呼和浩特市中考数学答案由刀豆文库小编整理，希望给你工作、学习、生活带来方便，猜你可能喜欢“呼和浩特市中考数学”。
2014年 呼 和 浩 特 市 中 考 试 卷
数学参考答案及评分标准
一、选择题
1．C2．D3．A4．B5．B6．C7．D8．C9．B10．C
二、填空题 11．160°12．1.613．63°或27°14．–y(3x–y)215．816．①
三、计算题
17．（1）解：原式=2 ×
32＋ 1–2
＋ 1
2··························································· 3分= 3–3＋2)＋ 1
2··································································· 4分
= –32
···························································································· 5分
（2）解：去分母得
3x2–6x–x2–2x = 0 ·························································································· 1分2x2 –8x = 0 ········································································································ 2分∴ x = 0或x = 4 ································································································ 3分 经检验：x = 0是增根
∴ x = 4是原方程的解 ······················································································ 5分 18．解：过点P作PD⊥AB于D ················································································· 1分
由题意知∠DPB = 45° 在RtΔPBD中，sin 45° = PDPB
∴ PB2PD ········································································································· 2分 ∵ 点A在P的北偏东65°方向上 ∴ ∠APD = 25°
数学试卷答案第1页（共6页）在RtΔPAD中 cos 25° =
PD
PA
∴ PD = PA cos 25° = 80 cos 25° ··········································································· 5分 ∴ PB = 802 cos 25°························································································· 6分
–2x＋3≥–3…………………①19．解：1(x–2a)1
2x
∴ 当a > 3时，不等式组的解集为x≤3 ······························································ 4分当a
由此可以估计初三学生60秒跳绳在120个以上的人数达到一半以上 ··············· 3分（2）x =
2×70＋10×90＋12×110＋13×130＋10×150＋3×170
≈121 ···· 6分
（3）记第一组的两名学生为A、B，第六组的三名学生为1、2、3 ················· 7分则从这5名学生中抽取两名学生有以下10种情况：AB，A1，A2，A3，B1，B2，B3，12，13，23 ∴ P =
410= 2
····································································································· 9分 21．证明：（1）∵ 四边形ABCD是矩形
∴ AD=BCAB=CD 又∵ AC是折痕
∴ BC = CE = AD································································································ 1分AB = AE = CD································································································ 2分 又DE = ED
数学试卷答案第2页（共6页）
AC
∴ ΔADE ≌ΔCED ····························································································· 3分（2）∵ ΔADE ≌ΔCED ∴ ∠EDC =∠DEA
又ΔACE与ΔACB关于AC所在直线对称 ∴ ∠OAC =∠CAB 而∠OCA =∠CAB
∴ ∠OAC =∠OCA ······························································································· 5分 ∴ 2∠OAC = 2∠DEA ·························································································· 6分 ∴ ∠OAC =∠DEA
∴ DE∥AC ··········································································································· 7分 22．解：设基本电价为x元/千瓦时，提高电价为y元/千瓦时 ·································· 1分
由题意得：
180x＋150y=213
180x＋60y =150 ······························································································ 3分 解之得：x=0.6y=0.7
··································································································· 4分
∴ 4月份的电费为：160×0.6=96元
5月份的电费为：180×0.6＋230×0.7 = 108＋161 = 269元
答：这位居民4、5月份的电费分别为96元和269元． ···································· 7分 23．解：（1）∵ y = k
x
过（1，4）点
∴ k = 4，反比例函数解析式为y = 4
x································································ 1分
（2）∵ B（m，n）A（1，4）
∴ AC = 4–n，BC = m–1，ON = n，OM = 1 ···················································· 2分 ∴AC4–ON= nn= 4n–1 而B（m，n）在y = 4x上
∴4
n
= m ∴ ON = m–1 而
BCm–1OM = 1
∴ ACON = BCOM······································································································ 4分 又∵ ∠ACB =∠NOM = 90°
∴ ΔACB∽ΔNOM ····························································································· 5分（3）∵ ΔACB与ΔNOM的相似比为2 ∴ m–1 = 2 ∴ m = 3
∴ B点坐标为（3，43）························································································ 6分
设AB所在直线的解析式为y = kx＋b 4∴ 3= 3k＋b 4 = k＋b∴ k = –43b = 16
3∴ 解析式为y = –43x＋16
3················································································ 8分
24．证明：（1）连接OC ···························································································· 1分
∵ AB为⊙O的直径 ∴ ∠ACB = 90°
∴ ∠ABC ＋∠BAC = 90° 又∵ CM是⊙O的切线 ∴ OC⊥CM
∴ ∠ACM ＋∠ACO = 90°·············································································· 2分 ∵ CO = AO ∴ ∠BAC =∠ACO
∴ ∠ACM =∠ABC ······························································································· 3分
（2）∵ BC = CD ∴ OC∥AD 又∵ OC⊥CE ∴ AD⊥CE
∴ ΔAEC是直角三角形
∴ ΔAEC的外接圆的直径为AC········································································· 4分 又∵ ∠ABC ＋∠BAC = 90° ∠ACM ＋∠ECD = 90° 而∠ABC =∠ACM ∴ ∠BAC =∠ECD 又∠CED =∠ACB = 90° ∴ ΔABC∽ΔCDE ∴ABBCCD= ED而⊙O的半径为3 ∴ AB = 6 ∴6BCCD= 2∴ BC2 = 12
∴ BC = 23········································································································ 6分 在RtΔABC中 ∴ AC =
36–12 6··················································································· 7分
∴ ΔAEC6········································································ 8分 25．解：（1）∵ y = ax2＋bx＋2经过点B、D
4a＋2b＋2 = 0
∴
a＋b＋2 = 5
4解之得：a =–114b =2
∴ y =–14 x2 –1x＋2 ·························································································· 2分
∵ A（m，0）在抛物线上 ∴ 0 =–14 m2 –1m＋2
解得：m =–4
∴ A（–4，0）···································································································· 3分 图像（略）············································································································ 4分（2）由题设知直线l的解析式为y = 1x–1
∴ S = 1AB·PF
= 1
×6·PF
= 314 x2 –12x＋2＋1–1x）································································ 5分
= –3
x2 –3x＋9
= –3
4（x＋2）2 ＋12 ··················································································· 6分
其中–4
∴ PB所在直线的解析式为y =–1x＋1 ·························································· 10分
设Q（a12a–1）是y = 1x–1上的任一点
则Q点关于x轴的对称点为（a，11
a）
将（a，1–12 a）代入y =–1x＋1显然成立 ····················································· 11分∴ 直线l上任意一点关于x轴的对称点一定在PB所在的直线上 ···················· 12分 注：本卷中各题如有不同解法，可依据情况酌情给分。
