = GI ， FE FI= ···································