文档：da2007年聊城市中考数学试题

聊城市二○○七年普通高中招生统一考试
数学题库（A ）参考答案及评分说明
一、选择题（每小题选对得4分，满分48分）
题号 1 2 3 4 5 6 7 8 9 10 11 12 答案
C
A
C
B
B
C
A
B
C
D
D
A
二、填空题（每小题填对得4分，满分20分） 13．4- 14．135° 15．1
2
16．32 17．80
三、解答题（满分82分） 18．（本题满分8分）
解：223111111a a a a a ⎛⎫⎛
⎫+÷- ⎪ ⎪+--⎝⎭⎝⎭
· 2221141111a a a a a +-=÷+--···················································································· 2分 22
21111141a a a a a +-=+--·· 21(1)(1)1
1(12)(12)1
a a a a a a a +-+=
+-+-·· ········································································· 4分 1
21
a =
-． ···································································································· 6分 当3
2
a =-
时，原式111
3214212a =
==--⎛⎫
⨯-- ⎪⎝⎭
． ············································· 8分
19．（本题满分8分）
解：（1）36292740437233
407
++++++=∵
， ·
············································· 2分 40301200⨯=∴．
即小辰家的轿车每月要行驶1200千米． ······························································ 4分 （2）4.7481200121005460.485500⨯⨯⨯÷=≈． ············································· 7分 即小辰家一年的汽油费用大约是5500元． ··························································· 8分 20．（本题满分10分） 解：（1）这个公式为2
2
2
()2a b a ab b +=++． ···················································· 2分 （2）ABC CDE ∵△≌△，BAC DCE ∠=∠∴．
90ACB DCE ACB BAC ∠+∠=∠+∠=∴°． ·
··················································· 4分 由于B C D ，，共线，
所以180()ACE ACB DCE ∠=-∠+∠°
1809090=-=°°°． ·
····················································································· 5分 （3）梯形ABDE 的面积为 2111
()()()()222
AB ED BD a b a b a b +=++=+·； ·
·············································· 7分 另一方面，梯形ABDE 可分成三个直角三角形，其面积又可以表示成 2111
222
ab ab c ++． ·
······················································································ 8分 所以，2
21111()2222a b ab ab c +=++． ····························································· 9分
即2
2
2
a b c +=． ·························································································· 10分 21．（本题满分10分）
解：根据题意，得2010200AC =⨯=． 过点A 作AD 垂直于直线BC ，垂足为D ． 在Rt ADC △中，
cos AD AC CAD =∠·200
cos301003==·°， ··················································································································· 2分
sin 200sin30100DC AC CAD =∠==··°． ·
······················································ 4分 在Rt ADB △中，tan 1003tan 75DB AD BAD =∠=·°． ···································· 7分 1003tan 75100CB DB DC =-=-∴°．
53tan 7552720
CB
=-≈∴
°． ·
······································································· 9分 即该游客自景点C 驶向景点B 约需27分钟． ····················································· 10分
22．（本题满分10分）
解：设这种计算器原来每个的进价为x 元， ························································· 1分 根据题意，得
4848(14)1005100(14)x x
x x
---⨯+=⨯-%%%%%． ···································· 5分 解这个方程，得40x =． ················································································· 8分
经检验，40x =是原方程的根． ········································································ 9分 答：这种计算器原来每个的进价是40元． ························································· 10分 23．（本题满分10分）
a
b c
c
A E D
C B b a
第20题图
75
30 C
B A
北 东
Ｄ 第21题图
解：（1）由题设可知，13.5 1.1514.65OA =+=，29
2
OD =
． 29(014.65) 1.152A C ⎛⎫
⎪⎝⎭
∴，，，． ········································································· 2分
设拱形抛物线的关系式为2
y ax c =+，则
22
14.650291.152a c a
c ⎧=+⎪⎨⎛⎫
=+⎪ ⎪⎝⎭⎩
·，
·． ······················································································ 4分 解得54
14.65841
a c =-=，． ·
··········································································· 5分 所以，所求函数的关系式为2
5414.65841
y x =-+． ·
·············································· 6分 （2）由20MN =米，设点N 的坐标为0(10)y ，，
代入关系式，得
2054
1014.658.229841
y =-⨯+≈． ·
······ 8分 0 1.158.229 1.15y -=-∴7.0797.08=≈．
即大幕的高度约为7.08米． ················ 10分 24．（本题满分12分） 解：（1）在BDE △和FDA △中，
12FB BD =
∵，12AE ED =，2
3
BD ED FD AD ==∴． ·
············································ 2分 又BDE FDA ∠=∠∵， BDE FDA ∴△∽△． ·
·················· 5分 （2）直线AF 与O 相切． ············· 6分 证明：连结OA OB OC ，，．
AB AC BO CO OA OA ===∵，，，
OAB OAC ∴△≌△． ·
···················· 7分 OAB OAC ∠=∠∴．
所以AO 是等腰三角形ABC 顶角BAC ∠的平分线．
AO BC ⊥∴． ·
····························································································· 9分 由BDE FDA △∽△，得EBD AFD ∠=∠．BE FA ∴∥． ································ 10分 由AO BE ⊥知，AO FA ⊥．∴直线FA 与O 相切． ······································· 12分 25．（本题满分14分）
解：（1）设公园A B ，需铺设草坪的面积分别为12S S ，，根据题意，得
16232622322221800S =⨯-⨯-⨯+⨯=． ······················································ 2分
13.5米
29米
第23题图
1.15米
y
x A
M
N
C
O B
D
第24题图
A C
D
E
O
B F
设图2中圆的半径为R ，由图形知，圆心到矩形较长一边的距离为252
， 所以25
cos302R =
°，有253
R =． 于是，2
21202512525
65252π2100836022
33S ⎛⎫=⨯-⨯⨯-⨯⨯⨯≈ ⎪⎝⎭． ························ 4分 所以公园A B ，需铺设草坪的面积分别为2
1800m 和10082m ． ································ 5分 （2）设总运费为y 元，公园A 向甲地购买草皮x 2m ，向乙地购买草皮(1800)x -2m ． ··················································································································· 6分 由于公园A B ，需要购买的草皮面积总数为180010082808+=（2m ）， 甲、乙两地出售的草皮面积总数为2
160812002808(m )+=． 所以，公园B 向甲地购买草皮2
(1608)m x -，
向乙地购买草皮2
1200(1800)(600)(m )x x --=-． ············································ 7分
于是，有01608018001200x x ⎧⎪⎨
-⎪⎩，
．
≤≤≤≤ ······································································ 9分
所以6001608x ≤≤． ··············································································· 10分 又由题意，得
300.25220.3(1800)320.25(1608)300.3(600)y x x x x =⨯+⨯-+⨯-+⨯-···
1.919344x =+． ·
························································································ 11分 因为函数 1.919344y x =+随x 的增大而增大，
所以，当600x =时，有最小值 1.96001934420484y =⨯+=（元）． ··················· 13分
因此，公园A 在甲地购买6002
m ，在乙地购买2
180********(m )-=；
公园B 在甲地购买16086001008-=（2
m ）．
此时，运送草皮的总运费最省． ······································································ 14分 说明：解答题各小题只给了一种解答及评分说明，其他解法只要步骤合理、解答正确，均应给出相应分数．。