2005年佛山升中数学试卷参考答案

2005年佛山市高中阶段学校招生考试数学试卷
参考答案及评分标准（课改实验区）
11．抽样调查 12．3
02
x <<
13．60 14．4 15．答案不唯一，如1y x =+，2y x =，（只要答案正确即可得满分）．
三、解答题答案及评分标准:
16．甲（或电动自行车）， 2， 乙（或汽车）， 2， 18， 90 注：此题每空1分，合计6分．
17．解法一：原式222
4(2)(2)(2)(2)x x x x x x x x
⎡⎤+--=-⎢⎥
-+-+⎣⎦（通分一个1分） ····· 2分 22444x x x -=-（合并2分、去括号1分） ·············································· 5分 4x
=． ······························································································· 6分 解法二：原式11(2)(2)()22x x x x x -+=--+ ··········································· 1分 1(2)(2)1(2)(2)
22x x x x x x x x -+-+=--+ ······························· 3分
224
x x x x x
+-=-=． ······························································ 6分 注：可能还有其它解法，依据各得分点酌情给分．
18．解：（写作法共4分）法一：
（1）以O 为圆心，任意长为半径画弧，分别交OA 、OB 于C 、D 两点； ····· 1分
（2）分别以C 、D 为圆心，大于
1
2
CD 的长为半径画弧，两弧交于E 点（不与O 点重合）
注：也可直接以A 、B 为圆心作图． ······················································ 2分 （3）射线OE 交弧AB 于F ； ······························································· 3分 则线段OF 将扇形AOB 二等分． ··························································· 4分 法二：（1）连接AB ；
（2）分别以A 、B 为圆心，大于
1
2
AB 的长为半径画弧，两弧交于C 点（不与O 点重合）；························································································· 2分 （3）连接OC 交弧AB 于D 点； ··························································· 3分 则线段OD 将扇形AOB 二等分． ··························································· 4分 （作图共2分）保留作图痕迹，作图准确． ·············································· 6分 19．解：根据题意，△ABC 是直角三角形， ···················· 1分
且 4.5BC =米，35ACB ∠=． ···
··························· 2分
tan 35AB
BC
=
∵， ················································ 4分 tan35AB BC =∴． 4.50.70 3.2≈⨯≈（米）． ······· 5分 第19题图
答：帐篷支撑竿的高约为3.2米． ····························· 6分 20．解法一：设口袋中有x 个白球，······························································ 1分
由题意，得
1050
10200
x =+， ··································································· 3分
解得30x =． ····················································································· 5分 答：口袋中大约有30个白球． ······························································· 6分 注：这里解分式方程是同解变形，可不检验，因而不给分． 解法二：P ∵（50次摸到红球）＝
501
2004
=， ·········································· 2分 1
10404
÷
=∴．401030-=∴． ·························································· 5分 答：口袋中大约有30个白球． ······························································· 6分 四、解答题答案及评分标准：
21．解：（1）1122PD P D ∵∥，∴△11PD O ∽△22P D O ，
······························· 2分 11
1222PD D O P D D O =∴
，即1122
b l b l =． ·
····························································· 4分 （2）1122
b l
b l =∵且1 3.2b =cm ，22b =cm ，18l =m ， ································ 5分
2
3.28
2l =∴．（注：可不进行单位换算） ·
················································ 6分 25l =∴m ． ·
······················································································ 7分 答：小“E ”的测试距离是25l =m ． ······················································· 8分
22．解法一：设三人普通房和双人普通房各住了x 、y 间， ······························· 1分
根据题意，得32501500.51400.51510x y x y +=⎧⎨⨯+⨯=⎩
，
·········································· 4分
解得813.
x y =⎧⎨=⎩，······················································································ 7分
答：三人间普通客房、双人间普通客房各住了8、13间． ···························· 8分 解法二：设三人普通房和双人普通房各住了x 、y 人． ······························· 1分
根据题意，得501500.51400.5151032x y x y
+=⎧⎪
⎨⨯⨯+⨯⨯=⎪⎩
，
···································· 4分 解得2426.x y =⎧⎨=⎩， ····················································································· 7分
且2483=（间），26132
=（间）． 答：三人间普通客房、双人间普通客房各住了8、13间． ···························· 8分 解法三：设三人普通房共住了x 人、则双人普通房共住了(50)x -人， ··········· 1分
根据题意，得501500.51400.5151032
x x
-⨯⨯
+⨯⨯=． ····························· ４分 解得24x =，5026x -= ····································································· 7分
且
2483=（间），26132
=（间）．答：三人间普通客房、双人间普通客房各住了8、13间． ························································································· 8分
23．解：（1）甲的中位数是94.5，乙的众数是99． ········································ 4分
注：此题因对数据要进行整理、分析，故每空给2分．
（2）学生的回答是多样的（只要学生说的有道理即可）例如：
甲考试成绩较稳定，因为方差、极差较小（或甲的平均数比乙的平均数高）；乙有潜力，因为乙的最好成绩比甲的最好成绩高等． ··········································· 7分 （3）10次测验，甲有8次不少于92分，而乙仅有6次，若想获奖可能性大，可选甲参赛；若想拿到更好的名次可选乙，因为乙有4次在99分以上． ·················· 9分 注：统计的解释和推断往往不是非此即彼的，从不同角度谈出合理的看法即可． 24．解：（1）1
25
a =-
，4c =，10EF =米．（2分＋2分＋3分） ·················· 7分 （2）误差估计如下：
解法一： 2.6 2.7∵ 2.65≈，10.6≈，
100.6≈∴． ································································ 8分 ∴差的近似值约为0.6米． ······················································· 9分
解法二：=∵在10到11之间，可得10.510.6<=<，
0.5100.6<<∴， ························································· 8分 ∴差的近似值约为0.5或0.6米． ··············································· 9分 注：答案应为0.5或0.6，只写一个答案的不扣分．没有误差估计过程不扣分． 五、解答题答案及评分标准：
25．解：（1）甲 √ 乙 × ················································ 2分
（2）证明（1）中对甲的判断： 连接EF 、FG 、GH 、HE ， ····························································· 3分 E ∵、F 分别是AB 、BC 的中点，∴EF 是△ABC 的中位线． ················ 4分
∴EF AC ∥，1
2
EF AC =
， ······························································· 5分 同理，HG AC ∥，1
2
HG AC =， ························································ 6分
∴EF HG ∥，EF HG =．∴四边形EFGH 是平行四边形． ···················· 7分 注：可反例说明（1）中对乙的判断：举矩形、菱形、等腰梯形等例子（用文字或图形说明，也给5分）．若将乙看作是正确的命题去证明，过程准确，给3分． （3）类似于（1）中的结论甲、乙都成立（只对一个给2分）． ··················· 10分 26．解：（1）设直线OM 的函数关系式为y kx =，1
()P a a ，，1()R b b
，． ············ 1分
则1
()M b a
，，11k b a ab
=
÷=∴． ··························································· 2分 ∴直线OM 的函数关系式为1
y x ab
=． ·················································· 3分
（2）Q ∵的坐标1()a b ，满足1
y x ab
=
，Q ∴点在直线OM 上． （或用几何证法，见《九年级上册》教师用书191页） ······························· 4分 四边形PQRM 是矩形，1
2
SP SQ SR SM PR ====
∴． SQR SRQ ∠=∠∴． ··········································································· 5分
∵2PR OP =，1
2
PS OP PR ==
∴．POS PSO ∠=∠∴． ······················· 6分 PSQ ∠∵是△SQR 的一个外角，
2PSQ SQR ∠=∠∴．2POS SQR ∠=∠∴ ·············································· 7分 QR OB ∵∥，SOB SQR ∠=∠∴． ·
····················································· 8分 2POS SOB ∠=∠∴． ········································································· 9分
1
3
SOB AOB ∠=∠∴． ········································································ 10分
（3）以下方法只要回答一种即可．
方法一：利用钝角的一半是锐角，然后利用上述结论把锐角三等分的方法即可． 方法二：也可把钝角减去一个直角得一个锐角，然后利用上述结论把锐角三等分后，
再将直角利用等边三角形（或其它方法）将其三等分即可． ············· 11分。